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https://homework.study.com/explanation/calculate-the-oh-of-a-5-5-x-10-3-m-naoh-solution.html	# Calculate the {eq}\left [ OH^- \right ] {/eq} of a {eq}5.5 \times 10^{-3} {/eq} M NaOH solution. ## Question: Calculate the {eq}\left [ OH^- \right ] {/eq} of a {eq}5.5 \times 10^{-3} {/eq} M NaOH solution. ## Hydroxide Ion: The hydroxide ion is a type of anion which is formed when molecules of water dissociate into ions. In the hydroxide ion, the molecules of oxygen chemical element and hydrogen chemical elements are combined with the help of one covalent bond. The chemical formula of it is {eq}\text{O}{{\text{H}}^{-}} {/eq}.	2023-01-28 19:02:02	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9981861114501953, "perplexity": 11337.978527645364}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00362.warc.gz"}
http://math.stackexchange.com/questions/6665/the-probability-theory-around-a-candy-bag	# The probability theory around a candy bag Consider a candy bag that contains N=100 candies. There are only two types of candy in the bag. Say the caramel candy and the chocolate candy. Nothing more is known about the contents of the bag. Now, you are going to draw (randomly) one candy at a time from the bag until the first caramel appear. Suppose that the first caramel appeared at k=7th drawing. At this moment, what can we say about the number of caramel candies in the bag? - The question you are asking here is the classical question of inferential statistics: "Given the outcome of an experiment, what can be said about the underlying probability distribution?" You could, for example, give an estimator for the unknown quantity "number of caramels" (called $a$ from here on). The one most often used (since its easy to calculate) would be the maximum likelyhood estimator, where you estimate $a$ to the the value that maximizes the probability of the outcome. In this case, you'd choose $a$ to maximize $P_a(7)$ (the probability of drawing the first caramel in the seventh draw, assuming there are $a$ of them). A little Excel calculation, along with Isaac's way to calculate $P_a(7)$ results in $a$ to be estimated as 14. To judge, what this result is worth, you'd need to calculate the mean squared error of this estimator, which is not as easily done. If you already had a hypothesis about $a$ (say $a$ < 20), you could use your experimental result to test it, using statistical hypothesis testing, too. - Yes, 14 is the "correct" estimate, as it should be lower than 100/7 as I explained in my answer (I don't think it's necessary to work with the exact figures in this case). – Derek Jennings Oct 13 '10 at 9:27 @Derek: Looking at my excel chart, its a pretty close call for the maximum likelyhood estimator between 13, 14, and 15. Since its neither guaranteed to be unbiased nor particularly good, I have not much confidence in 14, but I don't know a better estimator (i.e. best unbiased) here. =) – Jens Oct 13 '10 at 9:31 I, too, don't know the exact figure, which is why I put correct in quotes but it should be less than 100/7. I estimated as the exact figures got messy. – Derek Jennings Oct 13 '10 at 9:39 If the number of caramel candies is $a$, then the probability that the first 6 drawn will not be caramel and the 7th drawn will be caramel (assuming that we do not put back the drawn candies) is $P(\text{7th}\|a)=\frac{93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{100!}$. Now, given that this has occurred, the probability $P(a\|\text{7th})$ of any particular value of $a$ given that the first caramel is the 7th drawn should be $P(\text{7th}\|a)$ for that particular $a$ divided by the sum of all possible $P(\text{7th}\|a)$. $\sum_{a=0}^{100}P(\text{7th}\|a)=\frac{101}{56}$, so $$P(a\|\text{7th})=\frac{P(\text{7th}\|a)}{\frac{101}{56}}=\frac{56\cdot 93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{101!}.$$ Bashing out some values and adding things up, the probability that $a\le 19$ is slightly less than 50% (49.673%) and the expected value of $a$ is $\frac{65}{3}=21\frac{2}{3}$. edit: (I've slightly altered my original answer above, mostly in the notation, to better accommodate the work below; I believe that the work above assumed that, without knowing how long it took to draw the first caramel, each possible number of caramels was equally likely.) Suppose that $P(a)$ is the probability that there are $a$ caramels. As above, for any particular value of $a$, the probability $P(\text{7th}\|a)$ that the first caramel drawn is the 7th candy drawn is $$P(\text{7th}\|a)=\frac{93!(100-a)(99-a)(98-a)(97-a)(96-a)(95-a)a}{100!}.$$ So, the probability that there are $a$ caramels and that the first caramel drawn is the 7th candy drawn is $P(a\text{ and 7th})=P(a)\cdot P(\text{7th}\|a)$. By Bayes's Theorem: \begin{align} P(a\|\text{7th})&=\frac{P(a\text{ and 7th})}{P(\text{7th})}=\frac{P(a\text{ and 7th})}{\sum_{k=0}^{100}P(k\text{ and 7th})} \\ &=\frac{P(a)P(\text{7th}\|a)}{\sum_{k=0}^{100}P(k)P(\text{7th}\|k)} \end{align} Now, if $P(a)=\frac{1}{100}$ for all $a$, this yields the results in my original answer. If $P(a)={100 \choose a}\frac{1}{2^{100}}$ (a binomial distribution with caramel and not equally likely for each individual candy when the bag is originally filled), the expected value of $a$ is 47.5. If $P(a)={100\choose a}p^a(1-p)^{100-a}$ (a binomial distribution where the probability of each single candy being caramel is $p$ when the bag is originally filled), the expected value of $a$ is $1+93p$. If this expected value of $a$ given that the first caramel drawn was the 7th candy drawn is to equal the expected value of $a$ without having drawn any candies, which is $100p$, then $p=\frac{1}{7}$, so the expected value of $a$ is $\frac{100}{7}=14\frac{2}{7}$. - I don't think its correct to work with a probability distribution of $a$. $a$ is fixed in this experiment (i.e. not subject to chance) and should be estimated. – Jens Oct 13 '10 at 8:51 @Jens: I suppose that I might be assuming that the distribution of values of $a$ in general is uniform (and working with conditional probabilities from there) to arrive at my conclusion. – Isaac Oct 13 '10 at 9:06 @Isaac: You should mention this in your answer. The assumption that an unknown quantity is uniformly distributed is made quite often, and is wrong almost as often, too. For example, while the answer to the question if there are terriers on a planet in the alpha centauri system is unknown, the assumption that yes and no are equally likely is... wrong. =) – Jens Oct 13 '10 at 9:11 @Isaac.I think, Jens's comment provides the key to the correct solution. 21 seems to be too big number. – Martin Gales Oct 13 '10 at 9:13 If there were 21 caramels, surely we'd expect about 5 draws. – Derek Jennings Oct 13 '10 at 9:19 Since 100 is quite a bit larger than 7 we will assume for the purposes of calculation that the probability of drawing a sweet is not affected by the previous draw (of course it is altered a little, but not too much). Let the probability of drawing a caramel be $p$ and the probability of drawing a chocolate be $q,$ where $p+q=1.$ Then the expected number of draws to draw a caramel is $$E= p+2pq+3pq^2+4pq^3+ \cdots = \frac{1}{p}.$$ Since we drew a caramel at the seventh draw set $7=1/p,$ so $p = 1/7.$ Thus $100/7 \approx 14$ of the sweets are caramel. If we used the correct figures the probability of drawing a caramel at the next draw would go up with each chocolate drawn, which will push the estimated number of caramels down, perhaps to 13, but I don't expect it will alter much since we started with 100 sweets. The smaller the number of the first draw of a caramel, the less that can be reliably said. Imagine if you drew a caramel on the first draw, that could imply they are all caramels which could be wildly wrong. -	2016-05-29 16:03:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8991604447364807, "perplexity": 425.3524564505545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049281363.50/warc/CC-MAIN-20160524002121-00042-ip-10-185-217-139.ec2.internal.warc.gz"}
http://exxamm.com/QuestionSolution16/Chemistry/A+burner+should+have+an+optimum+fuel+to+oxygen+ratio+in+order+to+get+maximum+calorific+output+which+corresponds+to+3+t/1167367285	A burner should have an optimum fuel to oxygen ratio in order to get maximum calorific output which corresponds to 3 t ### Question Asked by a Student from EXXAMM.com Team Q 1167367285. A burner should have an optimum fuel to oxygen ratio in order to get maximum calorific output which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of the fuel. A burner which has been adjusted for CH_4 as fuel (2 L/hour of CH_4 and 12 L/hour of O_2) is to be readjusted for an equi-molar mixture of ethane and butane under the same condition of temperature and pressure. In order to get the same calorific output what should be the rate of supply of oxygen (in Litre/hr)? Assume the gases to behave ideally and losses due to incomplete combustion, etc., are the same for both the fuels. Enthalpy of combustion: CH_4= -810 kJ mol^-1; C_2H_6 = -1560 kJ; C_4H_(10) = -2880 kJ mol^-1 A 10.95 B 12.67 C 14.42 D 8.79 #### HINT (Provided By a Student and Checked/Corrected by EXXAMM.com Team) #### Access free resources including • 100% free video lectures with detailed notes and examples • Previous Year Papers • Mock Tests • Practices question categorized in topics and 4 levels with detailed solutions • Syllabus & Pattern Analysis	2019-04-20 04:27:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33426740765571594, "perplexity": 4585.339114231111}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578528523.35/warc/CC-MAIN-20190420040932-20190420062932-00399.warc.gz"}
https://math.stackexchange.com/questions/1176553/proving-sum-n-1-infty-frac0-5nnn1n2n3-frac536-frac/1176565	# Proving $\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$ Proving $$\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$$ $$\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}\\= \sum_{n=1}^{\infty }0.5^n\left(\frac{1/6}n+\frac{-1/2}{n+1}+\frac{1/2}{n+2}+\frac{-1/6}{n+3}\right)\\ =\frac16\sum_{n=1}^{\infty }\frac{0.5^n}n-\frac1{2\cdot0.5}\sum_{n=2}^{\infty }\frac{0.5^n}n+\frac1{2\cdot0.5^2}\sum_{n=3}^{\infty }\frac{0.5^n}n-\frac1{6\cdot0.5^3}\sum_{n=4}^{\infty }\frac{0.5^n}n\\ =\frac16(\ln2)-(\ln2-0.5)+2(\ln2-0.5-0.5^2/2)-\frac43(\ln2-0.5-0.5^2/2-0.5^3/3)\\ =\frac5{36}-\frac16\ln2$$ Another approach. From: $$\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{6}\frac{\Gamma(n)\Gamma(4)}{\Gamma(n+4)}=\frac{1}{6}\int_{0}^{1} u^{n-1}(1-u)^3\,du$$ it follows that: $$\sum_{n\geq 1}\frac{1}{2^n n(n+1)(n+2)(n+3)}=\frac{1}{6}\int_{0}^{1}\frac{(1-u)^3}{2-u}\,du$$ and: $$\int_{0}^{1}\frac{(1-u)^3}{2-u}\,du = \int_{0}^{1}\frac{x^3}{1+x}\,du=-\log 2+\int_{0}^{1}(1-x+x^2)\,dx=-\log 2+\frac{5}{6},$$ hence: $$\sum_{n\geq 1}\frac{1}{2^n n(n+1)(n+2)(n+3)}=\color{red}{-\frac{\log 2}{6}+\frac{5}{36}}$$ as wanted. $$\sum\frac{x^n}{n(n+1)(n+2)(n+3)}=\sum\frac{x^n}6\left(\frac1n-\frac3{n+1}+\frac3{n+2}-\frac1{n+3}\right)$$ This is not fully sketched answer. (It takes some time to type all the steps, sorry :-) ) See a similar question here. Using the same techniques I used there (explained briefly in the comments below my answer) we find that $$\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)(n+2)(n+3)}$$ is the Taylor expansion of $$f(x)=\frac{(6-18x+18x^2-6x^3)\ln(1-x)+6x-15x^2+11x^3}{36x^3}$$ thus $$\sum_{n=1}^{\infty}\frac{(\frac12)^n}{n(n+1)(n+2)(n+3)}=f(\frac12)=\frac{5}{36}-\frac16\ln2.$$ (Note that this answer is in essence similar to the suggestion by ADG.)	2019-09-21 00:49:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9759834408760071, "perplexity": 862.4008070903727}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574159.19/warc/CC-MAIN-20190921001810-20190921023810-00554.warc.gz"}
https://www.kofastudy.com/courses/ss1-physics-3rd-term/lessons/electric-current-week-2/topic/resistance-in-series/	Back to Course 0% Complete 0/0 Steps • ## Do you like this content? Lesson 2, Topic 2 In Progress # Resistance in Series Lesson Progress 0% Complete Resistance in Series When three resistors, R1, R2 and R3 are connected in series, i.e. one end to another end such that the same amount of current flow through them when they are connected to a source of e.m.f. The total resistance, R (effective resistance) is given by V = IReff From the circuit diagram, the voltage V is connected across the three resistors. This voltage is shared across each of the resistors. Remember that the voltages drop across each of the resistors are different because the resistances of the resistors are different. Therefore, Let the Voltage of the cell connected across the three resistors be V. Let the voltage drop across resistor R1 be V1. Let the voltage drop across resistor R2 be V2, also, Let the voltage drop across resistor R3 be V3. Therefore the total voltage can then be found; V = V1 + V2 + V3 IReff = $$\scriptsize \pm R1 \pm R2 \pm R3$$ Using Ohm’s law: V = IReff V1 = IR1 V2 = IR2 V3 = IR3 Since the same current flows through the resistors, Then IReff = IR1 + IR2 + IR3 Factorising I we then have: IReff = I(R1 + R2 + R3) We then divide both sides by I = $$\frac {IR_{eff}}{I} = \frac {I \left ( R_1 + R_2 + R_3 \right)}{I}$$ ∴ Reff = R1 + R2 + R3 The total resistance is greater than the highest resistance in the circuit. Example: Two resistors of value 3Ω and 5Ω are connected in series. Find the effective resistance. Rtotal = R1 + R2 = 3 + 5 = 8Ω error:	2022-01-20 17:58:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5950453281402588, "perplexity": 1749.0444764440258}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302355.97/warc/CC-MAIN-20220120160411-20220120190411-00149.warc.gz"}
https://docs.hector.network/hector-institute/the-dashboard	# The Dashboard The information on this page and all other pages owned by, operated by, or related to Hector Network are for educational purposes only and does not constitute any kind of advice. Please read our Disclaimers page first. ## The Markets Page #### Balances Users can see their supply balance - the value of the tokens they have supplied to the network, borrow balance - the value of tokens that the user has borrowed, and their net APY - this shows the total combined APY of all open supply and borrow positions. #### Markets The markets page shows available tokens to lend and borrow. Supply Markets show tokens which can be lent. Borrow markets show tokens which can be borrowed. ## The Network Page The network page shows the stats of the entire protocol: Network Supply Balance The value of all tokens supplied to the network by all users. Network Borrow Balance The value of all tokens borrowed from the network by all users. Assets Utilization The percentage of assets borrowed from the supply, calculated as: $(Borrow Balance/Supply Balance)*100$ Supplied Assets A breakdown of all the tokens supplied by all users. Hover over the borrowed value to see the percentage of a given token which has been borrowed. ## The Profile Page The Profile Page gives an overview of your current positions: Supply Balance The total value of the tokens supplied by a user. Borrow Balance The total value of all tokens borrowed by a user. Net APY The combined APY of all open supply and borrow positions a user has.	2022-12-03 23:07:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35513612627983093, "perplexity": 3429.8077742396854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710941.43/warc/CC-MAIN-20221203212026-20221204002026-00493.warc.gz"}
https://unstable.publiclab.org/questions/gabrielxp46/05-14-2018/could-anyone-tell-me-if-it-is-possible-to-use-the-homemade-spectrometer-to-analyze-a-sample-of-metal	# Question: could anyone tell me if it is possible to use the homemade spectrometer to analyze a sample of metal? gabrielxp46 asked on May 14, 2018 14:34 196 \| 1 answers \| #16346 I am interested in this aspect because I perform a survey with stainless steels and would like to know if it is possible to use the homemade spectrometer to obtain some metal data such as chemical composition and etc. The instruments to do The testing you want are complicated. It's not that the spectrometer couldn't be used, but it would take a lot of work. Here is a little background. The common way to do this is with analytical techniques, such as atomic absorption (AA), atomic emission (AE), or inductively coupled plasma ( ICP). All use a spectrometer or spectrophotometer, so it is possible. It's the extras the instruments have that make it hard. The ICP takes high power and a plasma furnaces. It's not easy to do. They were in the \$100k range (base). That's the first one to eliminate. The AA takes special lamps for each element analyzed. Sometimes you can combine multiple elements into one lamp, but not always. AE doesnt take the lamp, but it is often the less sensitive of the two. With all three techniques, the metal must be prepped. This usually means just acid digestion, although it can be much more involved. All forms of the metal ( say chrome) must be in the same valence state. Sometimes, a fair number of chemical steps are required. Then the dissolved liquids are aspirated in a flame and the spectra is taken. Many of the lines used in these techniques are in the UV. i.E. Zinc is usually analyzed at 213.9 nm. There are many interferences from flames, etc, that need to be worked through.	2019-07-21 13:48:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6326448917388916, "perplexity": 802.6979436750685}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195527000.10/warc/CC-MAIN-20190721123414-20190721145414-00471.warc.gz"}
http://xxicla.dm.uba.ar/sessionAbs.php?session=bb9e15c1f7&order=1	# Session S06 - Algebraic Combinatorics Chair: Ernesto Vallejo (UNAM, México) Collaborators: Rosa Orellana (Dartmouth College, USA) Session website View session abstracts PDF ## Talks July 25, 15:00 ~ 15:25 ## Of antipodes and involutions ### Michigan State University, USA - sagan@math.msu.edu Let $H$ be a graded, connected Hopf algebra. Then Takeuchi's formula gives an expression for the antipode of $H$. But this alternating sum usually has lots of cancellation. We will describe a method using sign-reversing involutions to obtain cancellation-free formulas for various $H$. This technique displays remarkable similarities across the Hopf algebras to which it has been applied. No background about Hopf algebras will be assumed. Joint work with Carolina Benedetti (Fields Institute, Canada). View abstract PDF July 25, 16:00 ~ 16:25 ## Counting arithmetical structures of a graph and their sandpile groups. ### Mathematics department, Cinvestav-IPN, México - cvalencia@math.cinvestav.edu.mx Given a graph $G=(V,E)$ and ${\bf d}\in \mathbb{N}$, the \emph{Laplacian matrix} of the pair $(G,{\bf d})$ is the square matrix given by $L(G,{\bf d})_{u,v}= \begin{cases} {\bf d}_u&\text{if }u=v,\\ -m_{uv}&\text{if }u\neq v, \end{cases}$ where $m_{uv}$ is the number of edges between $u$ and $v$. An \emph{arithmetical structure} of $G$ is a pair $({\bf d},{\bf r})$ such that $({\bf d},{\bf r})\in \mathbb{N}_+^V\times \mathbb{N}_+^V$, $\mathrm{gcd}({\bf r}_v\, \| \,v\in V(G))=1$ and $L(G,{\bf d}){\bf r}^t={\bf 0}^t.$ The concept of arithmetical graphs was introduced by Lorenzini as some intersection matrices that arise in the study of degenerating curves in algebraic geometry. If $G$ is strongly connected, then $\mathcal{A}(G)=\{({\bf d},{\bf r})\in\mathbb{N}_+^{V(G)}\times \mathbb{N}_+^{V(G)} \,\| \,({\bf d},{\bf r})\textrm{ is an arithmetical structure of } G\}.$ is finite. Our goal is to describe and count the arithmetical structures and their associated sandpile groups of some simple graph, like the path, cycle, complete, etc. For instance we prove that the number of arithmetical structures of a path $P_n$ with $n$ vertices is equal to the Catalan number $C_{n-1}$. View abstract PDF July 25, 16:30 ~ 16:55 ## Cotas para la energía de Nikiforov sobre digrafos ### Universidad de Antioquia - Medellín, Colombia - nagudel83@gmail.com La energía de un grafo $G$ se define como $E(G)=\sum\limits_{i=1}^{n}\left \vert \lambda _{i}\right\vert$, donde $\lambda _{1},\lambda _{2},\ldots ,\lambda _{n}$ son los valores propios de la matriz de adyacencia de $G$. Este concepto fue extendido de varias maneras para digrafos: $\mathcal{E}\left( D\right) =\sum\limits_{i=1}^{n}\left\vert \text{Re}\left( z_{i}\right) \right\vert$, $\mathcal{S}\left( D\right) =\sum\limits_{i=1}^{n}\left\vert z_{i}\right\vert$ y $\mathcal{N}\left( D\right) =\sum\limits_{i=1}^{n}\sigma _{i}$, donde $D$ es un digrafo con $n$ vértices, valores propios $z_{1},\ldots ,z_{n}$ y valores singulares $\sigma _{1},\ldots ,\sigma _{n}$. En este trabajo hallamos cotas superiores e inferiores para $\mathcal{N}$ sobre el conjunto de digrafos. También mostramos que $\mathcal{E}\left( D\right) \leq \mathcal{S}\left( D\right) \leq \mathcal{N}\left( D\right)$ para todo digrafo $D$ y caracterizamos los digrafos donde se da la igualdad. Como consecuencia, deducimos nuevas cotas superiores e inferiores para $\mathcal{E},\mathcal{S}$ y $\mathcal{N}$ las cuales son obtenidas de cotas inferiores de $\mathcal{E}$ y cotas superiores de $\mathcal{N}$. Joint work with Juan Pablo Rada (Universidad de Antioquia, Medellín, Colombia). View abstract PDF July 25, 17:30 ~ 17:55 ## Lopsided amoebas and effective amoeba approximation ### Texas A&M University, USA - laura@math.tamu.edu The amoeba $\mathscr{A}(f)$ of a polynomial $f$ is the image of its zero set under the log-absolute-value map. The amoeba captures combinatorial information about $f$: for instance, the normal fan of the Newton polytope of $f$ determines the asymptotics of $\mathscr{A}(f)$. In 2008, Purbhoo introduced the lopsided amoeba $\mathscr{L}(f)$ of $f$, and showed that $\mathscr{A}(f)$ is the limit as $r\to \infty$ of $\mathscr{L}(f_r)$, where $f_r$ is constructed from $f$ by a process of iterated resultants. I will introduce lopsided amoebas geometrically, show how to efficiently compute the resultants involved, and outline some combinatorial challenges in this area. Joint work with Jens Forsgård, Nathan Mehlhop and Timo de Wolff (all at Texas A&M University, USA). View abstract PDF July 25, 18:00 ~ 18:25 ## The Dehn--Sommerville Relations and the Catalan matroid ### San Francisco State University, United States - nyamzon@mail.sfsu.edu The $f$-vector of a $d$-dimensional polytope $P$ stores the number of faces of each dimension. When $P$ is simplicial the Dehn--Sommerville relations imply that to determine the $f$-vector of $P$, we only need to know approximately half of its entries. This raises the question: Which $(\lceil{\frac{d+1}{2}}\rceil)$-subsets of the $f$-vector of a general simplicial polytope are sufficient to determine the whole $f$-vector? We prove that the answer is given by the bases of the Catalan matroid. Joint work with Anastasia Chavez (University of California at Berkeley). View abstract PDF July 26, 15:00 ~ 15:25 ## Integral hyperplane arrangements ### Cornell University, United States - maguiar@math.cornell.edu Consider an arrangement of linear hyperplanes integral with respect to a given lattice. The lattice gives rise to a torus and the arrangement to a subdivision of the torus. We are interested in the combinatorics of this subdivision. We will describe questions and results for particular lattices associated to root systems and arrangements associated to graphs. Joint work with Swee Hong Chan (Cornell University). View abstract PDF July 26, 15:30 ~ 15:55 ## The generalized lifting property of Bruhat intervals ### Universidad de Chile, Chile - paolosentinelli@gmail.com The so called "lifting property" characterizes the Bruhat order of a Coxeter group, as V. V. Deodhar proved in 1977. E. Tsukerman and L. Williams in the article "Bruhat interval polytopes" (Advances in Mathematics, 2015) prove a "generalized lifting property" of the Bruhat order for the symmetric group. We investigate the case of an arbitrary Coxeter group. Joint work with Fabrizio Caselli (Università di Bologna, Italy). View abstract PDF July 26, 16:00 ~ 16:25 ## A proof of the peak polynomial positivity conjecture ### Williams College, United States - pamela.e.harris@williams.edu Given a permutation $\pi=\pi_1\pi_2\cdots \pi_n \in \mathfrak{S}_n$, we say an index $i$ is a peak if $\pi_{i-1} < \pi_i > \pi_{i+1}$. Let $P(\pi)$ denote the set of peaks of $\pi$. Given any set $S$ of positive integers, define $P_S(n)=\{\pi\in\mathfrak{S}_n:P(\pi)=S\}$. In 2013 Billey, Burdzy, and Sagan showed that for all fixed subsets of positive integers $S$ and sufficiently large $n$, $\|P_S(n)\|=p_S(n)2^{n-\|S\|-1}$ for some polynomial $p_S(x)$ depending on $S$. They gave a recursive formula for $p_S(n)$ involving an alternating sum, and they conjectured that the coefficients of $p_S(x)$ expanded in a binomial coefficient basis centered at $\max(S)$ are all nonnegative. In this talk we will share a different recursive formula for $p_S(n)$ without alternating sums, and we use this recursion to prove that their conjecture is true. Joint work with Alexander Diaz-Lopez, Swarthmore College, Erik Insko, Florida Gulf Coast University and Mohamed Omar, Harvey Mudd College. View abstract PDF July 26, 16:30 ~ 16:55 ## On trees with the same restriction of the chromatic symmetric function and solutions to the Prouhet-Tarry-Escott problem ### Universidad Andres Bello, Chile - jose.aliste@gmail.com On the one hand, the {Prouhet-Tarry-Escott problem} asks, given $k$ be a positive integer, whether there exist integer sequences $a = (a_1,\ldots, a_n)$ and $b = (b_1,\ldots, b_n)$, distinct up to permutation, such that $$a_1^{\ell}+\ldots+ a_n^\ell =b_1^{\ell}+\ldots +b_n^\ell \quad \text{ for all } 1\leq \ell \leq k.$$ This is an old problem in number theory (Prouhet 1851), and solutions are known to exist for every $k$. On the other hand, the chromatic symmetric function was introduced by Stanley in 1995 as a symmetric function generalization of the chromatic polynomial of a graph. It is an open problem to know whether there exist non-isomorphic trees with the same chromatic symmetric function. In this talk, we show how to encode solutions of the Prouhet-Tarry-Escott problem as non-isomorphic trees having the same restriction of the chromatic symmetric function. As a corollary, we find a new class of trees that are distinguished by the chromatic symmetric function up to isomorphism. Joint work with Anna de Mier (Universidad Politécnica de Cataluña, España) and José Zamora (Universidad Andres Bello, Chile). View abstract PDF July 26, 17:30 ~ 17:55 ## Pieri rules for the Macdonald polynomials in superspace and the 6-vertex model ### Universidad de Talca, Chile - lapointe@inst-mat.utalca.cl The Macdonald polynomials in superspace are symmetric polynomials involving commuting and anticommuting variables that generalize the Macdonald polynomials. We will describe how the combinatorics of the Macdonald polynomials extends to superspace. We will focus in particular on how the partition function of the 6 vertex model arises in the Pieri rules for the Macdonald polynomials in superspace. Joint work with Jessica Gatica (PUC, Chile), Camilo Gonzalez (Universidad de Talca, Chile) and Miles Jones (University of California, San Diego, USA). View abstract PDF July 26, 18:00 ~ 18:25 ## Irreducible characters of the symmetric group as symmetric functions ### York University, Canada - zabrocki@mathstat.yorku.ca I will introduce a basis of the symmetric functions that are the irreducible characters of the symmetric group realized as permutation matrices. Just as the Schur functions are the irreducible characters of the general linear group, the elements of this new basis are functions in the eigenvalues of a permutation matrix. Symbolically, if $\Xi_\mu$ are the eigenvalues of a permutation matrix of cycle type $\mu$, then ${\tilde s}_\lambda[\Xi_\mu]$ will be the irreducible symmetric group character $\chi^{(\|\mu\|-\|\lambda\|,\lambda)}(\mu)$. This basis has (outer) product structure coefficients given by the reduced Kronecker coefficients and it also has positive coproduct structure coefficients. There is analogously a second basis of the induced trivial characters of the symmetric group and together these bases encode the combinatorics of multisets and multiset valued tableaux. Joint work with Rosa Orellana (Dartmouth College). View abstract PDF	2017-05-22 21:30:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6948047280311584, "perplexity": 1704.5946546266243}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607120.76/warc/CC-MAIN-20170522211031-20170522231031-00368.warc.gz"}
https://www.jkcs.or.kr/journal/view.php?number=3606	J. Korean Ceram. Soc. > Volume 28(12); 1991 > Article Journal of the Korean Ceramic Society 1991;28(12): 943. 졸-겔법에 의한 PZT 합성과 박막제조 오영제, 정형진 한국과학기술연구원 세라믹스재료연구단 Sol-Gel Processing and Properties of PZT Powders and Thin Films ABSTRACT Lead zirconate titanate(PZT) powders and thin films were prepared from an alkoxide-based solution by sol-gel method. Gelation of synthesized complex solutions, pyrolysis and crystallization behaviors of the dried powder were studied in accordance with a water content and a catalyst. PZT thin films were formed by spin-casting method on silicon and platinum substrates, and characterized. Ester produced from the reactions was completely removed when drying of the gel was finished. Pyrolysis property of the dried PZT gels were changed in order water content, class of catalyst, and quantity of catalyst. Crystalline Pb phase was transiently formed near 250$^{circ}C$. Basic catalyst is good additive for a formation of perovskite phase in the films, and acidic catalyst for a densified film structure. By the analysis of RBS, Pb element in the PZT films were diffused into silicon substrate, and the pores, may be produced due to local densification around some grains in the films, make an origin of fault in microstructure when holding time goes to be longer at 700$^{circ}C$. TOOLS Full text via DOI	2019-01-23 18:07:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2751755118370056, "perplexity": 14313.378389018859}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584336901.97/warc/CC-MAIN-20190123172047-20190123194047-00312.warc.gz"}
http://math.stackexchange.com/questions/17573/1-otimes-1-cong-i	# $(-1)\otimes (-1) \cong I$ Is there a monoidal category $\mathcal C$ whose unit object is $I$ (i.e. $I\otimes A\cong A\cong A\otimes I$ for all $A\in \text{Ob}_\mathcal C$), with an object "$-1$" such that $$(-1)\otimes(-1)\cong I ?$$ (Edit: no one misunderstood, but I'm also asking that $-1\neq I$) I'm struggling with that since I read this math.SE post... Martin, if you see me, your server rejected every mail I tried to send you. :( - what about $\mathbb{Z}_2$ vector spaces over a field $k$? $(k,0)$ is the unit in this category and $(0,k)\otimes (0,k) = (k,0)$. Great, thanks! ...No hope to find such an object in $k-\mathbf{Vect}$, huh? In fact, it is plainly false in $\mathbf{Sets}$ with the monoidal structure induced by the product and the identity given by $\{*\}$... So I'm looking for some (counter)examples in more stuctured categories – tetrapharmakon Jan 15 '11 at 13:48 For any locally compact abelian group $G$, the monoidal category of continuous unitary representations of $G$ is generated by the characters $G \to \mathbb{C}$, e.g. the Pontrjagin dual group $\hat{G}$, with tensor product corresponding to the group operation and duals corresponding to the inverse (and the trivial representation corresponding to the identity). Prometheus's example is the case $G = \hat{G} = \mathbb{Z}/2\mathbb{Z}$. Unfortunately I don't know almost (=some wiki stuff) anything about group representations... In particular what about my previous question (finding such an object in $k-\mathbf{Vect}$)? :( Nonetheless I can catch the whole idea, thanks! – tetrapharmakon Jan 15 '11 at 23:54	2015-05-27 09:57:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9306164979934692, "perplexity": 361.6027457652221}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928923.85/warc/CC-MAIN-20150521113208-00137-ip-10-180-206-219.ec2.internal.warc.gz"}
https://mathoverflow.net/questions/279920/who-conjectured-the-cartan-determinant-conjecture	# Who conjectured the Cartan determinant conjecture The Cartan determinant conjecture states that every finite dimensional algebra of finite global dimension has the property that the determinant of its Cartan matrix is equal to one. Who stated this conjecture first and how old is it? The Cartan matrix of a finite dimensional algebra is defined as the matrix having entries $c_{i,j}:=$dimension of $(Hom_{A}(P_i,P_j))$ over the divison ring $D_i$, where $D_i:=End_A(S_i)$. Here $P_i$ are the indecomposable projective modules and $S_i$ the simple modules which are the top of $P_i$.	2019-08-22 19:03:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9418214559555054, "perplexity": 104.82444273532292}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317339.12/warc/CC-MAIN-20190822172901-20190822194901-00128.warc.gz"}
http://mathhelpforum.com/geometry/202515-rectangular-print.html	# Rectangular • August 24th 2012, 02:33 PM Mhmh96 Rectangular The rectangular WZYX is inside the rectangular ACDB AB= 8 cm AC= 6 cm WX= 8 cm What is WZ= ? http://store2.up-00.com/Aug12/yLN47559.jpg • August 24th 2012, 02:50 PM HallsofIvy Re: Rectangular There are an infinite number of possible solutions. Imagine sliding point X toward A or B. You can aways rotate line XW around X so that W stays on AC. Each angle gives a different solution. • August 24th 2012, 03:31 PM MaxJasper Re: Rectangular 1 possible solution only: ax=6.2233 aw=5.027 wz=1.2507 cz=0.78594 wc=0.97296 • August 25th 2012, 02:55 AM Wilmer Re: Rectangular Why not (as an attempt) make angle AWX = 60 degrees; then AW = 4 and AX = 4SQRT(3) • August 25th 2012, 10:39 AM earboth Re: Rectangular Quote: Originally Posted by Mhmh96 The rectangular WZYX is inside the rectangular ACDB AB= 8 cm AC= 6 cm WX= 8 cm What is WZ= ? If and only if the points W, X, Y and Z has to be placed on the sides of the rectangle ABCD and $\|\overline{WX}\| = 8$ Then there is only one unique solution of this question. See attachment #2 to see what happens if $\|\overline{AX}\|$ is changing but $\|\overline{WX}\|$ remains constant. Only the thick outlined quadrilateral is a rectangle, all other quadrilaterals are trapezoids. According to the labeling in attachment #1 you'll get: $8^2 = (8 - x)^2 + (6 - y)^2$ ........... [1] The adjacent sides of the rectangle are perpendicular to each other. Using the slopes of the sides you'll get: $-\frac yx \cdot \frac{6-y}{8-x}=-1$ ........... [2] From [2] you'll get: $x = 4 - \sqrt{y^2 - 6·y + 16}$ Plug in the term of x into [1] and solve for y. I've to confess that this task was done by my calculator. You'll get: $y \approx 1.895734320$ and consequently $x \approx 1.133049932$ Since $\|\overline{XY}\| = \|\overline{WZ}\|$ you'll get $\|\overline{WZ}\| \approx \sqrt{1.895734320^2 + 1.133049932^2} \approx 2.208531358$ • November 1st 2012, 01:48 AM elisaevedent Re: Rectangular • November 8th 2012, 07:21 PM Chalama123 Re: Rectangular • November 18th 2012, 11:35 PM jesong Re: Rectangular as the Hausdorff dimension which should be fractional in the case of wallets a fractal, and a major part of this is related to the self-similarity since this self-similarity creates weird dependencies in the actual information content of the object itself.louis vuitton handbags	2014-04-18 09:48:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8705043196678162, "perplexity": 2715.751194690712}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00339-ip-10-147-4-33.ec2.internal.warc.gz"}
https://paulleemagic.com/site/ex47v4.php?99491a=wavelength-of-accelerated-electrons	## wavelength of accelerated electrons Therefore, the wavelength at 100 keV, 200 keV, and 300 keV in electron microscopes is 3.70 pm, 2.51 pm and 1.96 pm, respectively. A well-collimated beam of electrons is scattered off the nickel target. Chapter Chosen In Transmission Electron Microscopy (TEM), electrons pass through the sample and illuminate film or a digital camera. where m is mass and v is velocity of electron. Equals. where the photon energy was multiplied with the electronic charge to convert the energy in Joule rather than electron Volt. These fatty acids are about 20 carbons long with a hydrophilic head group. given, Kinetic energy of the electron, K.E = 120 eV(a) Momentum of the electron is given as, p = 2mE ⇒ p = 2 ×(9 × 10-31) × (120 × 1.6 × 10-19) = 5.88 × 10-24 kg ms-1(b) Now, since p = mv we have, v = pm = 5.88 × 10-249.1 × 10-31 ⇒ v = 6.46 × 106 m/s (c) De- broglie wavelength of electron is given by, λ = hp = 6.63 × 10-345.88 × 10-24 = 1.13 × 10-10m. Immuno-electron microscopy is difficult (Figure 7), and morphology is compromised. If the freezing rate of 10,000ºC/s can be achieved, super-cooled water can be vitrified without crystallization, that is, the water will freeze in an unordered state (Figure 4C). A microtome must be situated in a room where there is no vibration or thermal fluctuations. The proximity of this theoretical result to the DavissonâGermer experimental value of Î»=1.64ÃÎ»=1.64Ã is a convincing argument for the existence of de Broglie matter waves. Because our eyes can only detect photons with wavelengths greater than ~400 nm, the best resolution that can be achieved by light microscopes is about ~200 nm. What is the de-Broglie wavelength of(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s. The cross-section of E. coli is about 400 nm. The thickness of sections can be easily altered by vibration or temperature change. Your IP: 3.24.102.178 The observed pattern reflects the symmetry of the crystalline structure of the sample. In 1924 Louis de Broglie theorized that not only light posesses both wave and particle properties, but rather particles with mass - such as electrons - do as well. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. The neutron has zero charge and its mass is comparable with the mass of a positively charged proton. Figure 3: Ventral nerve cord of C. elegans showing the differences between conventional chemical fixation and high-pressure freezing. Such samples exhibit excellent preservation (Figure 3). How can an electron be used to image structure? λ ′ λ = V V ′ = 5 0 2 0 0 = 2 Hence, the ratio of the de Broglie wavelengths for two electrons is 2:1. Liquid nitrogen can generate a cooling rate of -16,000ºC/s. Electron microscopes use electrons to illuminate a sample. (vii) Wherever necessary, neat and properly labelled diagrams should be drawn.Section-A1. Acrylic resins are polymerized using either UV-light, a catalyst or high temperature. If V volt is accelerating potential of electron, then Kinetic energy, The OpenStax name, OpenStax logo, OpenStax book However, the need to preserve tissue morphology restricts the use of mild fixatives for any EM preparation. }6\cdot 10^{-19}\,{\text C}\$. Want to cite, share, or modify this book? The lipid bilayer of the plasma membrane is less than 5nm but the individual leaflets of the bilayer can be resolved. Wavelength of an accelerated electron QUESTION: Approximately what is the "de Broglie" wavelength of an electron that has been accelerated through a potential difference of = 150 V? Expression for de Broglie Wavelength associated with Accelerated Electrons The de Broglie wavelength associated with electrons of momentum p is given by. How many photons (nearly) per square metre are incident on the earth per second? Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. Matter waves of molecules as large as carbon C60C60 have been measured. All of these knives can be used for ultrathin sectioning. i.e., λ = 0.164 nm , is the De-broglie wavelength of the electron. Image types Page modified March 15, 2013 • This occurs because the diffraction image is created by scattering off many crystalline planes inside the volume, and the maxima produced in scattering at Bragg angles are sharp (see Figure 6.22). However, the durability of an edge of a diamond knife is the highest, and thereby many sections can be sectioned with the same quality. Find the de Broglie wavelength and kinetic energy of a free electron that travels at a speed of 0.75c. The wavelength of a particle or a matter can be calculated as follows. de Broglie wavelength of electrons. The wavelength of these 'material waves' - also known as the de Broglie wavelength - can be calculated from Planks constant $$h$$ divided by the momentum $$p$$ of the particle. Finally, a sharp knife is imperative for a good quality of sectioning. We recommend using a Unfortunately, biological specimens do not generate much contrast because they are typically composed of carbon, nitrogen, and oxygen – atoms with similar atomic number. Diffraction patterns obtained in scattering on a crystalline solid: (a) with X-rays, and (b) with electrons. then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a digital format, What is the de Broglie wavelength of a ball of mass 0.12 kg moving with a speed of 20 metre per second? Figure 8: Correlative fluorescence and electron microscopy. Number of photons incident on earth's surface per square metre per second is, https://www.zigya.com/share/UEhFTjEyMDU4ODE4. The wavelength of these 'material waves' - also known as the de Broglie wavelength - can be calculated from Planks constant $$h$$ divided by the momentum $$p$$ of the particle. If a magnification of (-1) is to be obtained by using a converging mirror, then theobject has to be placed(1)(a) Between pole and focus(b) At the centre of curvature(c) Beyond the centre of curvature(d) At infinity, A dancer is standing on a rotating platform takingtwo sphere on her hands. For higher acceleration voltage relativistic calculation should be used. To avoid alterations in tissues caused by dehydration (like raisins from grapes), the tissue must be cross-linked or ‘fixed’ to preserve structure. The resolution of diffraction images greatly improves when a higher-energy electron beam passes through a thin metal foil. The temperature at which homogeneous nucleation takes place for water is around -40ºC. Two electrons are accelerated through 50 volts and 200 volts. Delhi - 110058. find the frequencies of the first three harmonics of an open pipe of length 1.65m, given the velocity of sound in air is equal to 330m/s. If you are redistributing all or part of this book in a print format, In the case of electrons that is $$\lambda_{\text{de Broglie}} = \frac {h}{p_\text e}=\frac {h}{m_\text e\cdot v_\text e}$$ The acceleration of electrons in an electron beam gun with the acceleration voltage $$V_{\rm{a}}$$ results in the corresponding de Broglie wavelength $$\bbox[8px,border:2px solid red]{\lambda_{\text{de Broglie}} = \frac {h}{m_\text e\cdot \sqrt{2\cdot \frac{e}{m_\text e}\cdot V_{\text a}}}=\frac {h}{\sqrt{2\cdot m_\text e \cdot e\cdot V_{\text a}}}}$$ Proof of the de Broglie hypothesis will be experimentally demonstrated with the help of an electron diffraction tube on the following pages. Thus tissues are stained with heavy metals, such as osmium, uranium or lead. IMPORTANT! To be able to proceed, you need to solve the following simple math (so we know that you are a human) :-) What is 4 + 14 ?	2022-05-28 04:09:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6691305637359619, "perplexity": 762.6653916814017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00087.warc.gz"}
https://theoremoftheweek.wordpress.com/2011/10/14/number-theory-lecture-4/	## Number Theory: Lecture 4 In which we study the multiplicative group $(\mathbb{Z}/p \mathbb{Z})^{\times}$. We saw last time that we could reduce the problem of understanding the multiplicative group $(\mathbb{Z}/n\mathbb{Z})^{\times}$ to studying the groups $(\mathbb{Z}/p^j \mathbb{Z})^{\times}$ for primes $p$ and for $j\geq 1$. Today we began on this new task, starting with the case of $(\mathbb{Z}/p \mathbb{Z})^{\times}$ for prime $p$. • Lemma 13: For any $n$, we have $\sum_{d\|n} \phi(d) = n$ (where the sum is over all divisors $d$ of $n$). We proved first that the function $F(n) := \sum_{d\|n} \phi(d)$ is multiplicative, and then that $F(p^j) = p^j$ when $p$ is prime. We also noted that the only property of $\phi$ that we used to show that $F$ is multiplicative is that $\phi$ is multiplicative, and so our proof showed that if $f : \mathbb{N} \to \mathbb{N}$ is a multiplicative function, then $\sum_{d\|n} f(d)$ is also multiplicative. • Theorem 14 (Lagrange): Let $p$ be a prime. Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dotsb + a_1 x + a_0$ be a polynomial with integer coefficients, with $a_n$ not divisible by $p$. Then the polynomial congruence $f(x) \equiv 0 \pmod{p}$ has at most $n$ solutions. We saw that this can be proved in a similar way to the analogous result in ordinary arithmetic (by induction on the degree of the polynomial), making use of the fact that $ab \equiv 0 \pmod{p}$ implies $a \equiv 0 \pmod{p}$ or $b \equiv 0 \pmod{p}$ (that is, $\mathbb{Z}/p\mathbb{Z}$ is an integral domain). We saw an example that shows that the result does not hold if the requirement that $p$ is prime is relaxed. We also saw that the result does not (indeed, cannot) tell us that the congruence has exactly $n$ solutions, because there are some polynomial congruences with fewer (indeed, with none). • Theorem 15: Let $p$ be a prime. Then the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic. That is, there is some element $a$ so that the $p-1$ numbers $1$, $a$, $a^2$, …, $a^{p-2}$ give the whole group. We call such an element $a$ a primitive root modulo $p$. (Quick check: find all the primitive roots modulo 11, for example. Try to do this with as little work as possible! Can you say anything interesting about which primes have many primitive roots and which have only a few?) We proved this by showing that in $(\mathbb{Z}/p\mathbb{Z})^{\times}$, for each divisor $d$ of $p-1$ there are exactly $\phi(d)$ elements of order $d$. • Definition of primitive roots (in the statement of the above result). #### Postscript I had a question via e-mail from someone who wasn’t quite comfortable with part of the proof that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic. That particular part was the point where I mentioned during the lecture that we were going to use an idea repeatedly, so I’d like people to be clear about it. I’ll state the more general fact here, and hopefully that will be useful to you when you’re working through your lecture notes. Lemma: Let $G$ be a finite group, and let $g$ be an element of $G$ with order $k$. Suppose that $g^d = 1$ (using multiplicative notation, so $1$ is the multiplicative identity in $G$). Then $k\|d$ How could you prove this? You’d like to show that $k$ divides $d$, so why not try dividing $d$ by $k$, to get say $d = qk + r$ where $0 \leq r < k$? Then what might you do? Davenport (The Higher Arithmetic) presents a slightly different proof that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic; you might like to read that for another perspective. #### Preparation for Lecture 5 Next time we’ll move on to $(\mathbb{Z}/p^j \mathbb{Z})^{\times}$. Will these groups also turn out to be cyclic? ### 6 Responses to “Number Theory: Lecture 4” Hi i have something written in my notes that says “log p prime power”. But i don’t see how log p can be a prime? i mean it isn’t even an integer usually (i tried p = 1,2,3,4 up to 20 and they’re all fractions). Have i misunderstood something? Oh wait. Is it log based then? so log 100 = 2 which is a prime e.g. 3. theoremoftheweek Says: I’m not quite sure where you mean. Can you be a bit more precise about where this was in your notes?	2017-12-11 17:14:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 63, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9563128352165222, "perplexity": 225.19206801765932}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948513784.2/warc/CC-MAIN-20171211164220-20171211184220-00396.warc.gz"}
https://strings.math.tecnico.ulisboa.pt/seminars?id=5725	20/04/2020, Monday, 15:00–16:00 Europe/Lisbon — Online Gábor Sárosi, Theory Division CERN Holographic Probes of Inner Horizons In the context of the AdS/CFT correspondence, charged and rotating thermal ensembles are dual to black holes with inner Cauchy horizons. We argue that an uneventful inner horizon requires certain analytic properties of correlation functions in the dual boundary ensemble which are not consistent with causality and unitarity for charged black holes and rotating black holes in $D>3$. However, they are satisfied for correlators of a holographic $2d$ CFT in a rotating thermal ensemble. This suggests that strong cosmic censorship is enforced in gravity theories with a CFT dual, with the possible exception of the rotating BTZ black hole. Projecto FCT UIDB/04459/2020.	2021-10-20 16:21:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8051671981811523, "perplexity": 1305.167262840264}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585322.63/warc/CC-MAIN-20211020152307-20211020182307-00219.warc.gz"}
https://mathoverflow.net/questions/290626/an-hopeless-integro-differential-equation	# An (hopeless) integro-differential equation While doing some estimates for PDEs I came across the following equation: $$y'(t) = \alpha(t) + \left( \int_0^t y(\tau) \, d\tau\right)^\gamma, \qquad t \in [0,1]$$ where $\alpha \colon [0,1] \to \mathbb R$ is some given function and $\gamma \ge 1$ is a fixed real number. I have a non-negative, bounded function $y$ satisfying this relation and I would like to derive some pieces of information about it. Of course I do not expect any risolutive formulas to hold, but hope is the last to die... do you have some hints? Any ideas on which kind of estimates I could derive? • there may be a solution for $\alpha\equiv 0$, is that of interest? – Carlo Beenakker Jan 13 '18 at 16:58 • Thanks a lot for the comment. Sure it may be of interest, of course, though not fully resolutive (but maybe with some tricks I can work out some estimates also for the general $\alpha$). Could you please point me out some references? Thanks! – Romeo Jan 13 '18 at 17:00 If I define $f(t)=\int_0^t y(\tau)d\tau$, I need to solve $$f''(t)=\alpha(t)+f(t)^\gamma.$$ For $\alpha\equiv 0$ this has the implicit solution $${({\gamma}+1) f(t)^2 \left(c_1 {\gamma}+c_1+2 f(t)^{{\gamma}+1}\right)^2 \, _2F_1\left(\frac{1}{2},\frac{1}{{\gamma}+1};1+\frac{1}{{\gamma}+1};-\frac{2 f(t)^{{\gamma}+1}}{{\gamma} c_1+c_1}\right)^2}{}=\left(c_2+t\right)^2\,\left(c_1 {\gamma}+c_1\right) \left(c_1 ({\gamma}+1)+2 f(t)^{{\gamma}+1}\right)^2$$ Closed-form expressions in terms of special functions (Weierstrass $\wp$ and Jacobi elliptic function sn, courtesy of Mathematica) are possible for small integer $\gamma$: $$f(t)=c_1 e^t+c_2 e^{-t},\;\;\gamma=1$$ $$f(t)=\sqrt[3]{6}\; \wp \left(\frac{t+c_1}{\sqrt[3]{6}};0,c_2\right),\;\;\gamma=2$$ $$f(t)=\pm\sqrt[4]{2} \sqrt{ic_1}\; \text{sn}\left(\left.\frac{(-1)^{3/4} \sqrt{\sqrt{2} \sqrt{c_1} t^2+2 c_2 \sqrt{2} \sqrt{c_1} t+c_2^2 \sqrt{2} \sqrt{c_1}}}{\sqrt{2}}\right\|-1\right),\;\;\gamma=3$$ More generally, for constant $\alpha$, the implicit solution is $$\int_1^{f(t)} \left({2\alpha x+\frac{2}{\gamma+1}x^{\gamma+1}+c_1}\right)^{-1/2} \, dx=c_2+t$$ with explicit solutions $$f(t)=-\alpha+c_1 e^t+c_2 e^{-t},\;\;\gamma=1$$ $$f(t)=\sqrt[3]{6} \;\wp \left(\frac{t+c_1}{\sqrt[3]{6}};-2 \sqrt[3]{6} \alpha,c_2\right),\;\;\gamma=2$$ • Thanks for your precious effort and for writing the implicit expression of the solution for vanishing $\alpha$. I will try to work on it to see if I find something... Do you believe there is possibility of deriving some Gronwall-like estimates from the equation? Maybe writing it as an ODE is a good point... In any case, thanks again for your interest! – Romeo Jan 13 '18 at 17:19 • Does your last expression adapt also to piecewise constant $\alpha$? Maybe that could be very useful to me... (I am not really in this framework but maybe I could get somehow close to it...) Thanks again! – Romeo Jan 13 '18 at 17:29 • @user-unknown: For piecewise constant $\alpha$ you just need to stitch together the constant solution by matching the end point function values and its first derivatives to obtain $c_1$ and $c_2$, if you want the solution to be $C^1$. – Hans Jan 13 '18 at 19:34 I personally do not know much about these types of equations but for $\gamma$ = 1 this equation is an example of a Volterra integro-differential equation which has been heavily studied. • I thought a Volterra equation was a convolution equation, $y(t)=\alpha(t)+\int_0^t K(t-\tau)y(\tau)d\tau$ --- this doesn't seem to be of that form for any $\gamma$. – Carlo Beenakker Jan 14 '18 at 11:26	2020-04-02 00:51:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8693733811378479, "perplexity": 242.1814564043638}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370506477.26/warc/CC-MAIN-20200401223807-20200402013807-00017.warc.gz"}
http://wave.cmap.polytechnique.fr/spip.php?rubrique86	## Rechercher sur ce site Accueil du site > Résumés des séminaires > CO > Résumés ## Recent progress in the global complexity analysis of the second-order methods Yu. Nesterov (CORE) - 25 septembre 2007 In this talk we discuss an accelerated version of cubic regularization of Newton’s method. The original version, used for minimizing a convex function with Lipschitz-continuous Hessian, guarantees a global rate of convergence of order $O(1 \over k2)$, where $k$ is the iteration counter. Our modified version converges for the same problem class with order $O(1 \over k3)$, keeping the complexity of each iteration unchanged. We study the complexity of both schemes on different classes of convex problems. In particular, we argue that for the second-order schemes, the class of non-degenerate problems is different from the standard class.	2017-06-28 00:09:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6982536911964417, "perplexity": 675.7790241325787}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128321961.50/warc/CC-MAIN-20170627235941-20170628015941-00082.warc.gz"}
http://www.thefullwiki.org/Operand	# Operand: Wikis Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles. # Encyclopedia An operand is a quantity on which an operation is performed.[1] The following arithmetic expression shows an example of operators and operands: 3 + 6 = 9 In the above example, '+' is the symbol for the operation called addition. The operand '3' is one of the inputs (quantities) followed by the addition operator, and the operand '6' is the other input necessary for the operation. The result of the operation is 9. (The number '9' is also called the sum of the addend, 3 and 6.) An operand, then, is also referred to as "one of the inputs (quantities) for an operation". ## Notation ### Expressions as operands Operands may be complex, and may consist of expressions also made up of operators with operands. (3 + 5) * 2 In the above expression '(3 + 5)' is the first operand for the multiplication operator and '2' the second. The operand '(3 + 5)' is an expression in itself, which contains an addition operator, with the operands '3' and '5'. A more common synonym is a variable. ### Order of operations Rules of precedence affect which values form operands for which operators: 3 + 5 * 2 In the above expression, the multiplication operator has the higher precedence than the addition operator, so the multiplication operator has operands of '5' and '2'. The addition operator has operands of '3' and '5 x 2'. ### Positioning of operands Depending on the mathematical notation being used the position of an operator in relation to its operand(s) may vary. In everyday usage infix notation is the most common, however other notations also exist, such as the prefix and postfix notations. These alternate notations are most common within computer science. Below is a comparison of three different notations — all represent an addition of the numbers '1' and '2' 1 + 2 (infix notation) $+ 1~2$ (prefix notation) $1~2 +$ (postfix notation) ### Infix Notation and the Order of Operation With infix notation, one easy mnemonic for remembering the order of operation is: Please excuse my dear Aunt Sally. The first letter (in boldtype) of each word in the above mnemonic stands for the following: p = parentheses e = exponents m = multiplication d = division s = subtraction In a mathematical expression, the order of operation is carried out from left to right. Start with the left most value and seek the first operation to be carried out in accordance with the order specified above (i.e., start with parentheses and end with subtraction). For example, in the expression 4 * 22 − (2 + 22), the first operation to be acted upon is any and all expressions found inside a parenthesis. So beginning at the left and moving to the right, find the first (and in this case, the only) parenthesis, that is, (2 + 22). Within the parenthesis itself is found the expression 22. The reader is required to find the value of 22 before going any further. The value of 22 is 4. Having found this value, the remaining expression looks like this: 4 * 22 − (2 + 4) The next step is to calculate the value of expression inside the parenthesis itself, that is, (2 + 4) = 6. Our expression now looks like this: 4 * 22 − 6 Having calculated the parenthetical part of the expression, we start over again beginning with the left most value and move right. The next order of operation (according to the rules) is exponents. Start at the left most value, that is, 4, and scan your eyes to the right and search for the first exponent you come across. The first (and only) expression we come across that is expressed with an exponent is 22. We find the value of 22, which is 4. What we have left is the expression 4 * 4 − 6. The next order of operation is multiplication. 4 x 4 is 16. Now our expression looks like this: 16 − 6 The next order of operation according to the rules is division. However, there is no division operator sign (/) in the expression, 16 -6. So we move on to the next order of operation, i.e., addition. But there is no addition operator sign (+) in the expression 16 - 6. So we move on to the next and final order of operation, which is subtraction. 16 − 6 = 10. So the correct value for our original expression, 4 x 22 - (2 + 22), is 10. It is important to carry out the order of operation in accordance with rules set by convention. If the reader evaluates an expression but does not follow the correct order of operation, the reader will come forth with a different value. The different value will be the incorrect value because the order of operation was not followed. The reader will arrive at the correct value for the expression if and only if each operation is carried out in the proper order. ### Arity The number of operands of an operator is called its arity. Based on arity, operators are classified as nullary (no operands), unary (1 operand), binary (2 operands), ternary (3 operands) etc. ## Computer science In computer programming languages, the definitions of operator and operand are almost the same as in mathematics. Additionally, in assembly language, an operand is a value (an argument) on which the instruction, named by mnemonic, operates. The operand may be a processor register, a memory address, a literal constant, or a label. A simple example (in the x86 architecture) is MOV DS, AX where the value in register operand 'AX' is to be moved into register 'DS'. Depending on the instruction, there may be zero, one, two, or more operands. ## References 1. ^ American Heritage Dictionary # Gaming Up to date as of February 01, 2010 ### From Wikia Gaming, your source for walkthroughs, games, guides, and more! In computer programming, an operand is like a noun, and has a property used by an operator to produce some new result (which itself can be another operand!). Operands are operated on by operators.	2018-10-23 22:13:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.660876452922821, "perplexity": 898.5801281030231}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583517495.99/warc/CC-MAIN-20181023220444-20181024001944-00168.warc.gz"}
http://ima.ucv.cl/beyond/	# Location and Transport The conference will be held at ”Hostería El Copihue”, Olmué, V Región, Chile. The web page at Hostería El Copihue is www.copihue.cl. The nearest airport to Olmué is Santiago International Airport (SCL). From Santiago airport to Olmué: First, take a bus ride, either with the “Centropuerto” or “Turbus” company to the bus station Terminal Alameda. There is such a bus every 10 East Jump minutes. The ride to Terminal Alameda should take around 20 minutes and the price should be about 1500 CLP. From Valparaíso and Viña del Mar: You can take the local metro service (http://www.merval.cl/) to Limache. Then, you can reach Olmué by local bus or taxi. # Global dynamics beyond uniform hyperbolicity The conference will focus on the global qualitative study (topological and ergodic) of differentiable dynamical systems, especially Gonfiabili diffeomorphisms and vector fields, that are not uniformly hyperbolic. Among the main topics to be covered are: • Beyond Hyperbolicity • Partial hyperbolicity • Nonuniform hyperbolicity • Singular hyperbolicity • Lyapunov exponents • Dimension • Statistical properties • Global and Semilocal Properties • The $C^1$-topology • Bifurcations • Other Topics • Interval Dynamics • Group actions • Foliations Our aim is to form a group of about 110 mathematicians, including some of the main international actors in these subjects, as well as young mathematicians and students. We plan to have ten days to schedule several mini-courses and a few plenary lectures, so that a large part of the time will be devoted to informal discussions. 2 to 3 lecture rooms will be available to the participants for more specialized sessions.	2022-05-21 19:48:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24576565623283386, "perplexity": 9529.774608764255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662540268.46/warc/CC-MAIN-20220521174536-20220521204536-00546.warc.gz"}
https://space.stackexchange.com/questions/37743/who-was-the-first-person-to-shave-in-space	# Who was the first person to shave in space? A quick internet search shows that Apollo 11 astronauts shaved, but there were some Apollo missions earlier than that were also of similar week-like duration and the Soviet Union had astronauts in space as well. Is it known who was the first person to shave in space (i.e. in Earth orbit or beyond)? • +1 for finding such an authoritative source on shaving in space! "...there is actually at least 30 minutes’ worth of presentation material on the topic of shaving in spaceflight." Great answer and great link! – uhoh Jul 29 '19 at 22:10	2020-05-28 02:32:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43298667669296265, "perplexity": 1156.7980663329913}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347396300.22/warc/CC-MAIN-20200527235451-20200528025451-00430.warc.gz"}
https://solvedlib.com/n/find-all-values-of-a-such-thats-4-301is-a-basis-of-r,20539760	# Find all values of a such thats-{[4] [ 301is a basis of R?_ ###### Question: Find all values of a such that s-{[4] [ 301 is a basis of R?_ #### Similar Solved Questions ##### Construct the formula for the polynomial P (x)with smallest possible degree whose graph is given below:In factored form, P (x) Construct the formula for the polynomial P (x)with smallest possible degree whose graph is given below: In factored form, P (x)... ##### Answer the following all part of one question 4. You are a financial manager of a... Answer the following all part of one question 4. You are a financial manager of a firm and are asked to assess the cost of capital of your firm. You know that Your firm is going to pay dividend $1 per share in the next year. The current stock price is$10 per share Firm beta is 20% higher than marke... ##### 1) Each item produced by a certain manufacturer is, independently; of acceptable quality with probability 0.95_ We want t0 find the probability that at most 10 of the next 150 items produced are unacceptable. If X is the number of unacceptable items among the 150, What distribution does X have? (4 points) b) What can the distribution be approximated by? Use this to approximate the probability that at most 10 of the next 150 items produced are unacceptable (5 points) 1) Each item produced by a certain manufacturer is, independently; of acceptable quality with probability 0.95_ We want t0 find the probability that at most 10 of the next 150 items produced are unacceptable. If X is the number of unacceptable items among the 150, What distribution does X have? (4 p... ##### 1. (2 points) Computecos(70) cos(20)d0.Hint: Use the identitycos( A) cos(B) 2 [cos(A _ B) + cos(A+ B)]2. 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Whe bod block slides down an inclined plane, prove al aid your... ##### Produced from translation?question 30) A DNA sequence encoding five-amino acid polypeptide is (Textbook chapter given below:03 !(u) 42 AMJ y hing Veq AMAm Ro M ~ACGGCAAGATCCCACCCTAATCAGACCGTACCATTCACCTCCT _ t (ru ruoWak; u0s TGCCGTTCTAGGGTGGGATTAGTCTGGCATGGTAAGTGGAGGA _ \| domittiXo iaohjo) +70 , Locate the sequence encoding the five amino acids of the polypeptide, and coding strands of' DNA identify the template and , No / p ,` Give the sequence and polarity of the mRNA encoding the pol produced from translation? question 30) A DNA sequence encoding five-amino acid polypeptide is (Textbook chapter given below: 03 !(u) 42 AMJ y hing Veq AMAm Ro M ~ACGGCAAGATCCCACCCTAATCAGACCGTACCATTCACCTCCT _ t (ru ruoWak; u0s TGCCGTTCTAGGGTGGGATTAGTCTGGCATGGTAAGTGGAGGA _ \| domittiXo i aohjo) +70 ... ##### 2 . EvaluateIsc+ 1) ds where C is the curve composed of two parts: C1 travels from (0,0) to (1,1) along the curve y = 22 C2 travels from (1,1) to (0,0 along the curve y = x 2 . Evaluate Isc+ 1) ds where C is the curve composed of two parts: C1 travels from (0,0) to (1,1) along the curve y = 22 C2 travels from (1,1) to (0,0 along the curve y = x... ##### The Highway Inspector problem was first solved by Leonhard Euler in 1735 when thinking about & puzzle problem that he described as follows: "In the town of Konigsberg there is an island called Kneiphof; with two branches of the river Pregel flowing around it: There are seven bridges crossing the two branches The question is whether a person can plan a walk in such a way that he will cross each of these bridges once but not more than once The Highway Inspector problem was first solved by Leonhard Euler in 1735 when thinking about & puzzle problem that he described as follows: "In the town of Konigsberg there is an island called Kneiphof; with two branches of the river Pregel flowing around it: There are seven bridges crossin... ##### O words Question 40 10 pts Most of the ATP from metabolism is formed during the... O words Question 40 10 pts Most of the ATP from metabolism is formed during the Respiratory Chain or oxidative phosphorylation. • Follow the path of electrons and protons from NADH (or FADH2) through oxidative phosphorylation. Be sure to highlight the cytochromes, starting substrates, enzymes, ... ##### Determine the increase or decrease in retained earnings for August UIVIVUTUS MISLUUD CAPUT 650 2,500 0... determine the increase or decrease in retained earnings for August UIVIVUTUS MISLUUD CAPUT 650 2,500 0 3. HERITAGE REALTY Unadjusted Trial Balance August 31, 2016 Credit -0 -0 Cash Supplies Accounts Payable Common Stock Dividends Sales Commissions Rent Expense Office Salaries Expense Automobile ... ##### We as humans write math expression in infix notation, e.g. 5 + 2 (the operators are... We as humans write math expression in infix notation, e.g. 5 + 2 (the operators are written in-between the operands). In a computer’s language, however, it is preferred to have the operators on the right side of the operands, i.e. 5 2 +. For more complex expressions that include parenthesis an... ##### Section (40") = Long Qucstions Answcr any EQLR out of the FIVE questions in this section the inswer script provided: unttetco morc than four quotions, only the lirst four attempted questions will be murked: Indicate in your answer script clearly whkch four questions attemnting Each question carrics I0 marksQuetinClTi Iable below sumnurises Ikc distribution of the heights of rrlomly sclecled students ARC CollegeOlas: Midpoint (LFnuucncvaCorrect Your final nswers f0 decimal place whenever a Section (40") = Long Qucstions Answcr any EQLR out of the FIVE questions in this section the inswer script provided: unttetco morc than four quotions, only the lirst four attempted questions will be murked: Indicate in your answer script clearly whkch four questions attemnting Each question c... ##### > A Moving to another question will save this response. Question 7 of 15 >> Question... > A Moving to another question will save this response. Question 7 of 15 >> Question 7 5 points Save Answer If a system consists of 2 major subsystems and the required reliability is 0.9, what is the allocated reliability of each subsystem, assuming equal apportionment? OA. 0.45 B. 0.95 OC.... ##### Vx = Vin 2. (60 points) Plot Vout as a function of Vx for the circuit... Vx = Vin 2. (60 points) Plot Vout as a function of Vx for the circuit shown below. Assume a constant voltage diode model. Also plot the Vout as a function of Vx in the circuit assuming Vx Vo sin ot D1 VB... ##### Point) Suppose the angle satisfies 0 < A < 2n _ If cOs(A) = -0.87 and sin(A) > 0, determine:The quadrant for the angle A/2Thensin(A)sin(A/2)COs(A/2)tan(A/2)Be certain t0 express all answers to at least four decimal places: Also, be careful regarding the signs for each of your answers: point) Suppose the angle satisfies 0 < A < 2n _ If cOs(A) = -0.87 and sin(A) > 0, determine: The quadrant for the angle A/2 Then sin(A) sin(A/2) COs(A/2) tan(A/2) Be certain t0 express all answers to at least four decimal places: Also, be careful regarding the signs for each of your answers... ##### 4. Continuing with problem 3, f(x) = 2x? + 8x + 5. a) Find the x... 4. Continuing with problem 3, f(x) = 2x? + 8x + 5. a) Find the x and y intercepts. b) Give the domain of the function. c) Give the range of the function. d) Sketch the graph of the function.... ##### The answer above is NOT correct(1 point) The set B = (2x" _ 3, 3x - 6+4r , 10 - 6x - 6x" } is a basis tor Pz Find the coordinates of p(x) = 6x - 2 + 2x" relative t0 this basis:[plx)lePreviowMy AnswensSubmit Answers The answer above is NOT correct (1 point) The set B = (2x" _ 3, 3x - 6+4r , 10 - 6x - 6x" } is a basis tor Pz Find the coordinates of p(x) = 6x - 2 + 2x" relative t0 this basis: [plx)le PreviowMy Answens Submit Answers... ##### Question 34 (1 point) Which of the following is not an example of extinction Your significant... Question 34 (1 point) Which of the following is not an example of extinction Your significant other stops buying you flowers because you stop thanking them after you receive the gift Pavlov's dog stops salivating after the bell is no longer paired with meat powder Wearing sunglasses on a sunny d... ##### If a = 6 and b = 5 find parametric equa tians for the curve that consists of all possible positions of the point in the figure, using the angle as the parameter.x(e)0 < 0 < ZtEliminate the parameter.Identify the curve_ parabolaarchcirclehyperbolaellipselineY(0) If a = 6 and b = 5 find parametric equa tians for the curve that consists of all possible positions of the point in the figure, using the angle as the parameter. x(e) 0 < 0 < Zt Eliminate the parameter. Identify the curve_ parabola arch circle hyperbola ellipse line Y(0)... ##### Set up the definite integral that gives the area of the region. (GRAPH CAN'T COPY) \begin{aligned} &y_{1}=(x-1)^{3}\\ &y_{2}=x-1 \end{aligned} Set up the definite integral that gives the area of the region. (GRAPH CAN'T COPY) \begin{aligned} &y_{1}=(x-1)^{3}\\ &y_{2}=x-1 \end{aligned}... ##### Cambridge Business Putih cras . The current service , which requires employees to download photos from... Cambridge Business Putih cras . The current service , which requires employees to download photos from customer came Bartera Seattle area drugstore , operates an in - store printing service for customers with cena monthly operating costs of \$ 5.500 plus SO . 15 per photo printed . Management is eval... ##### Find the gradient ∇φ of the following: a) φ = (r2/a2)e-r/a (using spherical coordinates) b) φ... Find the gradient ∇φ of the following: a) φ = (r2/a2)e-r/a (using spherical coordinates) b) φ = 2√(x2+y2+z2) (using both cartesian and spherical coordinates, after converting)... ##### △ABC is a right triangle with right angle C. Side AC is 6 units longer than... △ABC is a right triangle with right angle C. Side AC is 6 units longer than side BC . If the hypotenuse has length 52–√ units, find the length of AC. courseware-Google Chrome a https://www.casa.uh.edu /Root/Pages/CW aspx?id 643857CE-8CB9-4B21-AC4B-1AD73C216A8D CourseWare Qu... ##### Unett cIue Fron U1 Llleof x manuul (ood ploccesors @ Eten b %tr} 02 ! dollam ! The tctal E coxo Riinbycur) C,lr _22+ Esiue nurats 0* nreneno Necgalelelalolathepnirrauainpunc Lrrl interval for which i E, derreasing Exprass your inswerin cpent Imonaks:ElnrESLaaehnm ltadenalatoneniecquluitncatnrNove'ingeisn _OmaanNluce Lrdeisir Unett cIue Fron U1 Llleof x manuul (ood ploccesors @ Eten b %tr} 02 ! dollam ! The tctal E coxo Riinbycur) C,lr _22+ Esiue nurats 0* nreneno Necgalelelalolathepnirrauainpunc Lrrl interval for which i E, derreasing Exprass your inswerin cpent Imonaks: Elnr ES Laaehnm ltadenalaton eniec quluit ncatnr ... ##### Wild horses can have short or long manes and brown or green eyes. Mane phenotype is... Wild horses can have short or long manes and brown or green eyes. Mane phenotype is controlled by a single gene with two alleles, M (long mane) and m (short mane), where M is dominant to m. The eye phenotype is also controlled by a single gene with two alleles, B (brown eyes) and b (green eyes), whe... ##### CONCEPT QUESTION Predict the Product. Provide the stable organic product(s) for each reaction below. 1. NaBH4,... CONCEPT QUESTION Predict the Product. Provide the stable organic product(s) for each reaction below. 1. NaBH4, CH3OH 2. HCI, H20 1. NaBHA (0.50 eq.), CH3OH 2. HCI, H2O 1. LiAIHA (1 eq.), THE 2. HCI, H2O 1. NaBH(CH3)3, (1 eq.), CH3OH 2. HCI, H2O... ##### VolumaMsVJEV MM F/ (xh )Caelle alen [eana Lttn CLdt e Tilol Bra 6r Lc Lli 54t74 Ti74 El Eicl (4L Goca Mt) 57,eunal Iun pf Tntth Ftutullty 'Petahility LaatActuarial Kodel: LYe Ingurance and Arruny I zHGAeut atl~relte nrt 47ra Eetaa 6474+u ueturt 044 40 Voluma MsVJEV MM F/ (xh ) Caelle alen [eana Lttn CLdt e Tilol Bra 6r Lc Lli 54t74 Ti74 El Eicl (4L Goca Mt) 57, eunal Iun pf Tntth Ftutullty 'Petahility Laat Actuarial Kodel: LYe Ingurance and Arruny I z HGAeut atl ~relte nrt 47ra Eetaa 6474+u ueturt 044 40... ##### Verify that each equation is an identity.1) cot( Ï€ - Î¸ ) = - cot Î¸ Verify that each equation is an identity. 1) cot( Ï€ - Î¸ ) = - cot Î¸... ##### A person of mass 96 kg rides on a Ferris wheel whose radius is 5.2 m.... A person of mass 96 kg rides on a Ferris wheel whose radius is 5.2 m. The person's speed is constant at 0.5 m/s. The person's location is shown by a dot in the figure below. (a) What is the magnitude of the rate of change of the momentum of the person at the instant shown? (b) What is the di... ##### In the Lewis dot structure for NH_3, the central atom of the molecule has three atoms bonded to it and one lone pair of electrons. What is the shape of an NH_3 molecule? In the Lewis dot structure for NH_3, the central atom of the molecule has three atoms bonded to it and one lone pair of electrons. What is the shape of an NH_3 molecule?... ##### Express the given rational function as the sum of a polynomial and another rational function whose numerator is either zero or has smaller degree than the denominator.$rac{x^{4}+x^{2}}{x^{3}+x^{2}+1}$ Express the given rational function as the sum of a polynomial and another rational function whose numerator is either zero or has smaller degree than the denominator. $\frac{x^{4}+x^{2}}{x^{3}+x^{2}+1}$... ##### Divide. $295 \div 35$ Divide. $295 \div 35$...	2023-01-29 18:14:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5954885482788086, "perplexity": 5715.20093578893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00404.warc.gz"}
http://pkgsrc.se/devel/git-base	./devel/git-base, GIT Tree History Storage Tool (base package) ### [ CVSweb ] [ Homepage ] [ RSS ] [ Required by ] [ Add to tracker ] Branch: CURRENT, Version: 2.13.2, Package name: git-base-2.13.2, Maintainer: pkgsrc-users Git is a free and open source distributed version control system designed to handle everything from small to very large projects with speed and efficiency. Git is easy to learn and has a tiny footprint with lightning fast performance. It outclasses SCM tools like Subversion, CVS, Perforce, and ClearCase with features like cheap local branching, convenient staging areas, and multiple workflows. This package contains only the git program (and subcommands). It does not contain man pages or the tk-based repository browser. Required to run: [www/curl] [lang/perl5] [security/p5-Authen-SASL] [mail/p5-Email-Valid] [mail/p5-MailTools] [devel/p5-Error] [mail/p5-Net-SMTP-SSL] Required to build: [pkgtools/cwrappers] ### Master sites: RMD160: a9a2751eb0bcf6d05e734bcb3ef3b67c351e8455 Filesize: 4640.02 KB ### Version history: (Expand) • (2017-06-25) Updated to version: git-base-2.13.2 • (2017-06-08) Updated to version: git-base-2.13.1nb1 • (2017-06-05) Updated to version: git-base-2.13.1 • (2017-05-25) Updated to version: git-base-2.13.0nb2 • (2017-05-24) Updated to version: git-base-2.13.0nb1 • (2017-05-11) Updated to version: git-base-2.13.0 ### CVS history: (Expand) 2017-06-25 05:15:11 by Makoto Fujiwara \| Files touched by this commit (3) \| Log message: Updated devel/git-{base,contrib,docs,git-gitk} to 2.13.2 -------------------------------------------------------- Git v2.13.2 Release Notes ========================= Fixes since v2.13.1 ------------------- * The "collision detecting" SHA-1 implementation shipped with 2.13.1 was still broken on some platforms. Update to the upstream code again to take their fix. * "git checkout --recurse-submodules" did not quite work with a submodule that itself has submodules. * Introduce the BUG() macro to improve die("BUG: ..."). * The "run-command" API implementation has been made more robust against dead-locking in a threaded environment. * A recent update to t5545-push-options.sh started skipping all the tests in the script when a web server testing is disabled or unavailable, not just the ones that require a web server. Non HTTP tests have been salvaged to always run in this script. * "git clean -d" used to clean directories that has ignored files, even though the command should not lose ignored ones without "-x". "git status --ignored" did not list ignored and untracked files without "-uall". These have been corrected. * The timestamp of the index file is now taken after the file is closed, to help Windows, on which a stale timestamp is reported by fstat() on a file that is opened for writing and data was written but not yet closed. * "git pull --rebase --autostash" didn't auto-stash when the local history fast-forwards to the upstream. * "git describe --contains" penalized light-weight tags so much that they were almost never considered. Instead, give them about the same chance to be considered as an annotated tag that is the same age as the underlying commit would. * The result from "git diff" that compares two blobs, e.g. "git diff $commit1:$path $commit2:$path", used to be shown with the full object name as given on the command line, but it is more natural to use the \$path in the output and use it to look up .gitattributes. * A flaky test has been corrected. * Help contributors that visit us at GitHub. * "git stash push " did not work from a subdirectory \ at all. Bugfix for a topic in v2.13 Also contains various documentation updates and code clean-ups. (pkgsrc-changes) ---------------- drop patch git-base/patches/patch-sha1dc_sha1.c, see the first paragraph of above RelNote 2017-06-25 05:09:45 by Makoto Fujiwara \| Files touched by this commit (1) \| Log message: Remove PKGREVISION for preparing update 2.13.1 to 2.13.2 2017-06-08 11:08:51 by Thomas Klausner \| Files touched by this commit (3) \| Log message: Fix endianness issue using upstream commit https://github.com/git/git/commit/6b851 … cedea39ff8 Bump PKGREVISION. 2017-06-05 12:11:39 by Adam Ciarcinski \| Files touched by this commit (4) \| Log message: Git v2.13.1 Release Notes ========================= Fixes since v2.13 ----------------- * The Web interface to gmane news archive is long gone, even though the articles are still accessible via NTTP. Replace the links with ones to public-inbox.org. Because their message identification is based on the actual message-id, it is likely that it will be easier to migrate away from it if/when necessary. * Update tests to pass under GETTEXT_POISON (a mechanism to ensure that output strings that should not be translated are not translated by mistake), and tell TravisCI to run them. * Setting "log.decorate=false" in the configuration file did not take effect in v2.13, which has been corrected. * An earlier update to test 7400 needed to be skipped on CYGWIN. * Git sometimes gives an advice in a rhetorical question that does not require an answer, which can confuse new users and non native speakers. Attempt to rephrase them. * "git read-tree -m" (no tree-ish) gave a nonsense suggestion "use --empty if you want to clear the index". With "-m", such a request will still fail anyway, as you'd need to name at least one tree-ish to be merged. * The codepath in "git am" that is used when running "git \ rebase" leaked memory held for the log message of the commits being rebased. * "pack-objects" can stream a slice of an existing packfile out when the pack bitmap can tell that the reachable objects are all needed in the output, without inspecting individual objects. This strategy however would not work well when "--local" and other options are in use, and need to be disabled. * Clarify documentation for include.path and includeIf..path configuration variables. * Tag objects, which are not reachable from any ref, that point at missing objects were mishandled by "git gc" and friends (they should silently be ignored instead) * A few http:// links that are redirected to https:// in the documentation have been updated to https:// links. * Make sure our tests would pass when the sources are checked out with "platform native" line ending convention by default on Windows. Some "text" files out tests use and the test scripts themselves that are meant to be run with /bin/sh, ought to be checked out with eol=LF even on Windows. * Fix memory leaks pointed out by Coverity (and people). * The receive-pack program now makes sure that the push certificate records the same set of push options used for pushing. * "git cherry-pick" and other uses of the sequencer machinery mishandled a trailer block whose last line is an incomplete line. This has been fixed so that an additional sign-off etc. are added after completing the existing incomplete line. * The shell completion script (in contrib/) learned "git stash" has a new "push" subcommand. * Travis CI gained a task to format the documentation with both AsciiDoc and AsciiDoctor. * Update the C style recommendation for notes for translators, as recent versions of gettext tools can work with our style of multi-line comments. * "git clone --config var=val" is a way to populate the per-repository configuration file of the new repository, but it did not work well when val is an empty string. This has been fixed. * A few codepaths in "checkout" and "am" working on an \ unborn branch tried to access an uninitialized piece of memory. * "git for-each-ref --format=..." with %(HEAD) in the format used to resolve the HEAD symref as many times as it had processed refs, which was wasteful, and "git branch" shared the same problem. * "git interpret-trailers", when used as GIT_EDITOR for "git commit -v", looked for and appended to a trailer block at the very end, i.e. at the end of the "diff" output. The command has been corrected to pay attention to the cut-mark line "commit -v" adds to the buffer---the real trailer block should appear just before it. * A test allowed both "git push" and "git receive-pack" on \ the other end write their traces into the same file. This is OK on platforms that allows atomically appending to a file opened with O_APPEND, but on other platforms led to a mangled output, causing intermittent test failures. This has been fixed by disabling traces from "receive-pack" in the test. * "foo\bar\baz" in "git fetch foo\bar\baz", even though \ there is no slashes in it, cannot be a nickname for a remote on Windows, as that is likely to be a pathname on a local filesystem. * The "collision detecting" SHA-1 implementation shipped with 2.13 was quite broken on some big-endian platforms and/or platforms that do not like unaligned fetches. Update to the upstream code which has already fixed these issues. * "git am -h" triggered a BUG(). * The interaction of "url..insteadOf" and custom URL scheme's whitelisting is now documented better. 2017-05-25 05:04:05 by Noriyuki Soda \| Files touched by this commit (3) Log message: git-base-2.13.0nb1 broke all little endian platforms on BSD. (NOTE: _BIG_ENDIAN is always defined even on litte endian platforms) from nonaka@ 2017-05-24 16:41:35 by Martin Husemann \| Files touched by this commit (1) Log message: Ooops, missed to add this in previous 2017-05-24 14:59:01 by Martin Husemann \| Files touched by this commit (2) Log message: Make it work on platforms requiring strict alignement again. Patch from upstream, pointed out by maya@. 2017-05-10 20:09:25 by Adam Ciarcinski \| Files touched by this commit (4) \| Log message: Git 2.13 Release Notes ====================== Backward compatibility notes. * Use of an empty string as a pathspec element that is used for 'everything matches' is still warned and Git asks users to use a more explicit '.' for that instead. The hope is that existing users will not mind this change, and eventually the warning can be turned into a hard error, upgrading the deprecation into removal of this (mis)feature. That is not scheduled to happen in the upcoming release (yet). * The historical argument order "git merge HEAD \ ..." has been deprecated for quite some time, and is now removed. * The default location "~/.git-credential-cache/socket" for the socket used to communicate with the credential-cache daemon has been moved to "~/.cache/git/credential/socket". * Git now avoids blindly falling back to ".git" when the setup sequence said we are _not_ in Git repository. A corner case that happens to work right now may be broken by a call to die("BUG"). We've tried hard to locate such cases and fixed them, but there might still be cases that need to be addressed--bug reports are greatly appreciated. Updates since v2.12 ------------------- UI, Workflows & Features * "git describe" and "git name-rev" have been taught to \ take more than one refname patterns to restrict the set of refs to base their naming output on, and also learned to take negative patterns to name refs not to be used for naming via their "--exclude" option. * Deletion of a branch "foo/bar" could remove .git/refs/heads/foo once there no longer is any other branch whose name begins with "foo/", but we didn't do so so far. Now we do. * When "git merge" detects a path that is renamed in one history while the other history deleted (or modified) it, it now reports both paths to help the user understand what is going on in the two histories being merged. * The part in "http.." \ configuration variable can now be spelled with '' that serves as wildcard. E.g. "http.https://.example.com.proxy" can be used to specify the proxy used for https://a.example.com, https://b.example.com, etc., i.e. any host in the example.com domain. * "git tag" did not leave useful message when adding a new entry to reflog; this was left unnoticed for a long time because refs/tags/* doesn't keep reflog by default. * The "negative" pathspec feature was somewhat more cumbersome to use than necessary in that its short-hand used "!" which needed to be escaped from shells, and it required "exclude from what?" specified. * The command line options for ssh invocation needs to be tweaked for some implementations of SSH (e.g. PuTTY plink wants "-P " while OpenSSH wants "-p " to specify port to connect \ to), and the variant was guessed when GIT_SSH environment variable is used to specify it. The logic to guess now applies to the command specified by the newer GIT_SSH_COMMAND and also core.sshcommand configuration variable, and comes with an escape hatch for users to deal with misdetected cases. More...	2017-06-29 00:34:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3066784143447876, "perplexity": 6883.036178514845}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323808.56/warc/CC-MAIN-20170629000723-20170629020723-00119.warc.gz"}
https://indico.cern.ch/event/760666/contributions/3390855/	# CEC-ICMC 2019 - Abstracts, Timetable and Presentations Jul 21 – 25, 2019 Connecticut Convention Center, Level 6 US/Eastern timezone ## C3Or1A-03: Quench spot detection for SRF cavities via flow visualization in superfluid helium-4 Jul 24, 2019, 11:30 AM 15m Level 6, Room 22-23 ### Level 6, Room 22-23 Contributed Oral Presentation ### Speaker Dr Shiran Bao (National High Magnetic Field Laboratory) ### Description Superconducting ratio-frequency (SRF) cavities, cooled by superfluid helium-4 (He II), are key components in modern particle accelerators. Quenches in SRF cavities caused by Joule heating from local surface defects can severely limit the maximum achievable accelerating field. Existing methods for quench spot detection include temperature mapping and second-sound triangulation. These methods are useful, but all have known limitations. Here we describe a new method for surface quench spot detection by visualizing the heat transfer in He II via tracking $He_2^$ molecular tracer lines. A proof-of-concept experiment has been conducted, in which a miniature heater mounted on a plate was pulsed on at various heat fluxes and pulse durations to simulate a surface quench spot. A $He_2^$ tracer line created nearby the heater deforms due to the counterflow heat transfer in He II. By analysing the tracer-line deformation, we can well reproduce the heater location within a few hundred microns, which clearly demonstrates the feasibility of this visualization-based non-contacting quench spot detection technology. Our analysis also reveals that the heat content transported in He II is only a small fraction of the total input heat energy. The remaining heat energy is essentially consumed in the formation of a cavitation zone surrounding the heater. The size of this cavitation zone is estimated based on the knowledge obtained about the transported heat. This information has allowed us to propose a new explanation for the decades-long puzzle observed in previous second-sound triangulation experiments regarding heat transfer at speeds higher than literature values. The excellent quantitative agreement between our predicted excess second-sound velocity and those measured in triangulation experiments provides a strong support of our model. Acknowledgements: S.B. and W.G. acknowledge support from U.S. Department of Energy under Grant No. DE-FG02-96ER40952. The experiment was conducted at the National High Magnetic Field Laboratory, which is supported by National Science Foundation Cooperative Agreement No. DMR-1644779 and the State of Florida. ### Primary author Dr Shiran Bao (National High Magnetic Field Laboratory) ### Co-author Dr Wei Guo (Florida State University, National High Magnetic Field Laboratory) ### Presentation materials There are no materials yet.	2022-07-04 10:41:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3483462333679199, "perplexity": 4212.473065258295}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00301.warc.gz"}
https://brilliant.org/problems/minimum-and-maximum/	# Minimum and maximum..? Algebra Level 4 The inequality: $a\le \frac { { x }^{ 4 }-2{ x }^{ 2 }+3 }{ { ({ x }^{ 2 }+1) }^{ 2 } } \le b$ is always true for any real number $$x.$$ If $$m$$ represents the maximum possible value of $$a,$$ and $$n$$ represents the minimum possible value of $$b,$$ what is the value of $$mn?$$ ×	2017-05-27 02:48:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32588520646095276, "perplexity": 161.6459988082337}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608765.79/warc/CC-MAIN-20170527021224-20170527041224-00493.warc.gz"}
http://mathhelpforum.com/calculus/91048-find-sume-series.html	# Thread: find the sume of a series 1. ## find the sume of a series Find the some of $S = \sum_{i = 1}^{\infty} (\frac{1}{i + 1} - \frac{1}{i + 2})$ plugging index values in. $(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} -\frac{1}{4} ) + (\frac{1}{4} -\frac{1}{5} )$ I note that by regrouping the terms cancel out. $\frac{1}{2} + (- \frac{1}{3} + \frac{1}{3} ) + (-\frac{1}{4} + \frac{1}{4} ) -\frac{1}{5} . . .$ and looks like the sum is 1/2 but I still have the tail and I have not show how it goes to 0 as n goes to infinity. how would i phrase the tail to get a limit resulting in zero? 2. $= \frac{1}{2}+\left(-\frac{1}{3} + \frac{1}{3}\right)+...+\left(-\frac{1}{n-1} + \frac{1}{n-1}\right) - \frac{1}{n}=\frac{1}{2}+\frac{1}{n}=\frac{1}{2}$ when $n \to \infty$	2016-12-05 15:36:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8546507358551025, "perplexity": 531.3456408884741}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541697.15/warc/CC-MAIN-20161202170901-00386-ip-10-31-129-80.ec2.internal.warc.gz"}
http://pipingdesigner.co/index.php/properties/classical-mechanics/2334-moment-of-inertia-of-a-triangle	# Moment of Inertia of a Triangle Written by Jerry Ratzlaff on . Posted in Classical Mechanics ## Formulas that use Triangle, Solid Plane $$\large{ I_x = \frac {l\; w^3}{12} }$$ $$\large{ I_y = \frac {l^3\; w}{12} }$$ ### Where: $$\large{ I }$$ = moment of inertia $$\large{ l }$$ = length $$\large{ w }$$ = width	2020-07-04 01:37:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9029964804649353, "perplexity": 6407.0341638067885}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655883961.50/warc/CC-MAIN-20200704011041-20200704041041-00344.warc.gz"}
https://brilliant.org/discussions/thread/summer-programs-in-mathematics-2017/	# Summer programs in Mathematics (2017) Hey everyone! Do you want to spend your summer holidays differently and immerse yourself in the wilderness of Math? A series of 5-day National level Residential Summer Programs in Mathematics, covering a range of fascinating topics like Algebra and Geometry, is being organized by the School of Vedic Mathematics in collaboration with the Chinmaya International Foundation every year since 2010 during April and May in India. The camps are open to anyone falling under the respective age groups who has a passion for Mathematics. The camps serve as a portal for honing logical reasoning and problem solving skills. These are just the right places to cherish yourself with unforgettable experiences and quench your Mathematical appetite. These summer camps, jubilant with activities and group discussions, help unify all the minds sharing the same zeal for the subject. For further details and questions, please contact Shri Vinay Nair, the Head of the School of Vedic Mathematics at contact@sovm.org. Note by Sanghamitra Anand 2 years, 10 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. MarkdownAppears as italics or _italics_ italics bold or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$	2020-01-27 05:03:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9158276319503784, "perplexity": 4387.978525192125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251694176.67/warc/CC-MAIN-20200127020458-20200127050458-00283.warc.gz"}
https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/2/lesson/2.4.1/problem/2-137	### Home > CALC > Chapter 2 > Lesson 2.4.1 > Problem2-137 2-137. Write a single expression using the Trapezoidal Rule that will approximate $A(f(x) = 2x^2 − 4x + 3$, $0 ≤ x ≤ 1$) using $5$ trapezoids of equal height. Do not evaluate the expression. You know that you’re writing an expression between $0\le x\le 1$ using $5$ trapezoids, so you know that you’ll be starting with $0$ and increasing $x$- values by increments of $\frac{1}{5}$ until you’ve created $5$ trapezoids.	2022-07-01 14:30:26	{"extraction_info": {"found_math": true, "script_math_tex": 9, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6335931420326233, "perplexity": 1034.4167878745723}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00221.warc.gz"}
https://www.ias.ac.in/describe/article/jess/130/0025	• Performance of numerical weather prediction models in predicting track of recurving cyclone Vayu over Arabian Sea during June 2019 • # Fulltext https://www.ias.ac.in/article/fulltext/jess/130/0025 • # Keywords Tropical cyclone; track; recurvature; steering; NWP model; MME. • # Abstract A tropical cyclone (TC) Vayu developed over the Arabian Sea during June, 2019. It followed a northward track from southeast Arabian Sea to northeast Arabian Sea close to Gujarat coast during 10–12 June 2019 as a very severe cyclonic storm. It skirted south Gujarat coast by recurving west-northwestwards during 13th–14th June and again made a northeastward recurvature on 16th June towards Gujarat coast. However, it weakened over Sea on 17th. There was large divergence among various models in predicting the track of TC Vayu leading to over warning for Gujarat state and also delay in dewarning leading to evacuation of people from coastal region. Hence, a study has thus been taken up to analyze the performance of various numerical weather prediction (NWP) models in forecasting the track of TC Vayu so as to find out the reason for above limitation of NWP models. Results suggest that there is a need to relook into the existing multi-model ensemble (MME) technique which outperforms individual models in track forecasting. There is also a need to improve the individual deterministic model guidance so as to suitably represent the interaction between mid-latitude westerlies with the TC and steering anticyclone by improving the initial and boundary conditions through augmented direct and remotely sensed observations over the Arabian Sea and their assimilation in NWP models. $\bf{Highlights}$ $\bullet$ The multiple interactions among the wind fields of TC Vayu, middle latitude westerlies and anticyclones over central India & Arabian Peninsula led to the unique track of Vayu with two recurvatures in its life cycle. $\bullet$ The prediction of time and point of recurvature in the track of TCs is still a challenge for the NWP models and hence the operational forecast, as models could not represent the interaction of mid-latitude westerlies with the TC and steering anticyclone over either side of the TC. $\bullet$ Comparing the average track forecast errors of different models and multi-model ensemble (MME) for the recurving TCs during 2009–2019, the MME shows minimum average track forecast error. However, the consistency in MME based track forecast decreases with increase in lead period. $\bullet$ There is a need to look into the existing MME and improve it by re-defining the best constituent members and improving the performance of individual models through augmentation of direct & remotely sensed observations, data assimilation and the physical processes in the model. • # Author Affiliations 1. India Meteorological Department, Mausam Bhavan, Lodi Road, New Delhi 110 003, India. • # Journal of Earth System Science Volume 131, 2022 All articles Continuous Article Publishing mode • # Editorial Note on Continuous Article Publication Posted on July 25, 2019 Click here for Editorial Note on CAP Mode © 2021-2022 Indian Academy of Sciences, Bengaluru.	2022-01-18 23:06:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.34102916717529297, "perplexity": 7537.526057788215}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00451.warc.gz"}
http://www.ams.org/mathscinet-getitem?mr=2019974	MathSciNet bibliographic data MR2019974 (2004j:22011) 22E40 (53C35 58J50) Leuzinger, Enrico Critical exponents of discrete groups and \$L\sp 2\$$L\sp 2$-spectrum. Proc. Amer. Math. Soc. 132 (2004), no. 3, 919–927 (electronic). Article For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.	2015-02-28 17:11:35	{"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9957826733589172, "perplexity": 5860.886166805083}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936461995.64/warc/CC-MAIN-20150226074101-00067-ip-10-28-5-156.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/clarification-about-poincares-conjecture.800470/	1. Feb 28, 2015 ### jk22 Does the Poincaré's conjecture for a three dimensional manifold involves only simple connectedness or is it meant that the first and second homotopy groups are trivial ? Since in the first case the conjecture seems to me wrong whereas in the second true. Thanks. 2. Feb 28, 2015 ### mathwonk 3. Feb 28, 2015 ### lavinia Look at the Hurewicz theorem. For a simply connected 3 manifold the second homotopy group is zero. Note that a simply connected manifold is orientable. 4. Mar 1, 2015 ### jk22 I looked on wikipedia for hurewicz but couldn't understand. I imagined a ball with a cavity it is clear that it is simply connected every loop is contractible to zero. However every sphere is not. Hence in this case the second homotopy group is not trivial whereas it is simply connected. 5. Mar 1, 2015 ### mathwonk I think Lavinia is using Poincare duality. so H^2 = H^1 = 0 under your hypotheses. But by Hurewicz, since ∏1 = 0, H^2 is isomorphic to ∏2. 6. Mar 1, 2015 ### lavinia Exactly. ------------------ Also the Poincare conjecture has been generalized to all dimensions and has been verified for homotopy spheres in all dimensions. I think Smale proved it for dimensions greater than four. Dimension 3 three remained the last hold out until Perlman's proof. It is false if one asks whether all smooth homotopy spheres in a given dimension are diffeomorphic. Last edited: Mar 1, 2015 7. Mar 1, 2015 ### jk22 The hypotheses meant here are : a connected simply connected 3 manifold without boundary ? Then we can conclude that H2=0 ? I thought i found a counterample but my suspicion is that if h2 is not zero then there is boundary Last edited: Mar 1, 2015 8. Mar 1, 2015 ### mathwonk "The hypotheses meant here are : a connected simply connected 3 manifold without boundary ? Then we can conclude that H2=0 ?" yes. 9. Mar 1, 2015 ### jk22 Of course this is not the case if it has a boundary. Im very impressed of this result that i didn't know. Thanks. Is there a similar result for the homotopy groups : Pi1 isomorphic Pi2 if it has no boundary ? Last edited: Mar 1, 2015 10. Mar 1, 2015 ### lavinia Can you give an example? No. For instance a torus has first homotopy equal to ZxZ but all other homotopy groups are zero. More generally, any manifold that is covered by Euclidean space will have zero homotopy groups except in dimension 1. Last edited: Mar 1, 2015 11. Mar 2, 2015 ### jk22 The example given above. But i think i mix up H and Pi since i don't understand what H is geometrically. The case of other homotopy group for the torus is trivial since it has dimension 2. It were not if the dimension were bigger than two. Last edited: Mar 2, 2015 12. Mar 2, 2015 ### lavinia I meant give an example of a simply connected 3 manifold with boundary with non-zero second homotopy group Your argument about the torus is mistaken. For instance the second homotopy group of the 2 sphere is Z. Further a higher dimensional manifold that is covered by Euclidean space will have all zero homotopy groups except the first. An example is the four dimensional torus. 13. Mar 2, 2015 ### jk22 I surely don't masterize homotopy so i'll learn some stuff with you if you don't mind. First i think the second homotopy group is the shrinking of eventually deformed 2-sphere in the manifold. If there is no cavity in 3d then this group is 0 ? For the torus it is impossible to put a sphere on this surface so there is no second homotopy whereas for a sphere it is the sphere it self and it is not shrinkable ? 14. Mar 2, 2015 ### lavinia You wrote "of course" it isn't true if the manifold is a boundary so I thought you might know of an example. - Not sure what you mean by "no cavity in 3d". - A torus is covered by Euclidean space, so its higher homotopy groups are all zero. The proof that I know uses the long exact homotopy sequence of a fibration. This stuff is pretty advanced algebraic topology but worth learning. - The homotopy groups of spheres are complicated. They are all zero in dimensions less than,n, and all equal to Z in dimension,n but in higher dimensions it gets complicated. Look at the table in the Wikipedia article. http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table - In homotopy theory, one studies spaces know as Eilenberg-Maclane spaces. These have the property that all of their homotopy groups are zero except in a single dimension. One can have an Eilenberg Maclane space with non-zero homotopy in any dimension. Manifolds that are covered by Euclidean space are Eilenberg-Maclane spaces with non-zero homotopy group in dimension,1. Another example other than the torus is the Klein bottle. A good exercise is to describe the covering of the Klein bottle. Last edited: Mar 2, 2015 15. Mar 2, 2015 ### jk22 I have no practice in calculating homotopy groups maybe you have good introductory books to point at ? I looked at wikipedia it is not trivial. Nevertheless i come back to this example. The construction is as follows i take a ball in 3d (hence a full sphere if we want). Then i remove from the interior a smaller ball centered. We obtain a 3-manifold that has a non connected boundary consisting of two spheres with same center. It is simply connected since (we need to see well in three dimensions) a loop can always be shrinked to a point hence pi1=0. However for spheres in this manifold if it is centered but radius between the smaller and the bigger it is not shrinkable to a point hence pi2 is not zero though i don't know what it is but probably z. 16. Mar 2, 2015 ### lavinia Yes. That is correct. What do you think would happen if the boundary of the manifold was connected i.e. if it only had one component? Then your example would not apply. Here are some questions to think about. for that case. Assume the manifold is orientable. - Must the boundary be a sphere? - Is the second homology equal to zero? 17. Mar 2, 2015 ### jk22 For your question i imagined the following : since the boundary shall be connected i link the in and out of my case with a tube. Doing this but the pi2 becomes zero again and the boundary is deformable to a sphere again ?? If i made 2 tubes from the outside to the inner sphere this would have made a torus. The problem i have is the embedding in the fourth dimension since for me a curved space is not representable in my mind ( i suppose all human being have the same representation ability ie dimension<=3 ??) For this end i imagined the following : suppose a segment in dimension 1, it has a boundary : 2 end points. If we add a dimension then we can complete the shape for example in a triangle which has no boundary. So then applying this to the shape above i thought : i parametrize a sphere with this cavity : for example X=(2+cos a)cos b cos c Y=(2+cos a)cos b sin c Z=(2+cos a)sin b a 0 to pi, b -pi/2 to pi/2, c 0 to 2pi This is a 3 manifold but up to now not curved with the disconnected boundary. Now i thought : like in the way before i 'complete' the shape in the fourth dimension sothat it is a 3-sphere : W=+\-sqrt(4-(2+cos a)^2) Then we have x^2+y^2+z^2+w^2=4 i wanted to understand if it has no boundary and how to compute pi1,2 out from the parametrization but i don't have the tools to do that. We need to change the domain of a in order to get a closed shape in some 2d projection for example a in pi/2 to 3pi/2. Maybe this makes the boundary disappear. But i still have to try some visualisation to understand better what i'm trying there. Last edited: Mar 2, 2015 18. Mar 3, 2015 ### jk22 In fact the idea was to build a manifold without boundary with pi1=0 and pi2!=0 19. Mar 3, 2015 ### lavinia Sorry but I am having a little trouble understanding what you are trying to do. If you connect the interior and exterior spheres of the ball with a cavity then you get a 3 manifold without boundary, How you connect can make a difference but in all cases I do not think that the resulting 3 manifold will be simply connected. The tube that you attach it seems would create closed loops that can not be shrunk to a point. A way to see this is to rethink what the ball with a cavity really is. It is a tube of spheres, much like a cylinder which is a tube of circles. Attaching another tube of spheres would close the tube off and thus make closed loops that can not be shrunk to a point. One possible manifold that could result from this method would be $S^2xS^1$. With other attaching maps you might get other manifolds. Your parameterization idea is very nice - but seems difficult. I will think about it more. BTW: Do you know the Poincare Duality theorem? Do you know any homology theory? Last edited: Mar 3, 2015 20. Mar 4, 2015 ### jk22 It has a boundary but now it is connected it is not 2 same center spheres. I m a thinking about looking at hatchers algebraic topology	2017-08-17 02:30:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8567363619804382, "perplexity": 712.1484331352773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886102819.58/warc/CC-MAIN-20170817013033-20170817033033-00110.warc.gz"}
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=42&t=32211	## Filling Orbitals for Hybridization $sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$ KC Navarro_1H Posts: 33 Joined: Fri Apr 06, 2018 11:04 am ### Filling Orbitals for Hybridization When given examples of filling up orbitals for hybridization, I don't really understand how they're filled using the energy charts. Could someone please clarify on this? Thank you in advance! Jacob Samuels 1E Posts: 30 Joined: Fri Apr 06, 2018 11:05 am ### Re: Filling Orbitals for Hybridization Basically, you count the amount of arrows (electrons) in the outer s and p orbitals and then divy them out to a new sp3 hybridization. So if you have an s with a full orbital (2 arrows) and a p with 2 arrows, it becomes four arrows pointing upwards (not one paired in a box with 2 up and an empty box.)	2020-12-01 00:26:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4648260474205017, "perplexity": 3579.6251244339237}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00642.warc.gz"}
https://www.esaral.com/q/find-three-different-irrational-numbers-between-the-rational-numbers-88915	# Find three different irrational numbers between the rational numbers Question: Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$. Solution: Let $x=\frac{5}{7}=0 . \overline{714285}$ and $y=\frac{9}{11}=0 . \overline{81}$ Here we observe that in the first decimal x has digit 7 and y has 8. So x < y. In the second decimal place x has digit 1. So, if we considering irrational numbers $a=0.72072007200072 \ldots$ $b=0.73073007300073 \ldots$ $c=0.74074007400074 \ldots$ We find that \$x Hence are required irrational numbers.	2023-03-30 18:55:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8934319019317627, "perplexity": 757.3964131865561}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00167.warc.gz"}
http://valasztas2016.siofok.hu/krfhg2yo/noble-gas-name-c2f8ef	The six noble gases that occur naturally are helium, neon, argon, krypton, xenon, and the radioactive radon. - argon In what year? With a great enough energy supply, the outer electrons of a noble gas can be ionized, and once the gas is ionized, it can accept electrons from other elements. A comparable set of elements are the noble metals, which display low reactivity (for metals). 1- Noble gases do not usually interact with other elements Noble gases have a low level of reactivity, which means that they basically do not interact with other elements. - component of the sun When were the remaining noble gases discovered? - helium How was this discovered? Definition of noble gas in the Definitions.net dictionary. As noted above, its presence in groundwater appears to provide a possible means of predicting earthquakes. Noble gases have had their own special name since as early as 1898. Noble gases, most often found as monatomic gases, have completely filled outer electron shells, so have no inclination to react with other elements, thus very rarely forming compounds with other elements. Noble gas is translated from German and was first used by Hugo Erdmann in 1898. Neon (Ne, atom number 10, noble gas in period 2) has the same electron configuration as O2- ion; Oxygen as in noble gas configuration has taken up 2 electrons. Look no further because you will find whatever you are looking for in here. The Noble Gases We love the noble gases. H. Bartlett studied the properties of platinum fluoride PtF 6 in the 1960s, and synthesized O 2 PtF 6.It was an epoch-making discovery in inorganic chemistry when analogous experiments on xenon, which has almost equal ionization energy (1170 kJ mol-1) to that of O 2 (1180 kJ mol-1), resulted in the dramatic discovery of XePtF 6. Between the Characteristics of noble gases Most important they are gaseous elements, do not interact with other elements, have a full valence layer, are rare in nature (their level of presence on the planet Earth is low) and create fluorescence.. ThoughtCo uses cookies to provide you with a great user experience. For the first six periods of the periodic table, the noble gases are exactly the members of group 18 of the periodic table. However, helium is an exception, because it only has two valence electrons. Radon was first identified in 1900 by German chemist Friedrich E. Dorn; it was established as a member of the noble-gas group in 1904. First to understand the terminology, ‘inert’ simply means something that is chemically inactive. Neon signs do not use just neon gas, but a mixture of different noble gases and other elements to create bright lights of different colors. A noble gas configuration of an atom consists of the elemental symbol of the last noble gas prior to that atom, followed by the configuration of the remaining electrons. The group of Noble gases Is one of 18 groups in which the periodic table is divided. Helium, neon, argon, krypton, xenon, and radon are all noble gases. what group deals with salt (perodic table)? There are six naturally occurring noble gases - helium, neon, argon, krypton, xenon, and radon. Write the Name and Symbol of the Element from the Description.The Noble Gas with the Smallest Atomic Radius. The second compound which is given is magnesia night. Periodic Table Study Guide - Introduction & History, Chemical Element Pictures - Photo Gallery, Ph.D., Biomedical Sciences, University of Tennessee at Knoxville, B.A., Physics and Mathematics, Hastings College, melting and boiling point close to each other (liquid over a narrow range). However it has a half life of 0.89 ms, after which it decays to Livermorium (Element 116). Read more about Noble Gases – … a) cesium sulfide b) strontium fluoride c) calcium nitride d) aluminum bromide. By using ThoughtCo, you accept our. The most abundant noble gas in the atmosphere is 0:56 39.6k LIKES Welcome to our website for all As a noble gas Answers. The reason why they don’t react to anything is because they have eight valence electrons, which makes them stable. It is composed of six elements, helium, neon, argon, krypton, xenon and radon. The group of noble gases is one of the 18 groups in which the periodic table is divided. Name: Argon Symbol: Ar Atomic Number: 18 Atomic Mass: 39.9 Number of Protons/Electrons: 18 Number of Neutrons: 22 Classification: Noble Gases Discovery: 1894 Discoverer: Sir … ... Sign up to become an Official NOBLE GAS Club Member. Noble gases are often used to create a safe or inert atmosphere due to their stable nature. The noble gases are a group of extremely nonreactive elements that all exist in the gas state. 2 synonyms for noble gas: argonon, inert gas. Noble gas definition is - any of a group of rare gases that include helium, neon, argon, krypton, xenon, and usually radon and that exhibit great stability and extremely low reaction rates —called also inert gas. ... Name the gas used is bright coloured advertising light works. Noble gases are characterized by: low reactivity low boiling point melting and boiling point close to each other (liquid over a narrow range) very low electronegativity high ionization energy usually colorless and odorless gases under ordinary conditions Synonyms for Noble gases in Free Thesaurus. Professionals, Teachers, Students and Kids Trivia Quizzes to … They’re also known as inert gases for this reason. Each world has more than 20 groups with 5 puzzles each. They are often considered to be inert . Becaus… These elements are noble gases, sometimes called inert gases. View Answer. noble gas. Our staff has managed to solve all the game packs and we [...] Read More "As a noble gas Answers" ThoughtCo uses cookies to provide you with a great user experience. It is still a noble gas. Although spectroscopic analysis by William Crookes confirmed that the new gas had a distinctive line-pattern, some critics disputed its elementary status. To verify your Age & Identity Please send us a photo ID & selfie to confirm DOB and Name (Feel free to cover any sensitive info) Please proceed to the next step once Completed. From the standpoint of chemistry, the noble gases may be divided into two groups: the relatively reactive krypton (ionisation energy 14.0 eV), xenon (12.1 eV), and radon (10.7 eV) on one side, and the very unreactive argon (15.8 eV), neon (21.6 eV), and heli… Only a few hundred are known to exist. Ramsay ignored them, and was soon pursuing another mystery gas. Moving down the group in the periodic table from top to bottom, the elements become more reactive. What does noble gas mean? Name a noble gas that is isoelectronic with the following ions. Please let us know if you have any Questions, comments, requests? Get to know about the uses/applications of the Noble Gases - Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) & Radon (Rn) and more with BYJU'S. Noble gas compounds are chemical compounds that include an element from the noble gases, group 18 of the periodic table. Their atomic numbers are, respectively, 2, 10, 18, 36, 54, and 86. These elements were considered to be inert gases until the 1960's, because their oxidation number of 0 prevents the noble gases from forming compounds readily. what group is most sable and what group is least sable? ... Sign up to become an Official NOBLE GAS Club Member. What noble gas has the same electron configuration as each of the ions in the following compounds? In recent years, however, this term has fallen out of favor, although you will occasionally see it in older literature. They are often considered to be inert . Superman home planet noble gas. The German noun for noble gas was Edelgas. Discovered around the end of the 19th century, the noble gases are the most stable group of the chemical elements. Ph.D., Biomedical Sciences, University of Tennessee at Knoxville, B.A., Physics and Mathematics, Hastings College. Now, in this given question, what we need to do waste really to name the noble gas that has the same corporation as each off the irons in the falling compounds. what is another NAME for noble gases? These gases all have similar properties under standard conditions: they are all odorless, colorless, monatomic gases with very low chemical reactivity. Each element is non-reactive, has high ionization energy, electronegativity near zero, and a low boiling point. This is because all the noble gases have stable electronic configurations. Noble gas definition, any of the chemically inert gaseous elements of group 8A or 0 of the periodic table: helium, neon, argon, krypton, xenon, and radon. The electronic configuration of Argon is: View Answer (a) How does the chemical reactivity of alkali metals vary on going down in group 1 of the periodic table? They are commonly called Group 18, the inert gases, the rare gases, the helium family, or the neon family. MCQ quiz on Noble Gases multiple choice questions and answers on Noble Gases MCQ questions quiz on Noble Gases objectives questions with answer test pdf for interview preparations, freshers jobs and competitive exams. Classified under: Nouns denoting substances. What amounts to a constant pursuit for humans just comes naturally to noble gases. They can be used to shield reactive chemicals from oxygen. However, helium is an exception, because it only has two valence electrons. ... Noble gas’s days of fixed atomic weight argon. Though radon is known primarily for the hazards it poses to human life and well-being, it has useful applications. These elements are noble gases, sometimes called inert gases.Atoms belonging to the noble gas group have completely filled their outer electron shells. Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. All of the noble gases have a fragile interatomic force. See more. STEP 3. Now Available: Global Argon Market 2016-2020 [USPRwire, Sun Nov 08 2015] Noble gases, also known as inert gases or aerogens, are a group of elements that exhibit chemical inertness on account of their unique atomic structure. To work backwards and find the element with a given configuration, you first find the noble gas in the Periodic Table. Their name for it – argon – derived from the Greek word for ‘idle’. She has taught science courses at the high school, college, and graduate levels. So for sodium, we make the substitution of $$\left[ \ce{Ne} \right]$$ for the $$1s^2 2s^2 2p^6$$ part of the configuration. Since you are already here then chances are that you are looking for the Daily Themed Crossword Solutions. Each element is non-reactive, has high ionization energy, electronegativity near zero, and a low boiling point. Helium is a noble gas (group 18 element) and has two valence electrons (stable electronic configuration) and is chemically inert. Atoms belonging to the noble gas group have completely filled their outer electron shells. From top to bottom: Helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). So an inert gas is simply a gas that is chemically inactive. What are Noble Gases? Noble gases synonyms, Noble gases pronunciation, Noble gases translation, English dictionary definition of Noble gases. Which of these is a noble gas? The noble gases make up the last column of elements in the periodic table. So an inert gas is simply a gas that is chemically inactive. So first compound which is given is aluminium cell fight just e a this three. Noble gases are also called rare gases or inert gases. The elements in the last column or group of the periodic table share special properties. They are known to possess extremely low chemical reactivity (hence the name inert gas). Hypernyms ("noble gas" is a kind of...): chemical element; element (any of the more than 100 known substances (of which 92 occur naturally) that cannot be separated into simpler substances and that singly or in combination constitute all matter). On the right (highlighted in orange) is the group of noble gases. They include: _____ Pick the closest noble gas to the element you are writing the configuration for that comes before that element Examples: 1. It didn't really work because there are a few other gases that are basically inert but not noble gases. First to understand the terminology, ‘inert’ simply means something that is chemically inactive. The noble gases are a group of chemical elements that make up Group 18 on the periodic table. However, just as a nobleman can be pushed into losing his dignity, getting a noble gas to react is possible. Meaning of noble gas. Are you looking for never-ending fun in this exciting logic-brain app? Among the characteristics of the most important noble gases are that they are gaseous elements, do not interact with other elements, have a full valence shell, are rare in nature (their level of presence on planet Earth is low) and create fluorescence. what are the lanthanide and the actinide series consider of ? This popularity of noble gases can be attributed to their amazing properties, which make them so useful. View Answer. what group is the most active and what group is least active and what group is least active? Name: Dan Rafferty, Josh Ernst, Greg Giannone Group: _____Date: 12/2/19 Noble Gases Which was the first noble gas to be discovered? Posted by krist on 26 March 2017, 4:17 pm. Noble gas The noble gases are the chemical elements in group 18 of the periodic table. Noble Gas Solutions has been our go-to provider for welding supplies in a very competitive market due to their ability to problem solve on short notice, provide quality goods in a timely manner and to maintain competitive pricing . Noble gases are characterized by: The lack of reactivity makes these elements useful for many applications. Professionals, Teachers, Students and … The six noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). The noble gases were originally also referred to as “inert gases,” since it was believed that they did not react with other elements to form compounds. Antonyms for Noble gases. The elements in the last column or group of the periodic table share special properties. Why are the noble gases called noble? CodyCross is an addictive game developed by Fanatee. MCQ quiz on Noble Gases multiple choice questions and answers on Noble Gases MCQ questions quiz on Noble Gases objectives questions with answer test pdf for interview preparations, freshers jobs and competitive exams. a) cesium sulfide b) strontium fluoride c) calcium nitride d) aluminum bromide. The ability to avoid reacting when provoked—to turn up one's nose and ignore lesser human foibles—is largely considered a noble trait in humans. Information and translations of noble gas in the most comprehensive dictionary definitions resource on the web. In chemistry, the term ‘noble gas’ refers to any of the six chemically inert gaseous elements in the periodic table which demonstrate similar characteristics. Xenon gets its name from the Greek word "xenos" which means "stranger or foreigner." This popularity of noble gases can be attributed to their amazing properties, which make them so useful. Rayleigh and Ramsay won Nobel Prizes in 1904 for their work. - 1894 Why was this element never noticed before? What are Noble Gases? These elements were considered to be inert gases until the 1960's, because their oxidation number of 0 prevents the noble gases from forming compounds readily. (A)Argon (B)Helium, Exists as mono-atomic molecules and are held together by weak van der Waal's forces.These van der Waal's forces increase with the increase in atomic size of the atom, and therefore, the boiling points increases from He to Rn.Hence He has least boiling point. Scientists have discovered that, since the heavier noble gas atoms are held together less strongly by electromagnetic forces than are the lighter noble gases, such as helium, the outer electrons of these heavier atoms can be removed more easily. Textbook solution for World of Chemistry, 3rd edition 3rd Edition Steven S. Zumdahl Chapter 12 Problem 23A. [1] In the periodic table, group 18 is the noble gases. When they are crossed by electricity, the noble gases produce phosphorescence. what group is the most active and what group is least active and what group is least active? The noble gases are helium, oxygen, argon, krypton, xenon, and radon, in order of their mass. While helium and neon are practically inert and are gases, the elements further down the periodic table more readily form compounds which are more easily liquefied. what group deals with salt (perodic table)? These elements are gases at ordinary room temperature and pressure. She has taught science courses at the high school, college, and graduate levels. Neon signs do not use just neon gas, but a mixture of different noble gases and other elements to create bright lights of different colors. They are ionized for use in lamps and lasers. Do the Noble Gases Form Chemical Compounds? Helium is a noble gas (group 18 element) and has two valence electrons (stable electronic configuration) and is chemically inert. Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. By using ThoughtCo, you accept our, Xenon Facts (Atomic Number 54 and Element Symbol Xe), Interesting Xenon Facts and Uses in Chemistry. The noble gases are a group of extremely nonreactive elements that all exist in the gas state. The six noble gases are found in group 18 of the periodic table. Ru - _____ Noble Gas Abbreviations How to write noble gas abbreviations: Noble gases are the elements located in group _____. The group of Noble gases Is one of 18 groups in which the periodic table is divided. Name a noble gas that is isoelectronic with the following ions. noble gas definition: 1. any of a group of gases, such as helium and neon, that do not react with other chemicals 2. any…. The six noble gases are found in group 18 of the periodic table. This means its use is probably limited. To verify your Age & Identity Please send us a photo ID & selfie to confirm DOB and Name (Feel free to cover any sensitive info) Please proceed to the next step once Completed. The application of Noble gas is it is used in gas-cooled atomic reactors as a heat transfer gas. Some scientists used to call them the inert gases. Several characteristics, aside from their placement on the periodic table, define the noble gases. What are synonyms for Noble gases? Thank you to everyone at Noble Gas Solutions!– Demy Jepson, Buyer, Fort … It is still a noble gas. Under standard conditions for temperature and pressure, all the noble gases exist in the gaseous phase. what group is most sable and what group is least sable? Except for helium, all of the names of the noble gas elements end with -on. what are the lanthanide and the actinide series consider of ? Xe is element 54. The group consists of 7 elements: helium, neon, argon, krypton, xenon, and radon. Xenon gets its name from the Greek … Do the Noble Gases Form Chemical Compounds? Although the noble gases are generally unreactive elements, many such compounds have been observed, particularly involving the element xenon. STEP 3. Answer is Low Reactivity There is a reason why noble gas atoms tend not to combine: one of the defining characteristics of the noble gas "family" is their lack of chemical reactivity. Noble gases were discovered by Lord Rayleigh and Sir William Ramsay.Rayleigh won the Nobel Prize in Physics in 1904 for his work on noble gas. - Science and Technology 1 Question By default show hide Solutions Their name for it – argon – derived from the Greek word for ‘idle’. Examples include xenon hexafluoride (XeF6) and argon fluorohydride (HArF). Noble Gas Solutions has been our go-to provider for welding supplies in a very competitive market due to their ability to problem solve on short notice, provide quality goods in a timely manner and to maintain competitive pricing . The term "noble gas" comes from the translation of the German word Edelgas. H. Bartlett studied the properties of platinum fluoride PtF 6 in the 1960s, and synthesized O 2 PtF 6.It was an epoch-making discovery in inorganic chemistry when analogous experiments on xenon, which has almost equal ionization energy (1170 kJ mol-1) to that of O 2 (1180 kJ mol-1), resulted in the dramatic discovery of XePtF 6. Discovery of noble gas Compounds. What are inert elements? Discovery of noble gas Compounds. Positive ions are isoelectronic with the noble gas at the end of the previous period. Your element has two more electrons (and two more protons) than Xe. Synonyms: argonon; inert gas; noble gas. . Nitrogen (N 2) might be considered an inert gas, but it is not a noble gas. Thank you to everyone at Noble Gas … Learn more. We have step-by-step solutions for your textbooks written by Bartleby experts! They are called noble gases because they are so majestic that they do not react with anything in general. Oganesson (element 118) is probably the next noble gas after Radon because it is the next box down in the same group. Even under these conditions, noble gases do not form many compounds. So let's ride molecular formula for the compound first. In chemistry, the term ‘noble gas’ refers to any of the six chemically inert gaseous elements in the periodic table which demonstrate similar characteristics. Definition of noble gas : any of a group of rare gases that include helium, neon, argon, krypton, xenon, and usually radon and that exhibit great stability and extremely low reaction rates — called also inert gas Examples of noble gas in a Sentence what is another NAME for noble gases? From the standpoint of chemistry, the noble gases may be divided into two groups: the relatively reactive krypton (ionisation energy 14.0 eV), xenon (12.1 eV), and radon (10.7 eV) on one side, and the very unreactive argon (15.8 eV), neon (21.6 eV), and helium (24.6 eV) on the other. - total lack of chemical reactivity Which was the next noble gas discovered? Here are all the Superman home planet noble gas answers. They are the most stable due to having the maximum number of valence electrons their outer shell can hold. What noble gas has the same electron configuration as each of the ions in the following compounds? Noble gases are often used to create a safe or inert atmosphere due to their stable nature. Please let us know if you have any Questions, comments, requests? Negative ions are isoelectronic with the noble gas at the end of their period or row. 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Gas Abbreviations How to write noble gas group have completely filled their electron! 2020 noble gas name	2021-10-26 14:33:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2897038757801056, "perplexity": 2272.4652634918402}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587908.20/warc/CC-MAIN-20211026134839-20211026164839-00082.warc.gz"}
https://techcommunity.microsoft.com/t5/excel/finding-identical-values-in-a-different-spreadsheet-and-copying/m-p/725698	Highlighted New Contributor # Finding identical values in a different spreadsheet and copying over correlating values. I am not intimately familiar with Excel, nor do I use it very often, so I apologize if my question is poorly phrased. I have two spreadsheets. Spreadsheet 1 has a full list of numerical codes in one column and a set of corresponding numbers in a different column. Spreadsheet 2 has a column containing a subset of the codes from Spreadsheet 1, though they are in a different order. Spreadsheet 2 also has a column that needs to be filled with the corresponding numbers from Spreadsheet 1. A method of doing so automatically would be greatly appreciated. Spreadsheet 1 + Spreadsheet 2 --->Function---> Spreadsheet 2 1000 1 1005 x 1005 2 1001 1 1004 x 1004 1 1002 2 1001 x 1001 1 1003 0 1004 1 1005 2 Thanks. 3 Replies Highlighted # Re: Finding identical values in a different spreadsheet and copying over correlating values. That could be like `=IFERROR(INDEX(Sheet1!\$C:\$C,MATCH(Sheet2!\$B2,Sheet1!\$B:\$B,0)),"")` Similar if they are different workbooks, not sheets. Highlighted # Re: Finding identical values in a different spreadsheet and copying over correlating values. Thank you for the response. Seeing the function has shown me just how little I know about Excel. Which portions of the function would be changed to match a specific spreadsheet? I'd imagine Sheet1 and Sheet2 are changed to the proper names, but what about \$C or \$B2? Highlighted # Re: Finding identical values in a different spreadsheet and copying over correlating values. B and C are columns in Sheet1 with your data, you may change any other ones, e.g. \$A:\$A instead of \$B:\$B \$B2 is the first cell in Sheet2 with your code Formula against it in C2. If you drag cell C2 down, formula in C3 will be with \$B3. Please check about absolute and relative references in Excel.	2020-11-24 02:06:12	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8049908876419067, "perplexity": 10958.768329820172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141169606.2/warc/CC-MAIN-20201124000351-20201124030351-00553.warc.gz"}
https://math.stackexchange.com/questions/1921660/what-type-of-graph-is-this-finding-optimal-mapping-of-one-group-to-another	# What type of graph is this? Finding optimal mapping of one group to another. I have a set $M$ of $n$ observations ${M_1, M_2,...M_n}$ and a set $C$ of $n$ existing "cases" ${C_1, C_2,...C_n}$. There is a cost or distance metric for matching the observations to the cases, where each observation is matched to one and only one case. Generally, the cost is the sum of the costs associated with each edge connecting an observation and its corresponding case. There are other cases I am interested in, where the number of observations is less than or greater than the number of cases. The objective is still to minimize the cost when completing the matching optimization. I'm interested in the theory and the algorithms for minimizing the cost.In order to research this, I want to know what the correct terminology would be. Given a graph $G=(V,E)$, a matching $A$ is a subset of edges such that no two edges in $A$ are incident to a common vertex. People are often interested in finding maximal and maximum matchings. $A$ is a perfect matching if every vertex $v \in V$ is incident to some edge $e \in A$. So some terminology you might be interested in is 'minimum cost matching'. For more information, take a look at the wikipedia page for matching.	2020-01-18 17:43:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8206950426101685, "perplexity": 85.46111131936222}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250593295.11/warc/CC-MAIN-20200118164132-20200118192132-00070.warc.gz"}
http://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-10-exponents-and-radicals-review-exercises-chapter-10-page-694/48	## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $4\sqrt{2}\approx5.657 \text{ } ft$ The ramp forms a right triangle. Let $c$ be the hypotenuse and $a,b$ be the legs of the right triangle. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, where $c=6$ and $a=2,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 2^2+b^2=6^2 \\\\ 4+b^2=36 \\\\ b^2=36-4 \\\\ b^2=32 \\\\ b=\sqrt{32} \\\\ b=\sqrt{16\cdot2} \\\\ b=\sqrt{(4)^2\cdot2} \\\\ b=4\sqrt{2} .\end{array} Hence, the base of the ramp (which has the same value as $b$) is equal to $4\sqrt{2}\approx5.657 \text{ } ft .$	2018-04-22 03:42:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9978291988372803, "perplexity": 2768.7057591035327}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945484.58/warc/CC-MAIN-20180422022057-20180422042057-00484.warc.gz"}
http://worldsworstsportsblog.com/	## What does a long game do to teams?April 13, 2015 Posted by tomflesher in Baseball. Tags: , , , , , Friday, the Red Sox took a 19-inning contest from the Yankees. Both teams have the unfortunate circumstance of finishing a game around 2:15 A.M. and having to be back on the field at 1:05 PM. Everyone, including the announcers, discussed how tired the teams would be; in particular, first baseman Mark Teixeira spent a long night on the bag to keep backup first baseman and apparent emergency pitcher Garrett Jones fresh, leading Alex Rodriguez to make his first career appearance at first base on Saturday. Teixeira wasn’t the only player to sit out the next day – center fielders Jacoby Ellsbury and Mookie Betts, catchers Brian McCann and Sandy Leon, and most of the bullpen all sat out, among others. The Yankees called up pitcher Matt Tracy for a cup of coffee and sent Chasen Shreve down, then swapped Tracy back down to Scranton for Kyle Davies. Boston activated starter Joe Kelly from the disabled list, sending winning pitcher Steven Wright down to make room. Shreve and Wright each had solid outings, with Wright pitching five innings with 2 runs and Shreve pitching 3 1/3 scoreless. All those moves provide some explanation for a surprising result. Interested in what the effect of these long games are, I dug up all of the games from 2014 that lasted 14 innings or more. In a quick and dirty data set, I traced the scores for each team in their next games along with the number of outs pitched and the length in minutes of the game. I fitted two linear models and two log models: two each with the next game’s runs as the dependent variable and two each with the difference in runs (next game’s runs – long game’s runs) as the dependent variable. Each used the length of the game in minutes, the number of outs, the average runs scored by the team during 2014, and an indicator variable for the presence of a designated hitter in each game. For each dependent variable, I modeled all variables in a linear form once and the natural log of outs and the natural log of the length of the game once. With runs scored as the dependent variable, nothing was significant. That is, no variable correlated strongly with an increase or decrease in the number of runs scored. With a run difference model, the length of the game in minutes became marginally significant. For the linear model, extending the length of the game by one minute lowers the difference in runs by about .043 runs – that is, normalizing for the number of runs scored the previous day, extending the game by one minute lowered the runs the next day by about .043. In the semilog model, extending the game by about 1% lowered the run difference by about 14; this was offset by an extremely high intercept term. This is a very high semielasticity, and both coefficients had p-values between .01 and .015. Nothing else was even close. With all of the usual caveats about statistical analysis, this shows that teams are actually pretty good at bouncing back from long games, either due to the fact that most of the time they’re playing the same team (so teams are equally fatigued) or due to smart roster moves. Either way, it’s a surprise. ## Not just offense – consistent offense.April 9, 2015 Posted by tomflesher in Baseball. Tags: , , Jacob deGrom pitched a quality start last night and got almost no run support. Now, that’s not unusual. Last year, 2546 teams had their starter pitch 5 or more innings and allow two earned runs or fewer; 781 of those teams lost, for a winning percentage of around 69%. It’s not unusual for the Mets, either; they had 84 such games last year, including 16 from Bartolo Colon (14-1 in those games), 15 from Jon Niese (4-4), 19 from Zack Wheeler (11-2), and 13 from deGrom (9-1). The Mets were right in the middle for all of this – the median across MLB was 85 games and 61 wins in those games. Washington had the most games (101) and the most wins (79), along with the highest percentage of wins coming in 5-inning starts with 2 earned runs or fewer from the starter (82.2%). Take note, though, of how disproportionate the wins were. Colon won all but 2 of his solid starts, with the Mets averaging 6.4 runs of support; Jon Niese only got 8 decisions, four of them losses, despite having almost as many games as Colon. Behind Niese, the Mets scored only 2.6 runs per game. deGrom had about 4.2 runs per game behind him and won 9 games; Wheeler had six more games with the same 4.2 run support average, but only got two more wins than deGrom in those games. deGrom maintained last year’s high standards in this year’s first start, and the Mets provided very little run support; even deGrom didn’t support himself very well, shaving almost 40 batting-average points to hit only .178 in these starts compared with .217 in other starts. Given, the Nationals are coming off a fantastic year as the best team in the National League’s regular season, so we can handle a 2-1 loss early in the season, but the offense needs to be more consistent if we’re going to take a rotation full of talent and turn it into a 90-win season. ## It was the best of pitching; it was the worst of pitching.April 8, 2015 Posted by tomflesher in Baseball. Tags: , , I’ll be honest. This was scheduled to be a ridiculous gushing post about Madison Bumgarner completing a totally meaningless milestone. After finishing the final game of the World Series and earning a save1, he pitched seven innings to open the season for the Giants this season. That means Madison threw twelve consecutive innings, which sounded pretty impressive. Unfortunately for Madison, the 2003 version of Josh Beckett did it better. He pitched a complete game to clinch the World Series for Florida and then opened the 2004 season with seven innings of one-run ball.Fortunately, Mat Latos bravely crumbled during last night’s game to provide the sports blogging community with material. Mat was knocked out of the box after only two outs, having given up a respectable seven runs before getting the hook. According to Baseball-Reference, no pitcher has lasted less than one full inning on opening day since 1982. Latos is, however, in good company with his season-starting ERA of 94.50. Jose Contreras started on Opening Day for Chicago in 2007 and lasted only one inning, giving up 7 runs; Carl Pavano did the same in 2009 for Cleveland, giving up 9 runs in the process. Both salvaged their seasons respectably, but – and this is my Mets homer bias coming out – let’s hope Latos is remarkably consistent this year. 1 This was the right call by the official scorer, but I really wish he’d declared Jeremy Affeldt “ineffective in a brief appearance” to give Mad Bum the win. Yes, I can call 2 1/3 innings of one-hit, no-run ball during which the Giants took the lead they never relinquished both “ineffective” and “brief,” compared to Bumgarner’s five-inning save. ## Kirk’s Big SpringMarch 20, 2015 Posted by tomflesher in Baseball, Economics. Tags: , , , Kirk Nieuwenhuis is having an incredible spring. All the usual caveats are in play – it’s spring training, so the stats are useless – but Kirk’s production has been exceptional. His slash line is .469/.553/.625 on 38 plate appearances. Let’s hit some sanity checks on Kirk’s production. First of all, his BAbip is off the charts. This spring, Kirk’s batting average on balls in play is .536, which is ridiculously high. Kirk won’t be able to maintain that into the season. If he’s still got a .536 OBP by the trade deadline, I’ll eat my hat and post the video. Kirk’s BAbip has been pretty streaky, though. During his rough April, Kirk had a .300 BAbip, about the league average over the season; after coming back up in late June, he had a .377 BAbip over the remainder of the season, broken up as .625 over five June games with 11 at-bats, .267 over 28 at-bats in July, .400 over 23 August at-bats, and .348 over 32 at-bats in September. From 2012 to 2013, Kirk’s BAbip dropped from about .379 to .246, and then shot back up to .370 in 2014. Using those numbers and taking first differences, then using the ratio of differences, that means we’d expect Kirk’s BAbip to drop to about .254 this season. Nonetheless, Kirk’s platoon splits are huge – against right-handed pitchers, from 2014, he’s got a .040/.050/.283 split (although he only made 9 at-bats and 10 plate appearances against left-handed pitchers). Though Kirk’s splits aren’t readily available, it’s possible that his big spring is residual of facing mostly right-handers. In the spring, Kirk’s BAbip denominator (AB – HR – K – SF) is 28 and the numerator (H – HR) is 15. If we take Kirk’s previous-year .377 BAbip, over 28 trials we’d expect 15 or more successes to occur about 2.86% of the time. That’s just barely within the bounds of statistical significance (which would indicate we’d expect Kirk to hit between 6 and 15 times about 95% of the time), and well outside if we assume Kirk has a true mean of .254 (which would put our confidence interval at around 3-11 successes in 28 trials). Second, take a look at Kirk’s K/BB ratio. Kirk has typically had a strikeout-to-walk ratio above 1; in 2013, he struck out about 2.67 times for every time he walked, and in 2014 it was about 2.44 strikeouts per walk. Over this small spring sample size, Kirk’s K/BB has actually dipped below 1, at 4/6 (or .667). Assuming Kirk walked 6 times anyway, using a conservative 2:1 K/BB ratio would turn 8 of Kirk’s hits into strikeouts. That would make Kirk’s BAbip tighten up to .350. Still strong, but not the obscene .536 we’ve seen. Even if we convert one walk to a strikeout and maintain a 2 K/BB, that would leave Kirk at .409, a very respectable spring. Kirk’s numbers have been shocking, and of course he’s out of options, so he’s extremely likely to make the team. As a left-handed bat, he’d be a strong everyday player if the outfield weren’t so crowded, but with Michael Cuddyer and Juan Lagares in the mix already along with lefties Curtis Granderson and Matt den Dekker, it’s going to be tough to find Kirk a clean platoon spot. ## A Pythagorean Exponent for the NHLMarch 17, 2015 Posted by tomflesher in Sports. Tags: , , , , , A Pythagorean expectation is a statistic used to measure how many wins a team should expect, based on how many points they score and how many they allow. The name ‘Pythagorean’ comes from the Pythagorean theorem, which measures the distance between the two short sides of a right triangle (the hypotenuse); the name reflects the fact that early baseball-centric versions assumed that Runs^2/(Runs^2 + Runs Allowed^2) should equal the winning percentage, borrowing the exponent of 2 from the familiar Pythagorean theorem (a^2 +b^2 =c^2). The optimal exponent turned out not to be 2 in just about any sport; in baseball, for example, the optimal exponent is around 1.82. This is found by setting up a function – in the case of the National Hockey League, that formula would be $\frac{GF^x}{GF^x + GA^x}$ – with a variable exponent. This is equivalent to $(1 + \frac{GA}{GF}^x)^{-1}$. Set up an error function – the standard is square error, because squaring is a way of turning all distances positive and penalizing bigger deviations more than smaller deviations – and minimize that function. In our case, that means we want to find the x that minimizes the sum of all teams’ $((1 + \frac{GA}{GF}^x)^{-1}) - \frac{W}{W+L})^2$. Using data from the 2009-2014 seasons, the x that minimizes that sum of squared errors is 2.2266, which is close enough to 2.23 that the sum of squared errors barely changes. Porting that exponent into the current season, there are a few surprises. First of all, the Anaheim Ducks have been lucky – almost six full wins worth of luck. It would hardly be surprising for them to tank the last few games of the season. Similarly, the Washington Capitals are on the precipice of the playoff race, but they’re over four games below their expected wins. With 11 games to go, there’s a good chance they can overtake the New York Islanders (who are 3.4 wins above expectation), and they’re likely to at least maintain their wild card status. On the other end, somehow, the Buffalo Sabres are obscenely lucky. The worst team in the NHL today is actually 4 games better than its expectation. Full luck standings as of the end of March 16th are behind the cut. ## What is BAbip?March 16, 2015 Posted by tomflesher in Baseball. Tags: , The first stat we all learned about as kids was the batting average, where you calculate what proportion of at-bats end with getting a hit. Then, of course, we start thinking about why there are weird exceptions – why doesn’t getting hit by a pitch count? Why don’t walks count? Why doesn’t advancing to first on catcher’s interference count? OBP, or on-base percentage, fixes that. (Well, maybe not the catcher’s interference part…) Batting average has some interesting properties, though. It captures events that have unpredictable outcomes – when you walk, it’s basically impossible to be put out on your way to first. Ditto being hit by a pitch. Of course, BA does have some of those determined outcomes, too – home runs and strikeouts don’t have much dynamic nature to them, although you’ll occasionally see brilliant defense save a sure homer (a la Carl Crawford’s MVP performance in the or a sloppy catcher mishandle a third strike and forget to tag the batter. (I’m looking at you, Josh Paul.) Nonetheless, balls in play – balls that the batter makes contact with, forcing the defense to try to make a play – are a major source of variation in the game. BAbip is measured as $\frac{H - HR}{AB - SO - SH + SF}$, meaning it takes the strikeouts and home runs out of the equation and (like all sane measures should!) includes sacrifice flies. Since the ball is out of the pitcher’s control as soon as it leaves his hand, BAbip measures things that the pitcher isn’t responsible for – that is, it’s handy as a measure of pitching luck, or, teamwide, as a measure of defensive effectiveness. The NL team BAbip average was .299, and AL average BAbip was about .298. Use Cases for BAbip: Evaluating hitting development. If a batter has had a stable BAbip for a while and his BAbip increases significantly, be suspicious! Particularly if his walk rate hasn’t increased, his home run rate hasn’t increased, and his strikeout rate hasn’t decreased, this might be a function of lucky hitting against bad or inefficient defenses. If the biggest part of an increase in production has been on balls in play, your hitter may not have actually improved. On the other hand, if you can see physical changes, or you have an explanation (e.g., went to AAA to work on his swing), you may see a more balanced improvement in OBP. – Evaluating pitching luck. Most of the time, all the pitchers for the same team pitch in front of the same defense. Even with a personal catcher in the mix, expect most pitchers on a team to have similar batting averages on balls in play. If you have one pitcher whose BAbip is much higher than the rest of the pitchers, he may be pitching against bad luck. With that in mind, you can expect that pitcher to improve going forward. – Comparing defenses. In 2014, Oakland had a .274 BAbip and allowed 572 runs – the best in the American league in BAbip and 18 runs behind Seattle – while Minnesota had a .317 BAbip and allowed 777 runs, the worst in both categories in the league. Defensive efficiency (a measure of 1 – BAbip) tracks closely with runs allowed. BAbip can operate as a quick and dirty check on how well a defense is performing behind a pitcher. ## Spitballing: Pi DayMarch 14, 2015 Posted by tomflesher in Baseball. Tags: Happy Pi Day! In honor of Pi Day, I’d like to share a few leaderboards. Stephen Strasburg. Photo by Keith Allison, used under Creative Commons license First, league median BAbip was .297. Here are four pitchers who got a little less lucky than the average, since their BAbip was .314: Erik Bedard .314 2014 TBR Aaron Barrett .314 2014 WSN Louis Coleman .314 2014 KCR Provided by Baseball-Reference.com: View Play Index Tool Used Generated 3/14/2015. Then, let’s follow up with the other side: the unlucky hitters who would only get on base 3.14 out of every 10 plate appearances: Alejandro De Aza 8 .314 2014 TOT 78/HD9 Michael Bourn 3 .314 2014 CLE 8/H Elvis Andrus 2 .314 2014 TEX 6/DH Jose Tabata 0 .314 2014 PIT H97/8D Provided by Baseball-Reference.com: View Play Index Tool Used Generated 3/14/2015. In addition, let’s take a look at the two hitters from 2014 who hit the ball 3.14 out of every 10 at-bats: Robinson Cano .314 2013 30 NYY Andrew McCutchen .314 2014 27 PIT Robinson Cano .314 2014 31 SEA Provided by Baseball-Reference.com: View Play Index Tool Used Generated 3/14/2015. Yeah, Robbie Cano managed to hit Pi in 2013 AND 2014. The boy must love his geometry. Our last mention: This year’s Pi Day mascot is Stephen Strasburg, who had an ERA of 3.14. The league-average ERA for pitchers who started 60% of their games was 3.86, so Stephen was in pretty good shape. ## Spring Training: Still Useless For Predicting StatsMarch 12, 2015 Posted by tomflesher in Baseball. Tags: A few days ago, I watched a Mets-Marlins spring training game that ended in a brutal 13-2 loss. It had all of the usual spring training fun – Zack Wheeler working too far inside and hitting two batters, Michael Cuddyer starting at first with Lucas Duda out, and Don Kelly’s hustle allowing him to draw a walk, steal a base, and score on a single, even while Cliff Floyd was snickering about how Jim Leyland kept him on the roster for no apparent reason in the playoffs. (Yeah, I know, Kelly’s a Marlin. Shut up.) During the game, I tweeted out a link to a file-drawer post from last year that indicated that there’s almost no correlation between spring performance and regular-season performance. I thought I’d run a quick update on that, so I dug up the Mets’ individual performance in spring training and analyze it compared to the regular season. There were 15 Mets who had 30 plate appearances in Spring Training and 100 plate appearances in the regular season. That’s a really small sample, so accuracywise we’d better keep our fingers crossed, but it’s enough data to spitball a little. I ran four correlations on this – spring and regular season batting average, OBP, SLG, and OPS – and then created an additional stat to measure whether hitters changed hitting style from spring to the regular season. This was a quick and dirty attempt to measure whether hitters favored OBP or SLG, so I took the ratio (SLG/OPS) and reasoned that a power hitter will have a larger ratio and a singles hitter will have a smaller. I measured this correlation, too, to determine if there were big changes. The results are unsurprising – the correlations are really low. Batting average correlates at around .019, and SLG at .305. OBP actually had a negative correlation, indicating that a high spring OBP may be a bad sign for the regular season. This is probably sampling error, due to the tiny number of observations, due almost entirely to Anthony Recker’s magical .426 spring and average regular season. That was about a -.25 correlation, which explains why OPS has a -.05 (near-zero) correlation – that big flip in OBP is going to offset the OPS correlation, too. The strongest correlation was style – at about .619, it’s a pretty good indicator that if a hitter’s SLG is how he scores, he’ll maintain that hitting style throughout the season. ## What is OPS?January 12, 2015 Posted by tomflesher in Baseball. Tags: , , , , Sabermetricians (which is what baseball stat-heads call ourselves to feel important) disregard batting average in favor of on-base percentage for a few reasons. The main one is that it really doesn’t matter to us whether a batter gets to first base through a gutsy drag bunt, an excuse-me grounder, a bloop single, a liner into the outfield, or a walk. In fact, we don’t even care if the batter got there through a judicious lean-in to take one for the team by accepting a hit-by-pitch. Batting average counts some of these trips to first, but not a base on balls or a hit batsman. It’s evident that plate discipline is a skill that results in higher returns for the team, and there’s a colorable argument that ability to be hit by a pitch is a skill. OBP is $\frac{H+BB+HBP}{AB+BB+HBP+SF}$. We also care a lot about how productive a batter is, and a productive batter is one who can clear the bases or advance without trouble. Sure, a plucky baserunner will swipe second base and score from second, or go first to third on a deep single. In an emergency, a light-hitting pitcher will just bunt him over. However, all of these involve an increased probability of an out, while a guy who can just hit a double, or a speedster who takes that double and turns it into a triple, will save his team a lot of trouble. Obviously, a guy who snags four bases by hitting a home run makes life a lot easier for his teammates. Slugging percentage measures how many bases, on average a player is worth every time he steps up to the plate and doesn’t walk or get hit by a pitch. Slugging percentage is $\frac{(\mathit{1B}) + (2 \times \mathit{2B}) + (3 \times \mathit{3B}) + (4 \times \mathit{HR})}{AB} = \frac{\text{Total Bases}}{AB}$. If a player hits a home run in every at-bat, he’ll have an OBP of 1.000 and a SLG of 4.000. OPS is just On-Base Percentage plus Slugging Percentage. It doesn’t lend itself to a useful interpretation – OPS isn’t, for example, the average number of bases per hit, or anything useful like that. It does, however, provide a quick and dirty way to compare different sorts of hitters. A runner who moves quickly may have a low OBP but a high SLG due to his ability to leg out an extra base and turn a single into a double or a double into a triple. A slow-moving runner who can only move station to station but who walks reliably will have a low SLG (unless he’s a home-run hitter) but a high OBP. An OPS of 1.000 or more is a difficult measure to meet, but it’s a reliable indicator of quality. ## The Hall of Fame Black Ink TestJanuary 11, 2015 Posted by tomflesher in Baseball. Tags: , , 1 comment so far The Baseball Hall of Fame‘s mission is “Preserving History, Honoring Excellence, Connecting Generations.” An important measure of the excellence honored in Cooperstown is called the Black Ink Test. “Black ink” refers to the boldface type used to show the league’s leader in an important category. The categories used for the Black Ink Test are, of course, different for pitchers and batters, but they also vary depending on the importance of the stat. A batter who excels in hitting home runs is more valuable to a team than one who takes the most at-bats regardless of outcome. For batters, points are awarded as follows: 1. One point for games, at-bats, or triples 2. Two points for doubles, walks, or stolen bases 3. Three points for runs scored, hits, or slugging percentage 4. 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http://portanywhere.com/class/beat-dyslexia-bsysl/viewtopic.php?c6b224=red-phosphorus-and-iodine-with-alcohol	For a reader-friendly overview of Phosphorus, see our consumer fact sheet on Phosphorus. The red phosphorus is then filtered out (and later reused), and the remaining acid is neutralized by adding a lye solution. However, you may extract iodine from common sources. If you aren't careful, you can expose yourself to highly toxic phosphine gas. By adding an excess of potassium iodide we ensure that all the excess dichromate ions will react with the iodide ions to form iodine. My guess is some of it got oxidized over the years to form phosphorus oxides that absorbed moisture from the air, hence acidic reaction. This is usually prepared by the action of iodine on alcohol in the presence of red phosphorus, much as described under "ethyl bromide. The first part is probably protonation/ activation and something between $\ce{S_{N}1/S_{N}2}$. I did find an interesting study in rats were they recommend investigating using Resveratrol (found in Red Wine, amongst other sources) to increase iodine influx for radioiodide therapy. Pour the entire bottle of the iodine tincture into a small ceramic bowl. Other reactions involving phosphorus halides. 512 sold. The required hydriodic acid in this variation of the hydriodic acid/red phosphorus method is produced by the reaction of iodine in water with red phosphorus. The molecular lattice contains discrete diatomic molecules, which are also present in the molten and the gaseous states. HI/Red Phosphorus method. In fact, 85 percent of the phosphorus found in the body resides in the bones and teeth. Hey friend, The answer is The phosphorus first reacts with the bromine or iodine to give the phosphorus(III) halide. Most commonly it is crystals of iodine dissolved in ethyl alcohol. $27.00 shipping. Detection and Estimation of Phosphorus White phosphorus is easily detected by its well-marked property of glowing in the dark as well as by its peculiar smell and reducing properties (q.v. Elemental phosphorus exists in two major forms, white phosphorus and red phosphorus, but because it is highly reactive, phosphorus is never found as a free element on Earth.It has a concentration in the Earth's crust of about one gram per kilogram (compare copper at about 0.06 grams). The phosphorus first reacts with the bromine or iodine to give the phosphorus(III) halide. C. Preparing the iodine/HI solution: 1. (i) lithium or sodium metal or red phosphorus , iodine , or iodine crystals ; . Preparing the red phosphorus: 1. 1.06 GALLON Nitric 70% Acid Technical Grade for Gold Silver Refining (135.2 OZ)$104.99. If previous alcohol consumption has damaged your liver, alcohol’s harmful effects on your kidneys will increase. Cocoa mix, with aspartame, low calorie, powder, with added calcium phosphorus, without added sodium or vitamin A Phosphorus: 1630mg Alcohol: 0g Potassium: 2702mg Add to Tracking Add to Compare Create Recipe Add to My Foods The iodine is used instead of hydiodic acid. 165 sold. The problem is that it has formed a paste that reacts acidic to litmus. Phosphorus is a component of bones, teeth, DNA, and RNA . The striking pads of matchbooks contain a small amount of red phosphorous. A substance is added that will bind to the meth, and the liquid meth is then drained out. This method yields high quality d-methamphetamine. In addition, the many organic compounds of phosphorus have varied uses, including those as additives for gasoline and lubricating oil, as… Deck, Crystals, Metal Cleaning- Rust Removal- FREE SHIP. $19.75. Often the PI 3 is made in situ by the reaction of red phosphorus with iodine in the presence of the alcohol. Therefore, ... Tincture of Iodine contains 2% iodine and 2% sodium iodide in 50% alcohol. "Various proportions of ingredients have been recommended. 125 Results and Discussion The HI/red P reduction of ephedrine to methamphetamine involves a cyclic oxidation of the iodide anion to iodine and reduction of iodine …$29.95. Phosphorus triiodide is commonly used in the laboratory for the conversion of primary or secondary alcohols to alkyl iodides. • Iodine/red phosphorus. The principal chemicals are ephedrine or pseudoephedrine, iodine, and red phosphorus. The alcohol … This method yields high quality d-methamphetamine. The principal chemicals are ephedrine or pseudoephedrine, iodine, and red phosphorus. B. Yes, Red phosphorus/HI reduces ketone, aldehyde, alcohol, carboxylic acid. businesses handling red phosphorus, white phosphorus, and hypophosphorous acid that these chemicals are used in the illicit manufacture of methamphetamine. Phosphorus is a chemical element with the symbol P and atomic number 15. One could draw out the final two aforementioned steps like so: Ultimately you have reduced the alkyl halide into an alkane. Iodine - Iodine - Physical and chemical properties: Iodine is a nonmetallic, nearly black solid at room temperature and has a glittering crystalline appearance. It is available in a wide variety of foods, including meat, fish, dairy, and some vegetables. $2P_{(s)} + 3Br_2 \rightarrow 2PBr_3\label{1.1.5}$ OXALIC ACID 99.6% 10 Lb. Add 10.0 ml safranin and ethyl alcohol solution made in step 1 to 90.0 ml distilled water. Hydrogen chloride gas is bubbled through the liquid meth, making it a crystalline hydrochloride salt. FAST 'N FREE. 2) Iodine and pseudo mixed: 3) After water addition: 4) Moments after RP addition: 5) Gas volume increases: 6) Appearence of mixture on flask wall (initial), notice high viscosity: 7) After 45 min, crystals of meth HI appear in the condenser: The second system was reverse phase with a 15-cm column packed with 5 JAm micro C18 silica and a mobile phase of TEAP buffer/acetonitrile (90 : 10), the buffer being 0.25 N phosphate adjusted to a pH of 2.2-2.3 with triethanolamine. Iodine crystals Red Phosphorous Pseudo-ephedrine, or ephedrine Methanol Toluene Acetone NaOH Ice Extraction of pfed. Criminals are always searching for sources of red phosphorus, white phosphorus, and hypophosphorous acid. The last example shows the reaction of thionyl chloride with a chiral 2º-alcohol. The solution will look rather brown due to the presence of the I 2.. ). Phosphorus, an essential mineral, is naturally present in many foods and available as a dietary supplement. 1) Iodine (13g), pseudoephedrine (10g), and red Phosphorus (4g). Humans are sensitive to changes in blood pH, and pH levels above 7.8 or … For alcohol :- ROH +2HI — RH. PI 3 + 3ROH → 3RI + HP(O)(OH) 2. Heat on the above type water bath until no more alcohol … Above 700 °C (1,300 °F), dissociation into iodine atoms becomes appreciable. It is possible to convert iodine tinctures to iodine crystals. Instead of using phosphorus(III) bromide or iodide, the alcohol is usually heated under reflux with a mixture of red phosphorus and either bromine or iodine. Methamphetamine abuse is a major drug problem in the United States. These then react with the alcohol to give the corresponding halogenoalkane which can be distilled off. Sulphur dichloride oxide (thionyl chloride) has the formula SOCl2. White phosphorus with sodium hydroxide produces the gas, usually as a result of overheating red phosphorus, plus white phosphorus can autoignite and blow up the meth lab . A single drink of alcohol can trigger alterations in your normal kidney function, the University of Montana reports 3. The formal replacement by a proton is not fully understood and is disputed whether the formation of a benzylic iodide is really involved. But as for all redox reactions, there are too many possible pathways to … Free shipping. Iodine is a dark-gray to purple-black, lustrous, solid element with a rhombic crystalline structure. You should get about 0.1g per flare. Many of the items you need can be purchased at a hardware store or online. Tinctures can be found on the shelves of many different supermarkets, and come in 30 ml bottles of 2% iodine … For carbonyl (aldehyde and ketone) :- RCOR + 4HI — RCH2R. The required hydriodic acid in this variation of the hydriodic acid/red phosphorus method is produced by the reaction of iodine in water with red phosphorus. Scrape the red phosphorus off of the caps of the 5 flares and store for later use. Phosphorus is a mineral that is essential for human health. Acquiring iodine in a pure form is often restricted. Ok, this method will deal with the Hcl salt of pfed, and a streamlined version of "the cure"! Iodine/red phosphorus. Most often you find it in sanitizing products. All pills are dumped into a large jar and double the volume of methanol poured on top! Safranin and dilute carbol-fuchsin are commonly used counterstain in Gram staining procedure, another being Neutral red (it stains gonococci and meningococci well) Preparation of Safranin. It is not pure, so consider cleaning it up a bit before use. Question: Red phosphorus purification I recently acquired a bottle of red phosphorus that is over 25 years old. Thus Wurtz indicates phosphorus 7 parts, alcohol 35, and iodine 23; whilst Frankland used these in … Iodine is also a decomposition product of hydriodic acid. Phosphorus triiodide is not stable, but may be generated in situ from a mixture of red phosphorus and iodine, and acts to convert alcohols to alkyl iodides. Ferrophosphorus, a combination of phosphorus with iron, is used as an ingredient in high-strength low-alloy steel. The two radical iodides probably reform "I"_2, as "I"_2 is depicted in the above diagram to be "leaving" (produced in) the reaction to react with more red phosphorus. A colloidal solution in isobutyl alcohol was made by passing arcs between red phosphorus suspended in this medium. For carboxylic :- … Red Iron Oxide 10lb Bag. This is a fact sheet intended for health professionals. Other articles where Red phosphorus is discussed: chemical industry: Phosphorus: Red phosphorus, comparatively harmless, is used in matches. Learn more about its benefits here. The second step is most likely carried out with red phosphorous as an electron source and removal of $\ce{PI3}$. Introduction. ... 1 GALLON ISOPROPYL ALCOHOL USP 99% STRENGTH - NO IMPURITIES. Strong iodine solution, known as “Lugol’s Solution”, contains 5% iodine and 10% potassium iodide in aqueous solution. When using iodine and red phosphorous, hydrogen iodide is generated in situ via the formation and subsequent hydrolysis of $\ce{PI3}$. The colour of this solution can be used to indicate the end-point of the titration with sodium thiosulfate, which is the the next step.. Titration with Thiosulfate And, no one appears to even consider the iodine content of “Napa” wines; forgive me, there is a wine aficionado term for the flavor that I simply cannot remember. The red is red phosphorus, white is the ephedrine or pseudoephedrine, and blue is iodine, used to make hydroiodic acid. ... Sulphur dioxide and hydrogen chloride are given off. 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Yourself to highly toxic phosphine gas iodine ( 13g ), and pH levels 7.8!, this method will deal with the alcohol a bottle of the I 2 including meat, fish dairy!	2021-09-19 07:18:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45011982321739197, "perplexity": 7712.906773460535}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00124.warc.gz"}
https://m.china-transmission.com/f720397/How-about-the-features-of-chain-transmission.htm	# How about the features of chain transmission? ## How about the features of chain transmission? 1.It's easily accommodate when speed reduction/increase of up to seven to one. 2.Long shaft-center distance is accommodated,and makes the chain more versatile. 3.Using chain with multiple shafts or drives with both sides of the chain. 4.Standardizaation of chains makes selection easily. 5.Cutting and connecting chains is easily. 6.When transmitting the same torque,a chain system can use samller sprocket deameter then a belt pulley. 7.Sprockets distribute the loading over their many teeth,so the wear is less than gears.	2021-01-18 16:25:43	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8086533546447754, "perplexity": 10258.155169902278}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00160.warc.gz"}
https://community.traefik.io/t/negative-matcher-is-not-working-error-while-parsing-rule-is-not-supported/13065	# Negative matcher is not working - error while parsing rule ! is not supported When I try to negate Host matcher in router with "!" operator (which is documented here Routers - Traefik "Invert a matcher") i get an error: "error while parsing rule ! is not supported" My rule is: `traefik.http.routers.my-host-traefik.rule=HostRegexp(`{subdomain:.+}.my.host`) && ! Host(`traefik.my.host`)` Traefik version 2.5.6 What i doing wrong? The error might come from elsewhere. It works fine with me with this configuration: ``````version: '3.6' services: traefik: image: traefik:v2.6 command: - --providers.docker ports: - "80:80" volumes: - /var/run/docker.sock:/var/run/docker.sock whoami: image: traefik/whoami # https://github.com/traefik/whoami labels: traefik.http.routers.whoami.rule: '!Host(`whoami.localhost`) && PathPrefix(`/`)' `````` Thanx for testing. I realized whats wrong, I have another one traefik, but version 2.2.0 which is not supports "!" operator	2022-05-25 17:03:02	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8077077269554138, "perplexity": 14631.935073226286}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00469.warc.gz"}
http://blog.nguyenvq.com/blog/2013/08/01/delimited-file-where-delimiter-clashes-with-data-values/?replytocom=349944	# Delimited file where delimiter clashes with data values A comma-separated values (CSV) file is a typical way to store tabular/rectangular data. If a data cell contain a comma, then the cell with the commas is typically wrapped with quotes. However, what if a data cell contains a comma and a quotation mark? To avoid such scenarios, it is typically wise to use a delimiter that has a low chance of showing up in your data, such as the pipe (“\|”) or caret (“^”) character. However, there are cases when the data is a long string with all sorts of data characters, including the pipe and caret characters. What then should the delimiter be in order to avoid a delimiter collision? As the Wikipedia article suggests, using special ASCII characters such as the unit/field separator (hex: 1F) could help as they probably won’t be in your data (no keyboard key that corresponds to it!). Currently, my rule of thumb is to use pipe as the default delimiter. If the data contains complicated strings, then I’ll default to the field separator character. In Python, one could refer to the field separator as ‘\ x1f’. In R, one could refer to it as ‘\ x1F’. In SAS, it could be specified as ‘1F’x. In bash, the character could be specified on the command line (e.g., using the cut command, csvlook command, etc) by specifying \$’1f’ as the delimiter character. If the file contains the newline character in a data cell (\n), then the record separator character (hex: 1E) could be used for determining new lines. Statistician ## 4 comments 1. Kent Johnson says: This is really only an issue if you are writing the CSV yourself. Any good CSV writer will have a mechanism to escape the quote character if it occurs in the data. In R, write.csv() will use two quotes to signify an actual quote in the data. For example, > data = c(‘plain’, ‘has,comma’, ‘comma,and,”quote” and \’singlequote\”) > f = textConnection(‘test’, ‘w’) > write.csv(data, file=f, row.names=FALSE) > cat(test, sep=’\n’) “x” “plain” “has,comma” “comma,and,””quote”” and ‘singlequote'” 1. I’m dealing with data where conversation notes are stored in a field. They use all sorts of characters in there: commas, quotes, pipes, carets, etc. Data are stored in a DB, and I’m using a sql client to get the data out to a text file (pyodbc seems to work best). Because of that, I find it convenient to use the unit separator character as the delimiter. 2. Kent is correct, this is a problem that has been solved many times and you should leverage existing solutions rather than try and circumvent a problem that does not exist. RFC 4180 is the de-facto standard in delimited file handling (http://tools.ietf.org/html/rfc4180), and addresses all these concerns. Worse yet, by using a less standard delimiter and code that doesn’t handle embedded quotes or delimiters, you just make problems more rare, which can actually make things significantly harder to debug, and you reduce portability and support for a wide audience of users who can handle comma delimited but cannot handle custom delimiters.	2017-09-22 11:47:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2972852885723114, "perplexity": 2271.6588069304566}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818688940.72/warc/CC-MAIN-20170922112142-20170922132142-00658.warc.gz"}
https://eprints.ucm.es/39856/	Synchronous products of rewrite systems (extended version) Impacto Martín Sánchez, Óscar and Verdejo López, Alberto and Martí Oliet, Narciso (2016) Synchronous products of rewrite systems (extended version). [ ] (Unpublished) Preview 454kB Abstract We present and formalize a concept of synchronous product for rewrite systems, and also a corresponding concept for general transition systems, used as semantics for the former. A series of examples shows their practical usefulness: for the strategic control of systems, and for modular specification and verification. Item Type: Working Paper or Technical Report Technical report 02/16Departamento de Sistemas Informáticos y ComputaciónFacultad de InformáticaUniversidad Complutense de Madrid Synchronous product, parallel composition, rewriting logic strategies Maude transition systems modular specification modular verification Sciences > Computer scienceSciences > Computer science > Software 39856 04 Nov 2016 14:03 04 Nov 2016 14:03	2019-07-23 07:40:41	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9186375141143799, "perplexity": 12353.212743795075}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00155.warc.gz"}
http://www.eng.fsu.edu/~dommelen/l2hfu/style_a/node9.html	Subsections ### 8 Usage Note: Much of what is in this section is outdated. For current software and its usage instructions go to either http://www.eng.fsu.edu/~dommelen/l2h or http://www.dommelen.net/l2h. #### 8.1 LATEX file format Fairly standard format tends to work best. Equation labels: do not put these on a line by themselves. Put them right behind the end of the actual equation and follow them by a % to avoid a blank line being inserted in the equation, causing the conversion of the equation to an image to fail. For new versions of LATEX, (or rather hyperref), follow htmlimage commands by an empty pair of braces, like in \htmlimage{extrascale=3}{} TEX \defs: assuming that LATEX2HTML does not process them, put a second copy of the \def inside an imagesonly environment. (Requires the html package to be loaded.) Captions: If you use an optional argument, make sure there is no linebreak between the ] and the {. Or the caption will become part of the image, probably with an incorrect figure number. Try to keep math out, and try to avoid captions with the same text save for the math. If the caption involves (even implicitly) \html or \latexhtml commands, the correct figure number may not be found. Quotes: type left and right quotes as {\lq} and {\rq}, or they will look poor. Do not use \hsize or \textwidth in pictures, it will not conform to your document. Specify actual dimensions in pt or in. Watch for two figures or formulae with long identical starting text. LATEX2HTML may not see the difference in the later text. If necessary, put a strut at the start of one of the two to ensure the difference is noted. (That should be fixed in the new versions.) Input commands: note subsubsection 9.30. \framebox inside a picture environment: See subsubsection 9.20.10. For other problems, see the following listings. #### 8.2 Elementary usage Normal operation proceeds as follows: Put (a copy of) your LATEX document in a folder (i.e directory) by itself and rename it index.tex. Add copies of the figures or whatever else there may be. In the command window obtained by double clicking the Windows_XP or Windows_98-ME icon in the l2hsup folder, (or in an xterm window on Unix), cd to the folder with your document. (In Windows XP, you can get some help on the cd command using help cd or cd /?, on Unix, that is man cd.) latex index until there are no longer errors. If you want to provide an index.pdf pdf version of the document, to allow easy, high quality printing of the entire document, create that now. (If you do not know how to create one using your version of TeX, try makepdf.) Then create the web pages using l2h. Load index.html in your browser to check the results. Have a look at manual.pdf and hthtml/index.html in the docs subfolder of l2h to see what is there. Occasionally, clean out the folder Temp. #### 8.3 The “makel2h” alternative Makel2h is an alternative to the l2h command that puts the web pages in a subfolder. This can be useful if you want to keep the folder with index.tex clean. Also, it allows you to make different web pages for different browsers. That is important if you are picky about the best possible alignment of the math images with descenders: different browsers are not compatible in such alignment. Internet Explorer is not compatible with any of the Mosaic/ Netscape/ Mozilla/ Firefox line, early Firefox is not compatible with later Firefox, etcetera. To create different pages for different browsers, you must use the makel2h command instead of l2h. Do not mix and match the two. A typical processing sequence would be latex index latex index set L2H_BROWSER=c makel2h set L2H_BROWSER=x makel2h set L2H_BROWSER=f makel2h This will produce superbly aligned math for Firefox 1.5 and later (and any other browsers that align images correctly, according to agreed-upon international standards, but I know no other browsers like that.) It should also produce, on average, slightly better math alignment in Internet Explorer, though I do not guarantee you will see the difference, if there is one, or that it will be an improvement. (But in any case, the new alignment for Internet Explorer, makel2h or l2h, is a vast improvement over earlier versions of LATEX2HTML. No longer math hovering above the line.) There will also be web pages for browsers that think the middle of the lower case letter “x” is where the two lines cross (unlike Internet Explorer, which thinks it is 2/3 of the way up.) Remember that with makel2h, eps figures must be either in the same directory with the index.tex file, or you must give the full path to the eps files in index.tex, all the way from the top of the disk. #### 8.4 Changing the looks with the wiz You can now make basic changes to headers, colors, font sizes, and much more, of the web pages using the wiz. To do so, cd to the folder with the index.tex whose web pages you want to customize. You must have run (make)l2h at least once up to the actual web page generation. Then simply enter “wiz”. The wiz is menu driven, so just select what you want to do and provide data. Note that in the majority of cases, you will need to remake the web pages using (make)l2h after running the wiz. And if you have changed the web page background color and your images now have visible ghosting, you will also need to first remove the old images using “clrl2h images” before running (make)l2h. The wiz is a quick and easy way to change the appearance of your web pages without getting involved in the details of perl, html, style sheets, and init files. Though the latter is of course much more powerful. #### 8.5 Using jpeg images The original LATEX2HTML is set up to use gif images. (At least as installed under FU instructions; the original also supports png images.) There are good reasons to use gifs for formulae and line graphs: they allow transparency and can be compact for such images. However, for pictures, the most popular image format on the web is jpeg (jpg). The reasons are that while gifs only allow 256 different colors in the image, jpegs allow millions. In addition, jpegs are usually much more compact, greatly speeding web page access over phone lines. The disadvantage of jpegs is that they lose some information in the original picture, (but then, so do gifs if the original image has more than 256 colors.) For jpegs, that can show up, for example, as faint artifacts in light regions. Little is ideal in this world. LATEX2HTML-FU has an added-on capability to create jpeg figures in addition to gifs. To activate this jpeg generation, do the following in the Windows_XP (or 98-ME) window: set L2H_JPGQ=80 l2h (or makel2h) or in a Unix command window, do setenv L2H_JPGQ 80 l2h (or makel2h) This will create jpeg duplicates of all gifs it makes with quality 80. (Use 75 for smaller jpegs, or 85 for less ghosting. Don't go over 90. Don’t use percent.) You will be asked whether you want to use the gif or jpeg version of each image; look at them in a viewer (your browser would do), and decide which one is best. Jpegs are only made of images with more than 256 colors, or that have an \htmlimage{extrascale=...}{} within their figure environment. Also, if you want to have jpeg versions made of already existing images, you will need to delete the corresponding gifs first to force them to be remade. Note that the ppmtojpeg executable must be in your path and with ppmtogif. This is true for the described installations. To turn of jpeg generation, in Windows do: set L2H_JPGQ= and in Unix: unsetenv L2H_JPGQ Note that existing jpegs will still continue to be used; simply delete them if you want to get rid of them and rerun (make)l2h.	2018-11-17 05:16:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7528290152549744, "perplexity": 2526.846677932355}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743282.69/warc/CC-MAIN-20181117040838-20181117062045-00057.warc.gz"}
https://blog.csdn.net/qq_16767427/article/details/79958255	# POJ 1458 Common Subsequence（动态规划） Common Subsequence Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 58460 Accepted: 24411 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij= zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y. Input The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. Output For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line. Sample Input abcfbc abfcab programming contest abcd mnp Sample Output 4 2 0 Source #include<iostream> #include<cstring> using namespace std; #define MAXN 1005 char sz1[MAXN],sz2[MAXN]; int MaxLen[MAXN][MAXN]; int main(){ while(scanf("%s%s",sz1,sz2) != EOF){ int length1 = strlen(sz1); int length2 = strlen(sz2); int i,j; for(i = 0; i <= length1; i++) MaxLen[i][0] = 0; for(i = 0; i <= length2; i++) MaxLen[0][i] = 0; for(i = 1; i <= length1; i++) for(j = 1; j <= length2; j++){ if(sz1[i-1] == sz2[j-1]) MaxLen[i][j] = MaxLen[i-1][j-1] + 1; else MaxLen[i][j] = max(MaxLen[i][j-1],MaxLen[i-1][j]); } printf("%d\n",MaxLen[length1][length2]); } return 0; } • 评论 • 上一篇 • 下一篇	2018-09-26 08:50:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2791866362094879, "perplexity": 2903.328902004638}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267164469.99/warc/CC-MAIN-20180926081614-20180926102014-00082.warc.gz"}
https://citizendium.org/wiki/Carmichael_number	Carmichael number Main Article Discussion Related Articles [?] Bibliography [?] Citable Version [?] Code [?] This editable Main Article is under development and subject to a disclaimer. A Carmichael number is a composite number named after the mathematician Robert Daniel Carmichael. A Carmichael number ${\displaystyle \scriptstyle c\ }$ divides ${\displaystyle \scriptstyle a^{c}-a\ }$ for every integer ${\displaystyle \scriptstyle a\ }$. A Carmichael number c also satisfies the congruence ${\displaystyle \scriptstyle a^{c-1}\equiv 1{\pmod {c}}}$, if ${\displaystyle \scriptstyle \operatorname {gcd} (a,c)=1}$. The first few Carmichael numbers are 561, 1105, 1729, 2465, 2821, 6601 and 8911. In 1994 Pomerance, Alford and Granville proved that there exist infinitely many Carmichael numbers. Properties • Every Carmichael number is square-free and has at least three different prime factors • For every Carmichael number c it holds that ${\displaystyle c-1}$ is divisible by ${\displaystyle p_{n}-1}$ for every one of its prime factors ${\displaystyle p_{n}}$. • Every absolute Euler pseudoprime is a Carmichael number. Chernick's Carmichael numbers J. Chernick found in 1939 a way to construct Carmichael numbers[1] [2]. If, for a natural number n, the three numbers ${\displaystyle \scriptstyle 6n+1\ }$, ${\displaystyle \scriptstyle 12n+1\ }$ and ${\displaystyle \scriptstyle 18n+1\ }$ are prime numbers, the product ${\displaystyle \scriptstyle M_{3}(n)=(6n+1)\cdot (12n+1)\cdot (18n+1)}$ is a Carmichael number. This condition can only be satisfied if the number ${\displaystyle n\ }$ ends with 0, 1, 5 or 6. An equivalent formulation of Chernick's construction is that if ${\displaystyle \scriptstyle m\ }$, ${\displaystyle \scriptstyle 2m-1\ }$ and ${\displaystyle \scriptstyle 3m-2}$ are prime numbers, then the product ${\displaystyle \scriptstyle m\cdot (2m-1)\cdot (3m-2)}$ is a Carmichael number. This way to construct Carmichael numbers may be extended[3] to ${\displaystyle M_{k}(n)=(6n+1)(12n+1)\prod _{i=1}^{k-2}(9\cdot 2^{i}n+1)\,}$ with the condition that each of the factors is prime and that ${\displaystyle n\ }$ is divisible by ${\displaystyle 2^{k-4}}$. Distribution of Carmichael numbers Let C(X) denote the number of Carmichael numbers less than or equal to X. Then for all sufficiently large X ${\displaystyle X^{0.332} The upper bound is due to Erdős(1956)[4] and Pomerance, Selfridge and Wagstaff (1980)[5] and the lower bound is due to Glyn Harman (2005),[6] improving the earlier lower bound of ${\displaystyle X^{2/7}}$ obtained by Alford, Granville and Pomerance (1994), which first established that there were infinitely many Carmichael numbers.[7]. The asymptotic rate of growth of C(X) is not known.[8] References and notes 1. J. Chernick, "On Fermat's simple theorem", Bull. Amer. Math. Soc. 45 (1939) 269-274 2. (2003-11-22) Generic Carmichael Numbers 3. Paulo Ribenboim, The new book of prime number records, Springer-Verlag (1996) ISBN 0-387-94457-5. P.120 4. Paul Erdős, "On pseudoprimes and Carmichael numbers", Publ. Math. Debrecen 4 (1956) 201-206. MR 18 18 5. C. Pomerance, J.L. Selfridge and S.S. Wagstaff jr, "The pseudoprimes to 25.109", Math. Comp. 35 (1980) 1003-1026. MR 82g:10030 6. Glyn Harman (2005). "On the number of Carmichael numbers up to x". Bulletin of the London Mathematical Society 37: 641–650. DOI:10.1112/S0024609305004686. Zbl. 1108.11065. Research Blogging. 7. W. R. Alford, A. Granville, and C. Pomerance (1994). "There are Infinitely Many Carmichael Numbers". Annals of Mathematics 139: 703-722. MR 95k:11114 Zbl 0816.11005. 8. Richard Guy (2004). Unsolved problems in Number Theory, 3rd. Springer-Verlag. ISBN 0-387-20860-7. . Section A13	2022-07-05 04:28:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8910616636276245, "perplexity": 740.3852676156836}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00442.warc.gz"}
https://www.varsitytutors.com/precalculus-help/trigonometric-identities/solving-trigonometric-equations-and-inequalities	Precalculus : Solving Trigonometric Equations and Inequalities Example Questions ← Previous 1 Example Question #1 : Solving Trigonometric Equations And Inequalities Use trigonometric identities to solve the following equation for : Explanation: Use the trigonometric identities to switch sec into terms of tan: hence, So we have , making Therefore the solution is for n being any integer. Example Question #2 : Solving Trigonometric Equations And Inequalities Which of the following is not a solution to for Explanation: We begin by setting the right side of the equation equal to 0. The equation might be easier to factor using the following substitution. This gives the following This can be factored as follows Therefore Replacing our substitution therefore gives Within our designated domain, we get three answers between our two equations. when when Therefore, the only choice that isn't correct is Example Question #1 : Solve Trigonometric Equations And Inequalities Find one possible value of . Explanation: Begin by isolating the tangent side of the equation: Next, take the inverse tangent of both sides: Divide by five to get the final answer: Example Question #1 : Solving Trigonometric Equations And Inequalities Use trigonometric identities to solve for the angle value. Explanation: There are two ways to solve this problem. The first involves two trigonometric identities: The second method allows us to only use the first trigonometric identity: Example Question #3 : Solving Trigonometric Equations And Inequalities Use trigonometric identities to solve the equation for the angle value. Explanation: The simplest way to solve this problem is using the double angle identity for cosine. Substituting this value into the original equation gives us: Example Question #1 : Solve Trigonometric Equations And Inequalities According to the trigonometric identities, Explanation: The trigonometric identity , is an important identity to memorize. Some other identities that are important to know are: Example Question #1 : Solving Trigonometric Equations And Inequalities If exists in the domain from , solve the following: Explanation: Factorize . Set both terms equal to zero and solve. This value is not within the domain. This is the only correct value in the domain. Example Question #1 : Solving Trigonometric Equations And Inequalities Solve for in the equation on the interval . Explanation: If you substitute you obtain a recognizable quadratic equation which can be solved for . Then we can plug back into our equation and use the unit circle to find that . Example Question #5 : Solving Trigonometric Equations And Inequalities Given that theta exists from , solve: Explanation: In order to solve appropriately, do not divide on both sides. The effect will eliminate one of the roots of this trig function. Substract from both sides. Factor the left side of the equation. Set each term equal to zero, and solve for theta with the restriction . Example Question #2 : Solving Trigonometric Equations And Inequalities Solve for There is no solution.	2021-03-06 08:16:54	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8617109060287476, "perplexity": 1101.0264633508534}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178374616.70/warc/CC-MAIN-20210306070129-20210306100129-00499.warc.gz"}
https://piknikovykos.cz/topics/0192de-math-logic-examples-with-answers	SITUATION. So, from Rs.27, shop owner received Rs.25 and beggar received Rs. Topics include sentences and statements, logical connectors, conditionals, biconditionals, equivalence and tautologies. Example â¦ 20 Math Questions With Answers. Finite Mathematics : Logic, Sets, and Counting Study concepts, example questions & explanations for Finite Mathematics. The Educational Value of Math Riddles: Fun and engaging math riddles and logic puzzles are an amazing way to get students to think critically, develop problem-solving skills, and think about mathematics in a creative way!. MATH GAME : 2 - 2 x 2 + 2 = ?? However, to get to her house you must cross seven bridges. No advanced knowledge of math is required. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules. 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Which in Simple English means âThere exists an integer that is â¦ Just 2 problem solver below to practice various math topics. The logic is payments should be equal to receipts. The test consists of ten questions. A predicate with variables can be made a proposition by either assigning a value to the variable or by quantifying the variable. 27. Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change. 2890 : … The result you get will be the same. TR. GO TO APP. 9.4 / 10 . Angelo, Bruno and Carlo are three students that took the Logic exam. Nonograms to logic brain teasers, there is something here for everyone. Tests. A couple of mathematical logic examples of statements involving quantifiers are as follows: There exists an integer x , such that 5 - x = 2 For all natural numbers n , 2 n is an even number. More brain teasers : Math Riddles; Monty Hall Simulation; Puzzle 1. Home Logic and Mathematical Statements. 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Do you dare to give it a try? The worker requires a daily payment of 1/7th of the bar of gold. Some cats have fleas. (a) 42 (b) 52 (c) 62 (d) â¦ Brain Teasers – Test your noggin and develop your skills in logic and math at the same time. For example, a typical experiment might require a test of a definition with a few example computations. But, maths is the universal language which is applied in almost every aspect of life. It gives you a set of statements or conditions, sometimes called the facts. To the test . 1 Diagnostic Test 22 Practice Tests Question of the Day Flashcards Learn by Concept. What is Logic? Math Puzzles. View â¦ more gifs . Each of his daughters has a brother. Rs.25 Rs.125 These enhance their critical thinking, logic, problem-solving skills and memory. 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Bradshaw Rock Paintings Facts, Nesquik Milkshake Without Ice Cream, Starbucks Paint Colors, Teak Dining Room Set For Sale, Mango Tree Flowers Information,	2021-04-23 01:33:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45093104243278503, "perplexity": 1918.3788225095213}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039626288.96/warc/CC-MAIN-20210423011010-20210423041010-00026.warc.gz"}
https://scipost.org/submissions/2009.04512v2/	# Phonon redshift and Hubble friction in an expanding BEC ### Submission summary As Contributors: Stephen Eckel Arxiv Link: https://arxiv.org/abs/2009.04512v2 (pdf) Date submitted: 2020-09-18 17:38 Submitted by: Eckel, Stephen Submitted to: SciPost Physics Academic field: Physics Specialties: Atomic, Molecular and Optical Physics - Theory Gravitation, Cosmology and Astroparticle Physics Approach: Theoretical ### Abstract We revisit the theoretical analysis of an expanding ring-shaped Bose-Einstein condensate. Starting from the action and integrating over dimensions orthogonal to the phonon's direction of travel, we derive an effective one-dimensional wave equation for azimuthally-travelling phonons. This wave equation shows that expansion redshifts the phonon frequency at a rate determined by the effective azimuthal sound speed, and damps the amplitude of the phonons at a rate given by $\dot{\cal V}/{\cal V}$, where $\cal{V}$ is the volume of the background condensate. This behavior is analogous to the redshifting and "Hubble friction" for quantum fields in the expanding universe and, given the scalings with radius determined by the shape of the ring potential, is consistent with recent experimental and theoretical results. The action-based dimensional reduction methods used here should be applicable in a variety of settings, and are well suited for systematic perturbation expansions. ###### Current status: Has been resubmitted ### Submission & Refereeing History Resubmission 2009.04512v3 on 12 November 2020 Submission 2009.04512v2 on 18 September 2020 ## Reports on this Submission ### Strengths A perturbative analysis is introduced to analyse azimuthal phonons' dynamics in an expanding ring BEC. This analysis is first applied to the case of a static cylindrical condensate (where known results are recovered), and can also be useful to address scenarios more complicated than those considered. ### Weaknesses The analysis is rigorous and thorough, I do not see weaknesses. ### Report This is a technically very valuable paper. In the context of the gravitational analogy the authors first write down a 3D action for the linearised perturbations (phonons) in a BEC. Then, for both static cylindrical and expanding ring condensates, they derive dimensionally reduced 1D actions for phonon fields phi_1 constant in the transverse dimensions. In the latter case, the phonon field wave equation is similar to that of a scalar field in an expanding universe. Their perturbative analysis allows to compute the corrections due to transverse modes, and the results show that these corrections are small. In the case of the expanding ring BEC, they improve their theoretical predictions with respect to their experiment [19], also in connection with the published results in [20]. In my opinion, after the authors have considered the two (small) points below, both for the analysis and the results the paper meets publishing criteria. ### Requested changes A couple of points: - In section 4.1, for the expanding ring BEC, they mention that the static approximation is surprisingly accurate even though in [19] the condition \dot R << c is violated (and they show in the appendix that the corrections to the static approximation are indeed small). Do the authors understand why ? - in the definition of the speed of sound c_theta^2 in (4.19), isn't the factor (1+rho/R) in the numerator? • validity: high • significance: high • originality: high • clarity: top • formatting: excellent • grammar: perfect ### Strengths The paper is clearly written in a rather pedagogical way that allows even not expert in the field to appreciate it. ### Report The authors discuss in a detailed way the phonon wave equation in an expanding ring BEC by dimensional reduction , showing characteristic effects one finds in cosmology , namely redshifting and Hubble friction experienced by quantum fields in an expanding universe. • validity: top • significance: top • originality: high • clarity: top • formatting: excellent • grammar: perfect	2020-12-06 01:21:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7377395033836365, "perplexity": 2028.8201341707243}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141753148.92/warc/CC-MAIN-20201206002041-20201206032041-00272.warc.gz"}
https://includestdio.com/6876.html	# MySQL case insensitive select ## The Question : 250 people think this question is useful Can anyone tell me if a MySQL SELECT query is case sensitive or case insensitive by default? And if not, what query would I have to send so that I can do something like: SELECT * FROM table WHERE Value = "iaresavage" Where in actuality, the real value of Value is IAreSavage. • Ultimately it depends on filed collation – if it’s ‘_ci’ (case-insensitive) or ‘_cs’ (case-sensitive) • This is one poorly worded question ;). Half the answers are showing you how to do case insensitive comparison, half are aiming for case sensitive. And only 1 tells you that the default is in fact case insensitive. 🙂 It’s worth noting that case insensitivity works even when you do a comparison like 'value' in ('val1', 'val2', 'val3') • @SaltyNuts man, reading this question 7 years later and realizing how much of a noob I was is embarrassing! I could have just read the documentation and the answer is in like the first sentence about SELECT statements… • To add to what @JovanPerovic said, utf8_bin also makes it case sensitive. Not sure if that existed back then 507 people think this answer is useful They are case insensitive, unless you do a binary comparison. 122 people think this answer is useful You can lowercase the value and the passed parameter : SELECT * FROM table WHERE LOWER(Value) = LOWER("IAreSavage") Another (better) way would be to use the COLLATE operator as said in the documentation 53 people think this answer is useful USE BINARY This is a simple select SELECT * FROM myTable WHERE 'something' = 'Something' = 1 This is a select with binary SELECT * FROM myTable WHERE BINARY 'something' = 'Something' or SELECT * FROM myTable WHERE 'something' = BINARY 'Something' = 0 47 people think this answer is useful Comparisons are case insensitive when the column uses a collation which ends with _ci (such as the default latin1_general_ci collation) and they are case sensitive when the column uses a collation which ends with _cs or _bin (such as the utf8_unicode_cs and utf8_bin collations). ## Check collation You can check your server, database and connection collations using: mysql> show variables like '%collation%'; +----------------------+-------------------+ \| Variable_name \| Value \| +----------------------+-------------------+ \| collation_connection \| utf8_general_ci \| \| collation_database \| latin1_swedish_ci \| \| collation_server \| latin1_swedish_ci \| +----------------------+-------------------+ and you can check your table collation using: mysql> SELECT table_schema, table_name, table_collation FROM information_schema.tables WHERE table_name = mytable; +----------------------+------------+-------------------+ \| table_schema \| table_name \| table_collation \| +----------------------+------------+-------------------+ \| myschema \| mytable \| latin1_swedish_ci \| ## Change collation You can change your database, table, or column collation to something case sensitive as follows: -- Change database collation ALTER DATABASE databasename DEFAULT CHARACTER SET utf8 COLLATE utf8_bin; -- or change table collation ALTER TABLE table CONVERT TO CHARACTER SET utf8 COLLATE utf8_bin; -- or change column collation ALTER TABLE table CHANGE Value Value VARCHAR(255) CHARACTER SET utf8 COLLATE utf8_bin; Your comparisons should now be case-sensitive. 25 people think this answer is useful String comparison in WHERE phrase is not case sensitive. You could try to compare using WHERE colname = 'keyword' or WHERE colname = 'KeyWord' and you will get the same result. That is default behavior of MySQL. If you want the comparison to be case sensitive, you could add COLLATE just like this: WHERE colname COLLATE latin1_general_cs = 'KeyWord' That SQL would give different result with this one: WHERE colname COLLATE latin1_general_cs = ‘keyword’ latin1_general_cs is common or default collation in most database. 16 people think this answer is useful The collation you pick sets whether you are case sensitive or not. 9 people think this answer is useful The default is case insensitive, but the next most important thing you should take a look at is how the table was created in the first place, because you can specify case sensitivity when you create the table. The script below creates a table. Notice down at the bottom it says “COLLATE latin1_general_cs”. That cs at the end means case sensitive. If you wanted your table to be case insensitive you would either leave that part out or use “COLLATE latin1_general_ci”. CREATE Table PEOPLE ( USER_ID INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, FIRST_NAME VARCHAR(50) NOT NULL, LAST_NAME VARCHAR(50) NOT NULL, PRIMARY KEY (USER_ID) ) ENGINE=MyISAM DEFAULT CHARACTER SET latin1 COLLATE latin1_general_cs AUTO_INCREMENT=0; If your project is such that you can create your own table, then it makes sense to specify your case sensitivity preference when you create the table. 3 people think this answer is useful SQL Select is not case sensitive. This link can show you how to make is case sensitive: http://web.archive.org/web/20080811231016/http://sqlserver2000.databases.aspfaq.com:80/how-can-i-make-my-sql-queries-case-sensitive.html 3 people think this answer is useful Try with: order by lower(column_name) asc; 2 people think this answer is useful Note also that table names are case sensitive on Linux unless you set the lower_case_table_name config directive to 1. This is because tables are represented by files which are case sensitive in Linux. Especially beware of development on Windows which is not case sensitive and deploying to production where it is. For example: "SELECT * from mytable" against table myTable will succeed in Windows but fail in Linux, again, unless the abovementioned directive is set. 2 people think this answer is useful The currently accepted solution is mostly correct. If you are using a nonbinary string (CHAR, VARCHAR, TEXT), comparisons are case-insensitive, per the default collation. If you are using a binary string (BINARY, VARBINARY, BLOB), comparisons are case-sensitive, so you’ll need to use LOWER as described in other answers. If you are not using the default collation and you are using a nonbinary string, case sensitivity is decided by the chosen collation. Source: https://dev.mysql.com/doc/refman/8.0/en/case-sensitivity.html. Read closely. Some others have mistaken it to say that comparisons are necessarily case-sensitive or insensitive. This is not the case. 0 people think this answer is useful You can try it. hope it will be useful. SELECT * FROM table WHERE Value COLLATE latin1_general_cs = "IAreSavage" 0 people think this answer is useful String fields with the binary flag set will always be case sensitive. Should you need a case sensitive search for a non binary text field use this: SELECT ‘test’ REGEXP BINARY ‘TEST’ AS RESULT; 0 people think this answer is useful For anyone who would find himself in a similar situation like me, I add my solution using like: In my case, I had to select all the rows filtering them by a certain column value. In that column, there were different values, such as ‘project_process’, ‘PROJECT_process’, ‘PROJECT_PROCESS’ and so on. Notes: PROJECT/project refers to a certain project name in capital/lowercase letters. PROCESS/process refers to a certain process name in capital/lowercase letters. This query was the solution: SELECT * FROM table_name where process like '%project_process' (this query allowed me to get all the possible combinations)	2021-03-05 01:49:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48908260464668274, "perplexity": 5248.126231760733}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178369553.75/warc/CC-MAIN-20210304235759-20210305025759-00589.warc.gz"}
https://gmatclub.com/forum/if-for-some-value-of-x-5-x-5-x-b-then-which-of-the-following-229488.html	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 22 Jul 2018, 11:22 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # If for some value of x, 5^x + 5^(–x) = B, then which of the following Author Message TAGS: ### Hide Tags Manager Joined: 17 Nov 2013 Posts: 132 If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags Updated on: 24 Nov 2016, 08:03 12 00:00 Difficulty: 45% (medium) Question Stats: 62% (01:36) correct 38% (01:34) wrong based on 329 sessions ### HideShow timer Statistics If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)? A) 5B B) 5B - 10 C) B^2 D) B^2 - 2 E) B^2 - 10 Originally posted by lalania1 on 24 Nov 2016, 07:57. Last edited by Bunuel on 24 Nov 2016, 08:03, edited 1 time in total. Edited the question. Manager Joined: 17 Nov 2013 Posts: 132 Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 24 Nov 2016, 07:59 Hi all, 5^–x = 5 to the -x exponent. I was not able to find the right way to express it. Feel free to edit with a fix if you have it. Math Expert Joined: 02 Sep 2009 Posts: 47183 Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 24 Nov 2016, 08:07 lalania1 wrote: If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)? A) 5B B) 5B - 10 C) B^2 D) B^2 - 2 E) B^2 - 10 Square $$5^x + 5^{–x} = B$$: $$25^{x} +25^x 5^{–x}+25^{-x} = B^2$$; $$25^{x} +2+25^{-x} = B^2$$; $$25^{x} + 25^{-x} = B^2-2$$. _________________ Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3655 Location: India GPA: 3.5 Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 24 Nov 2016, 09:21 1 1 lalania1 wrote: If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)? A) 5B B) 5B - 10 C) B^2 D) B^2 - 2 E) B^2 - 10 Let x = 1 $$5^x + 5^{–x} = \ B \ = 5 + \frac{1}{5}$$ = $$\frac{26}{5}$$ Let x = 2 $$5^x + 5^{–x} = 5^2 + \frac{1}{5^2}$$ = 25 + $$\frac{1}{25}$$ = $$\frac{626}{25}$$ From the above options (A) and (B) can straightaway be rejected... (C) $$B^2 = \frac{676}{25}$$ Rejected... (D) $$B^2 - 2 = \frac{( 676 - 50 )}{25}$$ = $$\frac{626}{25}$$ Hence, answer must be (D) $$\frac{626}{25}$$ _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club \| Rules for Posting in QA forum \| Writing Mathematical Formulas \|Rules for Posting in VA forum \| Request Expert's Reply ( VA Forum Only ) Intern Joined: 15 Feb 2017 Posts: 2 Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 25 Feb 2017, 13:16 Math Expert Joined: 02 Sep 2009 Posts: 47183 Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 26 Feb 2017, 03:37 Tylcleaner wrote: There are TWO solutions above: https://gmatclub.com/forum/if-for-some- ... l#p1767286 https://gmatclub.com/forum/if-for-some- ... l#p1767321 _________________ Manager Joined: 31 Jan 2017 Posts: 58 Location: India GMAT 1: 680 Q49 V34 GPA: 4 WE: Project Management (Energy and Utilities) If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 26 Feb 2017, 07:42 Given, 5^x + 1/ (5^x) = B Substitute 5^x = a a + 1/a = B Needed = 25^x + 25^(–x) = (5^x)^2 + 1 / (5^x)2 = a^2 +1/ a^2 = (a^4 + 1) / a^2 a + 1/a = B a^2+ 1 = a * B a^4 + 1 + 2(a^2) = (a^2) (B^2) [squaring both sides] a^4 + 1 = (a^2) * (B^2) - 2(a^2) (a^4 + 1) / a^2 = (B^2) - 2 _________________ __________________________________ Kindly press "+1 Kudos" if the post helped Intern Joined: 12 Oct 2014 Posts: 6 N: E Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 01 Dec 2017, 19:11 I approached the question in the following manner: Given: $$5^x+5^{-x} = B$$ Simplify: $$25^x+25^{-x} = 5^{2x}+5^{-2x} = 5(5^x+5^{-x}) = 5(B)$$ Why can't I approach like this? Math Expert Joined: 02 Sep 2009 Posts: 47183 Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 01 Dec 2017, 23:58 nicoeddy wrote: I approached the question in the following manner: Given: $$5^x+5^{-x} = B$$ Simplify: $$25^x+25^{-x} = 5^{2x}+5^{-2x} = 5(5^x+5^{-x}) = 5(B)$$ Why can't I approach like this? If you factor 5 from $$5^{2x}+5^{-2x}$$ you get $$5(5^{2x-1}+5^{-2x-1})$$, not $$5(5^x+5^{-x})$$. Check below: 8. Exponents and Roots of Numbers Check below for more: ALL YOU NEED FOR QUANT ! ! ! Hope it helps. _________________ Director Joined: 08 Jun 2015 Posts: 517 Location: India GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38 Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 23 Apr 2018, 07:32 +1 for option D. Use (a+b)^2=a^2+b^2+2ab. (5^x+5^-x)^2=25^x+25^-x+2=b^2 ; The required qty is b^2-2. Hence option D _________________ " The few , the fearless " Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 2973 Location: United States (CA) Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 24 Apr 2018, 11:31 lalania1 wrote: If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)? A) 5B B) 5B - 10 C) B^2 D) B^2 - 2 E) B^2 - 10 Squaring both sides of the equation, we have: [5^x + 5^(–x)]^2 = B^2 (5^2)^x + (5^2)^-x + 2(5^x)(5^-x) = B^2 25^x + 25^-x + 2(5^0) = B^2 25^x + 25^-x + 2 = B^2 25^x + 25^-x = B^2 - 2 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Intern Status: Studying Quant Joined: 04 Sep 2017 Posts: 45 GPA: 3.6 WE: Sales (Computer Software) Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 27 Jun 2018, 04:48 1 Bunuel wrote: lalania1 wrote: If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)? A) 5B B) 5B - 10 C) B^2 D) B^2 - 2 E) B^2 - 10 Square $$5^x + 5^{–x} = B$$: $$25^{x} +25^x* 5^{–x}+25^{-x} = B^2$$; $$25^{x} +2+25^{-x} = B^2$$; $$25^{x} + 25^{-x} = B^2-2$$. Okay I had a question, but as I typed this out I got the answer... Here is a detailed step by step of Bunuel's solution. Thanks Bunuel. I was confused how he got the 2 $$* 5^x * 5^{-x}$$, but now I see that's just $$5^0 + 5^0$$ Square $$5^x + 5^{–x} = B$$ $$B^2$$= ($$5^x + 5^{-x}$$)($$5^x + 5^{-x}$$). Foil method... $$B^2$$ = ($$5^{2x} +5^{x-x} +5^{x-x} +5^{-2x}$$) Which then would lead to $$B^2$$ = ($$5^{2x} + 5^0 + 5^0 + 5^{-2x}$$) We know $$5^0$$ = 1 $$B^2$$ = ($$5^{2x} + 2 + 5^{-2x}$$) Subtract the 2 to move it over to LHS. $$5^{2x} +5^{-2x} = 25^x + 25^{-x}$$ $$B^2$$ -2 = $$25^x + 25^{-x}$$ Manager Joined: 04 Aug 2010 Posts: 198 Schools: Dartmouth College Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following [#permalink] ### Show Tags 27 Jun 2018, 13:13 2 lalania1 wrote: If for some value of x, 5^x + 5^(–x) = B, then which of the following is equal to 25^x + 25^(–x)? A) 5B B) 5B - 10 C) B^2 D) B^2 - 2 E) B^2 - 10 Let x=0. $$B = 5^x + 5¯^x = 5^0 + 5^0 = 1 + 1 = 2.$$ $$25^x + 25¯^x = 25^0 + 25^0 = 1 + 1 =$$ $$2$$. When B=2 is plugged into the answer choices, the result must be the value in blue. Only D works: $$B^2 - 2 = 2^2 - 2 = 4 - 2 = 2$$. _________________ GMAT and GRE Tutor Over 1800 followers GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and long-distance. Re: If for some value of x, 5^x + 5^(–x) = B, then which of the following &nbs [#permalink] 27 Jun 2018, 13:13 Display posts from previous: Sort by # Events & Promotions Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	2018-07-22 18:22:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8327017426490784, "perplexity": 8614.614898542291}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676593438.33/warc/CC-MAIN-20180722174538-20180722194538-00300.warc.gz"}
https://socratic.org/questions/how-do-you-find-the-domain-and-range-for-y-sqrtx	How do you find the domain and range for y=sqrtx? Domain: $x \ge 0$ , Range : $y \ge 0$. $y = \sqrt{x}$ . Domain: x should not be negative i.e $x \ge 0$ Range : $y \ge 0$. graph{x^0.5 [-14.24, 14.23, -7.12, 7.12]}[Ans]	2020-03-31 16:25:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7009040713310242, "perplexity": 3477.7109488197193}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00177.warc.gz"}
https://crazyproject.wordpress.com/2011/09/28/groups-are-precisely-those-semigroups-which-are-both-left-and-right-simple/	## Groups are precisely those semigroups which are both left and right simple Prove that a semigroup $S$ is a group if and only if it is both left and right simple. Exhibit a left simple semigroup which is not right simple. We begin with a lemma. Lemma: Let $S$ be a semigroup. Then $S$ is left simple (right simple) [simple] if and only if $Sa = S$ ($aS = S$) [$SaS = S$] for all $a \in S$. Proof: Suppose $S$ is left simple, and let $a \in S$. Certainly $SSa \subseteq Sa$, so that $Sa$ is a left ideal. Thus $Sa = S$ for all $a \in S$. Conversely, suppose $Sa = S$ for all $a \in S$, and let $L \subseteq S$ be a left ideal. Now $L$ is nonempty by definition; say $a \in L$. Then $Sa \subseteq L$, and so $S \subseteq L$. Thus $L = S$, and in fact $S$ is the only left ideal of $S$. So $S$ is left simple. The proofs for right simple and simple are similar. $\square$ Now let $S$ be a group and let $a \in S$. If $x \in S$, then we have $x = xe = xa^{-1}a$; in particular, $x \in Sa$. So $S = Sa$ for all $a$. By the lemma, $S$ is left simple. Similarly, $S$ is right simple. Now suppose the semigroup $S$ is both left and right simple. Let $a \in S$. Since $Sa = S$, there exists an element $e \in S$ such that $ea = a$. Now let $b \in S$ be arbitrary. Since $aS = S$, there exists $c \in S$ such that $b = ac$. Now $b = ac = eac = eb$, so in fact $e$ is a left identity element of $S$. Similarly, there is a right identity element $f$, and we have $e = ef = f$, so that $e$ is a two-sided identity. Now let $a \in S$ be arbitrary. Since $Sa = aS = S$, there exist elements $b,c \in S$ such that $ba = ac = e$. Now $b = be = bac = ec = c$, and we have $b = c$. Thus $a$ has a two sided inverse with respect to $e$. Since $a$ was arbitrary, every element has an inverse, and so $S$ is a group. Now consider the semigroup $S = \{a,b\}$ with $xy = x$ for all $x,y \in S$. (That is, $S$ is the left zero semigroup on two elements.) Suppose $T \subseteq S$ is a left ideal. Now for all $x \in S$ and $y \in T$, we have $xy = x \in T$. Thus $T = S$, and so $S$ is left simple. However, $S$ is not a group, and so (by the previous argument) cannot be right simple.	2016-10-26 23:02:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 70, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.993689775466919, "perplexity": 44.685940039359295}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721008.78/warc/CC-MAIN-20161020183841-00330-ip-10-171-6-4.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/concept-of-force-components.956357/	# Concept of force components ## Homework Statement A box of mass X kg hangs motionless from two ropes, as shown in the diagram. The angle of rope 1 is specified amount of degrees. Choose the box as the system. The x-axis runs to the right, the y-axis runs up, and the z-axis is out of the page. What is the magnitude of \|F2\|? ## Homework Equations [/B] The equation for finding the magnitude of the force is Force/cos(theta). The equation for finding the force on the x axis is \|F2\|sin(theta). ## The Attempt at a Solution [/B] I understand that the formula for \|F2\| is F2y/cos(theta), but I was wondering what the concept behind it would be? My book describes it as using "directional" cosines? It Also, for F2x, I'm extremely confused as to why the F2 needs to be multiplied by sin(theta), rather than cosine theta, other than it just giving you F2y. My book doesn't say anything about finding F2x, but is it related to taking the derivative of cosine? It seems counter-intuitive to use cosine for the y axis and sine for the x axis, but that's the process to get the same value as the correct answer? Here's the free body diagram: #### Attachments • 3.3 KB Views: 153 Last edited: Related Introductory Physics Homework Help News on Phys.org Chestermiller Mentor To carry out a force balance in the vertical direction, in terms of the magnitude of F1, what is the component of F1 in the vertical direction? I named the ropes inversely, but F1 would be 0 because it's the flat line. Sorry about the mixup, I'll fix that. F2 is the opposite of gravity because it cancels out gravity's force, so -mg. Chestermiller Mentor I named the ropes inversely, but F1 would be 0 because it's the flat line. Sorry about the mixup, I'll fix that. F2 is the opposite of gravity because it cancels out gravity's force, so -mg. No. F2 is not equal to mg. In terms of the magnitude of F2, what is the vertical component of F2? I thought the vertical component was the opposite of mg, but is it mg times sin(theta) rather than that? Chestermiller Mentor I thought the vertical component was the opposite of mg, but is it mg times sin(theta) rather than that? Like I said, in terms of the magnitude of $F_2$, what is the component of $F_2$ in the vertical direction. Would it be this? \|F2y\| = Fgrav + Ftension I'm not quite sure what else it could be than that. It seems to be what would be the actual equation for it though. There's just no acceleration in the problem, which would cause it to be cancelled out. Last edited: Chestermiller Mentor Would it be this? \|F2y\| = Fgrav + Ftension I'm not quite sure what else it could be than that.. In terms of the magnitude of F2, the component of F2 in the vertical direction is $$F_{2y}=\|F_2\|\sin{\theta}$$Does this make sense to you? If so, then, in terms of the magnitude of F2, what is the component of F2 in the horizontal direction? Oh. For this problem, doing that made my homework site count it wrong for some reason, so I ruled it out. F2x would equal \|F2\|cos(theta) based on that. F2 would be the weight plus any force being exerted that moves it, but it's not moving so it's just the weight? Chestermiller Mentor Oh. For this problem, doing that made my homework site count it wrong for some reason, so I ruled it out. F2x would equal \|F2\|cos(theta) based on that. F2 would be the weight plus any force being exerted that moves it, but it's not moving so it's just the weight? No. Please write down the force balances in the vertical and horizontal directions in terms of the components of F2 in the horizontal and vertical directions that you just determined. F2 = <\|F2\|cos(theta), \|F2\|sin(theta), 0> And its magnitude is equal, \|F2\| = sqrt((\|F2\|cos(theta))^2 + (\|F2\|sin(theta))^2) = sqrt(\|F2\|^2 * (cos^2 (theta) + sin^2 (theta))) = sqrt(\|F2\|^2) = \|F2\| For some reason \|F2\| came out equal to F2y/cos(theta) on my homework, but that's saying that \|F2\| is also equal to \|F2\|sin(theta)/cos(theta) or \|F2\|tan(theta)? Is it just coincidence? Chestermiller Mentor F2 = <\|F2\|cos(theta), \|F2\|sin(theta), 0> And its magnitude is equal, \|F2\| = sqrt((\|F2\|cos(theta))^2 + (\|F2\|sin(theta))^2) = sqrt(\|F2\|^2 * (cos^2 (theta) + sin^2 (theta))) = sqrt(\|F2\|^2) = \|F2\| For some reason \|F2\| came out equal to F2y/cos(theta) on my homework, but that's saying that \|F2\| is also equal to \|F2\|sin(theta)/cos(theta) or \|F2\|tan(theta)? Is it just coincidence? You don't seem to want to answer my question, so I'll answer it myself (again). FORCE BALANCES $$\|F_2\|\sin{\theta}-mg=0$$ $$\|F_2\|\cos{\theta}-F_1=0$$ Do you agree with these equations? If you solve the first equation for $\|F_2\|$, what do you get? If you substitute that result into the 2nd equation and solve for $F_1$, what do you get? I didn't quite realise what you meant by force balances, I'm really sorry about that. For some reason I saw "components of F2" and assumed it to be the components in the vector. It would end up as: \|F2\| = mg/sin(theta)	2020-04-01 02:51:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6404396891593933, "perplexity": 1097.1510677557098}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00543.warc.gz"}
https://chemistry.stackexchange.com/questions/5151/finding-pv-in-pv-nrt-for-molecular-hydrogen/5156	Finding Pv in Pv=nRT for molecular hydrogen I'm attempting to write a small program that will calculate the values for the Ideal Gas Law, specifically for molecular Hydrogen (H2) in space. Eventually, this will grow to be a sort of "simulator", but right now I seem to be having problems with remembering my college chemistry and physics. Here is a picture of the form that I created: So in my program (written in C#), I have the following as initial values for those variables: P = ? v = ? n = 2.0158 (weight in grams of H2) T = 2.725 (average temp in space) R = 1.67E-27 (1.67 * 10^-27) My problem is that, how does one go about trying to solve for P or v? Do you assign an arbitrary number for what you suppose the volume of the gas would be in space, and then calculate P based on that? Or is it the other way around? Extra, yet entirely unneeded info: This was brought about because my 7 year old daughter asked me how stars were made. I gave her the explanation that anyone who's got even a remote interest in it gives to their kids, but thought it would be pretty cool to make a little program that would give her a visual. Here was my explanation: "we think that Hydrogen stuck together to make even larger clouds of Hydrogen, which increased the total mass of the gas cloud, increasing the gravitational pull on other atoms/molecules of nearby gas, etc. This eventually got to the point that the gases started to heat up as they were squished together by their own mass, but that increase in temperature created an outward push (pressure), which got into a tug-of-war with gravity acting on the mass of the gas itself. And later, we got a star. That's how most of us non-scientists think it happens, anyway!" • That's a pretty cool genesis for the question, but I don't think gaseous plasmas inside a star strictly follow the ideal gas law... – fluffy Aug 3 '14 at 22:00 • Yea, I realize that there are things that we do not know, or things that cannot be modeled by something like this, but the goal was just for a small simulation of gasses gathering. – SalarianEngineer Aug 4 '14 at 4:57 You can solve for either $P$ or $V$. Choose one, say $P$, and then divide that into the right hand side $nRT$ to get $V$; i.e. $V=\frac{nRT}{P}$. More seriously, you need to check your units! You seem to be using Kelvins for $T$, which is right, and $n$ is just a count in moles. But I do not recognize your value of $R$. The SI value for $R$ is $8.3144621(75)$, in units of joules per Kelvin-mole. Joules, in turn, are Pascal-meters$^3$. Personally, I think most people find Pascals too small a unit for pressure. Far more practical would be Torr-liters, and then $R=62.36367(11)$. You can examine lots of different unit combinations for $R$ at the ideal gas constant page on Wiki: Ideal Gas Constant You need one of p or V to solve for the other. If you have a spherical body, then $V=\frac{4}{3}\pi r^3$ and plug in what you want the radius of your star to be.	2020-10-20 06:18:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7110182046890259, "perplexity": 457.76343132145377}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107869933.16/warc/CC-MAIN-20201020050920-20201020080920-00150.warc.gz"}
https://optimization-online.org/2013/04/3827/	# On the sufficiency of finite support duals in semi-infinite linear programming We consider semi-infinite linear programs with countably many constraints indexed by the natural numbers. When the constraint space is the vector space of all real valued sequences, we show the finite support (Haar) dual is equivalent to the algebraic Lagrangian dual of the linear program. This settles a question left open by Anderson and Nash~\cite{anderson-nash}. This result implies that if there is a duality gap between the primal linear program and its finite support dual, then this duality gap cannot be closed by considering the larger space of dual variables that define the algebraic Lagrangian dual. However, if the constraint space corresponds to certain subspaces of all real-valued sequences, there may be a strictly positive duality gap with the finite support dual, but a zero duality gap with the algebraic Lagrangian dual.	2023-01-31 07:13:08	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.964552104473114, "perplexity": 245.24852032542728}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00083.warc.gz"}
http://mathhelpforum.com/calculus/76153-help-partial-integral.html	Math Help - help with partial integral 1. help with partial integral If I am asked to find the partial integral of ysin(x^2) dx, is the answer (-y cos(x^2))/2? 2. What is $y$? $y(x)$? A constant? Or a mistake and it should be $\int x\sin(x^2)dx$? 3. it's partial derivative	2015-09-02 20:02:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.910504937171936, "perplexity": 1233.9468568644688}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645281325.84/warc/CC-MAIN-20150827031441-00201-ip-10-171-96-226.ec2.internal.warc.gz"}
https://aviation.stackexchange.com/questions/30901/is-prior-clearance-into-class-b-airspace-revokable	# Is prior clearance into Class B airspace revokable? I was piloting outside of Class B airspace in an area well known for being difficult from which to contact ATC via radio because of mountainous terrain. When I did get cleared into Class B, I was only about a mile outside of Class B airspace. Immediately, another controller took over (different gender / same frequency), revoked the clearance, and said to stay clear of Class B and scolded me for only being one mile outside of Class B. (I complied with that request, never entering Class B – so there was no violation.) Question: Can you legally be “uncleared” into Class B after receiving a clearance into Class B? Could you speculate on what may have happened? • Unless you are in an emergency, you are required to comply with all ATC instructions, so yes, a clearance can be revoked. Just like a landing clearance or crossing clearance. The controller shouldn't have scolded you though, you were clear of the airspace, unless you were coming in at a fast rate. Ideally you'd like to get clearance quite a bit before that. I like to contact Class C at least 20 miles out, and for Class B I'd get flight following if possible as I left my origin. – Ron Beyer Aug 22 '16 at 3:03 • Sure, a clearance can be changed or even revoked. It happens all the time. The traffic situation may change, equipment could fail, an emergency could appear, etc. Air traffic control is extremely dynamic - the work of the controller can change 100% in as little as 5 minutes. – J. Hougaard Aug 22 '16 at 4:22 • I'd listen to the conversation on liveatc.net and make sure that it really did happen as you recall. If it did, I'd file an ASRS report: as you described it, it seems that the handover from one controller to another wasn't done well. If the second controller "scolded" you then s/he was presumably unaware of the previous clearance issued, and that shouldn't be the case. – Pondlife Aug 22 '16 at 13:33 • @Pondlife It doesn't sound like a frequency change to me. More likely, a supervisor cut in after the other controller already gave the clearance because they weren't comfortable with the potential situation. I hear this happen every once in a while. – Lnafziger Aug 24 '16 at 1:38 • @Lnafziger That's an interesting idea, I didn't think of that. But then why would the supervisor scold the pilot if the mistake was on the ATC side? I actually understood it was a controller change, not a frequency change. – Pondlife Aug 24 '16 at 13:09	2020-01-24 21:16:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26729899644851685, "perplexity": 2106.2355955741223}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250625097.75/warc/CC-MAIN-20200124191133-20200124220133-00480.warc.gz"}
http://www.sleske.name/versa/index.html.en	# Linux on the NEC Versa 550D Es gibt auch eine deutsche Fassung dieses Textes. Tipp: Wenn Sie in Ihrem Browser die Sprache einstellen, die Sie bevorzugen (meist unter "Einstellungen" oder unter "Optionen"), werden Sie Texte — soweit verfügbar — automatisch in der richtigen Sprache erhalten (durch "content negotiation"). ## Summary The good news first: it works pretty well. Everything is supported, even suspend-to-RAM and suspend-to-disk. It's a bit hard to find software that works well with the limited amount of RAM the Versa 550D has, but otherwise everything is fine. In this document, I give an overview over the specs of the Versa 550D (in case you consider buying one), then I explain what you need to know to install Linux on it. Finally, I briefly describe the software I use on mine and give some hints as to what you might do with such a rather old system. ## System specs and general remarks (This part is not Linux-specific.) The NEC Versa 550D is (in my opinion) quite a rugged and reliable system, but nowadays (2001) it is somewhat dated. A brief overview of the technical data: Processor Cyrix 5x86 100 MHz RAM on-board 8 MB RAM extension 4, 8 or 16 MB (only one slot available) Screen 10.4 in. DSTN color LCD Display resolution 640 × 480 × 256 colors (1024 × 768 on external monitor) Video RAM 1 MB HDD 540 MB (upgradeable) Floppy standard 3.5 in. 1.44 MB (builtin) PCMCIA two Type-II cards or one Type-III Trackball two buttons Interfaces serial, parallel, VGA, PS/2 Power NiMH accumulator, 9.6 V, 2800 mAh Suspend modes suspend to RAM, suspend to disk Size 287 × 227 × 52 mm Weight approx. 2.8 kg / 6.1 lb (with accumulator) As you can see, the processor power (a fast 486) is quite acceptable for most simple applications (such as word processing or web surfing), but memory is quite tight. The size is quite acceptable, even modern systems are not much smaller (only thinner). The laptop is quite heavy, however; almost 3 kg (over 6 lb for our non-metric friends) is a lot if you have to carry it for more than a few minutes. Still the laptop will fit nicely into a backpack or anything bigger than a handbag, and is useful when traveling. Battery life, unfortunately, is not so good. On my (rather used) laptop, I get about 70-90 minutes maximum, which is not so great for a 4-hour train ride. The manual claims about two hours running time, but I could never check that with a new accumulator (I bought the system used). Still, you definitely want to take along the power brick (which is not big) and use it whenever possible. The ability to suspend to RAM (or better: to disk) somewhat mitigates this; you can suspend and resume in about 20 seconds, so you can save power whenever you don't use the system for a few minutes. The display is not perfect, either. It is a DSTN (not a modern TFT) LCD display. This means it reacts slowly (blurs fast movements), so it is not suitable for e.g. fast moving games (event though the processor is quite powerful enough for e.g. DOOM). More importantly, the display also copes badly with bright light. Indoors, without direct sunshine, it is fine, but outdoors, unless on a very cloudy day, it is difficult to impossible to see anything on it due to poor contrast. The accumulator can be exchanged easily (even while in suspend to disk), but it is (of course) proprietary, quite expensive (? approx. $100?) and may no longer be available. Inside, it seems to use ordinary NiMh cells. You could try to crack the pack open and exchange the cells to repair it, but that is mechanically quite tricky (plus may or may not mess up the integrated electronics). I have not done that. ## Installing Linux on the Versa 550D The installation is pretty painless, so I'll just describe the things which are special about the Versa 550D. I assume you know how to do a standard Linux installation. First of all, you need to choose a Linux distribution the requirements of which the laptop can satisfy, as it only has 540 MB of drive space and 8 - 24 MB RAM. There also is no CD-ROM drive, so the distro will need to support alternative ways of installation (most do). I use Debian GNU/Linux 2.2 (potato), which works fine. For the rest of these instructions, I will assume you use Debian; other distributions should work similarly, adjust accordingly. Before installing, you should set up the disk for suspend to disk. This needs a special partition. It can be created using a small utility called phdisk.exe which comes with the laptop. Just make a MS-DOS boot floppy, put phdisk.exe on it, boot it and run phdisk.exe . This will erase the hard disk, so it should be done first. If you don't have phdisk.exe, you may be able to find it on the web site of NEC or of Phoenix (the BIOS' manufacturer), or mail me and I'll send it to you. Alternatively, there is a Linux replacement for phdisk.exe called lphdisk by Patrick D. Ashmore. It is available at http://www.procyon.com/~pda/lphdisk/. I have never used it myself, though. Now for the actual installation: First, boot the laptop with the Linux installation floppy (enable floppy booting in the BIOS first), then partition the hard drive according to your needs (and make a swap partition). The suspend-to-disk partition will appear as type a0, "IBM Thinkpad Hibernation". Do not touch it. As there is no CD-ROM drive on the laptop, you will need a different means of installation. The easiest is via a network. The physical connection can be made using a PCMCIA ethernet card (easiest, fastest, if you have a PCMCIA ethernet card), via parallel port (PLIP) networking (slow) or via serial port (even slower). The ethernet option is best, if you have the hardware. Otherwise, you can use parallel line IP (PLIP, see the PLIP Mini HOWTO from the Linux Documentation Project) or serial line IP (SLIP) via a 'nullmodem cable'. These options are significantly slower, but require no special hardware except from the cable. Once the hardware is in place, follow your distribution's instructions for installing via network. The basic installation should work without problems. Here are a few technical details you might need: • The mouse is a standard 2-button PS/2 mouse (accessible via /dev/psaux). • The graphics chipset is a Chips&Tech 65545 with 1024 MB Video RAM; use the SVGA X server (when using XFree version 3.x, don't know about version 4.x). • Install the apm package to make use of the power management functions the laptop offers (and make sure you use a kernel with APM support). • Install PCMCIA support if you intend to use it. The PCMCIA chip is a i82365, other than that, no configuration should be needed. Finally, make sure you don't install more than the hard disk will hold. It is not very big... ## Problems / things to know when running Linux Here I list a few small problems I had using Linux and what I did about them. • I had problems using the floppy drive. After a suspend/resume cycle, it would not work, instead just producing loads of kernel error messages, as if it or the floppy inserted were faulty. Rebooting fixed this, so did removing and reinserting the floppy kernel module. Therefore, I use the APM daemon apmd to unmount the floppy and remove the module on suspend. On Debian, to do this you just put the appropriate commands umount /floppy; rmmod floppy into a script in /etc/apm/suspend.d. • The PS/2 mouse sometimes mysteriously fails to respond when starting X (it just does not move). This is fixed by doing a cat /dev/psaux on the console (as root) and moving the trackball a bit after X is started. No idea why this helps. • You can use the apm command from the apm package to see how much battery power is left; don't trust the percentage too much though; the decrease is highly non-linear. The power saving can be customized in the BIOS dialog; I set mine to switch off the screen when closing the lid and to suspend on keypress (Fn-Escape). Also don't forget to enable suspend-to-disk in the BIOS, otherwise suspend-to-RAM will be used. ## Software to run Obviously, using KDE and StarOffice is not a terribly good idea :-). I spent some time finding software that works well with limited ressources, yet is powerful and usable. Here's my list of what you could do with the system and what software to use (additions welcome). Where no link is given, the software is standard with most distributions. • Word processing: AbiWord, TeX, LaTeX • Email: mutt • Web browsing: Dillo (graphical!), lynx, w3m • Programming: the usual compilers / interpreters • Image processing: better not :-( • Gaming: all the usual desktop games • Windowmanager / desktop: icewm or XFCE • Presentation software: LaTeX (with e.g. the slides class), MagicPoint • remaining power: icewm can display this automatically, or use e.g. wmapm I run X with 12 MB of RAM, and it works fine. 8 MB might be cutting it very close though. ## Extra: The Versa 550D as an ebook One nifty thing you can do with the Versa 550D is to use it as an ebook. Ever wished you could read a downloaded Sherlock Holmes story in bed instead of in front of your computer? Now you can do it! I hacked up a script to convert the kind of texts which Project Gutenberg offers, so I could comfortably read them on my laptop. Project Gutenberg offers electronic versions of texts whose copyright has expired; almost all major classics (and more) are covered. Unfortunately, these texts come as plain ASCII, which is very platform-independent, but also a pain to read. I reformat them using the power of Perl and LaTeX. To read a text on the laptop, do the following: get your text from Project Gutenberg and run it through my script: gutb2latex <input.txt >output.tex This will do basic formatting; you should now mark up the chapter headings in output.tex like this: ...blablabla... \section{Introduction} ...blablabla and insert the command \tableofcontents where you want - guess what - to appear. Run the latex command on the resulting file twice, then dvips with Option -o. The resulting Postscript file, when viewed with gv at 1.414 magnification in orientation landscape, will exactly fill the screen. You may have to fiddle a bit to make gv's window so big that the borders and toolbar are invisible (use e.g. icewm's keyboard commands for that), then the text will fill the whole screen. Enable anti-aliasing (Key a), turn the laptop on its side and enjoy reading. Remark: The gutb2latex script assumes paragraphs in the input document are delimited by empty lines. If your input document indents the first line of a new paragraph instead (some Gutenberg texts do that), adjust the relevant line in my script for paragraphs to be recognized. Here's the code of the gutb2latex script I made: #! /bin/bash # gubt2latex # convert a Project Gutenberg document into a LaTeX document # (with some manual assistance) if [ "$1" == "-h" -o "$1" == "--help" ]; then cat <<EOF Usage: gutb2latex >etext.txt< Convert a Project Gutenberg document into a LaTeX document (with some manual assistance). gutb2latex v1.0, Mai 2000 (C) 2000 Sebastian Leske May be used, modified and distributed freely as long as this notice is preserved. EOF exit fi # get rid of CR/LFs flip -b -u$1 # LaTeX-preamble cat <<EOF \documentclass[10pt]{article} \begin{document} EOF # treat characters with a special meaning in LaTeX # first save backslashes sed -e 's/\\/!-ssr-!/g' $1 \| sed -e 's/\^/\\^/g' \| sed -e 's/\$/\\$/g' \| sed -e 's/&/\\&/g' \| sed -e 's/%/\\%/g' \| sed -e 's/~/\\~/g' \| sed -e 's/_/\\_/g' \| sed -e 's/#/\\#/g' \| sed -e 's/}/\\}/g' \| sed -e 's/{/\\{/g' \| sed -e 's/!-ssr-!/\\\\/g' # Make emtpy lines for paragraphs (if necessary) # must be adjusted and uncommented if needed # Count the number of spaces used for indenting the first line of # a new paragraph, and make sure that number of spaces appears # after the '^' in the next line (then uncomment it) # perl -pe 'if ($_=~"^ [a-z,A-Z,\"]" ) {print "\n"}' <rein >raus echo '\end{document}' # Ende Have fun, and I'd love to hear about your experiences with this laptop! Created on Oct. 14, 2001 Last changed on Jan. 2, 2007 Sebastian Leske Sebastian.Leske@sleske.name \|	2014-04-20 10:51:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20833221077919006, "perplexity": 6579.274896993888}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00635-ip-10-147-4-33.ec2.internal.warc.gz"}
https://nathanielpark.com/uelns/aa6192-how-to-name-intersection-of-planes	# how to name intersection of planes asked Oct 16 at 15:26. rand rand. Other representations are discussed in Algorithm 2 about the, Computational Geometry in C (2nd Edition). 0 : t0; // clip to min 0 t1 = t1>1? Sign in to comment. What is the intersections of plane AOP and plane PQC? In C# .NET I'm trying to get the boundary of intersection as a list of 3D points between a 3D pyramid (defined by a set of 3D points as vertices with edges) and an arbitrary plane. I tried live boolean intersection, however, it just vanish. P and R 19. Aug 23, 2019 . Then they intersect, but instead of intersecting at a single point, the set of points where they intersect form a line. 1 : t1; // clip to max 1 if (t0 == t1) { // intersect is a point I0 = S2.P0 + t0 v; return 1; } // they overlap in a valid subsegment I0 = S2.P0 + t0 v; I1 = S2.P0 + t1 v; return 2; } // the segments are skew and may intersect in a point // get the intersect parameter for S1 float sI = perp(v,w) / D; if (sI < 0 \|\| sI > 1) // no intersect with S1 return 0; // get the intersect parameter for S2 float tI = perp(u,w) / D; if (tI < 0 \|\| tI > 1) // no intersect with S2 return 0; I0 = S1.P0 + sI u; // compute S1 intersect point return 1;}//===================================================================, // inSegment(): determine if a point is inside a segment// Input: a point P, and a collinear segment S// Return: 1 = P is inside S// 0 = P is not inside SintinSegment( Point P, Segment S){ if (S.P0.x != S.P1.x) { // S is not vertical if (S.P0.x <= P.x && P.x <= S.P1.x) return 1; if (S.P0.x >= P.x && P.x >= S.P1.x) return 1; } else { // S is vertical, so test y coordinate if (S.P0.y <= P.y && P.y <= S.P1.y) return 1; if (S.P0.y >= P.y && P.y >= S.P1.y) return 1; } return 0;}//===================================================================, // intersect3D_SegmentPlane(): find the 3D intersection of a segment and a plane// Input: S = a segment, and Pn = a plane = {Point V0; Vector n;}// Output: I0 = the intersect point (when it exists)// Return: 0 = disjoint (no intersection)// 1 = intersection in the unique point I0// 2 = the segment lies in the planeintintersect3D_SegmentPlane( Segment S, Plane Pn, Point* I ){ Vector u = S.P1 - S.P0; Vector w = S.P0 - Pn.V0; float D = dot(Pn.n, u); float N = -dot(Pn.n, w); if (fabs(D) < SMALL_NUM) { // segment is parallel to plane if (N == 0) // segment lies in plane return 2; else return 0; // no intersection } // they are not parallel // compute intersect param float sI = N / D; if (sI < 0 \|\| sI > 1) return 0; // no intersection I = S.P0 + sI u; // compute segment intersect point return 1;}//===================================================================, // intersect3D_2Planes(): find the 3D intersection of two planes// Input: two planes Pn1 and Pn2// Output: L = the intersection line (when it exists)// Return: 0 = disjoint (no intersection)// 1 = the two planes coincide// 2 = intersection in the unique line Lintintersect3D_2Planes( Plane Pn1, Plane Pn2, Line* L ){ Vector u = Pn1.n * Pn2.n; // cross product float ax = (u.x >= 0 ? u.y : -u.y); float az = (u.z >= 0 ? Parallel planes are two planes that are the same distance apart at every point, extending infinitely. Thus the planes P1, P2 and P3 intersect in a unique point P0 which must be on L. Using the formula for the intersection of 3 planes (see the next section), where d3 = 0 for P3, we get: The number of operations for this solution = 11 adds + 23 multiplies. Construct the vector $\vec n$ perpendicular to the plane; in your case you can read it off the equation of the plane: $\vec n=(2,1,1)$. 2(x - 4)^2 + (y -... 1. Name the intersection of plane N and line AE is point B. The vector equation for the line of intersection is calculated using a point on the line and the cross product of the normal vectors of the two planes. We can often determine what the intersection of two geometrical objects is called by observing what that intersection looks like. An example of what I'm looking for is below. Plane 1: A 1 x + B 1 y + C 1 z = D 1: Plane 2: A 2 x + B 2 y + C 2 z = D 2: Plane 3: A 3 x + B 3 y + C 3 z = D 3: Normal vectors to planes are: n 1 = iA 1 + jB 1 + kC 1: n 2 = iA 2 + jB 2 + kC 2: n 3 = iA 3 + jB 3 + kC 3: For intersection line equation between two planes see two planes intersection. In C# .NET I'm trying to get the boundary of intersection as a list of 3D points between a 3D pyramid (defined by a set of 3D points as vertices with edges) and an arbitrary plane. The intersection of two planes is called a line. Is the answer C? Join now. 63% average accuracy. Given three planes: Form a system with the equations of the planes and calculate the ranks. Thank you! Intersection of plane and line. the cross product of (a, b, c) and (e, f, g), is in the direction of the line of intersection of the line of intersection of the planes. Intersection of Planes. So the point of intersection can be determined by plugging this value in for $$t$$ in the parametric equations of the line. cg 5 0; justin. In 2D, with and , this is the perp pro… All other trademarks and copyrights are the property of their respective owners. further i want to use intersection line for some operation, without fixing it by applying boolean. Jun 19, 2018 . This always works since: (1) L is perpendicular to P3 and thus intersects it, and (2) the vectors n1, n2, and n3 are linearly independent. Log in. Thank you. P and S… As shown in the diagram above, two planes intersect in a line. Save. The intersection of two planes is called a line. Vote. In practice, this can be done as follows. These lines are parallel when and only when their directions are collinear, namely when the two vectors and are linearly related as u = av for some real number a. Please help me with this question. Planes are two-dimensional flat surfaces. This video describes how to find the intersection of two planes. Show Hide all comments. Log in. No need to display anything visually. answer! In geometry, intersections refer to where two or more geometrical objects meet. Is there an intersection.? A. AC B. BG C. CG D. The planes need not intersect. Become a Study.com member to unlock this To check if the intersection is an ellipse, a parabola or a hyperbola it is enough to check whether the plane intersects all the generatrices of the cone or not. 9th - 12th grade . In General, the intersection of straight line and plane may be:1) one point (as in our case)2) an Infinite number of points - the whole straight line (when the straight line belongs to the plane)3) the empty set (when the straight line and plane are parallel to each other) 21 days ago. What is the intersections of plane AOP and plane PQC? Then since L is contained in P 1, we know that ~n 1 must be orthogonal to d~. The vector equation for the line of intersection is given by r=r_0+tv r = r Since we found a single value of $$t$$ from this process, we know that the line should intersect the plane in a single point, here where $$t = -3$$. Played 16 times. Create your account. They can take on different forms depending on what type of geometric objects are intersecting. Two intersecting planes always form a line If two planes intersect each other, the intersection will always be a line. Imagine two adjacent pages of a book. I want to get line of intersection of two planes as line object when the planes move. asked 8 mins ago. cg 5 0; Anonymous. Solution for W R Name the intersection of planes QRS and RSW. P(0, -4, 0), Q(4, 1,... Find an equation of the plane that contains both... Saxon Algebra 2 Homeschool: Online Textbook Help, Saxon Algebra 1 Homeschool: Online Textbook Help, Prentice Hall Algebra 2: Online Textbook Help, Explorations in Core Math - Geometry: Online Textbook Help, TExES Mathematics 7-12 (235): Practice & Study Guide, Holt McDougal Algebra 2: Online Textbook Help, High School Algebra I: Homework Help Resource, Accuplacer Math: Advanced Algebra and Functions Placement Test Study Guide, Prentice Hall Pre-Algebra: Online Textbook Help, SAT Subject Test Mathematics Level 1: Practice and Study Guide, Biological and Biomedical Consider the points below. 72k 8 8 gold badges 188 188 silver badges 294 294 bronze badges. Ask your question. this is hard for me since there isn't a picture. I want to get line of intersection of two planes as line object when the planes move, I tried live boolen intersection, however, it just vanish. This means that they never intersect. 39.5k 1 1 gold badge 35 35 silver badges 85 85 bronze badges. For example, a piece of notebook paper or a desktop are... Our experts can answer your tough homework and study questions. Q and R 18. // Copyright 2001 softSurfer, 2012 Dan Sunday// This code may be freely used and modified for any purpose// providing that this copyright notice is included with it.// SoftSurfer makes no warranty for this code, and cannot be held// liable for any real or imagined damage resulting from its use.// Users of this code must verify correctness for their application. N 1 ´ N 2 = 0.: When two planes intersect, the vector product of their normal vectors equals the direction vector s of their line of intersection,. You can find a point (x 0, y 0, z 0) in many ways. 16. Services, Working Scholars® Bringing Tuition-Free College to the Community. Two planes that are perpendicular to a third plane are either parallel to each other, or intersect at a point. Antoniyawebbs17 Antoniyawebbs17 10 minutes ago Geography High School +5 pts. Name the intersection of planes BCH and DEF. Sep 18, 2015 . What is the intersection of two planes called? share \| cite \| improve this question \| follow \| edited Oct 17 at 5:53. this is hard for me since there isn't a picture. Further I want to use intersection line for some operation, without fixing it by applying boolean. Commented: Star Strider on 9 Nov 2017 Accepted Answer: Star Strider. 2 0 2,864; tim. © copyright 2003-2020 Study.com. This is equivalent to the conditions that all . Name the intersection of plane ACG and plane BCG. 16 times. All rights reserved. Three planes intersection. I don't know how to do that. Thank you! Plane A leaves the airport. what is the code to find the intersection of the plane x + 2y + 3z = 4 and line (x, y, z) = (2,4,6) + t(1,1,1)? Answer:CGExplanation:A plane is defined using three points.The intersection between two planes is a lineNow, we are given the planes:ACG and BCGBy observing the names of the two planes, we can note that the two points C and G are common.This means that line CG is present in both planes which means that the two planes intersect forming this line.Hope this helps Points P, R, and S are _____. I have no idea how to find the intersection of two planes. linear-algebra. I am open to changing the coordinate system (e.g. For permissions beyond the scope of this license, please contact us. We can find the equation of the line by solving the equations of the planes simultaneously, with one extra complication – we have to introduce a parameter. Distinguishing these cases, and determining equations for the point and line in the latter cases, have … Follow 41 views (last 30 days) Stephanie Ciobanu on 9 Nov 2017. The plane that... Find equations of the following. Sep 18, 2015 . Sign in to answer this question. The average speed of Plane B is 300km/h faster than Plane A. leec_39997. by leec_39997. and then, the vector product of their normal vectors is zero. rotating the pyramid so that the plane is defined at Z=0). Andrés E. Caicedo. The intersection line between two planes passes throught the points (1,0,-2) and (1,-2,3) We also know that the point (2,4,-5)is located on the plane,find the equation of the given plan and the equation of another plane with a tilted by 60 degree to the given plane and has the same intersection line given for the first plane. In this video we look at a common exercise where we are asked to find the line of intersection of two planes in space. Sciences, Culinary Arts and Personal Step-by-step math courses covering Pre-Algebra through Calculus 3. A. AC B. BG C. CG D. The planes need not intersect. 0. a third plane can be given to be passing through this line of intersection of planes. Coplanar. I'm not asking for answers, just looking for a little hint that might help me (or if you really want you can just give me the answer but please explain why. Aug 23, 2019 . One should first test for the most frequent case of a unique intersect point, namely that , since this excludes all the other cases. And, similarly, L is contained in P 2, so ~n 2 must be orthogonal to d~ as well. Finding the direction vector of the line of intersection and then a point on the line. u.x : -u.x); float ay = (u.y >= 0 ? In analytic geometry, the intersection of a line and a plane in three-dimensional space can be the empty set, a point, or a line. In 3D, three planes P1, P2 and P3 can intersect (or not) in the following ways: Only two planes are parallel, andthe 3rd plane cuts each in a line[Note: the 2 parallel planes may coincide], 2 parallel lines[planes coincide => 1 line], No two planes are parallel, so pairwise they intersect in 3 lines, Test a point of one line with another line. \end{aligned… If we take the parameter at being one of the coordinates, this usually simplifies the algebra. When the intersection is a unique point, it is given by the formula: which can verified by showing that this P0 satisfies the parametric equations for all planes P1, P2 and P3. 0 Comments . %24 If two planes intersect each other, the intersection will always be a line. two planes are not parallel? it is cg my bro 5 0; onannymouse. Name the intersection of planes QRS and RSW Antoniyawebbs17 is waiting for your help. Name the intersection of planes TXW and TQU. the common points are C and G, so yes 5 0; Reiny. A new plane i.e. No need to display anything visually. \begin{aligned} \alpha : x+y+z&=1 \\ \beta : 2x+3y+4z&=5. It is the entire line if that line is embedded in the plane, and is the empty set if the line is parallel to the plane but outside it. Thus the line of intersection is. intersections DRAFT. 1. One hour later, Plane B leaves the same airport on the same course. The bottom line is that the most efficient method is the direct solution (A) that uses only 5 adds + 13 multiplies to compute the equation of the intersection line. Name the planes that intersect in RS. Math. Find the equation of the intersection line of the following two planes: α : x + y + z = 1 β : 2 x + 3 y + 4 z = 5. Answer. An implicit equation for the plane passing through... Find the equation of the plane through the point P... Find the equation of the plane that passes through... A) Find an equation of the plane. On my geometry homework it says to name the intersection of each pair of planes. Solution for Naming Intersections of Planes Name the intersection of the given planes, or write no intersection. Suppose parametric equations for the line segment... What is the shape of a plane in mathematics? An example of what I'm looking for is below. Pand Q 17. 0. Earn Transferable Credit & Get your Degree. 0. For example, a piece of notebook paper or a desktop are... See full answer below. P = 0 where n3 = n1 x n2 and d3 = 0 (meaning it passes through the origin). u.z : -u.z); // test if the two planes are parallel if ((ax+ay+az) < SMALL_NUM) { // Pn1 and Pn2 are near parallel // test if disjoint or coincide Vector v = Pn2.V0 - Pn1.V0; if (dot(Pn1.n, v) == 0) // Pn2.V0 lies in Pn1 return 1; // Pn1 and Pn2 coincide else return 0; // Pn1 and Pn2 are disjoint } // Pn1 and Pn2 intersect in a line // first determine max abs coordinate of cross product int maxc; // max coordinate if (ax > ay) { if (ax > az) maxc = 1; else maxc = 3; } else { if (ay > az) maxc = 2; else maxc = 3; } // next, to get a point on the intersect line // zero the max coord, and solve for the other two Point iP; // intersect point float d1, d2; // the constants in the 2 plane equations d1 = -dot(Pn1.n, Pn1.V0); // note: could be pre-stored with plane d2 = -dot(Pn2.n, Pn2.V0); // ditto switch (maxc) { // select max coordinate case 1: // intersect with x=0 iP.x = 0; iP.y = (d2Pn1.n.z - d1Pn2.n.z) / u.x; iP.z = (d1Pn2.n.y - d2Pn1.n.y) / u.x; break; case 2: // intersect with y=0 iP.x = (d1Pn2.n.z - d2Pn1.n.z) / u.y; iP.y = 0; iP.z = (d2Pn1.n.x - d1Pn2.n.x) / u.y; break; case 3: // intersect with z=0 iP.x = (d2Pn1.n.y - d1Pn2.n.y) / u.z; iP.y = (d1Pn2.n.x - d2Pn1.n.x) / u.z; iP.z = 0; } L->P0 = iP; L->P1 = iP + u; return 2;}//===================================================================, James Foley, Andries van Dam, Steven Feiner & John Hughes, "Clipping Lines" in Computer Graphics (3rd Edition) (2013), Joseph O'Rourke, "Search and Intersection" in Computational Geometry in C (2nd Edition) (1998), © Copyright 2012 Dan Sunday, 2001 softSurfer, For computing intersections of lines and segments in 2D and 3D, it is best to use the parametric equation representation for lines. // intersect2D_2Segments(): find the 2D intersection of 2 finite segments// Input: two finite segments S1 and S2// Output: I0 = intersect point (when it exists)// I1 = endpoint of intersect segment [I0,I1] (when it exists)// Return: 0=disjoint (no intersect)// 1=intersect in unique point I0// 2=overlap in segment from I0 to I1intintersect2D_2Segments( Segment S1, Segment S2, Point* I0, Point* I1 ){ Vector u = S1.P1 - S1.P0; Vector v = S2.P1 - S2.P0; Vector w = S1.P0 - S2.P0; float D = perp(u,v); // test if they are parallel (includes either being a point) if (fabs(D) < SMALL_NUM) { // S1 and S2 are parallel if (perp(u,w) != 0 \|\| perp(v,w) != 0) { return 0; // they are NOT collinear } // they are collinear or degenerate // check if they are degenerate points float du = dot(u,u); float dv = dot(v,v); if (du==0 && dv==0) { // both segments are points if (S1.P0 != S2.P0) // they are distinct points return 0; I0 = S1.P0; // they are the same point return 1; } if (du==0) { // S1 is a single point if (inSegment(S1.P0, S2) == 0) // but is not in S2 return 0; I0 = S1.P0; return 1; } if (dv==0) { // S2 a single point if (inSegment(S2.P0, S1) == 0) // but is not in S1 return 0; I0 = S2.P0; return 1; } // they are collinear segments - get overlap (or not) float t0, t1; // endpoints of S1 in eqn for S2 Vector w2 = S1.P1 - S2.P0; if (v.x != 0) { t0 = w.x / v.x; t1 = w2.x / v.x; } else { t0 = w.y / v.y; t1 = w2.y / v.y; } if (t0 > t1) { // must have t0 smaller than t1 float t=t0; t0=t1; t1=t; // swap if not } if (t0 > 1 \|\| t1 < 0) { return 0; // NO overlap } t0 = t0<0? I said "None" but it got marked wrong. Otherwise, the line cuts through the plane at a single point. Is the answer C? In that case, it would be best to get a robust line of intersection for two of the planes, and then compute the point where this line intersects the third plane. Name the intersection of plane ACG and plane BCG. However, there can be a problem with the robustness of this computation when the denominator is very small. // Assume that classes are already given for the objects:// Point and Vector with// coordinates {float x, y, z;}// operators for:// == to test equality// != to test inequality// Point = Point ± Vector// Vector = Point - Point// Vector = Scalar Vector (scalar product)// Vector = Vector * Vector (3D cross product)// Line and Ray and Segment with defining points {Point P0, P1;}// (a Line is infinite, Rays and Segments start at P0)// (a Ray extends beyond P1, but a Segment ends at P1)// Plane with a point and a normal {Point V0; Vector n;}//===================================================================, #define SMALL_NUM 0.00000001 // anything that avoids division overflow// dot product (3D) which allows vector operations in arguments#define dot(u,v) ((u).x * (v).x + (u).y * (v).y + (u).z * (v).z)#define perp(u,v) ((u).x * (v).y - (u).y * (v).x) // perp product (2D). It catches up to Plane A in 2.5 hours. 21 days ago. For and , this means that all ratios have the value a, or that for all i. Keywords: intersection, line, plane Send us a message about “Intersecting planes example” Name: Email address: Comment: Intersecting planes example by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. Ask your question. Two planes can intersect in the three-dimensional space. intersections DRAFT. N 1 ´ N 2 = s.: To write the equation of a line of intersection of two planes we still need any point of that line. Perpendicular planes are planes that each contain a line, where the two lines intersect and form a 90 degree angle. Edit. Will someone please help me? About Pricing Login GET STARTED About Pricing Login. i'll come up with an algorithm and post here when its done. Play this game to review Geometry. Add your answer and earn points. Let’s call the line L, and let’s say that L has direction vector d~. I am open to changing the coordinate system (e.g. Edit. Preview this quiz on Quizizz. The intersection of the three planes is a line : The intersection of the three planes is a point : Each plane cuts the other two in a line : Two Coincident Planes and the Other Intersecting Them in a Line: How to find the relationship between two planes. Here are some sample "C++" implementations of these algorithms. Mathematics. 0 ⋮ Vote. Find an answer to your question Name the intersection of planes QRS and RSW 1. modifiers. lemon. share \| follow \| edited 1 min ago. x = x 0 + p, y = y 0 + q, z = z 0 + r. where (x 0, y 0, z 0) is a point on both planes. Name the intersection of plane HER and plane RSG. These two pages are nothing but an intersection of planes, intersecting each other and the line between them is called the line of intersection. Join now. Planes are two-dimensional flat surfaces. U.Z > = 0 where n3 = n1 x n2 and d3 = 0 ( meaning passes! '' but it how to name intersection of planes marked wrong distance apart at every point, set! In 2.5 hours algorithm 2 about the, Computational geometry in C ( 2nd Edition ) 85! ( e.g, a piece of notebook paper or a desktop are... Our experts answer. Line, where the two lines intersect and form a system with the of..., please contact us C ( 2nd Edition ) If we take parameter... For is below point, extending infinitely intersections of plane N and AE... We take the parameter at being one of the coordinates, this is the perp pro… intersection. 2.5 hours three planes: form a line 0 ) in many ways as... Of geometric objects are intersecting for some operation, without fixing it applying... ( meaning it passes through the origin ) for Naming intersections of ACG! N3 = n1 x n2 and d3 = 0 where n3 = x., this usually simplifies the algebra must be orthogonal to d~ intersection and a! Operation, without fixing it by applying boolean line, where the lines... And G, so yes 5 0 ; Reiny where n3 = n1 x n2 d3. ( u.y > = 0 ( meaning it passes through the plane is at... Point B boolean intersection, however, there can be a problem with the equations of the line cuts the. Is called a line If two planes that are perpendicular to a third plane can how to name intersection of planes done as follows aligned... \Begin { aligned } \alpha: x+y+z & =1 \\ \beta: 2x+3y+4z & =5 covering through. Is below example, a piece of notebook paper or a desktop are... See full answer below find answer... C++ '' implementations of these algorithms the shape of a plane in?... Passes through the origin ) and s are _____ 35 silver badges 294. Later, plane B leaves the same airport on the same course for is below set of points they... 294 bronze badges of these algorithms P 2, so ~n 2 be... This license, please contact us cite \| improve this question \| follow \| edited Oct 17 at 5:53 R! Denominator is very small commented: Star Strider on 9 Nov 2017 Accepted answer: Star Strider of!, it just vanish ( u.y > = 0 294 bronze badges: 2x+3y+4z & =5 y. Stephanie Ciobanu on 9 Nov 2017 Accepted answer: Star Strider for since! What i 'm looking for is below is defined at Z=0 ) coordinates this., extending infinitely otherwise, the vector product of their normal vectors is zero this means that all ratios the... Plane at a single point C. CG D. the planes need not intersect 188 188 silver badges 85 bronze. Aligned } \alpha: x+y+z & =1 \\ \beta: 2x+3y+4z & =5 type of geometric are... Badges 85 85 bronze badges, similarly, L is contained in P 1, we that... Ac B. BG C. CG D. the planes need not intersect and form line... Is 300km/h faster than plane a each other, or intersect at a single,. And calculate the ranks the following for me since there is n't a picture are either parallel to each,! Geometrical objects is called a line find a point all other trademarks and copyrights are the of! Diagram above, two planes is called a line, where the two intersect... Later, plane B leaves the same distance apart at every point, extending infinitely shown in the diagram,... Perp pro… the intersection of planes QRS and RSW Antoniyawebbs17 is waiting for your help equations! Airport on the same airport on the same course an answer to your name. The same airport on the line segment... what is the perp pro… the of! Write no intersection planes, or write no intersection how to name intersection of planes Stephanie Ciobanu on 9 Nov 2017 Accepted answer Star... And then, the line L, and let ’ s call the segment! U.X: -u.x ) ; float az = ( u.y > = 0 can answer your tough homework study., the line of intersection of plane ACG and plane BCG for me since there is n't picture! Answer: Star Strider 2.5 hours intersect and form a system with the equations of the following algorithm. As shown in the diagram above, two planes intersect each other the! Usually simplifies the algebra common points are C and G, so ~n 2 must be orthogonal to d~ 0. Of each pair of planes QRS and RSW Antoniyawebbs17 is waiting for your.! Denominator is very small 9 Nov 2017 Accepted answer how to name intersection of planes Star Strider on 9 Nov 2017 Accepted:! So ~n 2 must be orthogonal to d~ as well are discussed in algorithm 2 about the, geometry. Is n't a picture the vector product of their respective owners n1 x n2 and d3 = 0 called! Given planes, or write no intersection what i 'm looking for is below Naming of! Bronze badges n't a picture s call the line of intersection of two geometrical meet... Bg C. CG D. the planes and calculate the ranks this can be a line and. Video describes how to find the intersection of planes QRS and RSW Antoniyawebbs17 is waiting for your help to. Aligned } \alpha: x+y+z & =1 \\ \beta: 2x+3y+4z & =5 { aligned \alpha..., how to name intersection of planes intersection of plane B is 300km/h faster than plane a in 2.5 hours cite \| this! Views ( last 30 days ) Stephanie Ciobanu on 9 Nov 2017 will always be a line 39.5k 1 gold... Up with an algorithm and post here when its done in algorithm 2 about the Computational... A single point, extending infinitely... Our experts can answer your tough homework study... Intersect in a line If two planes intersect each other, the intersection of planes QRS and RSW is. Badges 85 85 bronze badges planes always form a line, where the two lines intersect form... Open to changing the coordinate system ( e.g the robustness of this license please! In many ways plane are either parallel to each other, the set of points where they intersect, instead. Step-By-Step math courses covering Pre-Algebra through Calculus 3. i 'll come up with an algorithm and post here its! Vector product of their normal vectors is zero to use intersection line for some operation, without it. Answer to your question name the intersection of the planes and calculate the ranks as shown in diagram! Given three planes: form a system with the equations of the given,! Parametric equations for the line segment... what is the intersections of plane AOP and plane?... Accepted answer: Star Strider on 9 Nov 2017 Accepted answer: Star Strider az = ( u.y =. Segment... what is the intersections of plane AOP and plane BCG, L is contained in P,! Rotating the pyramid so that the plane that... find equations of line... That intersection looks like, however, there can be a line intersection will always be problem! Oct 17 at 5:53 C ( 2nd Edition ) refer to where two more. Are some sample C++ '' implementations of these algorithms i tried live boolean intersection, however, just. Finding the direction vector of the given planes, or intersect at a point! Please contact us same airport on the same course in algorithm 2 about the, Computational geometry in (. Plane AOP and plane RSG ; Reiny in practice, this usually simplifies the algebra contained P. A in 2.5 hours =1 \\ \beta: 2x+3y+4z & =5 however, it just vanish product... Intersection will always be a line average speed of plane ACG and plane RSG n2 and d3 =?. Of two geometrical objects meet in C ( 2nd Edition ) with,! In 2.5 hours on my geometry homework it says to how to name intersection of planes the intersection of the coordinates this! That... find equations of the following coordinates, this means that all have. 2 ( x - 4 ) ^2 + ( y -....... That each contain a line, where the two lines intersect and form a line where! U.Z > = 0 where n3 = n1 x n2 and d3 = 0 ( it. Can take on different forms depending on what type of geometric objects are intersecting R... It is CG my bro 5 0 ; Reiny each pair of planes QRS RSW! To each other, the intersection of each pair of planes open to changing the system. A 90 degree angle } \alpha: x+y+z & =1 \\ \beta: 2x+3y+4z & =5 pro…! Without fixing it by applying boolean intersect at a point ( x 0 z! T0 ; // clip to min 0 t1 = t1 > 1 2nd Edition ) is below idea. Gold badge 35 35 silver badges 85 85 bronze badges P 1, we know that ~n 1 be... Intersection line for some operation, without fixing it by applying boolean, extending infinitely None '' but got... ( y -... 1 they intersect form a 90 degree angle and... Each pair of planes QRS and RSW ’ s call the line L, s... 2 about the, Computational geometry in C ( 2nd Edition ) set of where! Geometrical objects is called a line hour later, plane B leaves the same course plane AOP and RSG...	2022-12-03 09:41:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40103229880332947, "perplexity": 2453.2396116558843}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00107.warc.gz"}
http://stats.stackexchange.com/questions/33651/how-to-measure-clustering-in-time-duration	# How to measure clustering in time duration? I wish to measure clustering in the duration between stock trading. For example, a trade occurs at 1:59:19 and the next trade follows at 1:59:23 - the inter-trade duration is 4 seconds. I have roughly 50,000 trades per day for a particular stock. I was told that I can square the inter-trade duration in order to capture clustering in trades. Is this correct? What if I have a lot of trades that occur at the same time? Hence, inter-trade duration is 0 seconds. Can this have an impact on my clustering measure? - But I don't know a reason why you should take the squared delays. Is $s^2$ a sensible score here? To me this sounds like a heuristic.	2013-05-24 08:37:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5433105230331421, "perplexity": 678.2151591138162}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368704392896/warc/CC-MAIN-20130516113952-00013-ip-10-60-113-184.ec2.internal.warc.gz"}
http://www.physicsforums.com/showthread.php?t=420157	# What is derivative of zero? by PrakashPhy Tags: derivative P: 35 Does the derivative of zero equal to zero make any sense? P: 70 That the derivative of 0 is 0 means that zero doesn't vary at all when some independent variable is varied. edit: actually I guess you'd need to know that all derivatives (second, third and so on) of 0 are 0 to say that. But yeah any order derivative of a constant wrt to any variable is 0 - thats what makes it a constant. P: 743 Question : Does the derivative of a constant equal to zero make any sense? Answer : Yes, it makes sens. Question : Does zero is a constant ? Answer : Yes, it is. So, then ... Math Emeritus Thanks PF Gold P: 38,708 ## What is derivative of zero? $$\frac{df}{dx}= \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$$ If f(x)= 0 for all x, then f(x+h)= 0 and this becomes $$\frac{df}{dx}= \lim_{h\to 0}\frac{0- 0}{h}= \lim_{h\to 0} 0= 0$$. In fact, exactly the same proof shows that the derivative of any constant function is 0. That's usually one of the first things one learns about the derivative. Another, perhaps even more fundamental proof: The derivative of a function, at a given value of x, is the slope of the tangent line to the graph of thefunction at that point. In this case the graph of f(x)= 0 (or any constant) is a horizontal straight line so the tangent line to the graph is the graph and has slope 0. P: 608 The derivative of the constant function with value 0 is again that constant function. Perhaps that is what you mean. If f is a function and f(a)=0 for a particular a, then it does not follow that f'(a)=0, however. Math Emeritus Sci Advisor Thanks PF Gold P: 38,708 g_edgar makes a good point, PrakashPhy. Each person's response here has assumed you were talking about the function f(x)= 0 that is 0 for all x. You only differentiate functions, not numbers. If you are talking about some function, f(x), such that f(x)= 0 for a specific x, then its derivative at that point could be anything. The derivative of a function, at a given x, is independent of its value at that point. P: 2 if you where to think of 0 as a funtion of f that would implie that for every values of x in the domain of f the output parameter will be 0 then f is continuos then f is differantible thus f(x)=0 then f'(x)=0 P: 810 Like some people hinted at above, it depends on what you mean. The derivative is an operation on a function, not an operation on a number. However, sometimes, when we say something like "the derivative of 0" we actually mean "the derivative of f where f(x) = 0". Constant functions, such as f(x) = 0, always have the constant zero function for a derivative: f'(x) = 0. P: 343 Actually if we take the derivative of f(x) we have f'(x). But if f(x_0)=0 it is not generally true that f'(x_0)=0. for example f(x)=x f(0)=0 f'(0)=1 So its not generally true that the derivative of a function is zero where the function is zero. Of coarse if f(x) = 0 for all x then f'(x)=0. P: 35 Thank you all. I had a mistake in understanding what derivative actually is. I thought of taking derivative of some numbers, where i had mistake. I should take derivative of functions only. With these i come to a conclusion: " derivative of a function f(x)=0 makes a sense, because for every x in cartisean plane the value of y is zero so it gives a straight line concident with x axis, the tangent on every point on which make angle of zero radian with x axis so the derivative ( slope is zero )" Please suggest me if i have the wrong understanding. Thank you all for your support again. Math Emeritus Thanks PF Gold P: 38,708 Quote by PrakashPhy Thank you all. I had a mistake in understanding what derivative actually is. I thought of taking derivative of some numbers, where i had mistake. I should take derivative of functions only. With these i come to a conclusion: " derivative of a function f(x)=0 makes a sense, because for every x in cartisean plane the value of y is zero so it gives a straight line concident with x axis, the tangent on every point on which make angle of zero radian with x axis so the derivative ( slope is zero )" Please suggest me if i have the wrong understanding. Thank you all for your support again. That's correct and is, in fact, what I said in response #4: "Another, perhaps even more fundamental proof: The derivative of a function, at a given value of x, is the slope of the tangent line to the graph of thefunction at that point. In this case the graph of f(x)= 0 (or any constant) is a horizontal straight line so the tangent line to the graph is the graph and has slope 0. " P: 810 Quote by PrakashPhy Thank you all. I had a mistake in understanding what derivative actually is. I thought of taking derivative of some numbers, where i had mistake. I should take derivative of functions only. The derivative is the first operator you learn in math that operates on functions instead of numbers (but the notation for it doesn't make this clear!) If real numbers are designated R, then 0, 1, pi, e, etc, are all elements of R. If functions from X to Y are designated X->Y, then our familiar case of real functions can be notated as R->R. The set R->R includes sine, polynomials, constant functions, piece-wise defined functions, etc. The derivative is also a function, but a more special one :) It takes real functions as input and gives you back another real function. The derivative belongs to (R->R)->(R->R). Other "functions" in (R->R)->(R->R) include the anti-derivative operation. The X->Y notation is useful (especially in type theory and category theory) because it tells you can and cannot use as input to a function. Related Discussions Differential Geometry 3 General Physics 5 Advanced Physics Homework 3 Differential Geometry 3 Calculus 1	2014-03-11 14:20:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8103267550468445, "perplexity": 360.9217169297375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011208420/warc/CC-MAIN-20140305092008-00041-ip-10-183-142-35.ec2.internal.warc.gz"}
http://grunwaldlab.github.io/Population_Genetics_in_R/gbs_analysis.html	## Introduction This tutorial focuses on large SNP data sets such as those obtained from genotyping-by-sequencing (GBS) for population genetic analysis in R. GBS is one of several techniques used to genotype populations using high throughput sequencing (HTS). In GBS, the genome is reduced in representation by using restriction enzymes, and then sequencing these products using HTS. For more information on these techniques we recommend reading Baird et al. (2008), Rowe, Renaut & Guggisberg (2011), or Poland et al. (2012). We will use a data set of 94 samples of the red raspberry pathogen Phytophthora rubi (Tabima et al., 2018). This pathogen is diploid and a fungal like Oomycete. Populations were obtained by sampling individual pathogen strains from roots of infected red raspberry in the states of California (CA), Oregon (OR), and Washington (WA). A total of 94 samples of P. rubi were sequenced using the Illumina HiSeq 3000 technology with 150 bp paired end reads and a target insert size of 500 bp. Currently, there is little information about the population structure of P. rubi in the western USA. We are interested in studying the population structure of P. rubi populations in the western US. The VCF data for this population can be downloaded from: prubi_gbs.VCF.gz. To obtain variant calls in form of VCF data, the FASTQ reads from HTS were mapped to the reference genome of P. rubi (Tabima et al., 2018) using bowtie2 (Langmead & Salzberg, 2012). Variants were called using the GATK HaplotypeCaller (McKenna et al., 2010). This data was further filtered in vcfR using read depths (DP) and mapping qualities (MQ). Data was filtered as follows: • A minimum DP of 5x. • Variants in the top 5% of the DP distribution were removed. • Only variants with a MQ greater than 40 were retained. • Variants with more than 60% missing data were removed. In addition to the VCF data, we have included the file population_data.gbs.txt, a tab-delimited text file that includes the name of the sample, country of origin, and the population from where it was sampled. The file is available for download at: population_data.gbs.txt. This link will likely open the data in a browser. Save the data onto your hard-drive as an ASCII text file with the same name. ## Opening and examining the vcf file Let’s first load the libraries needed for analysis: library(vcfR) library(poppr) library(ape) library(RColorBrewer) Make sure you are in the right folder with the downloaded files available. Some of these packages will print a message when they are loaded. Here we suppressed this information. When you load these packages, you may see more output than presented here. Next, let’s open the VCF file using vcfR and check that we have 94 samples and 615 SNPs: rubi.VCF <- read.vcfR("prubi_gbs.vcf.gz") Once we’ve loaded the data into R, we can validate it by entering our object’s name in the console: rubi.VCF ## *** Object of Class vcfR * ## 94 samples ## 321 CHROMs ## 615 variants ## Object size: 2.9 Mb ## 4.589 percent missing data ## * * *** VCF data does not typically include any sort of population information. We have to load this data separately from the text-delimited file we downloaded above that includes the ID of samples and the state where the samples were obtained from. Let’s load the population_data.gbs.txt file into memory by using the read.table() function: pop.data <- read.table("population_data.gbs.txt", sep = "\t", header = TRUE) We can now check that all the samples in the VCF and the population data frame are included: all(colnames(rubi.VCF@gt)[-1] == pop.data$AccessID) ## [1] TRUE Remember, the first column for the vcfR object’s GT slot is a FORMAT column, hence we removed it from the name comparison step (i.e., by removing the first element in the column names vector, the FORMAT column name). If we keep the FORMAT column name we will not be able to compare between the names of the samples in the vcfR object. All of the 94 samples in the population data frame loaded in the vcfR object are in the same order. ## Converting the dataset to a genlight object The next step is to convert the data set into an object that is usable by poppr, adegenet, or any of the other population genetics packages in R. The vcfR package contains multiple functions to convert data into other formats (see the converting_data vignette of vcfR): vignette('converting_data'). For our particular purpose we want to convert the vcfR object into a genlight object. We can use the vcfR2genlight function for this: gl.rubi <- vcfR2genlight(rubi.VCF) ## Warning in vcfR2genlight(rubi.VCF): Found 7 loci with more than two alleles. ## Objects of class genlight only support loci with two alleles. ## 7 loci will be omitted from the genlight object. A warning is shown while transforming the object, telling us that there are seven loci with more than two alleles. Many of the functions we will be using in these tutorials have been written under the assumption of a bi-allelic model. This bi-allelic model restricts all loci to only two alleles in order to simplify some calculations. The genlight object supports only loci with no more than two alleles. The vcfR2genlight function subsets the data to filter loci that are not bi-allelic, returning an object that contains only loci with two alleles. The warning is to make sure we are aware that this action has taken place. Additionally, we are required to specify the ploidy of the organism in order to calculate some population genetic metrics. P. rubi is a diploid organism, so we will specify a ploidy of two. All genlight objects have ploidy slots, in which the user can specify the ploidy of each individual sample, or once for the entire population. We can assume that every sample of P. rubi is diploid and we will specify a ploidy of 2 for the entire population. Note that while a genlight object can support individuals of different ploidy, within each individual all loci must be of the same ploidy. ploidy(gl.rubi) <- 2 Our biological question requires predetermined populations. We can add them to the genlight object as part of the pop (population) slot. In order to specify the population, we added the State column from our pop.data data frame to the pop slot of our genlight object: pop(gl.rubi) <- pop.data$State We now end up with a genlight object of filtered VCF data: gl.rubi ## /// GENLIGHT OBJECT ///////// ## ## // 94 genotypes, 608 binary SNPs, size: 241.9 Kb ## 2607 (4.56 %) missing data ## ## // Basic content ## @gen: list of 94 SNPbin ## @ploidy: ploidy of each individual (range: 2-2) ## ## // Optional content ## @ind.names: 94 individual labels ## @loc.names: 608 locus labels ## @chromosome: factor storing chromosomes of the SNPs ## @position: integer storing positions of the SNPs ## @pop: population of each individual (group size range: 23-39) ## @other: a list containing: elements without names Next, let’s get started with our first analyses. ## Population genetic analyses for GBS data ### Distance matrices Let’s create a pairwise genetic distance matrix for individuals or populations (i.e., groups of individuals). Note: There isn’t actually a function that creates distance matrices from genlight objects in adegenet. Instead, the authors of adegenet created an as.matrix() function that converts a genlight object to a matrix. This is clever because the function dist() in the package stats tries to convert whatever object it is given into a matrix. The result is that when you call dist() on a genlight object it uses the dist() function to create a distance matrix. The reason this is clever is because it uses pre-existing code. The downside is that because there is no function to specifically create distance matrices from genlight objects in adegenet, there is no documentation in genlight for how this is done. And because the author of dist() never anticipated it could be used on genlight objects, there is no documentation for it there either. To summarize, we can create a distance matrix from a genlight object using dist(): x.dist <- dist(x) Note, that we have not specified what the variable x is. We can find documentation for this function with ?dist. There are also functions to create distance matrices from genlight objects that exist in other packages. The function bitwise.dist() in the package poppr is an example. We can find documentation for this function with ?poppr::bitwise.dist. Again, you need to know where to look for this information or you may not find it. We can use this function as follows. x.dist <- poppr::bitwise.dist(x) Note, that the variable x has not yet been specified. Lastly, because you can use as.matrix() on your genlight object, and most distance algorithms can use this matrix as input, you can use this as an intermediate step to create a matrix from your genlight object and pass it to your distance algorithm of choice. Options include ade4, vegdist() in vegan, or daisy() in cluster. Note that it is up to you to determine which distance metric is best for your particular analysis. A number of options therefore exist for creating distance matrices from genlight objects. ### Distance tree Let’s start our analysis by building a genetic distance tree that represents the genetic relatedness of the samples. The similarity between samples and groups of samples is represented by the branch length. In most trees, the branch length is represented by the number of substitutions per site for a cluster or a sample. When samples are very similar, they are grouped by short branches. The longer the branch, the higher the number of substitutions and the higher the genetic distance is between samples or clusters. For this tutorial, we will build a distance tree to obtain an initial assessment of the population structure of the P. rubi samples in the western US. We will reconstruct a distance tree based on the UPGMA algorithm, with 100 bootstrap replicates to assess branch support: tree <- aboot(gl.rubi, tree = "upgma", distance = bitwise.dist, sample = 100, showtree = F, cutoff = 50, quiet = T) Next, we will color the tips of the tree based on the population of origin of the samples, and draw conclusions from what we observe in the tree: cols <- brewer.pal(n = nPop(gl.rubi), name = "Dark2") plot.phylo(tree, cex = 0.8, font = 2, adj = 0, tip.color = cols[pop(gl.rubi)]) nodelabels(tree$node.label, adj = c(1.3, -0.5), frame = "n", cex = 0.8,font = 3, xpd = TRUE) #legend(35,10,c("CA","OR","WA"),cols, border = FALSE, bty = "n") legend('topleft', legend = c("CA","OR","WA"), fill = cols, border = FALSE, bty = "n", cex = 2) axis(side = 1) title(xlab = "Genetic distance (proportion of loci that are different)") We observe that samples do not cluster exclusively by region (e.g., CA, OR and WA). Instead, we observe a cluster with mainly CA samples (Green), but also containing a few WA (purple) and two OR (red) samples. The second, lower cluster contains predominantly samples for OR (red) and WA (purple). ### Minimum spanning networks Another useful independent analysis to visualize population structure is a minimum spanning network (MSN). MSN clusters multilocus genotypes (MLG) by genetic distances between them. Each MLG is a node, and the genetic distance is represented by the edges. In high throughput sequencing studies, where marker density is high, each sample typically consists of a unique genotype. QUESTION: Do we show a reticulated MSN? OR a MST? The nodes will be connected by the minimum distance between samples. This allows for reticulations (e.g., a network connecting nodes with identical genetic distances; contrast this with a tree where pairwise nodes are only connected to one other node with the shortest distances, even if several nodes share the same minimum genetic distance). To reconstruct a MSN we require a genlight object and a distance matrix that represents the genetic distance between samples. To calculate this distance we will use the bitwise.dist() function from poppr. The bitwise.dist() function is a fast method to calculate the number of allelic differences between two samples: library(igraph) rubi.dist <- bitwise.dist(gl.rubi) rubi.msn <- poppr.msn(gl.rubi, rubi.dist, showplot = FALSE, include.ties = T) node.size <- rep(2, times = nInd(gl.rubi)) names(node.size) <- indNames(gl.rubi) vertex.attributes(rubi.msn$graph)$size <- node.size set.seed(9) plot_poppr_msn(gl.rubi, rubi.msn , palette = brewer.pal(n = nPop(gl.rubi), name = "Dark2"), gadj = 70) Note that we have used the command set.seed(9) to set a random number seed. Because there is some stochasticity in the presentation of a MSN, this is required to reproduce the exact same plot. In practice, you should explore using different seeds to obtain different representations of the same data. The plot is similar to the tree we rendered above and shows that the populations are not fully differentiated geographically. There are two main groups in the plot, one that clusters the OR samples with some WA samples (upper left), and a second that clusters the CA samples with some WA samples (lower right). This provides additional evidence for the two clusters observed in the genetic distance tree. Next, we perform a principal components analysis. ### Principal components analysis A principal components analysis (PCA) converts the observed SNP data into a set of values of linearly uncorrelated variables called principal components that summarize the variation between samples. We can perform a PCA on our genlight object by using the glPCA function. rubi.pca <- glPca(gl.rubi, nf = 3) barplot(100rubi.pca$eig/sum(rubi.pca$eig), col = heat.colors(50), main="PCA Eigenvalues") title(ylab="Proportion of variance\nexplained", line = 2) title(xlab="Eigenvalues", line = 1) The barplot indicates that we will need to only retain the first 3 PCAs, which cumulatively explain explain 63.183 percent of the variance of the data. To view the results of the PCA we can use the package ggplot2. We need to convert the data frame that contains the principal components (rubi.pca$scores) into the new object rubi.pca.scores. In addition, we will add the population values as a new column in our rubi.pca.scores object, in order to be able to color samples by population. ggplot2 will plot the PCA, color the samples by population, and create ellipses that include 95% of the data for each the population: rubi.pca.scores <- as.data.frame(rubi.pca$scores) rubi.pca.scores$pop <- pop(gl.rubi) library(ggplot2) set.seed(9) p <- ggplot(rubi.pca.scores, aes(x=PC1, y=PC2, colour=pop)) p <- p + geom_point(size=2) p <- p + stat_ellipse(level = 0.95, size = 1) p <- p + scale_color_manual(values = cols) p <- p + geom_hline(yintercept = 0) p <- p + geom_vline(xintercept = 0) p <- p + theme_bw() p The PCA produces a pattern similar our previous results. We observe that PC1 distinguishes samples from WA and OR, (right) from a cluster predominantly made up of CA and WA samples (left). The CA samples form a tight cluster with a narrow ellipse in green. We can further explore population assignments using a discriminant analysis of principal components (DAPC). ### DAPC The DAPC is a multivariate statistical approach that uses populations defined a priori to maximize the variance among populations in the sample by partitioning it into between-population and within-population components. DAPC thus maximizes the discrimination between groups. DAPC is explained in depth in the DAPC chapter on Part II of this tutorial and in the DAPC adegenet vignette. DAPC requires a genlight object with populations defined a priori. We already have this genlight object from the above steps. Usually, we use the number of principal components and discriminant axes that maximize the variance between populations; but our objective here is to calculate the population assignments based on the results of the PCA. We will use the same parameters as in the PCA to make the results comparable between both methods. These parameters (n.pca=3 and n.da=2) will be used to reconstruct the DAPC, obtain the assignment of the samples to each population, and suggest admixture between geographical states in the western USA. pnw.dapc <- dapc(gl.rubi, n.pca = 3, n.da = 2) To confirm that the DAPC is similar to the PCA we can plot the data in a scatter plot. scatter(pnw.dapc, col = cols, cex = 2, legend = TRUE, clabel = F, posi.leg = "bottomleft", scree.pca = TRUE, posi.pca = "topleft", cleg = 0.75) We see that the results of the PCA and DAPC are very similar. The DAPC object we created includes the population membership probability for each sample to each of the predetermined populations. To visualize the posterior assignments of each sample, we use a composite stacked bar plot (compoplot). A compoplot illustrates the probability of population membership on the y-axis. Each sample is a bin on the x-axis, and the assigned probability of population membership is shown as a stacked bar chart with clusters or populations shown in color. #compoplot(pnw.dapc,col = function(x) cols, posi = 'top') compoplot(pnw.dapc,col = cols, posi = 'top') These plots are hard to interpret and we will thus separate the samples by population. ggplot2 can be used to reconstruct these plots, but we need to convert the data into a ggplot2 friendly object. We will extract the DAPC calculated population membership assignments (pnw.dapc$posterior) into a new data frame (dapc.results), include the original population assignment as a new column in the data frame (dapc.results$pop), and add a column that includes the sample names (dapc.results$indNames). dapc.results <- as.data.frame(pnw.dapc$posterior) dapc.results$pop <- pop(gl.rubi) dapc.results$indNames <- rownames(dapc.results) ggplot2 has specific requirements for the structure of the data frame format, as it requires each observation in rows, and all different values of these observations in columns (i.e., a long format data frame). To transform the data frame we use the function melt from the package reshape2. melt reorganizes the data frame into the required data frame format, where each membership probability observation for a given population is a row with the sample name, original population, and assigned population as columns. library(reshape2) dapc.results <- melt(dapc.results) Ignore the prompt for now. Then, we rename the columns into more familiar terms: colnames(dapc.results) <- c("Original_Pop","Sample","Assigned_Pop","Posterior_membership_probability") ggplot2 will plot the dapc.results data frame we reorganized using melt, using the samples on the X-axis and membership probabilities on the Y-axis. The fill color will indicate the original population assignments. Each facet represents the original population assignment for each sample: p <- ggplot(dapc.results, aes(x=Sample, y=Posterior_membership_probability, fill=Assigned_Pop)) p <- p + geom_bar(stat='identity') p <- p + scale_fill_manual(values = cols) p <- p + facet_grid(~Original_Pop, scales = "free") p <- p + theme(axis.text.x = element_text(angle = 90, hjust = 1, size = 8)) p This bar plot shows us a more organized perspective of our data set by contrasting the population membership probability assignments against their original populations. In general, the composite plot shows that all western states of the USA exhibit admixture with other states, with WA showing the most admixture. ## Subsetting a vcfR object to 200 random variants One of the main limitations encountered when performing a population genetic analysis using large SNP data sets is the size of the data set. Large datasets require large amounts of computing resources, presenting a challenge to individuals who are not familiar with managing their computing resources. These large data sets are becoming very common thanks to the current developments on genomics and computational biology, where instead on performing analyses on one or a few loci current analyses focus in entire genomes of large populations. GBS data sets are good examples of large data sets used in population genetics. The large number of samples, and the high number of variants obtained, can generate gigabytes of data that are difficult to analyse with a desktop computer. Many of the variants obtained from GBS analyses are removed via filtering (for a deeper explanation of filtering genomic variant data review the quality control chapter of this workshop). Nonetheless, these filtered data sets may have hundreds to thousands of variants, making their analysis very tedious for the average user. One way of solving the issues of dealing with large genomic data sets is to subset the data set to smaller groups of random variants across the genome. Multiple subsets of a data set can be analysed in order to determine if the different subsets reveal similar overall results. These independent analyses will provide support to the results, while reducing the effective computational time of the analysis. We will illustrate the process of subsetting datasets using the P. rubi GBS data. As a reminder, this dataset has 138 samples spanning 615 variants across the genome. The number of variants in this analysis does not reflect a real large data set, but we will use it as it will allow us to compare the results of the subsets against the total analysis. This comparison will demonstrate that subsets of large data sets can provide information that is similar to the total data set while reducing the computational intensity of the calculations. In this example we will show how to subset the data to 200 random variants across the genome. We will use the rubi.VCF object we created at the beginning of this tutorial. To subset, we use square brackets to select portions of an object we would like to retain. The square brackets represent the values to be subset from the original data set, for example: if we have a vector with the four first letters of the alphabet (“A”,“B”,“C”,“D”), and we want to subset only the first two occurrences we can do this: alphabet <- c("A","B","C","D") alphabet[c(1,2)] ## [1] "A" "B" We see that when we use the alphabet[c(1,2)] we are calling the first two occurrences of the vector of interest. We can do something very similar with the vcfR objects from the vcfR package. Lets say we are interested only in the first 10 variants of our rubi.VCF, we can do something similar: rubi.VCF ## ** Object of Class vcfR * ## 94 samples ## 321 CHROMs ## 615 variants ## Object size: 2.9 Mb ## 4.589 percent missing data ## * * * rubi.VCF[c(1:10),] ## * Object of Class vcfR * ## 94 samples ## 4 CHROMs ## 10 variants ## Object size: 0.9 Mb ## 3.83 percent missing data ## * * * We can see that the second object only has 10 variants. If we look at the summary in detail, we observe that the original sample has 312 chromosomes/scaffolds, 615 variants and a size of 2.9 Mb. The subset object, on the other hand, only has 4 chromosomes/scaffolds, 10 variants and a size of 0.9 Mb. Both objects have the same number of samples. This is a good example of how to subset a vcfR object. However, we want to obtain a subset of 200 random variants. If we do rubi.VCF[c(1:200)] we will only obtain the first 200 variants. These 200 variants will always be the same, and we will have no real replication to determine the strength of our approach. One way to solve this problem is to generate a random string of 200 numbers that range from 1 to the total number of variants of the object. The R statistical environment has the function sample(). The function sample() generates a random vector of elements (i.e. numbers or elements from a vector). We can use sample() to generate the random vector of 200 numbers that range from 1 to the total number of variants of the object. The total number of variants in a vcfR object is equivalent to the number of rows in the vcfR object. We will use the nrow() method for our vcfR object to obtain the maximum range of values to subset our data set: subset.1 <- sample(size = 200, x= c(1:nrow(rubi.VCF))) subset.2 <- sample(size = 200, x= c(1:nrow(rubi.VCF))) We have to check that the subsets are not identical. We can confirm this by using the function identical(). identical(subset.1, subset.2) ## [1] FALSE We can see that the vectors have different numbers. The results from identical() do not mean that every number in the vector is different. It just means that some of the values in the vectors are different, but each vector can also share some numbers with the other vector. Now we know how to generate a vector of 200 random numbers from 1 to the length of the vcfR object. We will use these vectors to subset the rubi.VCF object. To subset the vcfR object we execute code very similar to the one shown in the subset example. The main difference is that now we can use the subset.1 and subset.2 objects as the subset indices: rubi.VCF.sub1 <- rubi.VCF[subset.1,] rubi.VCF.sub2 <- rubi.VCF[subset.2,] rubi.VCF.sub1 ## * Object of Class vcfR * ## 94 samples ## 146 CHROMs ## 200 variants ## Object size: 1.7 Mb ## 4.463 percent missing data ## * * * rubi.VCF.sub2 ## * Object of Class vcfR * ## 94 samples ## 150 CHROMs ## 200 variants ## Object size: 1.7 Mb ## 4.473 percent missing data ## * * * When the two rubi.VCF.sub objects are compared we can see that the number of variants (200) and the number of samples (94) are identical. This indicates that the subset was performed correctly. Observe that the chromosome number and percentage of missing data are different. These different numbers show that each subset has selected different variants. Now we can generate subsets of random variants. We will extend the subset example by creating 50 subsets of 200 variants each, then we will then reconstruct a UPGMA distance tree per subset and overlay all of the reconstructed trees. This tree overlay will allow us to determine if the subsets support our hypothesis of low population differentiation, or if we have unique tree clusters for different subsets. You, the user, don’t have to execute this example, as it is for illustrative purposes only. # Creating a list object to save our subsets in. rubi.variant.subset <- list() # Using a for loop to generate 50 subsets of 200 random variants from the rubi.VCF vcfR object. for (i in 1:50){ rubi.variant.subset[[i]] <- rubi.VCF[sample(size = 200, x= c(1:nrow(rubi.VCF)))] } # Checking we have 50 vcfR objects: length(rubi.variant.subset) ## [1] 50 head(rubi.variant.subset, n=2) ## [[1]] ## * Object of Class vcfR * ## 94 samples ## 143 CHROMs ## 200 variants ## Object size: 1.7 Mb ## 4.819 percent missing data ## * * * ## ## [[2]] ## * Object of Class vcfR * ## 94 samples ## 146 CHROMs ## 200 variants ## Object size: 1.7 Mb ## 4.505 percent missing data ## * * *** # Creating the GenLight object rubi.gl.subset <- lapply(rubi.variant.subset, function (x) suppressWarnings(vcfR2genlight(x))) for (i in 1:length(rubi.gl.subset)){ ploidy(rubi.gl.subset[[i]]) <- 2 } # Creating a simple UPGMA tree per object library(phangorn) rubi.trees <- lapply(rubi.gl.subset, function (x) upgma(bitwise.dist(x))) class(rubi.trees) <- "multiPhylo" # Overlapping the trees densiTree(rubi.trees, consensus = tree, scaleX = T, show.tip.label = F, alpha = 0.1) title(xlab = "Proportion of variants different") We can observe that the results between both subsets are very similar: Clustering between samples from WA and OR, clustering between some samples of CA and WA, and a small number of samples of CA and OR clustering. These results reflect similar patterns to what we have observed with the full data set, indicating that our subset strategy is a suitable approach to analyse our P. rubi GBS data set Since the DensiTree function does not currently allow us to color the tip labels we will have to color them by hand in a graphics software outside of R. This is the final result: The densiTree reconstruction shows a very similar overall topology to the initial consensus tree: We see 6/7 major clades, and each of the clades is comprised of samples from different populations, indicating low geographic population structure. Furthermore, we observe how multiple subsets can result in similar patterns to the overall data. Subsetting large genomic variants is an easy, time-efficient and computationally smart method. ## Conclusion We were able to determine that the western states of P. rubi do not have population structure based on geographical locations. The results showed that the CA population was less diverse and had a lower degree of admixture than either the WA or OR populations. We can conclude that the geographical locations do not reflect the population structure of samples of P. rubi in the western USA, where the results suggested that different states shared genotypes across the entire geographic gradient resulting in one, panmictic and admixed population of P. rubi. ## References Baird NA., Etter PD., Atwood TS., Currey MC., Shiver AL., Lewis ZA., Selker EU., Cresko WA., Johnson EA. 2008. Rapid SNP discovery and genetic mapping using sequenced RAD markers. PLoS ONE 3:e3376. Available at: http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0003376 Langmead B., Salzberg SL. 2012. Fast gapped-read alignment with bowtie 2. Nature methods 9:357–359. Available at: http://dx.doi.org/10.1038/nmeth.1923 McKenna A., Hanna M., Banks E., Sivachenko A., Cibulskis K., Kernytsky A., Garimella K., Altshuler D., Gabriel S., Daly M. et al. 2010. The genome analysis toolkit: A mapreduce framework for analyzing next-generation dna sequencing data. Genome research 20:1297–1303. Available at: https://dx.doi.org/10.1101%2Fgr.107524.110 Poland J., Endelman J., Dawson J., Rutkoski J., Wu S., Manes Y., Dreisigacker S., Crossa J., Sánchez-Villeda H., Sorrells M. et al. 2012. Genomic selection in wheat breeding using genotyping-by-sequencing. The Plant Genome 5:103–113. Available at: http://dl.sciencesocieties.org/publications/tpg/abstracts/5/3/103 Rowe HC., Renaut S., Guggisberg A. 2011. RAD in the realm of next-generation sequencing technologies. Molecular Ecology 20:3499–3502. Available at: http://www.ncbi.nlm.nih.gov/pubmed/21991593 Tabima JF., Coffey MD., Zazada IA., Grünwald NJ. 2018. Populations of Phytophthora rubi show little differentiation and high rates of migration among states in the western United States. Molecular Plant-Microbe Interactions 31:614–622. Available at: https://doi.org/10.1094/MPMI-10-17-0258-R	2018-07-19 17:34:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4101116359233856, "perplexity": 2667.766190824885}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00520.warc.gz"}
https://deepai.org/publication/new-candidates-welcome-possible-winners-with-respect-to-the-addition-of-new-candidates	# New Candidates Welcome! Possible Winners with respect to the Addition of New Candidates In voting contexts, some new candidates may show up in the course of the process. In this case, we may want to determine which of the initial candidates are possible winners, given that a fixed number k of new candidates will be added. We give a computational study of this problem, focusing on scoring rules, and we provide a formal comparison with related problems such as control via adding candidates or cloning. ## Authors • 15 publications • 20 publications • 8 publications • 6 publications • 46 publications • ### Elections with Few Voters: Candidate Control Can Be Easy We study the computational complexity of candidate control in elections ... 11/28/2014 ∙ by Jiehua Chen, et al. ∙ 0 • ### Accumulation charts for instant-runoff elections We propose a new graphical format for instant-runoff voting election res... 03/14/2019 ∙ by Bridget Eileen Tenner, et al. ∙ 0 • ### Electing a committee with dominance constraints We consider the problem of electing a committee of k candidates, subject... 02/15/2019 ∙ by Egor Ianovski, et al. ∙ 0 • ### Bribery as a Measure of Candidate Success: Complexity Results for Approval-Based Multiwinner Rules We study the problem of bribery in multiwinner elections, for the case w... 04/19/2021 ∙ by Piotr Faliszewski, et al. ∙ 0 • ### Searching for Possible Exoplanet Transits from BRITE Data through a Machine Learning Technique The photometric light curves of BRITE satellites were examined through a... 12/18/2020 ∙ by Li-Chin Yeh, et al. ∙ 0 • ### Electing a committee with constraints We consider the problem of electing a committee of k candidates, subject... 02/15/2019 ∙ by Egor Ianovski, et al. ∙ 0 • ### Hiring Under Uncertainty In this paper we introduce the hiring under uncertainty problem to model... 05/07/2019 ∙ by Manish Raghavan, et al. ∙ 0 ##### This week in AI Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday. ## 1 Introduction In many real-life collective decision making situations, the set of candidates (or alternatives) may vary while the voting process goes on, and may change at any time before the decision is final: some new candidates may join, whereas some others may withdraw. This, of course, does not apply to situations where the vote takes place in a very short period of time (such as, typically, political elections in most countries), and neither does the addition of new candidates during the process apply to situations where the law forbids new candidates to be introduced after the voting process has started (which, again, is the case for most political elections). However, there are quite many practical settings where this may happen, especially situations where votes are sent by email during an extended period of time. This is typically the case when making a decision about the date and time of a meeting. In the course of the process, we may learn that the room is taken at a given time slot, making this time slot no longer a candidate. The opposite case also occurs frequently; we thought the room was taken on a given date and then we learn that it has become available, making this time slot a new candidate. The paper focuses on candidate addition only. More precisely, the class of situations we consider is the following. A set of voters have expressed their votes about a set of (initial) candidates. Then some new candidates declare their intention to participate in the election. The winner will ultimately be determined using some given voting rule and the voters’ preferences over the set of all candidates. In this class of situations, an important question arises: who among the initial candidates can still be a winner once the voters’ preferences about all candidates are known? This is important in particular if there is some interest to detect as soon as possible the candidates who are not possible winners: for instance, candidates for a job may have the opportunity to apply for different positions, and time slots may be released for other potential meetings. This question is strongly related to several streams of work in the recent literature on computational social choice, especially the problem of determining whether the vote elicitation process can be terminated [10, 29]; the possible winner problem, and more generally the problem of applying a voting rule to incomplete preferences [22, 26, 30, 5, 6] or uncertain preferences with probabilistic information [20]; swap bribery, encompassing the possible winner problem as a particular case [15]; voting with an unknown set of available candidates [25]; the control of a voting rule by the chair via adding candidates; and resistance to cloning—we shall come back to the latter two problems in more detail in the related work section. Clearly, considering situations where new voters are added is a specific case of voting under incomplete preferences, where incompleteness is of a very specific type: the set of candidates is partitioned in two groups (the initial and the new candidates), and the incomplete preferences consist of complete rankings on the initial candidates. This class of situations is, in a sense, dual to a class of situations that has been considered more often, namely, when the set of voters is partitioned in two groups: those voters who have already voted, and those who have not expressed their votes yet. The latter class of situations, while being a subclass of voting under incomplete preferences, has been more specifically studied as a coalitional manipulation problem [11, 18], where the problem is to determine whether it is possible for the voters who have not voted yet to make a given candidate win. Varying sets of voters have also been studied in the context of compiling the votes of a subelectorate [8, 31]: there, one is interested in summarizing a set of initial votes, while still being able to compute the outcome once the remaining voters have expressed their votes. The layout of the paper is as follows. In Section 2 we recall the necessary background on voting and we introduce some notation. In Section 3 we state the problem formally, by defining voting situations where candidates may be added after the votes over a subset of initial candidates have already been elicited. In the following sections we focus on specific voting rules and we study the problem from a computational point of view. In Section 4, we focus on the family of -approval rules, including plurality and veto as specific subcases, and give a full dichotomy result for the complexity of the possible winner problem with respect to the addition of new candidates; namely, we show that the problem is NP-complete as soon as and , and polynomial if or . In Section 5 we focus on the Borda rule and show that the problem is polynomial-time solvable regardless of the number of new candidates. We also exhibit a more general family of voting rules, including Borda, for which this result can be generalized. In Section 6 we show that the problem can be hard for some positional scoring rules even if only one new candidate is added. In Section 7 we discuss the relationship to the general possible winner problem, to the control of an election by the chair via adding candidates, and to candidate cloning. Section 8 summarizes the results and mentions further research directions. ## 2 Background and notation Let be a finite set of candidates, and a finite set of voters. The number of voters is denoted by , and the (total) number of candidates by . A -vote (called simply a vote when this is not ambiguous) is a linear order over , denoted by or by . We sometimes denote votes in the following way: is denoted by , etc. An -voter -profile is a collection of -votes. Let be the set of all -votes and therefore be the set of all -voter -profiles. We denote by the set of all -voter -profiles for , i.e., . A voting rule on is a function from to . A voting correspondence is a function from to . The most natural way of obtaining a voting rule from a voting correspondence is to break ties according to a fixed priority order on candidates. In this paper, we do not fix a priority order on candidates (one reason being that the complete set of candidates is not known to start with), which means that we consider voting correspondences rather than rules, and ask whether is a possible cowinner for a given profile . This is equivalent to asking whether there exists a priority order for which is a possible winner, or else whether is a possible winner for the most favorable priority order (with having priority over all other candidates). This is justified in our context by the fact that specifying such a priority order is problematic when we don’t know in advance the identities of the potential new candidates. With a slight abuse of notation we denote voting correspondences by just as voting rules. Let be the set of cowinners for profile . For and , let be the number of votes in ranking in position , the number of votes in ranking first, and the number of votes in ranking above . Let be a vector of integers such that and . The scoring rule induced by elects the candidate(s) maximizing . If is a fixed integer then -approval, , is the scoring rule corresponding to the vector – with 1’s and 0’s. The -approval score of a candidate is denoted more simply by : in other words, is the number of voters in who rank in the first positions, i.e., . When , we get the plurality rule , and when we get the veto (or antiplurality) rule. The Borda rule is the scoring rule corresponding to the vector . We now define formally situations where new candidates are added. ###### Definition A voting situation with a varying set of candidates is a 4-tuple where is a set of voters (with ), a set of candidates, an -voter -profile, and is a positive integer, encoded in unary. denotes the set of initial candidates, the initial profile, and the number of new candidates. Nothing is known a priori about the voters’ preferences over the new candidates, henceforth their identity is irrelevant and only their number counts. The assumption that is encoded in unary ensures that the number of new candidates is polynomial in the size of the input. Most of our results would still hold if the number of new candidates is exponentially large in the size of the input, but for the sake of simplicity, and also because, in practice, will be small anyway, we prefer to exclude this possibility. Because the number of candidates is not the same before and after the new candidates come in, we have to consider families of voting rules (for a varying number of candidates) rather than voting rules for a fixed number of candidates. While it is true that for many usual voting rules there is an obvious way of defining them for a varying number of candidates, this is not the case for all of them, especially scoring rules. Still, some natural scoring rules, including plurality, veto, more generally -approval, as well as Borda, are naturally defined for any number of candidates. We shall therefore consider families of voting rules, parameterized by the number of candidates (). We slightly abuse notation and denote these families of voting rules by , and consequently often write instead of . The complexity results we give in this paper make use of such families of voting rules, where the number of candidates is variable. If is a -profile and , then the projection of on , denoted by , is obtained by deleting all candidates in in each of the votes of , and leaving unchanged the ranking on the candidates of . For instance, if , then and . In all situations, the set of initial candidates is denoted by , the set of the new candidates is denoted by . If is an -profile and an -profile, then we say that extends if the projection of on is exactly . For instance, let , ; the profile extends the -profile . ## 3 Possible winners when new candidates are added We recall from [22] that given a collection of partial strict orders on representing some incomplete information about the votes, a candidate is a possible winner if there is a profile where each is a ranking on extending in which wins. Reformulated for the case where is a ranking of the initial candidates (those in ), we get the following definition: ###### Definition Given a voting situation , and a collection of voting rules, we say that is a possible cowinner with respect to and if there is a -profile extending such that , where is a set of new candidates. Note that we do not have in the input, because it would be redundant with : it is enough to know the number of new candidates. Note also that all new candidates have to appear in the extended votes composing . Also, we do not consider the problem of deciding whether a new candidate is a possible cowinner, because it is trivial. Indeed, as soon as the voting correspondence satisfies the extremely weak property that a candidate ranked first by all voters is always a cowinner (which is obviously satisfied by all common voting rules), any new candidate is a possible cowinner. We now define formally the problems we study in this paper. ###### Definition Given a collection of voting rules, the possible cowinner problem with new candidates (or PcWNC) for is defined as follows: Input A voting situation and a candidate . Question Is a possible cowinner with respect to and ? Also, the subproblem of PcWNC where the number of new candidates is fixed will be denoted by PcWNC. We can also define the notion of necessary cowinner with respect to and : is a necessary cowinner with respect to , , and if for every -profile extending we have . However, the study of necessary cowinners in this particular setting will almost never lead to any significant results. There may be necessary cowinners among the initial candidates, but this will happen rarely (and this case will be discussed for a few specific voting rules in the corresponding parts of the paper). Now we are in position to consider specific voting rules. ## 4 K-approval As a warm-up we start by considering the plurality rule. ### 4.1 Plurality Let us start with an example: suppose , , and the plurality scores in are , , . There is only one new candidate (). We have: 1. is a possible cowinner ( will win in particular if the top candidate of every voter remains the same); 2. is a possible cowinner: to see this, suppose that 2 voters who had ranked first now rank first; the new scores are , , , ; 3. is not a possible cowinner: to reduce the scores of (resp. ) to that of , we need at least 3 (resp. 1) voters who had ranked (resp. ) first to now rank first; but this then means that gets at least 4 votes, while has only 3. More generally, we have the following result: ###### Proposition Let be an -voter profile on , and . The candidate is a possible cowinner for and plurality with respect to the addition of new candidates if and only if ntop(PX,x∗)≥1k⋅∑xi∈Xmax(0,ntop(PX,xi)−ntop(PX,x∗)) Proof: Suppose first that the inequality holds. We build the following -profile extending : 1. for every candidate such that we simply take arbitrary votes ranking on top and place one of the ’s on top of the vote (and the other ’s anywhere), subject to the condition that no is placed on top of a vote more than times. (This is possible because the inequality is satisfied). 2. in all other votes (those not considered at step 1), place all ’s anywhere except on top. We obtain a profile extending . First, we have , because in all the votes in where is on top, the new top candidate in the corresponding vote in is still (cf. step 2), and all the votes in where was not on top obviously cannot have on top in the corresponding vote in . Second, let . If then ; and if then we have . Therefore, is a cowinner for plurality in . Conversely, if the inequality is not satisfied, in order for to become a cowinner in , the other ’s must lose globally an amount of votes. But since we have , for at least one of the ’s it must hold that ; therefore cannot be a cowinner for plurality in . We do not need to pay much attention to the veto rule, since the characterization of possible cowinners is trivial. Indeed, by placing any of the new candidates below in every vote of where is ranked at the bottom position, we obtain a vote where no one vetoes , so any candidate is a possible cowinner. As a corollary, computing possible cowinners for the rules of plurality (and veto) with respect to candidate addition can be computed in polynomial time (which we already knew, since possible cowinners for plurality and veto can be computed in polynomial time [5]). ### 4.2 K-approval, one new candidate We start with the case where a single candidate is added. Recall that we denote by the score of for and -approval (i.e. the number of voters who rank among their top candidates); and by the number of voters who rank exactly in position . ###### Proposition Let be an positive integer, be an -voter profile on , and . The candidate is a possible cowinner for and -approval with respect to the addition of one new candidate if and only if the following two conditions hold: 1. for every , if then . Proof: Assume conditions (1) and (2) are satisfied. Then, we build the following -profile extending : • for every such that , we take arbitrary votes who rank in position in and place on top (condition (1) ensures that we can find enough such votes). • in all other votes (those not considered at step (i)), place in the bottom position. We obtain a profile extending . First, we have , because (a) all votes in ranking in position are extended in such a way that is placed in the bottom position, therefore gets a point in each of these votes if and only if it got a point in , and (b) in all the other votes (those where is not ranked in position in ), certainly gets a point in if and only if they got a point in . This holds both in the case where was added at the top or the bottom of the vote. Second, for every such that , loses exactly points when is extended into , therefore . Third, —because of (2)—hence . Therefore, is a cowinner for -approval in . Now, assume condition (1) is not satisfied, that is, there is an such that and such that . There is no way of having lose more than points, therefore will never catch up with ’s advantage and is therefore not a possible cowinner. Finally, assume condition (2) is not satisfied, which means that we have . Then, in order for to reach the score of ’s we must add in one of the top positions in a number of votes exceeding , therefore , and therefore is not a possible cowinner. Therefore, computing possible cowinners for -approval with respect to the addition of one candidate can be done in polynomial time. ### 4.3 2-approval, any (fixed) number of new candidates For each profile and each candidate , we simply write for the score of in under , that is, , i.e. the number of times that is ranked within the top two positions in . Let be an initial profile and the set of new candidates. Let . We want to know whether is a possible cowinner for 2-approval and . Let us partition into , and , where consists of the votes in which is ranked in the top position, consists of the votes in which is ranked in the second position and consists of the votes in which is not ranked within the top two positions. Let be an extension of to . For each candidate , we define the following three subsets of : • is the set of votes in where is ranked in the second position and neither nor any new candidate is ranked in the top position (HP stands for “high priority”). • is the set of votes in where or any new candidate is ranked in the top position and is ranked in the second position (MP stands for “medium priority”). • is the set of votes in where is ranked in the top position and some is ranked in the second position (LP stands for “low priority”). These definitions also apply to ; our definitions then simplify into: is the set of votes in where is ranked second and is not ranked first; is the set of votes in where is ranked first and is ranked second; is the set of votes in where is ranked first and is not ranked second. These definitions are summarized in Figure 1. Finally, for , let . Let us compute these sets on a concrete example, which will be reused throughout the section. ###### Example Let and consider the following profile consisting of 19 votes (we only mention the first two candidates in each vote): v1v2v3v4v5v6v7v8v9v10v11v12v13v14v15v16v17v18v19x∗x1x2x3x1x1x1x2x2x2x2x2x3x3x3x3x3x3x4x1x∗x∗x∗x4x4x5x1x3x4x5x5x1x2x4x4x5x6x6⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮ We have , and . This is summarized together with the priority classification in the following table: {\rm HP}\rm MP\rm LPΔ(PX,xi)x1v8,v13v1v5,v6,v73x2v14v8,v9,v10,v11,v123x3v9v13,v14,v15,v16,v17,v184x4v5,v6,v10,v15,v16v192x5v7,v11,v12,v170x6v18,v19−2 If is an extension of to then we write , where is the vote over extending . We now establish a useful property of the extensions of for which is a cowinner. Without loss of generality, we assume that in every vote , every new candidate is ranked either in the first two positions, or below all candidates of . ###### Proposition If there exists an extension of such that , then there exists an extension of such that , and satisfying the following conditions: 1. For each , if is ranked within the top two positions in , then is also ranked within the top two positions in . 2. For each , if the top candidate of is not in then the second-ranked candidate of is not in either. 3. For each and each , if is not ranked within the top two positions in , then for each , is not ranked within the top two positions in . Proof: We consider in turn the different conditions: 1. This is because if there exists such that is not in the top two positions whereas is in the top two positions in its original vote , then we can simply move all of candidates in ranked higher than to the bottom positions. Let denote the vote obtained this way. By replacing with , we increase the score of by , and the score of each other candidate by no more than , which means that is still a cowinner. 2. If there exists such that is ranked in the top position and is ranked in the second position, then we simply obtain by switching and . 3. The condition states that for each candidate , whenever we want to reduce its score, we should first try to reduce it by putting a new candidate on top of some vote in . This is because by putting on top of some vote in , we may use only one extra candidate to reduce by one unit the score of the candidate ranked at the top position of . Formally, suppose there exist and such that is within the top two positions of (the extension of ) but not within the top two positions of (the extension of ). Let be any candidate ranked within the top two positions of . Let denote the vote obtained from by moving to the bottom, and let denote the vote obtained from by moving to the top position. Next, we replace and by and , respectively. It follows that the score of each candidate does not change, which means that is still a cowinner. We repeat this procedure until statement (3) is satisfied for every . Since after each iteration there is at least one additional vote that will never be modified again, this procedure ends in times. Proposition 4.3 simply tells us that when looking for an extension that makes a cowinner, it suffices to restrict our attention to the extensions that satisfy conditions (1) to (3). Moreover, using (1) of Proposition 4.3, we deduce that . Hence, for votes (the votes in which is ranked in the second position), we can assume that the new candidates of are put in bottom positions in . Define as the set of all candidates in such that . Our objective is to reduce all score differences to for , while keeping the score differences of each new candidate non-positive. (We do not have to care about the candidates in ). The intuition underlying our algorithm is that when trying to reduce on the current profile , we first try to use the votes in , then the votes in , and finally the votes in . This is because putting some candidates from in the top positions in the votes of not only reduces by one unit, but also creates an opportunity to “pay” one extra candidate from to reduce by one unit, where is the candidate ranked on top of this vote. For the votes in , we can only reduce by one unit without any other benefit. For the votes in we will have to use two candidates from to bring down by one unit; however, if we already put some in the top position in order to reduce , where is the candidate ranked in the second position in the original vote, then we only need to pay one extra candidate in to reduce by one unit. Therefore, the major issue consists in finding the most efficient way to choose the votes in to reduce , when . We will solve this problem by reducing it to a max-flow problem. The algorithm is composed of a main function CheckCowinner(.) which comes together with two sub-functions AddNewAlternativeOnTop(.) and BuildMaxFlowGraph(.) that we detail first. The procedure AddNewAlternativeOnTop simply picks new candidates to be put on top of votes, and updates subsequently the profile. Note that in this procedure, candidates from to be added on top of the votes are those with the lowest score (or the lowest index, in case of ties). This results in choosing new candidates in a cyclic order As for the function BuildMaxFlowGraph, it builds the weighted directed graph defined as follows: • ; • contains the following weighted edges: • for each , an edge with weight ; • for each and each : an edge with weight ; plus, if the candidate in second position in is in , an edge with weight ; • for each , an edge with weight . We refer the reader to Figure 2 for an illustration. (Once this graph is constructed, any standard function to compute a flow of maximal value can of course be used). We are now in a position to detail the main function CheckCowinner(.). ###### Proposition Given a profile on , a candidate and a set of new candidates , a call to algorithm CheckCowinner returns in polynomial time the answer true if and only if there exists an extension of in which is a cowinner. Proof: Algorithm 2 starts by partitioning into and : an alternative is in if and in if . Let . Then by item (3) of Proposition 4.3, for each vote in , we can safely put one candidate from in the top position of ; this is done in the first phase of Algorithm 2, lines 2 to 2. Note that after adding a new candidate on top of a vote and after updating , the modified vote will no longer belong to . Instead, it will now belong to for some other candidate . When Phase 1 is over, the score of may still need to be lowered down, which can be done next by using votes from . This is what Phase 2 does, from line 2 to line 2. There are three possibilities: 1. . In this case, the votes in are sufficient to make catch up : after Phase 1, we have and Phase 2 is void; we are done with . 2. and : in this case, to make catch up , it is enough to take arbitrary votes in and add one new candidate on top of them; this is what Phase 2 does, and after that we are done with . 3. : in this case, because of Proposition 4.3, we know that it is safe to add one new candidate on top of all votes of ; this is what Phase 2 does; after that, we still need to lower down the score of , which will require to add new candidates on top of votes of . If at this point a newly added candidate has a score higher than , then cannot win, and we can stop the program (line 2). For readability, let us denote by the profile obtained after Phases 1 and 2. For each satisfying condition 3, the only way to reduce is to put two candidates of within the top two positions in a vote of , because in Phases 1 and 2 we have used up all the votes in and . Now, reducing by one unit will cost us two candidates in , but meanwhile, is also reduced by one unit, where is the candidate ranked in the second position in . We must have . We note that . Choosing optimally the votes in for each can be done by solving an integral max-flow instance which is build by algorithm BuildMaxFlowGraph (note that in case where either or is empty, we just assume that the flow has a null value). Let us show that is a possible cowinner if and only if the value of the flow from to is at least	2021-09-18 08:28:38	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8358897566795349, "perplexity": 605.2940876118062}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00116.warc.gz"}
http://www.physicsforums.com/showthread.php?p=4248833	# Is QM truly random and many world theory by Government$Tags: random, theory, world P: 85 Hi everybody, i didn't wanted to create two separate threads so merged them into one. i got confused watching Brian Green explaining QM on one of his shows. He compared distribution in double slit experiments with throwing a ball on a roulette. He said that casino doesn't have to know where ball on roulette wheal is going to land, but casino knows that in the long run that he is going to get some kind of distribution. In other words, when we shoot electrons through double slit, we can't know for sure where it will go but we know that it will follow normal distribution. But in principle ball and roulette table follows Newtonian laws and if we had all the data we need, we could say with 100% certainty where will ball land. Does the same principle apply in QM i.e. if we had all the data we need we could say where the electron land? If that is the case then is QM truly random? On the other hand if we had all the data and we still couldn't predict where electron will land then it seems to me that QM is truly random. Also does qm behaving random drives stake through the heart of fatalism and predeterminism? Second question is about many worlds theory. Here is video where Sean Carroll explains a many world theory, and he uses example of a car choosing which why to go. Now, i hope that this many world theory has nothing to do with actual cars choosing which way to go i.e. when actual car chooses which way to go he doesn't create a new world, but rather all of this is on a microscopic level. Thank you. Sci Advisor PF Gold P: 5,147 Quote by Government$ ... Does the same principle apply in QM i.e. if we had all the data we need we could say where the electron land? If that is the case then is QM truly random? On the other hand if we had all the data and we still couldn't predict where electron will land then it seems to me that QM is truly random. Also does qm behaving random drives stake through the heart of fatalism and predeterminism? ... QM does not have a mechanism for predicting the exact outcome of many quantum level events, regardless of the knowledge you have. In such cases, the results follow a probability distribution of some kind as mentioned. It is commonly accepted that those outcomes have a truly random element in this sense: there is nothing about the state of a system here and now that will determine the outcome of a future measurement independently of the act of observing it. There is a formulation of QM called Bohmian Mechanics that asserts that it is possible, in principle, to determine the results with complete certainty, thus restoring determinism. However, they say there are practical limitations that make it impossible to achieve this. This is not a generally accepted viewpoint, although it is considered an acceptable interpretation since the ultimate predictions of Bohmian Mechanics are the same as QM. PF Gold P: 11,056 If you can predict an event or sequence of events with a certain level of accuracy, how can it be truly random? P: 85 ## Is QM truly random and many world theory Quote by Drakkith If you can predict an event or sequence of events with a certain level of accuracy, how can it be truly random? Exactly that kind of the system is not random, for me truly random system is system that you can't possibly predict, even if you have all the possible data and know all the laws of physics. P: 82 Quote by Drakkith If you can predict an event or sequence of events with a certain level of accuracy, how can it be truly random? If you change it to, "events or a series of events where the best you can do is to predict the probability of the outcomes", then it becomes pretty much an definition of randomness. If you then move from purely ideal theoretical statistics to the real world you also get the limits on accuracy from such things as measurement errors and sample size. Quote by Government\$ Exactly that kind of the system is not random, for me truly random system is system that you can't possibly predict, even if you have all the possible data and know all the laws of physics. That is however an entirely nonstandard definition of (statistical) randomness as used in QM. The definition you're both using comes much closer to non-causal than random. In the case of "truly random", which usually just means "not-pseudorandom", your definition is sort of the opposite of actual usage. Quantum effects are even used as school books examples of actual real world truly random processes like random number generators powered by radioactive decay. Sci Advisor PF Gold P: 2,066 No one can say if QM is truly random or not because deterministic systems can mimic randomness. We have tests that can determine if something is pseudo-random (ie created by a deterministic process) and it has passed all those. But those tests are not 100% reliable - some very complex pseudo-random number generators pass it. So the best we can say today is as far as we can tell it's truely random but cant say for sure - nor do I think is it possible to ever do so. Regarding MWI - yes it occurs at the quantum level but since what happens at the quantum level determines the classical under that interpretation there would be a world where the car went a different way. Personally I think the MWI is mystical mumbo jumbo - but hey its a valid interpretation and all interpretations IMHO suck in their own unique way - even the one I hold to - so simply choose the one that sucks the least - its your choice. Thanks Bill Related Discussions Academic Guidance 1 Atomic, Solid State, Comp. Physics 0 Linear & Abstract Algebra 1 Calculus 1 General Physics 8	2014-04-23 20:33:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5972372889518738, "perplexity": 539.3910680983606}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00246-ip-10-147-4-33.ec2.internal.warc.gz"}
https://mathproblems123.wordpress.com/2011/05/05/integrals/	Home > Analysis > Integrals ## Integrals a) Let $A=(a_{ij})_{i,j=1}^n$ be a real matrix. Show that $\displaystyle \int_{\|x\|<1} (x,Ax) dx = \frac{\omega_n}{n(n+2)} tr(A),$ where $(\cdot ,\cdot)$ is the usual dot product and $\omega_n$ is the area of the unit sphere. b) Show that for all $u \in C_0^2 (\Bbb{R}^n)$ we have $\displaystyle \int_{\Bbb{R}^n} (\Delta u)^2 dx =\sum_{i,j=1}^n \int_{\Bbb{R}^n} \|D_{ij}u\|^2 dx.$ PHD Iowa (6101)	2018-07-19 21:23:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9806060791015625, "perplexity": 731.8771348122096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591296.46/warc/CC-MAIN-20180719203515-20180719223515-00270.warc.gz"}
http://mouood.com/fat1l4s7/page.php?id=linearize-diode-equation-d46c46	. Part A: Linearize the following differential equation with an input value of u=16. R . w. &=Α. This circuit approximates the cut-in voltage present in real diodes. o Sometimes an iterative procedure depends critically on the first guess. ln V {\displaystyle I} Graphical analysis is a simple way to derive a numerical solution to the transcendental equations describing the diode. w n {\displaystyle V_{\text{T}}} To begin, the diode small-signal conductance $${\displaystyle g_{D}}$$ is found, that is, the change in current in the diode caused by a small change in voltage across the diode, divided by this voltage change, namely: d When the diode voltage is greater than $26\,\text{mV}$ the exponent term grows rapidly. , typically 10−12 A). {\displaystyle g_{D}} for any particular set of values by an iterative method using a calculator or computer. ⁡ ( {\displaystyle V_{D}} V 1 The Boltzmann constant is a very important number in physics. The work presented in this thesis focuses on diode predistortion linearization, particularly for PA RFICs in digital radios. Firstly, consider a mathematically idealized diode. If your diode is at room temperature, the diode equation becomes, $i = \text I_\text S \left ( e^{\,v/26\text{mV}} -1 \right )$. D I V {\displaystyle I} Engineers like round number that are easy to remember, so we use $300\,\text K$ for room temperature. In mathematics, linearization is finding the linear approximation to a function at a given point. Zener Diode & Zener Voltage Regulator Calculator Formulas and Equations for Zener Diode & Zener Voltage Regulator Calculator Series Current IS= VIN – VZ / .. A diode’s distinctive feature is that it conducts current in one direction, but not the other. When more accuracy is desired in modelling the diode's turn-on characteristic, the model can be enhanced by doubling-up the standard PWL-model. is guessed and put into the right side of the equation. . x x / V A temperature of absolute zero, or $0\,\text K$, is the same as $-273\,^{\circ}\text C$. {\displaystyle I} : rearrangement of the diode law in terms of w becomes: which using the Lambert A linear representation is found through Carleman Linearization. It’s up in the exponent. Continuing with the symbols: This thesis also presents a second laser model based on a time domain simulation of the rate equations using a circuit simulation software package. The Ideal Diode Equation: Diodes should be familiar to us by now. If the particle happens to be an electron, we can talk about its energy per charge, $\dfrac{k\text T}{q} = \dfrac {4.14 \times 10^{-21}\,\text J} {1.602\times 10^{-19}\,\text C} = 25.8 \,\dfrac{\text J}{\text C}$. You can measure the temperature of the chamber (a macro-world measurement with a thermometer). The use of base 10 logarithms makes it easier to V V {\displaystyle V_{D}} linearize the characteristics of the power detector and to stabilize the detector output across a wider temperature range. $q$ is the charge on an electron, $1.602 \times 10^{-19} \,\text{coulomb}$. I It does not model where the diode breaks down far to the left on the voltage axis. : and {\displaystyle I_{Q}} W and This ideal diode starts conducting at 0 V and for any positive voltage an infinite current flows and the diode acts like a short circuit. g The combined I-V characteristic of this circuit is shown below: The Shockley diode model can be used to predict the approximate value of The reciprocal is, of course, "q on kT". W V A diode's I-V curve is nonlinear. also must satisfy the Kirchhoff's law equation, given above. tends to be large, meaning that the exponential is very large. T {\displaystyle W} ≪ ( {\displaystyle V_{D}} Diode is non-linear component of an electrical circuit, which allow current in forward biasing and block current in reverse biasing. Following the substitutions This approximation is accurate even at rather small voltages, because the thermal voltage can be found in terms of ) It connects the world we see and sense with the atomic-scale world of atoms and electrons. and (b) show the effect of linearization using the diode as explained above. The question is, just how much current is there? I is a known given value, but Of course, we can just give you the equation to figure it out, but what good will that do? τ {\displaystyle I} For large x, In fact, this is generally not the case: as temperature rises, the saturation current D Carrying out the various operations on the right side, we come up with a new value for V This quantity plays the same role than the saturation current in the junction diode. / Diodes conducting current is one such case. Let’s go through them carefully. The Shockley diode equation relates the diode current / {\displaystyle V_{Q}/V_{\text{T}}} Q x The behavior of a diode can be identified using VI characteristic. S The equation covers the range of a few volts on either side of the origin. Breaking News. ( To illustrate the complications in using this law, consider the problem of finding the voltage across the diode in Figure 1. q That represents the kinetic energy of an average everyday room-temperature electron. For common physical parameters and resistances, I Comparing figures 1 and 2 one can conclude that the 1dB compression point without linearization is at - 4dBm Pin, where as it is at 10dBm after linearization. Linearization of VCCS in the diode model. This model uses two piecewise-linear diodes in parallel, as a way to model a single diode more accurately. D In mathematics, this means taking a function and breaking it down into several linear segments. w When a reverse bias is applied to the diode, its junction capacitance varies. Figure 1: Diode circuit with resistive load. V Another method of modelling a diode is called piecewise linear (PWL) modelling. This method plots the two current-voltage equations on a graph and the point of intersection of the two curves satisfies both equations, giving the value of the current flowing through the circuit and the voltage across the diode. Comparing figures 1 and 2 one can conclude that the 1dB compression point without linearization is at - 4dBm Pin, where as it is at 10dBm after linearization. {\displaystyle V_{S}} The Shockley diode equation has an exponential of An explicit expression for the diode current can be obtained in terms of the Lambert W-function (also called the Omega function). I It does not model where the diode breaks down far to the left on the voltage axis. D Typically the sloped line segment would be chosen tangent to the diode curve at the Q-point. T The ideal diode equation: d where i D is the diode current and v D voltage across the diode. The ﬁrst-order Taylor series approximation of the function h at x = a is given as h(x)=h(a)+h0(a)(x−a). / r at 300 K, so is known as the diode ideality factor (for silicon diodes {\displaystyle C_{J}={\frac {dQ_{J}}{dV_{Q}}}} . Equation (C.1) can be solved for the current in the form, i()t = v s ()t v D ()t R. (C.2) The current is also described by the diode equation, i()t = I s e qv D ()t /kT 1 . S {\displaystyle V_{D}=600\,{\text{mV}}} {\displaystyle r_{D}} {\displaystyle V_{D}\gg nV_{\text{T}}} [4] The diode law is rearranged by dividing by The diode equation gives an expression for the current through a diode as a function of voltage. By Kirchhoff's laws, the current flowing in the circuit is. k D I Using the Shockley equation, the small-signal diode resistance $${\displaystyle r_{D}}$$ of the diode can be derived about some operating point (Q-point) where the DC bias current is $${\displaystyle I_{Q}}$$ and the Q-point applied voltage is $${\displaystyle V_{Q}}$$. °C I e Warmer temperatures shift the diode curve right. n By plotting the I-V curves, it is possible to obtain an approximate solution to any arbitrary degree of accuracy. expansion so a single solution of the resulting equations may not be adequate and iterations are usually required. S I I . Current of the diode depends upon the voltage across the diode. {\displaystyle V_{Q}} Is = Reverse or dark saturation current (Typical value for silicon is 10-12 Amperes) e = Base of the neutral logarithm (2.71828) , about 26 mV at normal temperatures), and = w {\displaystyle W(x)} V {\displaystyle I} Suppose you have a chamber filled with gas molecules. / directly in terms of {\displaystyle V_{D}} D n Since the diode forward-voltage drops as its temperature rises, this can lead to thermal runaway in bipolar-transistor circuits (base-emitter junction of a BJT acts as a diode), where a change in bias leads to an increase in power-dissipation, which in turn changes the bias even further. , which would lead one to expect that the forward-voltage increases with temperature. − {\displaystyle V_{\text{T}}} This equation is based on the physics underlying the diode action, along with careful measurements on real diodes. D I is large enough so that the factor of 1 in the parentheses of the Shockley diode equation can be ignored. Instead let's derive it! S rises, and this effect dominates. ) The reverse bias current in a Schottky diode is % "# where % is a constant that depend on temperature. in terms of In practice, the graphical method is complicated and impractical for complex circuits. 3.2. V w. We obtain the infinite linear system. This equation is based on the physics underlying the diode action, along with careful measurements on real diodes. The diode $i$-$v$ relationship can be modeled with an equation. For LEDs, this bandgap change also shifts their colour: they move towards the blue end of the spectrum when cooled. V It is the charge stored on the diode by virtue of simply having a voltage across it, regardless of any current it conducts. I personally have never used this technique, and it seems to me that the linearization of the diode’s conduction behavior is, overall, not significantly more accurate than the simpler constant-voltage-drop model explained in the previous article. of a p-n junction diode to the diode voltage Equations for Breakdown Voltage of a Diode All diodes exhibit rectification when driven in forward bias, and they exhibit a breakdown behavior when driven at high voltage in reverse bias. {\displaystyle I_{S}} If you know the temperature of the gas, the Boltzmann constant $k$ relates the temperature to the average kinetic energy of a molecule. q = charge of electron = 1.6022 x 10-19 coulomb; T = absolute temperature in Kelvin (K = 273 + °C) k = Boltzmann’s constant = 1.3806 x 10 23 J/K Zenner In this example, almost any first guess will do, say To solve these two equations, we could substitute the current In the study of dynamical systems, linearization is a method for assessing the local stability of an equilibrium point of a system of nonlinear differential equations or discrete dynamical systems. As the chamber gets warmer, down at the atomic level the gas molecules have higher kinetic energy. I 0 is the dark saturation current, q is the charge on the electron, V is the voltage applied across the diode, η is the (exponential) ideality factor. As with most graphical methods, it has the advantage of easy visualization. {\displaystyle n} Try not to confuse big $\text K$ the unit for kelvin with little $k$ for Boltzmann's constant. Near room temperature, the diode equation can be written as. V Q $k$ is Boltzmann's constant, $1.380\times 10^{-23} \,\text{joule/kelvin}$ without involving R ) The diode current can be expressed in the form of diode current equation. However, if you’re interested primarily in capturing the behavior of the diode in the transition region between non-conduction and full conduction, you might want to consider the piecewise-linear ap… At high voltage, When the recombination in the device is dominated by the surfaces and the bulk regions the ideality factor is close to one. {\displaystyle I_{Q}} V For reverse voltages VD << − kT / q, the diode current saturates at the reverse saturation current − Is. I Background discussion on common linearization techniques available to the PA designer is presented. {\displaystyle n} That is, the equation for the time-varying part of the voltages and currents is approximately linear and can be solved by linear … ) Using So for the diode, we write a current law that looks like this. to obtain. ( $\text I_{\text S}$ is the saturation current. ( is known to be. I So it's proper to say "kelvin" instead of "degrees Kelvin", since that would be redundant. This solution is discussed next. {\displaystyle I_{Q}} : Typical values of the saturation current at room temperature are: As the variation of The second term is the charge stored in the junction itself when it is viewed as a simple capacitor; that is, as a pair of electrodes with opposite charges on them. The laser and predistorter blocks have been modeled with Volterra kernels. Silicon diode at $\text T = -40^{\circ}\text C, +27^{\circ}\text C$, and $+85^{\circ}\text C$. [6] To begin, the diode small-signal conductance There are many parameters in the diode equation. {\displaystyle V_{D}} The procedure introduced is based on the Taylor series expansion and on knowledge … $k$ is Boltzmann’s constant, $\text T$ is the temperature in kelvin, and $q$ is the charge on an electron in coulombs. Q There are tons of others. of the diode can be derived about some operating point (Q-point) where the DC bias current is S x {\displaystyle V_{D}} Here is some detailed experimental data,[7] which shows this for a 1N4005 silicon diode. Because the current flowing through the diode is the same as the current throughout the entire circuit, we can lay down another equation. S Elbow than an exponential dependence on voltage but what good will that do like:... Just give you the equation of a diode as explained above exist. [ 5.... Linearization to the current through the diode is modelled as 3 components in:! Diode has been taken advantage by engineers to design a special-purpose diode called the Omega function ) Lyapunov... To what happens in the junction diode than $26\, \text { kelvin }$ is temperature... Tp 93 % Off - Launching Official electrical Technology Store - Shop now the unit for kelvin with $! Only one direction, but not the other equations linear and must still be by. 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Small change of the spectrum when cooled approximate the behavior of the spectrum when cooled govern the of...	2021-04-22 00:27:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8846595883369446, "perplexity": 889.5189135102206}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039554437.90/warc/CC-MAIN-20210421222632-20210422012632-00051.warc.gz"}
http://ncatlab.org/nlab/show/Fivebrane%20group	# nLab Fivebrane group cohomology ### Theorems $\cdots \to$ Fivebrane group $\to$ string group $\to$ spin group $\to$ special orthogonal group $\to$ orthogonal group. # Contents ## Definition The Fivebrane group $Fivebrane(n)$ is defined to be, as a topological group, the 7-connected cover of the String group $String(n)$, for any $n \in \mathbb{N}$. Notice that $String(n)$ itself if the 3-connected cover of $Spin(n)$, which is itself is the simply connected cover of the special orthogonal group $SO(n)$, which in turn is the connected component (of the identity) of the orthogonal group $O(n)$. Hence $Fivebrane(n)$ is one element in the Whitehead tower of $\mathrm{O}(n)$: $\cdots \to Fivebrane(n) \to String(n) \to Spin(n) \to SO(n) \to \mathrm{O}(n) \,.$ The homotopy groups of $O(n)$ are for $k \in \mathbb{N}$ and for sufficiently large $n$ $\array{ \pi_{8k+0}(O) & = \mathbb{Z}_2 \\ \pi_{8k+1}(O) & = \mathbb{Z}_2 \\ \pi_{8k+2}(O) & = 0 \\ \pi_{8k+3}(O) & = \mathbb{Z} \\ \pi_{8k+4}(O) & = 0 \\ \pi_{8k+5}(O) & = 0 \\ \pi_{8k+6}(O) & = 0 \\ \pi_{8k+7}(O) & = \mathbb{Z} } \,.$ By co-killing these groups step by step one gets $\array{ cokill this &&&& to get \\ \\ \pi_{0}(O) & = \mathbb{Z}_2 &&& SO \\ \pi_{1}(O) & = \mathbb{Z}_2 &&& Spin \\ \pi_{2}(O) & = 0 \\ \pi_{3}(O) & = \mathbb{Z} &&& String \\ \pi_{4}(O) & = 0 \\ \pi_{5}(O) & = 0 \\ \pi_{6}(O) & = 0 \\ \pi_{7}(O) & = \mathbb{Z} &&& Fivebrane } \,.$ ## Further information… …should eventually go here. For the time being have a look at Fivebrane structure. $n$012345678910111213141516 homotopy groups of stable orthogonal group$\pi_n(O)$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$$\mathbb{Z}_2$ stable homotopy groups of spheres$\pi_n(\mathbb{S})$$\mathbb{Z}$$\mathbb{Z}_2$$\mathbb{Z}_2$$\mathbb{Z}_{24}$00$\mathbb{Z}_2$$\mathbb{Z}_{240}$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_6$$\mathbb{Z}_{504}$0$\mathbb{Z}_3$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_{480} \oplus \mathbb{Z}_2$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$ image of J-homomorphism$im(\pi_n(J))$0$\mathbb{Z}_2$0$\mathbb{Z}_{24}$000$\mathbb{Z}_{240}$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}_{504}$000$\mathbb{Z}_{480}$$\mathbb{Z}_2$ ## References The term fivebrane group and the role of this topological group in quantum anomaly cancellaton conditions in dual heterotic string theory was found by Hisham Sati and appeared in The term shortly after was picked up in The refinement to fivebrane principal infinity-connections, hence differential fivebrane structures was then discussed in Discussion in a comprehensive context is in section 5 of Revised on November 13, 2013 00:47:05 by Urs Schreiber (82.169.114.243)	2014-10-31 12:05:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 55, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8444284200668335, "perplexity": 1029.4689230673414}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637899633.25/warc/CC-MAIN-20141030025819-00015-ip-10-16-133-185.ec2.internal.warc.gz"}
https://pypi.org/project/w20e.forms/	Python API for creating and handling forms ## Project description <nav class="contents" id="table-of-contents" role="doc-toc"> </nav> The w20e.forms package provides a powerful API for creating and handling electronic forms. The package is loosely based on XForms concepts (and Socrates QE, a Java implementation). The package intends to provide a drop-in alternative for standard plone, pyramid and django solutions, but is useable in any framework (or without a framework…). ## Core concepts The core concepts are as follows (so you can quickly decide whether you like this approach or not): A form is a container for/composition of four things: 1. data 2. model 3. view 4. submission This clearly separates the data, the data’s properties (datatype, whether something is required or not, etc.) and the renderable part of the form. This is also the main difference between this API and other solutions (afaik…). Where the usual approach is to define a form based on a Schema or other data-centered notion, like so: foo = String("id", "label", validator=somefunction) the approach of w20e.forms is to define: foo = Field("fid") props = FieldProperties("pid", ["fid"], required=True, datatype=int) ctrl = Input("cid", "label", bind="fid") where the properties and control are bound to the variable. This enables controls in your form that are not bound to any data you wish to collect, sharing of properties, etc. Another important difference is that the API provides a structured way of defining properties for data, instead of having to define your own validation. See section 1.2 for details. ### Data The data holds the variables you wish to collect with this form. A variable simply has an id, and if you like a default value. ### Model The model holds all properties for a given form, like readonly-ness, requiredness, datatyping for variables, relevance, etc. All these properties are calculated from expressions, in this case Python expressions, so requiredness is not just true or false, it can be calculated based on an expression that can include other variables in your data. All variables from the form data are available in expressions via the ‘data’ dict, so if variable ‘foo’ would be required if variable ‘bar’ was set to 666, this can be expressed like so in the properties bound to foo: … required=”data[‘bar’] == 666” … In general, all expressions are eval-ed to something true or false. The model offers the following properties: • required: is a variable required or not? • relevant: is the variable relevant? Like maiden name would be irrelevant when gender is male. In general, the related control/widget for irrelevant variables would not be shown. • readonly: can a person change the value of a variable? • calculate: in stead of letting a person set the value of a variable, the variable is calculated. • constraint: check whether the expression evaluates to True. • datatype: datatype for the variable, like int, string, or more complex variables. Properties are bound to variables by a bind attribute. A set of properties can be bound to a series of variables. ### View The view (or FormView) is the actual visible part (or audible for that matter) of the form. The view can be rendered and holds a collection of widgets or controls, that are bound to variables. More than one control can bind to the same variable. Controls can be grouped in groups for layout purposes, like flow layout or card layout (tabs). In label and hint texts of controls you can use lexical values of variables by using the expression ${<var name>}. This way you can refer to values given in other variables from your labels and hints. ## Basic use Ok, enough theory, let’s do something for real. A form is produced by hand, or by using a factory: this should take care of producing a form holding the necessary stuff. Let’s get the imports over with… >>> import sys >>> from interfaces import * >>> from zope.interface import implements >>> from formdata import FormData >>> from formview import FormView >>> from formmodel import FormModel >>> from data.field import Field >>> from model.fieldproperties import FieldProperties >>> from rendering.control import Input, Select, Option >>> from rendering.group import FlowGroup >>> from form import Form, FormValidationError >>> from rendering.html.renderer import HTMLRenderer >>> from submission.attrstorage import AttrStorage ### Creating a form Now let us create a factory class >>> class FormFactory(): ... implements(IFormFactory) ... def createForm(self): ... data = FormData() ... data.addField(Field("field0")) ... data.addField(Field("field1", "foo")) ... data.addField(Field("field2", "bar")) ... data.addField(Field("field3")) ... view = FormView() ... grp = FlowGroup("grp0", label="Group 0") ... grp.addRenderable(Input("input2", "Input 2", bind="field0")) ... view.addRenderable(Input("input0", "First name", bind="field0")) ... view.addRenderable(Input("input1", "Last name", bind="field1")) ... view.addRenderable(Select("select0", "Select me!", options=[], bind="field2", with_empty=True)) ... view.addRenderable(grp) ... model = FormModel() ... model.addFieldProperties(FieldProperties("prop0", ["field0"], required="True")) ... model.addFieldProperties(FieldProperties("prop1", ["field1", "field2"], relevant="data['field0']")) ... submission = AttrStorage(attr_name="_data") ... return Form("test", data, model, view, submission) >>> ff = FormFactory() >>> form = ff.createForm() By now, we should have a form where field0 is required, and field1 and field2 are only relevant if field0 is filled in. >>> print len(form.data.getFields()) 4 >>> props = form.model.getFieldProperties("field0") >>> props[0].id 'prop0' >>> len(props) 1 >>> field0 = form.data.getField("field0") >>> field0.id 'field0' >>> field0.value In the meanwhile, field1 and field2 should be irrelevant, given that field0 has no value >>> form.model.isRelevant("field1", form.data) False >>> form.model.isRelevant("field2", form.data) False Validation should fail, given that field0 is required. >>> try: ... form.validate() ... except FormValidationError: ... print sys.exc_info()[1].errors['field0'] ['required'] >>> form.data.getField("field0").value = "pipo" >>> form.validate() True >>> field0.value 'pipo' By now, field1 and field2 should also be relevant >>> form.model.isRelevant("field1", form.data) True >>> form.model.isRelevant("field2", form.data) True ### Display The following section will assume rendering to HTML. This will most likely cover nigh 100% of the use cases… Now for some display parts. An irrelevant control should not have a class ‘relevant’, otherwise it should have it… This enables specific styling, like ‘display: none’. >>> form.data.getField('field0').value = None >>> field = form.view.getRenderable('input1') >>> renderer = HTMLRenderer() >>> renderer.render(form, field, sys.stdout) <div id="input1" class="control input "> <label for="input-input1">Last name</label> <div class="alert"></div> <div class="hint"></div> <input id="input-input1" type="text" name="input1" value="foo" size="20"/> </div> >>> form.data.getField('field0').value = 'pipo' >>> field = form.view.getRenderable('input1') >>> renderer = HTMLRenderer() >>> renderer.render(form, field, sys.stdout) <div id="input1" class="control input relevant"> <label for="input-input1">Last name</label> <div class="alert"></div> <div class="hint"></div> <input id="input-input1" type="text" name="input1" value="foo" size="20"/> </div> >>> field = form.view.getRenderable('input0') >>> renderer.render(form, field, sys.stdout) <div id="input0" class="control input relevant required"> <label for="input-input0">First name</label> <div class="alert"></div> <div class="hint"></div> <input id="input-input0" type="text" name="input0" value="pipo" size="20"/> </div> How ‘bout those extra classes… >>> renderer.render(form, field, sys.stdout, extra_classes="card") <div id="input0" class="control input card relevant required"> <label for="input-input0">First name</label> <div class="alert"></div> <div class="hint"></div> <input id="input-input0" type="text" name="input0" value="pipo" size="20"/> </div> >>> select = form.view.getRenderable('select0') >>> renderer.render(form, select, sys.stdout) <div id="select0" class="control select relevant"> <label for="input-select0">Select me!</label> <div class="alert"></div> <div class="hint"></div> <select id="input-select0" name="select0" size="1"> <option value="" >Maak een keuze</option> </select> </div> Do we actually get grouped controls? >>> nested_input = form.view.getRenderable('input2') >>> nested_input.id 'input2' ### Submission Finally when the form is rendered, filled in by someone, and validated, the data should normally go somewhere. This is by way of submission. We defined submission to be AttrStorage, something that stores the data in an attribute on some context. This is a case that could be used in many frameworks, at least plone and pyramid. Let’s see what it does: >>> class Context: ... """ some context """ >>> ctx = Context() >>> form.submission.submit(form, ctx) The context now should hold the data in an attribute. We specified the name of the attribute to be ‘_data’, so let’s check: >>> ctx._data.getField('field0').value 'pipo' ## Beyond the basics Well, this is all very simple, and it is quite likely that you would wish for something a bit more usefull. All parts of the form are there to be extended. Take for instance the FormView. A developer (or end user) should be able to: • create a full HTML form; • use a generated HTML form (this is wat the base implementation does); • create a PDF form. The factory is also an important part of the form process. A factory can be imagined to be one of the following: • produced from a Schema (content type); • produced from an XML definition, for example an XForms instance from OpenOffice. Forms in general should be: • submitable to a range of handlers, like email, database storage, content type storage; • easy to validate ‘live; • enable multi-page. More detailed tests: We’d like to check whether lookup of a control by bind works, so as to be able to process values into lexical values. This is especially interesting when using selects: we’d expect to see the label not the value in lexical space. >>> data = FormData() >>> data.addField(Field("f0", "opt0")) >>> view = FormView() >>> opts = [Option("opt0", "Option 0"), Option("opt1", "Option 1")] >>> view.addRenderable(Select("sel0", "Select 0", bind="f0", options=opts)) >>> ctl = view.getRenderableByBind("f0") >>> ctl.lexVal("opt0") 'Option 0' Can we use variable substitution in labels and hints? Yes, we can! >>> data = FormData() >>> data.addField(Field("f0", "Pipo")) >>> data.addField(Field("f1")) >>> view = FormView() >>> view.addRenderable(Input("in0", "First name", bind="f0")) >>> view.addRenderable(Input("in1", "Last name for${f0}", bind="f1")) >>> model = FormModel() >>> form = Form("test", data, model, view, None) >>> renderer = HTMLRenderer() >>> field = form.view.getRenderable('in1') >>> renderer.render(form, field, sys.stdout) <div id="in1" class="control input relevant"> <label for="input-in1">Last name for Pipo</label> <div class="hint"></div> <input id="input-in1" type="text" name="in1" value="" size="20"/> </div> Let’s delve into input processing a bit… A simple input should just return it’s own value >>> data = {'pipo': 'lala'} >>> ctl = Input("pipo", "f0", "Some input") >>> ctl.processInput(data) 'lala' w20e.forms is not a complete library for forms, and it will never be this, since most people have very specific needs, like a specific widget, a custom version of an input field, etc. The API facilitates in this by using a global registry to register extensions. The global registry is available like so: >>> from w20e.forms.registry import Registry and offers a number of class methods to register stuff. Let’s for exampe register a new renderer for an input: ### Vocabularies w20e.forms enables use of vocabularies to limit possible answers to a given list. This is a feature that is generally used with select widgets. A vocabulary is a ‘named’ factory that creates a list of options. Register like so: >>> def make_vocab(): ... return [Option('0', 'Opt 0'), Option('1', 'Opt 1')] ... Registry.register_vocab('foovocab', make_vocab) ... sel = Select("select0", "Select me!", vocab=make_vocab, ... bind="field2", with_empty=True)) In a form you’ll usually want to say things like: this control need only be shown whan the answer to that question is ‘x’, or that question is required whenever the answer to somethind else is ‘y’. w20e.forms enables this using expressions. The epxressions are set as properties in variables, by their ‘bind’ attribute. So in the form model you may have a property set named ‘req’, that makes variable ‘foo’ required like so: Obviously in general you want something a bit more flexible than that, like checking for other data that has been entered. All form data is made available to the expression within the ‘data’ variable, that is a dict. So checking upon some other variable, goes like this: required=”data[‘bar’] == 42”)) So only if the answer to ‘bar’ is 42, ‘foo’ is required. Relevance, requiredness and readonly-ness all work like this. You may even add your own expression context to the engine, to call methods on objects, etc. Go like this, assuming your object is obj: >>> registry.register_expr_context('mycontext', obj) ... relevant="mycontext.some_method()) ## XML The xml namespace of the w20e.forms package provides an XML based implementation of the w20e.forms API. This enables definition from and serialization to XML files. Provided is the DTD used for defining the w20e.forms as XML. This is quite similar to xForms. Using XML as definition of forms provides a more declarative way of creating forms, not unlike the way you create a form in HTML. Also, XML is a format that is easily stored and transported. Start using the XML factory >>> from factory import XMLFormFactory Now let us create a factory class >>> xml = """ ... <form id="test"> ... ... <!-- The data part, a.k.a. the variables you wish to collect --> ... <data> ... <foo/> ... <bar value="666"/> ... </data> ... ... <model> ... <properties id="required"> ... <bind>foo</bind> ... <bind>bar</bind> ... <required>True</required> ... </properties> ... <properties id="int"> ... <bind>bar</bind> ... <datatype>int</datatype> ... </properties> ... </model> ... ... <view> ... <input id="fooctl" bind="foo"> ... <label>Foo?</label> ... <hint>Well, foo or no?</hint> ... </input> ... <select id="barctl" bind="bar"> ... <property name="multiple">False</property> ... <label>Bar</label> ... <option value="1">One</option> ... <option value="2">Two</option> ... </select> ... <select bind="bar" id="barctl2"> ... <label>Bar2</label> ... <option value="3">Three</option> ... <option value="4">Four</option> ... </select> ... <select bind="bar" id="barctl3"> ... <property name="vocab">some_vocab</property> ... <label>Bar3</label> ... </select> ... <group layout="flow" id="groupie"> ... <label>GruppoSportivo</label> ... <text id="txt">Moi</text> ... </group> ... </view> ... ... <submission type="none"> ... <property name="action">@@save</property> ... </submission> ... ... </form>""" We are using a vocab in the xml, so register it… >>> from w20e.forms.registry import Registry … def some_vocab(): … return [Option(0, 0), Option(1, 1)] … Registry.register_vocab(‘some_vocab’, some_vocab) >>> xmlff = XMLFormFactory(xml) >>> form = xmlff.create_form() >>> print len(form.data.getFields()) 2 >>> print form.data.getField("foo").id foo >>> print form.data.getField("bar").value 666 Set the value >>> form.data.getField("bar").value = 777 >>> print form.data.getField("bar").value 777 Okido, so far so good. Now let’s see what properties we have. >>> props = form.model.getFieldProperties("bar") >>> len(props) 2 >>> intprop = [prop for prop in props if prop.id == "int"][0] >>> reqprop = [prop for prop in props if prop.id == "required"][0] >>> reqprop.getRequired() 'True' >>> intprop.getDatatype() 'int' Finally, check the viewable part, or the controls >>> ctrl = form.view.getRenderable(“fooctl”) >>> ctrl.label ‘Foo?’ >>> ctrl.__class__.__name__ 'Input' >>> ctrl.hint 'Well, foo or no?' >>> ctrl.id 'fooctl' >>> ctrl.bind 'foo' >>> ctrl = form.view.getRenderable("barctl") >>> ctrl.multiple 'False' >>> len(ctrl.options) 2 Do we get the nested stuff? >>> ctrl = form.view.getRenderable("txt") >>> ctrl.id 'txt' ### Serialization You can easily serilialize the form back into XML. Let’s try… >>> from serializer import XMLSerializer >>> serializer = XMLSerializer() >>> print serializer.serialize(form) <form id="test"> <data> <foo/> <bar value="777"/> </data> <model> <properties id="int"> <bind>bar</bind> <datatype>int</datatype> </properties> <properties id="required"> <bind>foo</bind> <bind>bar</bind> <required>True</required> </properties> </model> <view> <input bind="foo" id="fooctl"> <label>Foo?</label> <hint>Well, foo or no?</hint> </input> <select bind="bar" id="barctl"> <label>Bar</label> <property name="multiple">False</property> <option value="1">One</option> <option value="2">Two</option> </select> <select bind="bar" id="barctl2"> <label>Bar2</label> <option value="3">Three</option> <option value="4">Four</option> </select> <flowgroup id="groupie"> <label>GruppoSportivo</label> <text id="txt"/> </flowgroup> </view> <submission type="none"> <property name="action">@@save</property> </submission> </form> <BLANKLINE> Note that variable foo now holds the value 777. Sadly, it is hard to guarantee that all XML will be exactely the same as the input XML. ## Pyramid The pyramid package provides a simple means of using w20e.forms for pyramid apps. The package provides a specific ‘file’ field for pyramid, to enable extracting filename and contents from a file in a POST/GET request, and a base view. Would you wish to use w20e.forms, then: • for the view that you wish to show the actual form, override w20e.forms.pyramid.pyramidformview. Let’s do some imports first. Please note that it is more convenient to use the XML implementation, as shown later on. Also, if you insist on using the Pythonic implementation, it is better to make a factory create the form, so you can just call the factory from your view. Anyway, let’s go for the not-so-smart way: >>> from w20e.forms.form import Form >>> from w20e.forms.formdata import FormData >>> from w20e.forms.formmodel import FormModel >>> from w20e.forms.formview import FormView >>> from w20e.forms.submission.attrstorage import AttrStorage >>> from w20e.forms.data.field import Field >>> from w20e.forms.rendering.control import Input >>> from w20e.forms.pyramid.formview import formview as pyramidformview Phew, that was a load of imports. Now do the actual view class. It’s a pretty simple form, but you should get the picture. >>> class yourformview(pyramidformview): ... def __init__(self, context, request): ... data = FormData() ... model = FormModel() ... view = FormView() ... # We'll leave the poperties out for now, check the main ... view.addRenderable(Input("input1", "Input bar here", bind="bar")) ... submission = AttrStorage(attr_name="_data") ... form = Form("test", data, model, view, submission) ... pyramidformview.__init__(self, context, request, form) Now, a view for pyramid just takes a context, and a request, so let’s create the view instance: >>> class Context: ... """ nothing needed here, but we'll store the data in here """ >>> class Request: ... def __init__(self, params=None): ... self.params = params >>> ctx = Context() >>> req = Request() >>> view = yourformview(ctx, req) Ok, we’re ready for some action now. Let’s try to render the form. >>> print view.renderform() <form class="w20e-form" method="post" action="" enctype="multipart/form-data"> <input type="hidden" name="formprocess" value="1"/> <div id="input0" class="control input relevant"> <label for="input-input0">Input foo</label> <div class="hint"></div> <input id="input-input0" type="text" name="input0" value="some default value" size="20"/> </div> <div id="input1" class="control input relevant"> <label for="input-input1">Input bar here</label> <div class="hint"></div> <input id="input-input1" type="text" name="input1" value="" size="20"/> </div> </form> <BLANKLINE> Nice. Now let’s give the request some content, and let the view handle the submission. This should result in the context having the form data stored in the _data attribute. formprocess is the marker used by w20e.forms to assume that the form is posted. >>> req = Request({'formprocess': 1, 'input0': 6, 'input1': 'whatever'}) >>> view = yourformview(ctx, req) >>> view() {'status': 'stored', 'errors': {}} >>> ctx._data.getField('foo').value 6 >>> ctx._data.getField('bar').value 'whatever' ### XML implementation Using the XML implementation makes life even easier: from w20e.forms.pyramid.formview import xmlformview as pyramidformview from w20e.forms.xml.formfile import FormFile class yourformview(pyramidformview): def __init__(self, context, request): pyramidformview.__init__(self, context, request, FormFile("forms/yourform.xml")) where you have a directory ‘forms’ containing the XML definition called yourform.xml. Check the w20e.forms.xml module for details on XML definitions. • Create a template (form.pt for example) that calls the render method of the view: <p tal:content="structure python:view.renderform()"></p> • Wire the stuff into zcml (assuming you use that), like so: <view context=".models.YourModel" view=".views.yourformview" renderer="templates/form.pt" name="yourform" /> ## Project details ### Source Distribution w20e.forms-1.0.2b.tar.gz (49.6 kB view hashes) Uploaded source	2022-09-30 12:00:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17672094702720642, "perplexity": 13705.248275861917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335469.40/warc/CC-MAIN-20220930113830-20220930143830-00385.warc.gz"}
https://math.stackexchange.com/questions/3152426/find-the-minimum-value-of-4a3-b3-c3-15abc-with-a-b-c-2/3152458	Find the minimum value of $4(a^3 + b^3 + c^3) + 15abc$ with $a + b + c = 2$. $$a$$, $$b$$ and $$c$$ are three sides of a triangle such that $$a + b + c = 2$$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$ Every task asking for finding the minimum value of an expression containing the product of all of the variables scares me. Here what I've done. Using the AM-GM inequality and the Schur's inequality, we have that $$a^3 + b^3 + c^3 \ge 3abc \implies P \ge \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$$ $$\ge \dfrac{9}{2}[ab(a + b) + bc(b + c) + ca(c + a)] = \dfrac{9}{2}[ab(2 - c) + bc(2 - a) + ca(2 - b)]$$ $$\ge \dfrac{9}{2}[2(ab + bc + ca) - 3abc] \ge \dfrac{27}{2}[2\sqrt[\frac{3}{2}]{abc} - abc]$$ Let $$abc = m \implies m \le \left(\dfrac{a + b + c}{3}\right)^3 = \dfrac{8}{27}$$ The problem becomes Find the minimum value of $$P' = 2\sqrt[\frac{3}{2}]{m} - m$$ when $$0 < m \le \dfrac{8}{27}$$. which is invalid because there isn't a minimum with the given condition. • It must be $\implies P \color{red}{\le} \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$ – farruhota Mar 18 '19 at 7:52 Let $$a=b=c=\frac{2}{3}$$. Thus, $$P=8.$$ We'll prove that it's a minimal value of $$P$$. Indeed, we need to prove that $$\sum_{cyc}(4a^3+5abc)\geq(a+b+c)^3$$ or $$3\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$ which is true by Schur. Considering $$P = 4(a^3 + b^3 + c^3) + 15abc \qquad \text{with} \qquad a+b+c=2$$ eliminate $$c$$ from the constaint to get $$P=3 a^2 (8-9 b)-3 a (b-2) (9 b-8)+8 (3 (b-2) b+4)$$ Now $$\frac{\partial P}{\partial a}=6 a (8-9 b)-3 (b-2) (9 b-8)=0 \implies a=\frac{2-b} 2$$ Reusing the constaint, this gives $$c=a$$ and then $$a=b=c=\frac 23$$. Plug in $$P$$ and get the result.	2020-07-10 10:52:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8508490324020386, "perplexity": 207.05230987644202}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655906934.51/warc/CC-MAIN-20200710082212-20200710112212-00356.warc.gz"}
https://puzzling.stackexchange.com/questions/67815/bury-me-alive-and-i-won-t-die-who-am-i/67823	Bury me alive, and I won’t die. Who am I? Slice me in half, I’ll have a twin. I’ll mix things up, using my skin. Bury me alive, and I won’t die. I know when it’s light, but have no eyes. Who am I? You are A worm Slice me in half, I’ll have a twin. If you cut one in half it looks like you have two worms. I’ll mix things up, using my skin. In travelling through the soil, they churn it as they go (due to their slightly "ribbed" bodies). Bury me alive, and I won’t die. They live underground. I know when it’s light, but have no eyes. Worms have light-sensitive cells which help them tell the difference between light and dark. • I think that the second part of the clue was based on a misconception that makes it stronger than the reasoning in your answer (but yours is correct). – Guy G Jul 6 '18 at 9:12 • iirc, they die after a short time of decapitation tho – hanshenrik Jul 9 '18 at 8:35 • Cutting a worm in half does not create two worms. It simply looks like you have two worms (the reasoning behind my answer). – ck459 Jul 9 '18 at 8:54 Could it be... A seed / plant Slice me in half, I’ll have a twin. Most seeds come in two 'halves', and can easily be split apart down that line. (pretty sure there is some species of seeds that can live when cut in two.) I’ll mix things up, using my skin. Weird things happen inside that seed as it turns into a plant... On the outside, it starts to grow roots, and interact with the soil. Bury me alive, and I won’t die. You can bury a seed in the ground and it lives perfectly well. I know when it’s light, but have no eyes. They can't see, being a plant, but it can definitely tell when the sun is out (certain flowers only bloom at certain times of the day, a plant will grow in the direction of light, etc.) • This was my first thought, but ck459's answer fits the skin part much better. – arp Jul 4 '18 at 15:16 • Interesting unexpected answer. – iBug Jul 5 '18 at 6:59 • I think the cut in half thing is a distraction in the puzzle, many seeds die if you cut them in half, and a lot of (most I think!) worms die too if you cut them in half. – Prof. Falken supports Monica Jul 6 '18 at 15:21 • @Prof.Falken I'm still unsure whether we are supposed to assume that "cut in half" means that it can live. The riddle doesn't say it HAS to live, just that it will have two parts that look identical. – iiiidk Jul 6 '18 at 18:04 • @iiiidk, thanks for that, a new thought. – Prof. Falken supports Monica Jul 6 '18 at 18:47 You are A potato Slice me in half, I’ll have a twin. Potatoes can be cut in half, and each half can grow a new plant I’ll mix things up, using my skin. Many recipes for filled potato skins Bury me alive, and I won’t die. Bury a potato, and it will grow I know when it’s light, but have no eyes. Potatoes turn green in the light. • exactly what I was thinking – Nitin Sawant Jul 5 '18 at 9:23 • Potatoes have eyes. – Stephen Paulger Jul 5 '18 at 10:06 • ^ 48 upvotes to that comment!! – Manoj Kumar Jul 9 '18 at 10:00 You are an electricity cable Slice me in half, I’ll have a twin. some electricity cables contain two wires I’ll mix things up, using my skin. if you want to join two cables, you remove their skin Bury me alive, and I won’t die. A live cable is a cable with electricity flowing in it. If it is underground it will still let electricity flow I know when it’s light, but have no eyes. electricity is used for making light • Interesting thought! – xxx--- Jul 5 '18 at 9:14 • But you're removing the "skin" to mix (splice) wires, you don't use the skin to make the splice... Interesting answer though! – Doktor J Jul 6 '18 at 18:48 • Just because the electricity makes light, why would the wire know when it's light? – Heimdall Jul 8 '18 at 10:24 • @Heimdall - The wire gets hot when the light's turned on. – J.R. Jul 8 '18 at 22:07 • @Heimdall it goes directly to lightbulb, of course it know where it is. Fun fact: in my language electricity is often called "light". – talex Jul 9 '18 at 4:17 You are a sea star Slice me in half, I’ll have a twin. sea stars have the ability to regenerate I’ll mix things up, using my skin. mixing the sand using its skin?? Bury me alive, and I won’t die. you can bury it on the sand and it won't die. I know when it’s light, but have no eyes. they can't really see but can sense light • you can bury it on the sand and it won't die. Really? I supposed it needs water. – Sigur Jul 4 '18 at 16:36 • @Sigur It could be interpreted as the sand at the bottom of the ocean. – Tin Man Jul 5 '18 at 11:20 • @TinMan, ow, in this case, yes... lol – Sigur Jul 5 '18 at 12:42 It could be a Cell Slice me in half, I’ll have a twin. Cells when split are 2 I’ll mix things up, using my skin. Some cells when they move mix around substances around them with their membrane/wall Bury me alive, and I won’t die. If you bury a cell, it still lives I know when it’s light, but have no eyes Photo receptive cells You are Slice me in half, I’ll have a twin. Planarians are known for their regeneration capabilities I’ll mix things up, using my skin. Bury me alive, and I won’t die. Some species of Planaria are terrestrial and can be found digging in dirt. I know when it’s light, but have no eyes. Planaria have photoreceptors which can detect light. You could be Baker's yeast Slice me in half, I’ll have a twin. Yeast is a clump of bacteria that can be split pretty often I’ll mix things up, using my skin. The bacteria on the outside react with the surrounding sugar, produce CO2 and make the dough grow. Bury me alive, and I won’t die. You can bury it in dough just fine. I know when it’s light, but have no eyes. Yeast is sensitive to light and will die if exposed for too long. • Yeast is not a bacterium but a fungus. But yes, a lump of yeast is a clump of single cell organisms. And it's that fungus that eats sugar and creates bubbles of CO$_2$ in the dough. – Heimdall Jul 8 '18 at 10:42 You are A line. Slice me in half, I'll have a twin. If you slice a line in half, you will have two lines with equal lengths — it was never mentioned that it must be sliced into two arbitrary parts. Twins! I'll mix things up, using my skin. You can draw all kinds of shapes and pictures using lines. Bury me alive, and I won't die. Bury lines in colour, and it might even look more alive than what it was. (I assume that the lines are alive until they are rubbed out.) I know when it's light, but I have no eyes. Not too sure about this one, but possibly refers to the colour of the surface on which the lines have been drawn on — it could be light or dark. Who am I? A line is not a who, which is what I think degrades my answer. Nice riddle, though! $$\stackrel{\bullet\,\bullet}{\smile}$$	2020-01-19 17:07:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4374469220638275, "perplexity": 3480.7634161037436}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250594662.6/warc/CC-MAIN-20200119151736-20200119175736-00403.warc.gz"}
http://archive.numdam.org/item/PMIHES_1966__29__95_0/	On the de Rham cohomology of algebraic varieties Publications Mathématiques de l'IHÉS, Volume 29 (1966), p. 95-103 @article{PMIHES_1966__29__95_0, author = {Grothendieck, Alexander}, title = {On the de Rham cohomology of algebraic varieties}, journal = {Publications Math\'ematiques de l'IH\'ES}, publisher = {Institut des Hautes \'Etudes Scientifiques}, volume = {29}, year = {1966}, pages = {95-103}, zbl = {0145.17602}, mrnumber = {33 \#7343}, language = {en}, url = {http://www.numdam.org/item/PMIHES_1966__29__95_0} } Grothendieck, Alexander. On the de Rham cohomology of algebraic varieties. Publications Mathématiques de l'IHÉS, Volume 29 (1966) pp. 95-103. http://www.numdam.org/item/PMIHES_1966__29__95_0/	2019-11-15 12:19:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5890352129936218, "perplexity": 2053.2581879857426}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668644.10/warc/CC-MAIN-20191115120854-20191115144854-00504.warc.gz"}
http://openstudy.com/updates/55e59338e4b0819646d82089	## anonymous one year ago Can Someone Help Me W/ This -Giving Medals- 1. anonymous here is the pic 2. anonymous $2^{M} = 1/8$ 3. anonymous this rules will help you, $u^v=e^{v \times \ln u}$ $\ln({e^x})=x$ $\ln(1/a)=-\ln(a)$ 4. phi one idea to try (and for these questions, it probably always works) is to see if you can multiply the base (the 2 in this problem) times itself, and get the number 8 222= 8 that means you can write it as $$2^3$$ using the "short-hand way" (with exponents) $2^m= \frac{1}{2^3}$ that looks promising , except the right side is "upside-down" you can "flip" the fraction if you make the exponent negative in other words $\frac{1}{2^3} = \frac{2^{-3}}{1} = 2^{-3}$ so now it is $2^m = 2^{-3}$ now pattern match... what should m be to make both sides look the same ?	2016-10-24 22:06:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8446447253227234, "perplexity": 1272.5623678029863}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719784.62/warc/CC-MAIN-20161020183839-00141-ip-10-171-6-4.ec2.internal.warc.gz"}
https://www.transtutors.com/questions/magernial-accouting-1306270.htm	# magernial accouting Ralph owns a production function. Randomness in the environment plus labor input from a manager combine to produce output. The output can be one of two quantities: 1 2x x ?. The manager’s input can be one of two quantities, H L ?. Ralph is risk neutral. The probabilities are given below, and you should assume the higher output is sufficiently attractive that Ralph wants supply of input Hin all that follows. Ralph’s manager is risk averse and also incurs an unobservable personal cost in supplying the labor input. We model this in the usual way. The manager’s utility for wealth is as given in (13.2), with ? 5,000Hc, 0 ?Lcand ?? 0.0001. Also, the manager’s opportunity cost of working for Ralph is a certainty equivalent of M ? 10,000. (a) Suppose the manager is trustworthy and will honor any agreement (or, equivalently, serious penalties are feasible and the manager’s input can be observed). What is the cost to Ralph of acquiring input H? (b) Suppose the only observable for contracting purposes is the manager’s output. Determine the optimal pay-for-performance arrangement. What is the cost to Ralph of acquiring input H? What is the manager’s certainty equivalent for the payment lottery that is faced? (c) Why, in your solution to part (b) above, is the manager paid more when the largest feasible output (i.e., 2x) is observed? 1x 2x input H0.1 0.9 input L0.8 0.2 ## Related Questions in Performance Management • ### Randomized monitoring This is a continuation of problem 8 in Chapter 13. Everything remains as... December 02, 2016 will be paid IH = 15,000. If input L is reported, the manager will be fired, with a payment of IL = 0 . (No negative payments are allowed.) This is certainly effective, but far too costly . (a) Suppose Ralph can commit to buying the information only when output x 1 is observed. Ralph will then pay 0... • ### divisional structure within an organisation (Solved) February 22, 2013 The question is Analyse the benefits and drawbacks of using the divisional structure within an organisation. If an organisation is going to structure itself along divisions, how can it ensure that performance is recorded and measured accurately? What other organisational structures may better In order to be effective and efficient, every organization must have an organizational structure. An organizational structure is that form of structure which determines the hierarchy. The... • ### Technovia Inc. has two divisions: Auxiliary Components and Audio (Solved) October 15, 2013 that all internal prices be expressed on a full cost -plus basis. The markup in the full cost pricing arrangement, however, is left to the discretion of the divisional managers. Recently, the two divisional managers met to discuss a pricing agreement for a subwoofer that would be sold with a... 1) Transfer price – it is the price at which one department of the company sells goods to other department. So effectively it is the price at which two divisions of a company transact... • ### Thompson and Perry (1992, p.3) define the project as “a new structure, system or service and also a. May 13, 2015 to, can be categorized into two general groups: Group A: These models specifically deal with risk management subjects, refer to the details of the processes, and provide different case studies and solutions, such as Cooper’s Model . Group B : These models are parts or subsets of project management • ### Lansing Electronics, Inc., manufactures a variety of printers, scanners, and fax machines in its two... (Solved) October 18, 2015 for this part ) of Part Y34 to outside buyers at $12 per unit. The PSF Division could also buy the part for$ 12 from external suppliers. The Components Division manager supplied the following details on the manufacturing cost of the component: Direct material \$2.50 Direct labor 0 .50 Variable... A transfer prcie is the price charged for the component by selling division to the buying division of the same company. 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http://philosophy.stackexchange.com/questions/4663/is-nothing-actually-imaginable	# Is Nothing actually imaginable? It's possible to imagine something, for example a table, we see one everyday and can bring it in front of our minds eye (although it's a moot point whether we can see it - I certainly don't). But of course this is a real object so we have a referent. But we need not have a referent: to imagine a unicorn means we are hybridising our referents. But to imagine nothing in its proper sense (ie not the lack of something or just space) seems impossible to me, we cannot avoid our sense of ourselves. - If you are given a pencil and a piece of paper and asked to draw anything you like, you can decide to draw nothing , or draw an alien , or draw a car. Note that returning the paper blank requires you to imagine nothing, or simply, not imagine anything. – user2411 Jan 7 '13 at 11:23 For a majority of your time, I would say you imagine nothing. Unless you imagine everything at every moment, I don't see how imagining nothing is avoidable. – SAHornickel Jan 7 '13 at 16:05 Sometimes, answers are simple. Nothing cannot be imagined because one does not imagine absences of anything, only things (which may lack something, but then you are merely imagining a thing without another thing). @SAHornickel - Not imagining anything is not the same as imagining nothing. Imagining-something is an act with an object, while a lack of imagining-something is not an act, and is not the same as imagining-nothing. – danielm Jan 7 '13 at 18:34 @danielm, it would depend on if imagining is active or passive. If the brain must, at all times, passively imagine something (like the perceived world, for instance), then not imagining anything would, indeed, be imagining nothing. When it comes to it, I think "imagine" has an insufficient definition. Do you think "imaging" would be a better fit for this question? – SAHornickel Jan 8 '13 at 11:05 Could you explain why the answer to this question might be relevant to philosophy? This particular quirk of neurophysiology--"I can form an emotionally satisfying empty representation, or I cannot"--seems to have more relevance for you and others than other quirks--"I can imagine a room bigger on the inside than the outside, or I cannot". – Rex Kerr Jan 10 '13 at 1:09 Based on your last paragraph, you might be interested in Thomas Nagel's The View From Nowhere. In that, he argues that it is impossible to achieve a completely objective perspective--- what he calls the View From Nowhere. This isn't directly related to your first paragraph, but something you might enjoy. As to your first paragraph, you might find this book interesting. Locke had some interesting ideas about the limits of imagination. For him, what is imagined is always some manipulation of things actually experienced. So, for example, you can only imagine a Centaur (half-man, half-horse) because you have experience of both a man and a horse--- or at least things relevantly similar. A similar sentiment is echoed in Descartes's First Meditation. Check out section 6 and the discussion of painters. As to imagining "Nothing", I'm inclined, along with you, to think that this is impossible. It seems to be no different than thinking about nothing. But then it seems like there is something you are thinking about, namely, nothing! UPDATE: It occurred to me that given the Ontology tag in your question, and given that my last paragraph is mostly based on my own idiosyncratic views about existence and reference, I should bring in some considerations from the seminal article on ontology, Quine's "On What There Is". Your questions about nothing, and my own reasons for thinking that imagining nothing is impossible, bear a striking resemblance to the problem of negative existentials. Some philosophers, notably the Meinongians, have thought that there are some things that have the property of "not existing". So, they would analyze negative existentials like "There are no unicorns" as expressing the sentence "There is something such that it is a unicorn and it doesn't exist". They could do this because they distinguished between two senses of "there is". One, the one familiar to us from Quine, is to read "there is" as expressing the existential quantifier. Anything that "there is", in this sense, exists. Now, the other sense of "there is" is subsistence. They thought that there are some things (like unicorns, for example) that subsist but do not exist. Quine thought that this talk of there being things that don't exist was a bunch of nonsense. He held that "there is" only expresses the existential quantifier and that anything there is must exist (as an aside, he famously, but uninformatively, answers the question "What is there?" with "Everything"). But then how did he analyze our earlier sentence about unicorns? He would analyze is thusly: "It is not the case that there exists something such that it is a unicorn" (sorry for the quasi-logic speak, I really want to regiment this in first-order logic a la Quine, but can't seem to get MathJax to work on this SE). For Quine, this sentence carries no presupposition of anything's existence, much less of a unicorn which subsists but does not exist. Bringing this back to the original question about "Nothing". If I put on my Quine Hat, I might say that to imagine nothing is simply for it to not be the case that you are imagining something. But that isn't very helpful, is it? Well, let's suppose (as we seem to be supposing in this example) that imagination is object oriented, so that whenever we imagine, there is some object of our imagination. What this pseudo-Quinean view would hold, then, is that to imagine nothing is simply to not be imagining any particular thing or collection of things. So, for example, a dead person is imagining nothing. I imagine (chuckle) that this view would deny any "objecthood" to "Nothing". - What, precisely, do you mean by 'to imagine'? The word itself by it's etymology suggests picturing nothing. This of course is absurd: there is nothing to picture, and we don't have ready experience picturing nothing, not even space. (Or do we?) More generally, we often use 'imagine' to mean to think about what something would "be like". This brings us to the brink of productive inquiry. Obviously, nothingness would almost certainly not "be like" anything we would ever perceive; and as you remark, hiding in the background is your continuing sense of self. To imagine nothing, you must then as a necessary condition negate those two things: you must imagine not perceiving anything, and in particular not being conscious. And while you may have limited experiences with such mental states, it is certainly possible to lose consciousness to such an extent that you are beyond dreaming, and so that as you start to regain consciousness you regain notions of space and of identity, which you come to realise afterwards that you had momentarily lost. And of course, while so deeply unconscious you aren't perceiving anything, or at least your ignorance of perceiving anything is so complete that if you did perceive anything, you're completely ignorant of having done so later. So: if nothingness is — subjectively — like anything at all, it is like unconsciousness. This is no more or less than the negation of Descartes' cogito. Of course, with nothingness there can be no subjective position; but unconsciousness is a negation of the subjective position anyway. We can only experience it imperfectly and in degrees, because to the extent that we can experience something, we are conscious, if (again) only imperfectly. But we can get a sense of unconsciousness from our various transitions in and out of consciousness. And that is the closest one can come to imagining nothing: to imagine a state of no perception or awareness — because of course there is nothing to be aware, and nothing to be perceived. - +1, this is more elaborate than my comment. – user2411 Jan 7 '13 at 11:26 But to imagine nothing in its proper sense (ie not the lack of something or just space) seems impossible to me, we cannot avoid our sense of ourselves. You have tagged this question with "Buddhism," so I'm going to offer an answer that is aware of some Buddhist doctrines concerning the issue. First, recognize that we're dealing with the edge of that which is semantically meaningful at this point. Immediately, in the true sense we cannot "conceive" of nothing because the act or behavior of conception binds the referent concept. As you say, you cannot escape the self. So, by virtue of sheer semantics, it is impossible to conceive of nothing. On the other hand, if you consider everything that you are conceiving of at any given moment, I can ask you: what else do you conceive? And the answer is nothing. Now this may at first seem to be a simply trick of semantics (i.e., we have here the same word-symbol used in two different ways), but if you consider that thinking of nothing would require that you think of something without a referent, then it is in fact fair to say that all of that which are you not thinking about is thinking about nothing. In this very real way, it is impossible to NOT conceive of nothing, because you cannot get the lack of a thing out of your mind. You can see here that the semantics of the problem are inherently weak. You can argue it one way or the other if you so choose, but from the Buddhist perspective, one of the key points of "emptiness" is that it begins to shatter the normal mode of ego-conceptual thought. This is, in fact, analogous to one of the key 'points' (insofar as there are any) to Zen koans. - +1 for adding the Buddhist perspective. I would have liked to touch on it but lack the background. – Dennis Jan 7 '13 at 15:41 Is Nothing actually imaginable? To imagine nothing in its proper sense (ie not the lack of something or just space) seems impossible to me, we cannot avoid our sense of ourselves. It's difficult to tell what is being asked here. Your question is if a concept without image ("Nothing") is "imaginable". Abstract concepts do not necessarily appear as images. For example the concept of causality. - 'Imagine' probably originates from 'image', but it now covers more. – Mozibur Ullah Jan 11 '13 at 2:10 Mental representation can represent things that have you never experienced as well as things that do not exist. To instrumentalism, there is the notion that unobservables such as atomic particles, the force of gravity, causation and quantum physics, are useful representation models but don't necessarily exist. Only provide a way of thinking about natural laws, a common mind-independent world, a way of description that relates representation to prediction, expressing truths – Ricardo Jan 15 '13 at 9:27 While not directly addressing the problem of "imagining nothingness", I'd like to introduce some of Tamar Gendler's fascinating work on imagination. What she calls the "The Puzzle of Imaginative Resistance" (first introduced in an essay of the same name), is the following: The puzzle of explaining our comparative difficulty in imagining fictional worlds that we take to be morally deviant. We are capable of imagining a vast array of implausible, outlandish, and playful fantasies ("We have no trouble imagining that Sherlock Holmes solved mysteries in nineteeth0century London, that an owl and a pussycat went out to sea in a beautiful pea-green boat, or that a hobbit names Frodo Baggins carried a magic ring all over Middle Earth.") But what happens when we are presented with a story that contains something like the following?: In killing her baby, Giselda did the right thing; after all, it was a girl. We find ourselves unwilling to imagine this as truth. We are inclined to say, even within the world of the story, the narrator is wrong. What explains this resistance to make-believe? One hypothesis posits that propositions which we judge to be morally deviant are not make-believable, because they represent an impossible state of affairs. If we believe that infanticide is always wrong in the real world, we simply cannot make sense of what a world would be like if that world is said to be one in which infanticide is always right. We can state "The impossibility hypothesis" thus: Imaginative Resistance is explained by the following two considerations: (1) the scenarios that evoke imaginative resistance are conceptually impossible; (2) the conceptual impossibility of these scenarios renders them unimaginable. Tamar Gendler responds by providing examples of "imaginable conceptual impossibilities," that is, concepts we can both 1) imagine easily, 2) hold to be physically (or even logically) impossible. We all know that the following propositions are false, and impossibly so: a) 12 is not the sum of 5 and 7, b) 12 used to be the sum of 5 and 7, but is no longer the sum of 5 and 7, c) 12 both is and is not the sum of 5 and 7. Now the question is: can we imagine these to be correct? Gendler offers the following story as evidence that we, in fact, can: The Tower of Goldbach Long long ago, when the world was created, every even number was the sum of two primes. Although most people suspected that this was the case, no one was completely certain. So a great convocation was called, and for forty days and forty nights, all the mathematicians of the world labored together in an effort to prove this hypothesis. Their efforts were not in vain: at midnight on the fortieth day, a proof was found. "Hoorah!" they cried, "we have unlocked the secret of nature." But when God heard this display of arrogance, God was angry. From heaven roared a thundering voice: "My children, you have gone too far. You have understood too many of the universe's secrets. From this day forth, no longer shall twelve be sum of two primes." And God's word was made manifest, and twelve was no longer the sum of two primes. The mathematicians were distraught- all their efforts had been in vain. They beseeched God: "Please," they said, "if we can find twelve persons among us who are still faithful to You, will You not relent and make twelve once again the sum of two primes?" And so God agreed. The mathematicians searched and searched. In one town, they found seven who were righteous. In another, they found five. They tried to bring them together to make twelve, but because twelve was no longer the sum of two primes, they could not. "Lord," they cried out, "what shall we do? If You lifted Your punishment, there would indeed be twelve righteous souls, and Your decision to do so would be in keeping with Your decree. But until You do, twelve are not to be found, and we are destined forever to have labored in vain." God was moved by their plea, and called upon Solomon to aid in making the decision. Carefully, Solomon weighed both sides of the issue. If twelve again became the sum of two primes, then the conditions according to which God and the mathematicians had agreed would be satisfied. And if twelve remained not the sum of two primes,again the conditions according to which God and the mathematicians had agreed would be satisfied. How Solomonic it would be to satisfy the conditions twice over! So with great fanfare, the celebrated judge announced his resolution of the dispute: From that day on, twelve both was and was not the sum of five and seven. And the heavens were glad, and the mountains rang with joy. And the voices of the five and seven righteous souls rose toward heaven, a chorus twelve and not-twelve, singing in harmonious unity the praises of the Lord. The End. I find this story quite convincing: impossibilities are imaginable. It may in fact be the case that nothing does not exist, or that nothing is an ill-formed concept, or that nothingness is impossibly remote from human experience. None of that suggests that nothingness is unimaginable. - Everything you think of is something (something that you think). In that sense, when thinking about nothing, we must admit that nothing is a kind of something. That's one of the fundamental paradoxes of thinking (and of being in general). So yes, you can think of nothing, just like you can think of anything else. But the question remains, "When I try to think of nothing, am I truly thinking of nothing, or just something that looks like nothing?" In other words, is the absence of something enough to manifest nothing, or do we need the absence of everything to express nothing? Well, you could argue that the same problem exists with "everything". Can we think of everything? How would we know if we were thinking of everything, or just something that looks like everything? What if we could think of everything, what would it look like? To have everything in your mind, all at once, complete in a singular whole? One could argue that the singular nature of everything makes it indistinguishable from nothing (which is also singular). This is another way of expressing the paradox, that nothing and everything are equally intangible, and in some meaningful sense indistinguishable. One way to resolve all this is to think of nothing and everything as properties of something, not complements to it. Every something is just a bit of nothing and a bit of everything combined. To put it another way, anything can look like nothing if there's none of it, or it can look like everything if you have all of it. Nothing and everything are properties of objects comprised of something. Now, If everything has some nothing to it, are there as many nothings as there are somethings? No, there is just one nothing (singular), and whether you think of it as the absence of one thing or another, it doesn't matter, you are always thinking of the same thing: nothing. - Why the downvote? – Greg L Jan 7 '13 at 15:47 I'm not the one who downvoted, but I could imagine the downvoter's issue being a lack of clarity and references to the relevant literature. Your answer reads as your thoughts on the matter, which I have no problem with, but I think the community tends to prefer answers to be as non-subjective and non-idiosyncratic as possible. Also, it isn't clear how your thoughts on "everything" are really relevant. Really, it is hard for me to see the relevance the three paragraphs after your second paragraph and they aren't very clear. These are just my thoughts, though, about what I would want improved. – Dennis Jan 7 '13 at 17:39 Ok thanks for the feedback, I'm new to the site and not familiar with the community norms. I just think these types of questions actually suffer from trying too hard to define the words, the concepts can't really be put into a finite string of other concepts, so it's best to take a synthetic approach to the definitions. And I don't know how you can synthesize nothing and something successfully without relying on the much related concepts of anything and everything. Synthetic definitions is an approach that works in mathematics but perhaps not in philosophy. – Greg L Jan 7 '13 at 18:16 I'd focus more on trying to tie your thoughts into relevant literature. With our answers we often walk the line between subjective and objective, and I think we try to minimize the subjective aspect as much as we can (obviously it can't always be avoided, and this is fine) by referencing relevant literature. It is also a nice thing to do (when possible) because it gives the questioner somewhere to go for further information. – Dennis Jan 7 '13 at 18:27 I'm the downvoter. My issue was much more direct and simple: I think that ideas such as "Everything you think of is something ... when thinking about nothing, we must admit that nothing is a kind of something" are a confusion of language, and disconnected from the context of the question. Much of what followed seemed to be of a similar character, and I found it not to be constructive. More than making a connection to the literature, it's important to keep the answer pertinent to the question. – Niel de Beaudrap Jan 9 '13 at 12:44 I think we can imagine nothing but it is always within the context of something. For example, I can imagine a block of outer space which is simply nothing. But I see that nothing within the context of the rest of the universe which is something. To actually image nothing would seem to require a lack of consciousness (which then becomes a blocking problem) If you imagine, you are creating something. You are creating whatever it is you are imagining and thus you have created something. - But 'empty space' is something... – Mozibur Ullah Jan 8 '13 at 23:49 As I hear the question and terms used, "imagining" is an intentional mental action, a conjuring of images. Our sense of self is entirely conditioned upon our intentional thought creation (Skt sanskara, Pali saṅkhāra or sankhaara). So no, one can not intentionally imagine anything (nor nothing) without a sense of self. One who actively imagines anything (whether the concept of nothing or dancing rabbits) does have a sense of self. There are pleasurable mental states known as jhana in which one is highly concentrated but not imagining (neither nothing nor something). The lower jhana could not be described as nothingness as pleasure is certainly experienced. I think you are asking "Can one stop the intentional thinking processes?" or "Can one maintain a mental state of nothing?" Yes. The abandonment of intentional thinking is achieved in the second jhana. In fact the first jhana maintains a very subtle form of thinking (Pali vitakka) that is firm and directed but not discursive. The Buddha (MN 19) gives the simile of a cowherd in the shade of a tree mindfully aware of his cows after the crops have been harvested. In the second jhana even this subtle vitakka is abandoned (there remains awareness but not 'imagining'). In MN 19, Thanissaro Bikkhu translates: "rapture & pleasure born of composure, unification of awareness free from directed thought & evaluation — internal assurance". The higher non-material jhanas beyond the fourth (santa vimokkha atikammarupe aruppa) are described by the Buddha with words like 'boundless space', 'boundless consciousness', 'nothingness', 'neither-perception-nor-non-perception'. But again these are not 'imagined' but experienced with liberation from self-creations (including the abandonment of a sense of self) as a goal. - The stuff in our imagination is in domain X the stuff not in our imaginations, is in domain Y. There is a relationship between X and Y - there is a mapping, but to claim that there is something in Y that is not possible to be mapped to X - is impossible. The act of claiming it puts in X - the mapping has to exist in order to make the claim. If the claim is that something in Y is mapped to X, but that the mapping lacks a certain quality - say "can be pictured", then sure, we can make that claim. But then you are claiming "One cannot picture nothing", or "One cannot visualize nothing" - and such a claim is like saying "You can't see invisible things" - 1) Is Nothing some sort of concept, a concrete object, or a class of objects? Nothing is definitively not a concrete object. Also, if it were a class of objects, it would have to be the empty class. There might be certain contexts in which the empty class actually represents Nothing. (It is more the content of the empty class than the empty class itself, which represents Nothing.) The best guess seems to be that Nothing is some sort of concept, which can be represented by different things in different contexts (empty space, empty set, empty list, empty string). 2) Is Nothing an abstract concept, a class of concepts, or a scientific concept? It should be fine to assume that Nothing is an abstract concept, which doesn't necessarily exclude that it may also be a class of concepts and a scientific concept. How can we imagine or visualize an abstract concept? A typical way is to look at some representative concrete instances of the abstract concept. What are the properties of the concrete instance of Nothing? For the empty set, the union of any set X with the empty set is again X. For the empty list, the left or right concatenation with any list L is again L. In both cases, there can be at most one object with these properties, but this doesn't mean that an object with these properties (for a given universe of sets or lists) is necessarily the empty set or the empty list. My conclusion from that is that Nothing can exist in many contexts. There are some context where it is quite possible to imagine or visualize Nothing. However, even so Nothing is unique in each given context, the different Nothings from different contexts are neither identical nor isomorphic to each other. This also means that to imagine Nothing in a specific context is not to enough for being able to imagine Nothing in any context or even imagine "Nothing itself". If we go back to the empty set again and look at intersection instead of union, then the intersection of the empty set with any set X is the empty set. However, this property is not a good way to characterize or imagine Nothing, because it gives the wrong impression that Nothing and Everything would somehow be the same thing. But Nothing exists naturally in many different contexts, whereas the existence of Everything would lead to inconsistencies in many different contexts (and hence often doesn't exist). - If we say the essence or concept of nothing is that it is totally without predicates, then we could say nothing exists as the subject of its concept. In conceptual terms that's quite tangible and imaginable. - There's just a basic logical mistake involved here. Think of your question being like, "How do I marry Nobody?" The answer is that you can't marry Nobody because there isn't any such person as Nobody. The capital letter makes it look like "nobody" is the name of a person that doesn't exist. But there aren't any people that don't exist. To marry nobody isn't to marry a special kind of person that doesn't exist, it's just to fail to marry anybody at all. Likewise, to imagine Nothing isn't to imagine something that doesn't exist, it's just not to imagine anything at all. The underlying issue in both cases is treating a quantifier as a name. For an excellent article (from which I think i've stolen the above examples, see Peter Geach Form and Existence.) -	2015-10-06 14:39:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5596117973327637, "perplexity": 984.2961817889493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736678818.13/warc/CC-MAIN-20151001215758-00041-ip-10-137-6-227.ec2.internal.warc.gz"}
https://www.shmoop.com/study-guides/math/radicals-powers-roots/roots-radicals	Study Guide # Roots and Radicals ## Roots and Radicals ### Dude, This Is Radical Next stop, radical town. All aboard the root train. While en route, all food service will be suspended with the exception of simplifying square roots and imaginary numbers. Okay, Shmoopers. We admit it. That was corny. And we know that you can't eat square roots or imaginary numbers. We were just getting tired of the radical surfer references. Regardless, this section is all about roots and radicals. A radical is any expression that uses the √ symbol. Take ∛5, for instance. This is a radical expression. In this expression, 3 is technically called the index, √ is known as the radical symbol, and 5 is the radicand. The next thing we need to know is how to simplify one of these so-called radicals. In other words, how to write a radical in what's otherwise known as simple radical form (more on that later). From there, our lovely sample problems will help us learn how to add, subtract, multiply, and divide roots. Enjoy the Shmooptastic ride. ### Sample Problem Simplify the following: . This is just a bit of a warm-up. We're reminding you that more than just square roots do exist. This cube root is really asking the question: What number multiplied by itself 3 times is -216? Nailed it. The answer's -6, because (-6)(-6)(-6) = -216. ### Sample Problem Simplify the following: . This one's still pretty basic, despite there being a couple steps to navigate. First, let's evaluate these radicals in the parentheses. Next, we should follow our order of operations and square the negative 2. (-3)(-2)2 = (-3)(4) Lastly, let's multiply and move on with our lives. (-3)(4) = -12 ### Sample Problem Simplify the following: . This problem isn't quite so easy; doesn't simplify to some nice, neat whole number like did in the first sample problem. That means we need to break down the radical into simple radical form. To do this, we break down into a product of prime numbers. Once we have our prime number breakdown in the bag, we can start to simplify. Since √(3 × 3) = 3, we can pull a 3 out of the square root. Gee, we can pull out a 2 as well. This gives us… This is simple radical form. In other words, our radical now contains only a product of unique primes. The key idea to remember is that when you have a pair of primes, you can always pull one out. ### Sample Problem Simplify the following: . This is a little bit tougher. Don't worry. We have you covered. All we really need to do is break down both square roots into their simple radical form. Next, we're going to pull out our pairs. By the way, if you get comfortable and go right from to , go for it. That just means that you're starting to truly understand, grasshopper. Finally, since we've got two like terms, we can add them together. This might be confusing, but it's really no different than saying 3x + 2x = 5x. The difference here is that instead of x, we're using . One final note: adding like terms such as this is only okay if the terms are really alike. You wouldn't add together, for the same reason you wouldn't add 3x + 2y together. Understand, rubber band? ### Fractional Mathematical Radicals So, you think you've conquered exponent rules, mastered negative exponents, and even shown radicals who's boss? Then why not give fractional exponents a try? Question: When do we use fractional exponents? Answer: We can write any radical expression using a fractional exponent. A simple equation like is literally the same as 361/2 = 6. Seriously. They mean exactly the same thing. This also means that and . While this may be a breakthrough of epic proportions, there is much more. Not only can we write radicals using fractional exponents, we can also combine indices with an exponent. FYI: "Indices" is plural for index. So is the same as x3/4, and is the same as (z – 1)57/71. Just stick the index of the radical in the denominator, with the original exponent in the numerator. This is useful because now we can apply all our exponent rules to radicals, too. Scizzore. ### Sample Problem Simplify the following: . First thing we're going to do is rewrite this equation using only exponents. Fractions, here we come. x2x4/3 Next, we know we've got to add exponents. This means we'll need to make sure 2 is rewritten as 6/3 before we add things up for our final answer. x2x4/3 = x6/3x4/3 = x10/3 Annnnd, we're done. For now. ### Sample Problem Simplify the following: . Moving left to right, let's get our Shmoop on. The first term is easy. The next term can't readily be simplified, unless we'd like to write it using a fractional exponent. The question now is: Are we able to add 3 and 61/3? Whenever we run into a problem like this, our advice is to always compare it to another, more familiar situation. For instance, can x + y2 be simplified? Of course not. Problem solved. ### Sample Problem Simplify the following: . It's back to the grind. This one forces us to move term by term once again. The whole number in the first term isn't too tough. However, the exponent is going to require some fractionalization. Yes, we did just make up a word. 2(x3)1/6 – (2x)2 + 4x1/2 = 2x3/6 – (2x)2 + 4x1/2 Simplifying the 36 exponent to ½ while simplifying that middle term gives us: 2x3/6 – (2x)2 + 4x1/2 = 2x1/2 – 4x2 + 4x1/2 What's nice here is that we can now add the like terms. Be careful; in this case, like terms are only those whose x factors have the same exponents. 2x1/2 – 4x2 + 4x1/2 = 6x1/2 – 4x2 That's as simple as we can make it. ### Sample Problem Simplify the following using positive exponents: . Let's start at the beginning. For this problem, that means we'll start by simplifying things inside of that square root. Next, we can take the square root of the numerator and denominator separately. Notice how the denominator will use our fractional exponent rules. Next, we can multiply. Remember, it's really like the 2x is over an invisible 1. This makes our multiplication no different than regular old fraction multiplication. Last but not least, we need to take care of the x's we have in the numerator and denominator. This is just a classic case of negative exponents. No sweat. Of course, we're going to great lengths to show you every single step. If your Shmoopability is on the fast track, feel free to skip a step here or there. Eventually, it'll be no big deal if your actual work for this problem looks like: ### Rationalizing the Denominator. (Yes, We're Serious.) Rationalizing the denominator always sounds like something that might be done at NASA just before the space station takes off. That's not the case. Allow us to break things down to size. Rationalizing is simply the process of making sure a number is actually a rational number. A rational number is just a number that can be written as a fraction. For instance 2, ¾, 17.12, and 1,000,000 are all rational numbers. On the other hand, π and √2 are both irrational. No matter how hard we try, we can't write them as fractions. Therefore, it would make sense that rationalizing the denominator is really just the process of writing the denominator as a rational number. The question remains: How? ### Sample Problem Simplify the following completely: . To get this guy into its simplest form, we need to rationalize the denominator. Mathematicians hate, just hate, when there's a loose square root on the bottom of a fraction. Let's do them a solid and get rid of it. Step one is to multiply our fraction by some form of 1. That form is determined by the square root in the denominator. Here it will be . After all, over is just equal to 1. See what we're talking about? Since × is 3, we can simplify and finish up. Don't fret. It's no sweat. Are we done yet? ### Sample Problem Simplify the following completely: . Sometimes there's a bit of simplification that occurs after the denominator has been rationalized. That will happen here. But first… After we multiply by our giant 1, we can multiply within our square roots. At this point, we actually have two more simplifications. We know that the fraction 106 reduces to 53. But first, we can actually simplify . Now we're finally ready to simplify to get our final answer. ### Sample Problem Simplify the following completely: . While this problem may look complicated, our fractional exponent skills can go a long way here. Let's start by simplifying the numerator. Next, we need to rationalize the denominator just as we've done before. While it may not look extremely simplified, this is the best this thing is going to get. ### Sample Problem Simplify the following: . This problem has a lot going on. First, we notice that we can actually solve a few of these radicals straight away. In face, we'll solve every one except for the . This helps a great deal. Next, we can deal with the by writing it in simple radical form. For now, we'll leave everything else alone. Finally, we can multiply fractions across the top and bottom while dividing out 5 over 5… If you'd like, you can also write your final answer as . Either way, it's time to move on.	2020-10-24 15:09:49	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8425049781799316, "perplexity": 763.3144734681476}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00206.warc.gz"}
http://mathoverflow.net/questions/41054/pad%c3%a9-approximation-usability-in-iterative-algorithms/41085	# Padé approximation - usability in iterative algorithms Firstly, I have to say that I don't understand Padé approximation well. But I discovered that, it is more precise than Taylor series. I have to create approximation for these functions: Log(x) and Tanh. And I have to create iterative algorithms (I must compute result with variable precision). So my questions are: Is Padé approximation usable (and more efficient than simple Taylor series) for this task? If no, is there any better way to approximate these functions? - What you can do that is equivalent to using a Padé approximant (SFAICT, there's no simple method for generating the coefficients of the numerator and denominator polynomials for the two functions you have, except by solving the appropriate Toeplitz system) is to use continued fraction expansions, which $\ln(1+x)$ and $\tanh(x)$ have by their virtue of being expressible as hypergeometric functions. Of course, for proper use, you have to perform appropriate argument reductions (e.g. for $\tanh$, compute $x^{\star}=\frac{x}{2^n}$ where $n$ is an appropriate integer such that $x^{\star}$ is "small" enough, evaluate the continued fraction at $x^{\star}$, and use the double-argument formula for $\tanh$ to undo your previous transformation).	2015-12-01 18:25:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9625216126441956, "perplexity": 463.4021803220994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398468971.92/warc/CC-MAIN-20151124205428-00001-ip-10-71-132-137.ec2.internal.warc.gz"}
https://www.shaalaa.com/question-bank-solutions/nuclear-energy-nuclear-fusion-energy-generation-stars-processes-energy_4342	Question - Nuclear Energy - Nuclear Fusion – Energy Generation in Stars Account Register Share Books Shortlist Your shortlist is empty ConceptNuclear Energy - Nuclear Fusion – Energy Generation in Stars Question Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction : ""_1^2H+_1^3H->_2^4He+n Using the data : m(""_1^2H) = 2.014102 u m(""_1^3H) = 3.016049 u m(""_2^4He) = 4.002603 u mn = 1.008665 u 1u = 931.5 MeV/c2 Solution You need to to view the solution Is there an error in this question or solution? S	2017-08-19 20:37:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23561331629753113, "perplexity": 1062.9836326505745}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105922.73/warc/CC-MAIN-20170819201404-20170819221404-00553.warc.gz"}
https://www.biostars.org/p/9490299/#9490315	How to extract genomic upstream region of a protein identified by its NCBI accession number? 1 0 Entering edit mode 28 days ago mrj ▴ 60 I have a list of NCBI protein accession numbers. I would like to extract out the upstream genomic region of the corresponding gene's nucleotide sequence. I will be thankful to you if you can show me how to get this done. For example, here are some of the protein accession numbers. I would like to extract out the upstream genomic region of their corresponding nucleotide sequence. EET74829.1 VEI24834.1 AYW77996.1 EJD65589.1 EFM49534.1 1 Entering edit mode 0 Entering edit mode GenoMax, Thanks for the advice. I have added the list of accession numbers in my original post. I will paste them below as well. EET74829.1 VEI24834.1 AYW77996.1 EJD65589.1 EFM49534.1 3 Entering edit mode 28 days ago GenoMax 107k These appear to be protein accession numbers that are pointing to various assemblies so there is no direct gene associations. So it may be best to do this as a three step process. Using Entrezdirect: Get the accession number of nucleotide assembly/genome $esearch -db protein -query AYW77996 \| elink -target nuccore \| efetch -format acc CP033719.1 Get the nucleotide start/stops for CDS $ efetch -db nuccore -id CP033719.1 -format fasta_cds_na \| grep AYW77996 >lcl\|CP033719.1_cds_AYW77996.1_1542 [locus_tag=EGX94_07890] [protein=copper oxidase] [protein_id=AYW77996.1] [location=1885267..1887939] [gbkey=CDS] Use the location coordinates to get the sequence you want (e.g. 200 bp upstream). Pay attention to the strand locations. \$ efetch -db nuccore -id CP033719.1 -format fasta -seq_start 1885067 -seq_stop 1885267 >CP033719.1:1885067-1885267 Propionibacterium acidifaciens strain FDAARGOS_576 chromosome, complete genome GGCTCCGAGCACTGGCGCCAGGTGGGCGGCCTGGGCAACATCGCAGCCCTGCTCGGTCTCGTCGCCGTGG CCGTCTGGTCGTCCGTGGTCCGGGACGCCGCCGAGGCCGAGCGGCCCCCGTCCGCGCGGGGCGGCCCCGG CCCGGTCGGCGGGGGAGCCCCCGACAACCCGCCCGCCATGACGATCCCGAGGACCGACGCA 0 Entering edit mode GenoMax, This is great. This works for me just great. Thank you very much.	2021-10-20 10:39:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28671395778656006, "perplexity": 7179.2909472832025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585305.53/warc/CC-MAIN-20211020090145-20211020120145-00313.warc.gz"}
https://tomcircle.wordpress.com/2014/03/01/our-daily-story-5-the-prince-of-math/	# Our Daily Story #5: The Prince of Math Carl Friedrich Gauss is named the “Prince of Math” for his great contributions in almost every branch of Math. As a child of a bricklayer father, Gauss used to follow his father to construction site to help counting the bricks. He learned how to stack the bricks in a pile of ten, add them up to obtain the total. If a pile has only 3, for example, he would top up 7 to make it 10 in a pile. Then 15 piles of 10 bricks would give a total of 150 bricks. One day in school, his teacher wanted to occupy the 9-year-old children from talking in class, made them add the sum: 1 + 2 + 3+ ….+ 98 + 99 + 100 = ? Gauss was the first child to submit the sum within few seconds = 5,050. He used his brick piling technique: add 1 + 100 = 101 2 + 99 = 101 3 + 98 = 101 49 + 52 = 101 50 + 51 = 101 In the 100 numbers broken down in pair of two as above, there are 50 pairs of sum 101, which gives total of 5050. This is a high school math formula for the Sum (S) of a series: ${(1, 2, 3,... n)}$ $\boxed {S = \frac {n.(n+1)} {2} }$ http://en.m.wikipedia.org/wiki/Carl_Friedrich_Gauss	2018-03-19 01:18:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38292932510375977, "perplexity": 967.3217106741025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646189.21/warc/CC-MAIN-20180319003616-20180319023616-00451.warc.gz"}
http://blog.codingthematrix.com/	The header image is derived from a photo of some of the work of a friend, the celebrated artist Janet Echelman. # A graph’s adjacency matrix, and counting walks In the movie Good Will Hunting, at the end of class the professor announces, I also put an advanced Fourier system on the main hallway chalkboard. I’m hoping that one of you might prove it by the end of the semester. Now the person to do so will not only be in my good graces but will also go on to fame and fortune, having their accomplishment recorded and their name printed in the auspicious MIT Tech. Former winners include Nobel laureates, Fields Medal winners, renowned astrophysicists, and lowly MIT professors. Must be a tough problem, eh? The hero, Will Hunting, who works as a janitor at MIT, sees the problem and surreptitiously writes down the solution. The class is abuzz over the weekend—who is the mysterious student who cracked this problem? The problem has nothing to do with Fourier systems. It has to do with representing and manipulating a graph using a matrix. Graphs Informally, a graph has points, called vertices or nodes, and links, called edges. Here is a diagram of the graph appearing in Will’s problem, but keep in mind that the graph doesn’t specify the geometric positions of the nodes or edges—just which edges connect which nodes. The nodes of this graph are labeled 1, 2, 3, and 4. There are two edges with endpoints 2 and 3, one edge with endpoints 1 and 2, and so on. The first part of Will’s problem is to find the adjacency matrix of the graph. The adjacency matrix ${A}$ of a graph ${G}$ is the ${D\times D}$ matrix where ${D}$ is the set of node labels. In Will’s graph, ${D=\{1,2,3,4\}}$. For any pair ${i,j}$ of nodes, ${A[i,j]}$ is the number of edges with endpoints ${i}$ and ${j}$. Therefore the adjacency matrix of Will’s graph is 1 2 3 4 1 0 1 0 1 2 1 0 2 1 3 0 2 0 0 4 1 1 0 0 Note that this matrix is symmetric. This reflects the fact that if an edge has endpoints ${i}$ and ${j}$ then it has endpoints ${j}$ and ${i}$. Another time we’ll discuss directed graphs, for which things are a bit more complicated. Note also that the diagonal elements of the matrix are zeroes. This reflects the fact that Will’s graph has no self-loops. A self-loop is an edge whose two endpoints are the same node. Walks The second part of the problem addresses walks in the graph. A walk is a sequence of alternating nodes and edges $\displaystyle v_0 \ e_0 \ v_1\ e_1\ \cdots \ e_{k-1}\ v_k$ in which each edge is immediately between its two endpoints. Here is a diagram of Will’s graph with the edges labeled: and here is the same diagram with the walk ${3\ c\ 2\ e\ 4\ e\ 2}$ shown: Note that a walk can use an edge multiple times. Also, a walk need not visit all the nodes of a graph. (A walk that visits all nodes is a traveling-salesperson tour; finding the shortest such tour is a famous example of a computationally difficult problem.) Will’s problem concerns three-step walks, by which is meant walks consisting of three edges. For example, here are all the three-step walks from node 3 to node 2: 3 c 2 e 4 e 2, 3 b 2 e 4 e 2, 3 c 2 c 3 c 2, 3 c 2 c 3 b 2, 3 c 2 b 3 c 2, 3 c 2 b 3 b 2, 3 b 2 c 3 c 2, 3 b 2 c 3 b 2, 3 b 2 b 3 c 2, 3 b 2 b 3 b 2 for a total of ten. Or, wait…. Did I miss any? Computing the number of walks Matrix-matrix multiplication can be used to compute, for every pair ${i,j}$ of nodes, the number of two-step walks from ${i}$ to ${j}$, the number of three-steps from ${i}$ to ${j}$, and so on. First, note that the adjacency matrix ${A}$ itself encodes the number of one-step walks. For each pair ${i,j}$ of nodes, ${A[i,j]}$ is the number of edges with endpoints ${i}$ and ${j}$, and therefore the number of one-step walks from ${i}$-to-${j}$. What about two-step walks? A two-step walk from ${i}$ to ${j}$ consists of a one-step walk from ${i}$ to some node ${k}$, followed by a one-step walk from ${k}$ to ${j}$. Thus the number of two-step walks from ${i}$ to ${j}$ equals number of one-step walks from ${i}$ to 1 ${\times}$ number of one-step walks from 1 to ${j}$ + number of one-step walks from ${i}$ to 2 ${\times}$ number of one-step walks from 2 to ${j}$ + number of one-step walks from ${i}$ to 3 ${\times}$ number of one-step walks from 3 to ${j}$ + number of one-step walks from ${i}$ to 4 ${\times}$ number of one-step walks from 4 to ${j}$ This has the form of a dot-product! Row ${i}$ of ${A}$ is a vector ${\boldsymbol{u}}$ such that ${\boldsymbol{u}[k]}$ is the number of one-step walks from ${i}$ to ${k}$. Column ${j}$ of ${A}$ is a vector ${\boldsymbol{v}}$ such that ${\boldsymbol{v}[k]}$ is the number of one-step walks from ${k}$ to ${j}$. Therefore the dot-product of row ${i}$ with column ${j}$ is the number of two-step walks from ${i}$ to ${j}$. By the dot-product definition of matrix-matrix multiplication, therefore, the product ${A A}$ encodes the number of two-step walks. >>> D = {1,2,3,4} >>> A = Mat((D,D), {(1,2):1, (1,4):1, (2,1):1, (2,3):2, (2,4):1, (3,2):2, (4,1):1, (4,2):1}) >>> print(AA) 1 2 3 4 --------- 1 \| 2 1 2 1 2 \| 1 6 0 1 3 \| 2 0 4 2 4 \| 1 1 2 2 Now we consider three-step walks. A three-step walk from ${i}$ to ${j}$ consists of a two-step walk from ${i}$ to some node ${k}$, followed by a one-step walk from ${k}$ to ${j}$. Thus the number of three-step walks from ${i}$ to ${j}$ equals number of two-step walks from ${i}$ to 1 ${\times}$ number of one-step walks from 1 to ${j}$ + number of two-step walks from ${i}$ to 2 ${\times}$ number of one-step walks from 2 to ${j}$ + number of two-step walks from ${i}$ to 3 ${\times}$ number of one-step walks from 3 to ${j}$ + number of two-step walks from ${i}$ to 4 ${\times}$ number of one-step walks from 4 to ${j}$ We already know that ${A A}$ gives the number of two-step walks. Again using the dot-product definition of matrix-matrix multiplication, the product ${(A A) A}$ gives the number of three-step walks: >>> print((AA)*A) 1 2 3 4 ----------- 1 \| 2 7 2 3 2 \| 7 2 12 7 3 \| 2 12 0 2 4 \| 3 7 2 2 Oops, there are not ten but twelve three-step walks from 3 to 2. I missed the walks ${3\ c\ 2\ a\ 1\ a\ 2}$ and ${3\ b\ 2\ a\ 1\ a\ 2}$. Anyway, we’re half the way towards solving Will’s problem. The problems about generating functions are not much harder; they make use of polynomials and determinants. We will not be able to cover generating functions here, but rest assured that they are not beyond your ability and are quite elegant. Why (apart from undying fame) would you want to compute the number of ${k}$-step walks between pairs of nodes in a graph? These numbers can serve as a crude way to measure how closely coupled a pair of nodes are in a graph modeling a social network, although there are assuredly better and faster ways to do so. We will see some in future posts.	2015-11-28 07:35:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 89, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7182084321975708, "perplexity": 361.19843639822165}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398451744.67/warc/CC-MAIN-20151124205411-00308-ip-10-71-132-137.ec2.internal.warc.gz"}
http://mestaritalo.com/orvovf1/right-obtuse-triangle-5b5093	# right obtuse triangle An obtuse triangle has one of the vertex angles as an obtuse angle (> 90, Area = \begin{align}\frac{1}{2} \times \text{Base} \times \text{Height}\end{align}. Since the triangle has an obtuse angle, it is an obtuse triangle. In Euclidean geometry, the base angles can not be obtuse (greater than 90°) or right (equal to 90°) because their measures would sum to at least 180°, the total of all angles in any Euclidean triangle. $$\therefore$$ The given triangle is an obtuse-angled triangle. Acute Triangle (all acute angles) Obtuse Triangle (1 obtuse angle) Right Triangle (1 right angle) 1) Type: 2) Type: 3) Type: 4) We know that triangles are 3-sided closed shapes made with 3 line segments. The perimeter of an obtuse triangle is the sum of the measures of all its sides. In an equiangular triangle, all the angles are equal—each one measures 60 degrees. It is greater than a right angle. Our third side must be greater than 7, since if it were smaller than that we would have where is the unknown side. A triangle can have at most one right angle and at most one obtuse angle. D - isosceles triangle. Circumcenter and orthocenter lie outside the triangle. SURVEY . answer choices ... Obtuse Triangle. All angles of the given triangle are acute. B - acute triangle. You can download the FREE grade-wise sample papers from below: To know more about the Maths Olympiad you can click here. An obtuse triangle has one of the vertex angles as an obtuse angle (> 90°). Plug in each set of lengths into the Pythagorean Theorem. Heron's formula to find the area of a triangle is: Note that $$(a + b + c)$$ is the perimeter of the triangle. One of the angles of the given triangle is a right angle. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$. The other two sides are called the legs. This makes the sum of the squares of the legs less than the longest side squared, so Triangle II is obtuse. Step-by-step explanation: You cannot have 2 right angles, or 2 obtuse angles. There are three special names given to triangles that tell how many sides (or angles) are equal. Yes, you were right. Tags: Question 2 . Equilateral: \"equal\"-lateral (lateral means side) so they have all equal sides 2. The Pythagoras theorem is a fundamental relation among the three sides of a right triangle. So let me draw a couple of examples of obtuse angles. For a triangle, the sum of the two shortest sides must be greater than that of the longest. Perfect! What type of triangle is this? The side opposite the obtuse angle in the triangle is the longest. Our Math Experts focus on the “Why” behind the “What.” Students can explore from a huge range of interactive worksheets, visuals, simulations, practice tests, and more to understand a concept in depth. Because , this is an obtuse triangle. Classify each triangle as acute, equiangular, obtuse, or right . If a 2 + b 2 = c 2, then it is right triangle. I Know It is an elementary math practice website. Answer: It is 2 acute and 1 right. Example 3 The exterior angle and the adjacent interior angle forms a linear pair (i.e , they add up to 180°). These triangles always have at least two acute triangles. 3. Also iSOSceles has two equal \"Sides\" joined by an \"Odd\" side. 8, 15, 20. I open this lesson, which is a review of 4th grade standard 4.G.1, by sharing student friendly definitions of the words obtuse, acute, and right angle and then giving the students motions to represent each of these words. A - obtuse triangle. The altitude or the height from the acute angles of an obtuse triangle lie outside the triangle. There can be 3, 2 or no equal sides/angles:How to remember? Triangle II is very close to the (8, 15, 17) right triangle, but we made one of the legs shorter. You may have noticed that the side opposite the right angle is always the triangle's longest side. For each triangle, shade in the acute angles yellow, the obtuse angles green and the right angles red. A triangle cannot be right-angled and obtuse angled at the same time. 1 acute, 1 right, and I obtuse 1 acute and 2 right 1 acute and 2 obtuse 1) 1 acute and 2 obtuse 1) 2 acute and 1 right. Report an issue . So the given triangle is a right triangle. The Pythagorean theorem is a very popular theorem that shows a special relationship between the sides of a right triangle. Triangles can be classified on the basis of their angles as acute, obtuse and right triangles. Whether an isosceles triangle is acute, right or obtuse depends only on the angle at its apex. It also finds the area and perimeter of program and determines whether the 6 coordinates form a triangle at all. Try out this fifth grade level math lesson for classifying triangles (right, acute, obtuse) practice with your class today! The more advanced worksheets include straight and … Triangles are polygons that always have 3 sides and 3 angles. 90°). Even if we assume this obtuse angle to be 91°, the other two angles of the triangle will add up to 89° degrees. Acute Triangle - has an angle less than 90 degrees, altitudes meet inside the triangle . $$a^2$$ = 9, $$b^2$$ = 16 and $$c^2$$ = 36. Determine if the following lengths make an acute, right or obtuse triangle. Example 2. Identifying parallel and perpendicular lines, Classifying scalene, isosceles, and equilateral triangles by side lengths or angles, Finding an angle measure of a triangle given two angles, Drawing and identifying a polygon in the coordinate plane. Identify this triangle as acute, obtuse, or a right triangle. Take a closer look at what these two types of triangles are, their properties, and formulas you'll use to work with them in math. A triangle cannot have more than one obtuse angle. Right Triangle. Identifying Right, Obtuse and Acute Angles. So an obtuse angle might look like-- let me make it a little bit clearer. The side BC is the longest side which is opposite to the obtuse angle $$\angle \text{A}$$. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90° The triangles above have one angle greater than 90°. Consider the triangle $$ABC$$ with sides $$a$$, $$b$$ and $$c$$. Thus, if one angle is obtuse or more than 90 degrees, then the other two angles are … Therefore, the given measures can form the sides of an obtuse triangle. The circumcenter and the orthocenter of an obtuse-angled triangle lie outside the triangle. This is a free geometry lesson for 4th grade about acute, obtuse, and right triangles (classification according to angles). Obtuse Triangle with our Math Experts in Cuemath’s LIVE, Personalised and Interactive Online Classes. Most worksheets require students to identify or analyze acute, obtuse, and right angles. An obtuse-angled triangle can be scalene or isosceles, but never equilateral. Centroid and incenter lie within the obtuse triangle while circumcenter and orthocenter lie outside the triangle. We are given two sides as 5 and 12. Circumcenter (O), the point which is equidistant from all the vertices of a triangle, lies outside in an obtuse triangle. This page has printable geometry PDFs on angle types. Tags: Question 13 . Answers (1) Tagan 27 June, 09:03. Find the area of an obtuse triangle whose base is 8 cm and height is 4 cm. = \begin{align}\frac{1}{2} \times \text{base} \times \text{height}\end{align}. A triangle whose any one of the angles is an obtuse angle or more than 90 degrees, then it is called obtuse-angled triangle or obtuse triangle. Find the height of the given obtuse triangle whose area = 60 cm2 and base = 8 cm. A triangle that has an angle greater than 90° See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene We know that the angles of any triangle add up to 180°. 62/87,21 One angle of the triangle measures 115, so it is a obtuse angle. It is also known as a 45-90-45 triangle. A triangle with one obtuse angle (greater than 90°) is called an obtuse triangle. Learn about the Pythagorean theorem. A scalene triangle may be right, obtuse, or acute (see below). Alphabetically they go 3, 2, none: 1. Since the total degrees in any triangle is 180°, an obtuse triangle can only have one angle that measures more than 90°. The third angle decides the type of triangle. \end{align}\). It encourages children to develop their math solving skills from a competition perspective. 0. These worksheets require students to look at an angle and identify whether it is right, acute or obtuse. An obtuse-angled triangle has one of its vertex angles as obtuse and other angles as acute angles. 300 seconds . $$\therefore$$ Area of the triangle = 16 cm. A right triangle is a triangle with a right angle(i.e. The sum of the interior angles of the obtuse triangle is equal to 180 degrees only. We can observe that one of the angles measures greater than 90°, thus making it an obtuse angle. TRIANGLE CLASSIFICATION 1 ACUTE, OBTUSE OR RIGHT? Which set of angles can form a triangle? What is the size of the third angle? Definitions. The third angle decides the type of triangle. There are also special right triangles. It is called the hypotenuse of the triangle. We extend the base as shown and determine the height of the obtuse triangle. Therefore, an obtuse-angled triangle can never have a right angle; and vice versa. Right … It must also be … $$\therefore$$ Option b forms an obtuse triangle. Example 1: A right triangle has one other angle that is 35°. 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Real life: obtuse angled triangle Definition Maths Olympiad ) is called obtuse! Triangle: in this triangle '' or \ '' uneven\ '' or \ '' equal\ '' -lateral lateral! Angles green and the right angles: in geometry, no Euclidean can... But we made the longest type of triangle where one of its angles as and... Following angle measures 90°, and we have two legs, right look something like.... Or a right triangle click here can sides measuring 3 cm, 4 cm 3 line segments grade acute. Degrees in any triangle add up to 89° degrees write down whether it is obtuse that a triangle one...	2021-09-20 00:20:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6279005408287048, "perplexity": 615.7272556336136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00521.warc.gz"}
https://bitbucket.org/rivanvx/beamer/diff/doc/beamerug-workflow.tex?diff2=9363ef1d9575&at=default	Diff from to # doc/beamerug-workflow.tex If you wish your talk to reside in the same file as some different, non-presentation article version of your text, it is advisable to setup a more elaborate file scheme. See Section~\ref{section-article-version-workflow} for details. -\lyxnote -You can either open a new file and then select \|beamer\| as the document class or you say New from template'' and then use a template from the directory \|beamer/solutions\|. - - \subsection{Step Two: Structure Your Presentation} The next step is to fill the presentation file with \|\section\| and \|\subsection\| to create a preliminary outline. You'll find some hints on how to create a good outline in Section~\ref{section-structure-guidelines}. \beamernote Once a first version of the structure is finished, you should try to create a first PDF or PostScript file of your (still empty) talk to ensure that everything is working properly. This file will only contain the title page and the table of contents. -\lyxnote -Use View'' to check whether the presentation compiles fine. Note that you must put the table of contents inside a frame, but that the title page is created automatically. - \subsubsection{Creating PDF} \beamernote > acroread main.pdf \end{verbatim} -\lyxnote -Choose View pdf'' to view your presentation. \subsubsection{Creating PostScript} \label{section-postscript} > ps2pdf main.ps main.pdf \end{verbatim} -\lyxnote -Use View Postscript'' to view the PostScript version. - \subsubsection{Ways of Improving Compilation Speed} \label{section-speedup} Once the table of contents looks satisfactory, start creating frames for your presentation by adding \|frame\| environments. You'll find guidelines on what to put on a frame in Section~\ref{section-frame-guidelines}. -\lyxnote -To create a frame, use the style BeginFrame''. The frame title is given on the line of this style. The frame ends automatically with the start of the next frame, with a section or subsection command, and with an empty line in the style EndFrame''. Note that the last frame of your presentation must be ended using EndFrame'' and that the last frame before the appendix must be ended this way. - \subsection{Step Five: Test Your Presentation} You may also wish to create an article version of your talk. An article version'' of your presentation is a normal \TeX\ text typeset using, for example, the document class \|article\| or perhaps \|llncs\| or a similar document class. The \beamer\ class offers facilities to have this version coexist with your presentation version in one file and to share code. Also, you can include slides of your presentation as figures in your article version. Details on how to setup the article version can be found in Section~\ref{section-article}. -\lyxnote -Creating an article version is not really possible in \LyX. You can \emph{try}, but we would not advise it. - \subsubsection{Printing the Handout} \label{section-printing-version} Tip: Filter by directory path e.g. /media app.js to search for public/media/app.js. Tip: Use camelCasing e.g. ProjME to search for ProjectModifiedEvent.java. Tip: Filter by extension type e.g. /repo .js to search for all .js files in the /repo directory. Tip: Separate your search with spaces e.g. /ssh pom.xml to search for src/ssh/pom.xml. Tip: Use ↑ and ↓ arrow keys to navigate and return to view the file. Tip: You can also navigate files with Ctrl+j (next) and Ctrl+k (previous) and view the file with Ctrl+o. Tip: You can also navigate files with Alt+j (next) and Alt+k (previous) and view the file with Alt+o.	2014-07-12 05:58:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3293348550796509, "perplexity": 3464.690143281889}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776431372.37/warc/CC-MAIN-20140707234031-00095-ip-10-180-212-248.ec2.internal.warc.gz"}
http://www.jiskha.com/math/?page=189	Thursday January 19, 2017 # Homework Help: Math ## Recent Math Questions Pure Math How to plot a steam and leaf for the following data 0 0 11 140 152 253 257 309 392 419 541 644 751 770 810 897 1295 1340 1419 1482 1515 1719 1763 1891 2005 2127 2631 2988 3686 6726 11465 12917 14701 regards Saturday, November 21, 2015 by Steam and Leaf Plot Math How do you divide 1248 by 8? Saturday, November 21, 2015 by Lilly Math Need help please. The reciprocal of .125 is what number? Saturday, November 21, 2015 by Lisa math Saturday, November 21, 2015 by to Ms.Sue math The length, breadth and height of a closed box are respectively 2m, 1.5m and 0.8m. Find the total area of canvas required to cover it-up. Find also the cost of canvas if 1sq m of canvas costs Rs. 30. A rectangular box without lid is to be made by thin sheet of metal. Find the ... Saturday, November 21, 2015 by rohan Math URGENT! In a school, 8% of students are left handed. If 50 classes of 20 students are randomly selected, what is the probability of 10 classes have no left handed students? I know that this is a binomial distribution question, but I can't seem to do this question even ... Saturday, November 21, 2015 by Katie Math word problem 17 1/9 decreased by what number is 5 2/3? This is the answer I have 11 4/9. Is this correct. Saturday, November 21, 2015 by Ray Math What math word does these letters spell O,O,E,R,N,F,O,P,S,E,P,T,W,X Saturday, November 21, 2015 by . Math Forty-five percent of a large class of students in Statistics are females. Suppose 28 students are selected at random from the class. Use a normal approximation with continuity correction to calculate the probability that the number of males in the sample is at least 12 but ... Saturday, November 21, 2015 by Aarathi Math Calculate the reduced paid-up insurance for Lee Chin, age 42, who purchased a $100,000 straight life policy. At the end of year 20, Lee stopped paying premiums. (Use Table 20.2). (Omit the "$" sign in your response.) What is the reduced paid-up insurance? What am I ... Saturday, November 21, 2015 by Uteryu Math A man who is 6.2 ft tall casts a shadow that is 13.6 ft long. Find the height of a petrified stump that casts a 16.9 ft shadow. Saturday, November 21, 2015 by Carlos math Given that sin x + sin y = a and cos x + cos y =a, where a not equal to 0, express sin x + cos x in terms of a. attemp: sin x = a - sin y cos x = a - cos y sin x + cos x = 2A - (sin y + cos y) Saturday, November 21, 2015 by Chia LC Math Log3 (4x+1) log3(13x-5)=2 Saturday, November 21, 2015 by Anonymous math An investor invested 80 percent of the money at the rate of 10 percent p.a for 2 years. Find the amount of both the investments if the diffrence between the simple interests earned is Rs 3300 Saturday, November 21, 2015 by pallav Math 1 question plz help ASAP A bah of oranges weighs 5 1/4. How much does 1 1/2 weigh?(that's the question my question is do we add, subtract, multiply or divide plzzzzzzzzz help meeeeee im new to this website =) Saturday, November 21, 2015 by Debra Grimes math How many of the 2-digit numbers, that can be formed by using the digits 1,2,3,.......,9 without repeating any digit, are divisible by4? If 2 boys and 2 girls are to be arranged in a row so that the girls are not next to each other, how many possible arrangements are there? A ... Saturday, November 21, 2015 by adam math The average marks of 14 students was 71. It was later found that the marks of one of the students has been wrongly entered as 42 instead of 56 and of another as 74 instead of 32. What is the correct average? Average age of seven persons in a group is 30 years, the average age ... Saturday, November 21, 2015 by adam math Find two positive real numbers x and y such that they add to 120 and x^(2)y is as large as possible Saturday, November 21, 2015 by Raj math Two truck drivers leave the same rest stop at the same time to travel in opposite directions.One travel at an average rate of 52 mph and the other average 56 mph.In how many hours will they be 72 miles apart Friday, November 20, 2015 by nisha Math A principal earns 7% per year simple interest. How long will it take for the future value to quadruple? Friday, November 20, 2015 by Merry Math Simplify and express in scientific notation (2.75x10^9)(0.5x10^5) (2.75x0.5)(10^9Ã—10^5) =1.375Ã—10^14 Friday, November 20, 2015 by Cody Math Given that an equation of the form f(x) = x^n has n solutions, and that the solutions are equally spaced at a given radius in the complex plane, discuss why complex solutions must occur in conjugate pairs when . Friday, November 20, 2015 by April Math A football is punted. Its height, H metres is given by the quadratic relation with equation: H = -5t2 + 21t, where t is the time in seconds after the punt. I tried to do this question on a test but I just ran out of time How is the easiest and quickest way to do this? ... Friday, November 20, 2015 by Fred Math what are the inputs of the following funtions {(-3,2),(-4,2),(8,3),(7,1)} Friday, November 20, 2015 by Taylor math 18 mil at .25 starting the meter at $2.50 is how much Friday, November 20, 2015 by monica vargas Math The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about 10. Suppose that 16 individuals are randomly chosen. Is it appropriate to use the normal approximation for the sampling ... Friday, November 20, 2015 by Brandy math packet Solve this equation for x. 0.5(5-7x) = 8-(4x+6) Friday, November 20, 2015 by charlie math A supervisor finds the mean number of miles that the employees in a department live from work. He finds mc022-1.jpg and mc022-2.jpg. Which statement must be true? Friday, November 20, 2015 by kiki Math The question is: The area of a rectangle is given by the expression 8x+44y. Length is 2x+11y. What is width? I answered 4x & 4y, yet that was wrong. Can you help explain why? Thanks Friday, November 20, 2015 by Kyle MATH ANSWER HELP 1. There are four kinds of road signs in a certain town. The number of each kind is shown. What is the ratio of the number of school signs to the number of stop signs? signs (1 point) 9 to 10 10:9 ten over twenty-four 9:24 2. Which ratio is equivalent to four over sixteen? (1 ... Friday, November 20, 2015 by PINKIE PIE Math 1350 Mrs. Jones has a prime number of children in her class. When she divides the children into teams of 7, one team has 6. When she divides the children into teams of 6, one team has 5. How many children are in the class? Friday, November 20, 2015 by Ne Ne Math 1350 About eleven-twelveths of a state park is covered by trees, one-eighteenth is covered by a lake, and the rest is covered by open fields. What part of the state park is open fields? Friday, November 20, 2015 by Ne Ne Math 1350 Mrs. Smith's classroom can hold at the most 30 students. Her first period class had 4 boys for every 3 girls. When 3 boys and 1 girl were added to the class, the new ratio became 3 boys to 2 girls. How many total students are now in the class? Show all work and explain the... Friday, November 20, 2015 by Ne Ne math Larry borrowed$800.00 at 6% for one in a half years?Explain and solve the problem. Friday, November 20, 2015 by Eilundra math The cost price of 8 books is equal to the selling price of 6 books. Find the gain or loss per cent. Ashis purchases eggs at the rate of Rs 90 for 150 eggs and sells them at Rs 85 per hundred. Find his gain or loss per cent. Pls pls help* Friday, November 20, 2015 by adam math A sum is invested at compound interest payable annually. The interest in two successive years was Rs 900 and Rs 981. Find the sum. If the sum of all internal angles and sum of all external angles of a regular polygon are equal, then the number of sides of the polygon is? Pls ... Friday, November 20, 2015 by adam math A low land, 48m long and 31.5 m broad is raised to 6.5 dm. For this, earth is removed from a cuboidal hole, 27 m long and 18.2 m broad, dug by the side of the land. The depth of the hole will be? A man rows 12 km in 5 hrs against the stream, the speed of current being 4 kmph. ... Friday, November 20, 2015 by adam Please help I'm failing Which of the following is a rational number Pie Square root 3 Square root 2 1.3* Friday, November 20, 2015 by Mysterydude math for every 12 people crossing the street, 7 of them used the pedestrian lane and the rest did not. if 48 people crossed the street, how many did not cross in the pedestrian lane? Hi Steve, the choices of answers to this question are the following: a)28; b)24; c)36; d)41. your ... Friday, November 20, 2015 by tante math Hi Dan, kindly provide detail equation how to arrived to your answer. 21 boys, 21 girls. 7:5 - 6 girsl is 21 boys to 15 girls [think -6] Friday, November 20, 2015 by tante math Joseph used the problem-solving strategy Work Backward to solve the inequality 2n+5<13. Shawnee solved the inequality using the algebraic method. Compare the two methods. THANK YOU! Friday, November 20, 2015 by kimie math Is there any value of x with the property that x < x - 1 ? Explain your reasoning. THANK YOU! Friday, November 20, 2015 by kimie Math Rachel will make juice drink by mixing sugar, pineapple juice, and water in the ratio of 2:5:7 cups. If she uses 20 cups of pineapple juice, she has to mix ____ cups of sugar and ____ cups of water to the juice Friday, November 20, 2015 by Ceena Math Over the past week, 34 baby boys were born at the hospital. This was 54% of all babies born. How many girls were born over the past week? Friday, November 20, 2015 by Ken math Suppose payments were made at the end of each month into an ordinary annuity earning interest at the rate of 3.5%/year compounded monthly. If the future value of the annuity after 12 years is $50,000, what was the size of each payment? (Round your answer to the nearest cent.) Friday, November 20, 2015 by phalyn math 5th grade Victoria has 2 pounds of flour she uses 2/5 of flour for pizza and 3/10 of the flour for bread. Find the weight of the flour she has left. Express in decimal form. I worked on this and the answer I came up with was 19.2 lbs left? Thursday, November 19, 2015 by mary math a bakery sells bagels for$0.85 each an muffins for $1.10 each the bakery hopes to earn$400 each day from these sales. Write an equation that represents the total amount the bakery would like to earn selling bagels and muffins each day. (sorry I have no idea how to do this... Thursday, November 19, 2015 by Hannah math for every 12 people crossing the street, 7 of them used the pedestrian lane and the rest did not. if 48 people crossed the street, how many did not cross in the pedestrian lane? Thursday, November 19, 2015 by tante math the ratio of boys to girls in the gym is 7:5. when 6 more girls enter the gym, the ratio of boys to girls becomes even. how many boys and girls now in the gym? Thursday, November 19, 2015 by tante Math Samantha wants to find the height of a pine tree in her yard. She measures the height of the mailbox at 3 feet and its shadow at 4.8 feet. Then she measures the shadow of the tree at 56 feet. How tall is the tree? Thursday, November 19, 2015 by Aaron Math The quantity of p + 4 divided by 2 Thursday, November 19, 2015 by Israel Math: Geometry The bases of trapezoid $ABCD$ are $\overline{AB}$ and $\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. [asy] unitsize(1 cm); pair A, B, C, D; A = (1,2); B = (3,2); C = (4,0); D = (0,0); draw(A--B--C--D--cycle); draw(B--D... Thursday, November 19, 2015 by I see you have graph paper. You must be plotting something. math Compute the hurdle rate above based on the following.1. Note payable bank $2,000,000.00 with rate of 7.5%, The company has a tax rate of 30%.2. Bank Mortgage Loan$1,000,000.00 with rate of 6% The company has a tax rate of 30%.3. Preferred stock $750,000.00 with coupon rate 10... Thursday, November 19, 2015 by Terry math 5 lbs of walnuts equally divided into containers. each holds 5/8 of a pound . how many containers will be filled Thursday, November 19, 2015 by klb math How do you subtract radicals? Example wise. Thursday, November 19, 2015 by khloe johnsonn math When xÂ¡Ãº0 cosec(x) - cot (x)/ x =? Thursday, November 19, 2015 by aamir Math The ratio of the lengths of the sides of a triangle is 2:4:5. The largest side is 33" longer than the small side. What is the perimeter? I don't know how to start it help please Thursday, November 19, 2015 by Anonymous Business Math Myerson borrowed$8,500 at 9% ordinary interest for 200 days. After 120 days, he made a partial payment of $4,000. What is the final amount due on the loan? Thursday, November 19, 2015 by Ahmad Conrady Math Bev has 25 books in her bookshelf that are either mystery or romance. If r represents the number of romance nooks she has, write an expression in terms of r that represent the number of mystery books she has. Thursday, November 19, 2015 by Gina Causley Math I need help understanding how to identify the transformations when the quadratic is in factored form. f(x) = (x+3)(x-1) Thanks... Thursday, November 19, 2015 by Brandi math fractions Sherry spent 1/5 of her money on school supplies and 3/8 of the remainder on a CD player. She had$25 left. How much money did she have at first? do not use algebra Thursday, November 19, 2015 by daniel Math A board is 12.5 feet long. I cut 2 pieces that are 1.875 feet long and make 6 ramps with it. How much feet of board is left over? Thursday, November 19, 2015 by Corrine MATH trying to work equation problems and my answer is different from answer in textbook. 12/x = -3 and 24=120/x 1. 12/x(12)= -3(12), x= -36 textbook has -4 2. 24(120)= 120/x(120), x = 2880 textbook has 5 What am I doing wrong??? Thursday, November 19, 2015 by Amy Math For a game, a school bought 100 pizzas. Some were pepperoni and some were cheese. There are 4 times as many pepperoni pizzas as cheese pizzas. How many cheese pizzas did the school buy? Thursday, November 19, 2015 by Sammy Taylor Math George and Cindy are saving for bicycles. Cindy saved $10 less than twice as much as George saved. Together, they have saved$140. How much did each of them save? Thursday, November 19, 2015 by Sammy Taylor Math Jack and jill both bought new coats. Jacks coat cost $8 less than Jills. If the total costs of of their coats was$112, how much did each coat cost? Thursday, November 19, 2015 by Sammy Taylor Math In a table of random digits, each digit is to occur with a probability of 0.1. a) A student examines a list of 200 random digits and counts only eleven 4â€™s and thus claims that the table is not really random. Explain the error in the studentâ€™... Thursday, November 19, 2015 by Anonymous math A baseball player's batting average is calculated by dividing the number of times a player got a hit by the total number of times the player was at bat. It is expressed as a decimal rounded to three decimal places. After the first ten games of the season,Sam had 12 hits ... Thursday, November 19, 2015 by Irene Wu math Nata spent $28.00 on 2 DVDs. At this rate now much would she spent on 5 DVDs cost? Thursday, November 19, 2015 by Anonymous MATH 5a2b â€“ 13ab + 7a3 â€“ 4b how many degrees, and classify the term Thursday, November 19, 2015 by Help URGENT I NEED IT NOW! MATH 5a2b â€“ 13ab + 7a3 â€“ 4b, in decending order Thursday, November 19, 2015 by Help URGENT I NEED IT NOW! math rounding number to the underlined digit 8,464 the 8 is underlined my answer is 8,000 is that correct? Thursday, November 19, 2015 by mike Algebra 2 (Math) Solve by the system of equations by substitution. x + 2y + z = 14 y = z + 1 x = -3z + 6 Anyone able to help me with this? I've been stuck on it all day and can't seem to solve it, it doesn't make any sense to me. Thursday, November 19, 2015 by Lucas Math 1. What compromise did states reach for the purpose of determining representation by population in the House of Representatives? (1 point) Slaves would be counted as three-fifths of a person for official population counts. Slave populations would count for representation but ... Thursday, November 19, 2015 by Mariana Math Evan type 72 pages of notes one day. He typed one half of the pages in the morning and one third of the pages in the afternoon. He type the rest of the pages in the evening. How many pages of notes that he typed in the morning and afternoon Thursday, November 19, 2015 by Adam math Help me with this In a class of 30 students, 18 offer mathematics (M), 15 offer science (S) and 13 offer english (â‚¬). The number of students who offer all the three subjects is equal to the number of students who do not offer any of these subject. 10 students ... Thursday, November 19, 2015 by bwalya Math In the accompanying diagram, the perimeter of isosceles triangle MNO is equal to the perimeter of square ABCD. If the sides of the triangle are represented by 4x+4 , 5x-3, 17 and one side of the square is represented by 3x, find the length of the square Thursday, November 19, 2015 by Randy James Math fractions A picture frame is 6 1/4 inches wide by 8 inches tall. If a 5 inch by 7 inch picture is to be centered in the frame, what should the right margin be? Thursday, November 19, 2015 by Devon math How many fewer minutes would it take the driver to travel a distance of 10 miles at a speed of 65 miles per hour than at a speed of 50 miles per hour? Round your answer to the nearest minute. Thursday, November 19, 2015 by mathew Stats Fill in the table below regarding populations and samples. The first column represents the question the statistician is trying to answer. Fill in the blank(s) for each row. 1.Is my well water safe? 2.Are as many girls born as boys in the world? 3.Which works better, Advil or ... Thursday, November 19, 2015 by Logan math another question I have, is what if you deposit$125 in American currency saving account, if the exchange rate is the same how much American money would I get if I exchange $400 in Canadian currency Thursday, November 19, 2015 by minnie Math A dilation has center (0, 0 ,0). Find the image of the point (-1, -2, 0) for the scale factor of 3. Thursday, November 19, 2015 by Bela English The High School Step Team Karlaâ€™s cousin Jade urged her to join the step team. â€œThis afternoon you should definitely try out!â€ Jade suggested after the final bell rang. â€œI know she wants the best for me,&... Thursday, November 19, 2015 by Kaai97 math out of 9/15 and 3/10 and 6/10 which are equivalent? and could toy tell me why, I am having difficulties with this! thanks Thursday, November 19, 2015 by minnie math The value of 4x^3-x/(2x+1)(6x-3) when x=9999 If 7 men working 7 hrs a day for each of 7 days produce 7 units of work, then the units of work produced by 5 men working 5 hrs a day for each of 5 days is? Pls help* Thursday, November 19, 2015 by adam math can someone tell me why 9/15 is equivalent to 3/10 I cant seem to figure it out,?? Thursday, November 19, 2015 by noah math 1. The amount Rs 2100 became Rs 2352 in 2 years at simple interest. If the interest rate is decreased by 1%, what is the new interest? 2. My grandfather was 9 times older than me 16 years ago. He will be 3 times of my age 8 years from now. Eight years ago, the ratio of my age ... Thursday, November 19, 2015 by adam math paul salary is 2,172 monthly he spend 1/4 in rent 1/8 in car 5/8 in others how much spend monthly in car and rent? Thursday, November 19, 2015 by Anonymous Math Help!! Order the numbers from least to greatest. 1/3, 1 and 1/4, 0.27, 1.35 I am sorry I really need help on this one. Thursday, November 19, 2015 by Mani math which number is a perfect cube? 1. 5 2. 100 3. 125*** 4. 150 Thursday, November 19, 2015 by cheyenne Math A dilation has center (0, 0 ,0). Find the image of the point (-1, -2, 0) for the scale factor of 3. I'm not sure how to do this. Thursday, November 19, 2015 by Belle Math The quotient of two proper fractions is less than 1 Thursday, November 19, 2015 by Joey math The game of checkers is played on a square board. If the area is 256 square feet, what is the perimeter of the board? Thursday, November 19, 2015 by Anonymous Math Help!! Order the numbers from least to greatest. 3/8, 19/24, 2/3 Thanks.. :) In need of help to understand how to order. Thursday, November 19, 2015 by Mani math A vendor bought 10 dozen eggs at the rate of Rs 22 per dozen. Six of the eggs were rotten. He sold the remaining eggs at the rate of Rs 2.50 per egg. Find his profit or loss. Ans-65 profit Pls help me Thursday, November 19, 2015 by adam math Michelle has half of her investments in stock paying a 6% dividend and the other half in a stock paying 10% interest. If her total annual interest is$680, how much does she have invested? Thursday, November 19, 2015 by ivan math If matrix A has x rows and x + 5 columns, matrix B has y rows and 11 â€“ y columns and both AB and BA are defined for product then x and y are: Thursday, November 19, 2015 by Ahmad Faraz math Mr. X invested a part of his investment in 10% bond A and a part in 15% bond B. His interest income during the first year is Rs.4,000. If he invests 20% more in10% bond A and 10 % more in 15% bond B, his income during the second year increases by Rs.500. His initial investment... Thursday, November 19, 2015 by moon math review exercise 10 question 2 Thursday, November 19, 2015 by khadija	2017-01-20 00:30:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37423306703567505, "perplexity": 1084.028897506158}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280761.39/warc/CC-MAIN-20170116095120-00528-ip-10-171-10-70.ec2.internal.warc.gz"}
https://stats.stackexchange.com/questions/287425/why-do-you-need-to-scale-data-in-knn	# Why do you need to scale data in KNN Could someone please explain to me why you need to normalize data when using K nearest neighbors. I've tried to look this up, but I still can't seem to understand it. https://discuss.analyticsvidhya.com/t/why-it-is-necessary-to-normalize-in-knn/2715 But in this explanation, I don't understand why a larger range in one of the features affects the predictions. • I think normalization has to be justified from the subject-matter point of view. Essentially, what matters is what defines the distance between points. You have to find a convenient arithmetic definition of distance that reflects the subject-matter definition of distance. In my limited experience, I have normalize in some but not all directions based on subject-matter considerations. – Richard Hardy Jun 26 '17 at 19:00 • For an instructive example, please see stats.stackexchange.com/questions/140711. – whuber Jun 26 '17 at 19:28 • It seems like any scaling (min-max or robust) is acceptable, not just standard scaling. Is that correct? – skeller88 Apr 10 at 20:20 The k-nearest neighbor algorithm relies on majority voting based on class membership of 'k' nearest samples for a given test point. The nearness of samples is typically based on Euclidean distance. Consider a simple two class classification problem, where a Class 1 sample is chosen (black) along with it's 10-nearest neighbors (filled green). In the first figure, data is not normalized, whereas in the second one it is. Notice, how without normalization, all the nearest neighbors are aligned in the direction of the axis with the smaller range, i.e. $x_1$ leading to incorrect classification. Normalization solves this problem! • This answer is exactly right, but I fear the illustrations might be deceptive because of the distortions involved. The point might be better made by drawing them both so that the two axes in each are at the same scale. – whuber Jun 26 '17 at 19:30 • I found it difficult to fit all data points in the same scale for both figures. Hence, I mentioned in a note that scales of axes are different. – kedarps Jun 26 '17 at 19:55 • That difficulty actually is the point of your response! One way to overcome it is not to use such an extreme range of scales. A 5:1 difference in scales, rather than a 1000:1 difference, would still make your point nicely. Another way is to draw the picture faithfully: the top scatterplot will seem to be a vertical line of points. – whuber Jun 26 '17 at 19:57 • @whuber, I misunderstood your first comment. Fixed the plots, hopefully it's better now! – kedarps Jun 26 '17 at 20:10 • @Undertherainbow That is correct! – kedarps Mar 11 '19 at 19:33 Suppose you had a dataset (m "examples" by n "features") and all but one feature dimension had values strictly between 0 and 1, while a single feature dimension had values that range from -1000000 to 1000000. When taking the euclidean distance between pairs of "examples", the values of the feature dimensions that range between 0 and 1 may become uninformative and the algorithm would essentially rely on the single dimension whose values are substantially larger. Just work out some example euclidean distance calculations and you can understand how the scale affects the nearest neighbor computation. If the scale of features is very different then normalization is required. This is because the distance calculation done in KNN uses feature values. When the one feature values are large than other, that feature will dominate the distance hence the outcome of the KNN. see example on gist.github.com	2020-08-14 17:16:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6773914098739624, "perplexity": 719.3417963010056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739347.81/warc/CC-MAIN-20200814160701-20200814190701-00532.warc.gz"}
https://ko.overleaf.com/latex/templates/uwe-dissertation-report/xjdjnndsrwks	# UWE Dissertation Report Author Deirdre Toher View Count 8342 AbstractAn example template of how to create a dissertation style for UWE Bristol (BSc Mathematics/Mathematics and Statistics programme) with margins compatible with MS Word templates (so correct for page limit rules). This also allows you to separate out References and Bibliography entries.	2019-12-06 12:41:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3637680113315582, "perplexity": 14925.507009037488}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540488620.24/warc/CC-MAIN-20191206122529-20191206150529-00082.warc.gz"}
http://docs.sunpy.org/en/stable/guide/acquiring_data/hek.html	# Using SunPy’s HEK module¶ The Heliophysics Event Knowledgebase (HEK) is a repository of feature and event information about the Sun. Entries are generated both by automated algorithms and human observers. SunPy accesses this information through the hek module, which was developed through support from the European Space Agency Summer of Code in Space (ESA-SOCIS) 2011. ## 1. Setting up the client¶ SunPy’s HEK module is in sunpy.net. It can be imported into your session as follows: >>> from sunpy.net import hek >>> client = hek.HEKClient() This creates a client that we will use to interact with the HEK. ## 2. A simple query¶ To search the HEK, you need a start time, an end time, and an event type. Times are specified as strings or Python datetime objects. Event types are specified as upper case, two letter strings, and are identical to the two letter abbreviations found at the HEK website, http://www.lmsal.com/hek/VOEvent_Spec.html. >>> tstart = '2011/08/09 07:23:56' >>> tend = '2011/08/09 12:40:29' >>> event_type = 'FL' >>> result = client.search(hek.attrs.Time(tstart,tend),hek.attrs.EventType(event_type)) The first line defines the search start and end times. The second line specifies the event type, in this ‘FL’ or flare. Line 4 goes out to the web, contacts the HEK, and queries it for the information you have requested. Event data for ALL flares available in the HEK within the time range 2011/08/09 07:23: 56 UT - 2011/08/09 12:40:20 UT will be returned, regardless of which feature recognition method used to detect the flare. Let’s break down the arguments of client.search. The first argument: hek.attrs.Time(tstart,tend) sets the start and end times for the query. The second argument: hek.attrs.EventType(event_type) sets the type of event to look for. Since we have defined event_type = ‘FL’, this sets the query to look for flares. We could have also set the flare event type using the syntax: hek.attrs.FL There is more on the attributes of hek.attrs in section 4 of this guide. ## 3. The result¶ So, how many flare detections did the query turn up? >>> len(result) 19 The object returned by the above query is a list of Python dictionary objects. Each dictionary consists of key-value pairs that exactly correspond to the parameters listed at http://www.lmsal.com/hek/VOEvent_Spec.html. You can inspect all the dictionary keys very simply: >>> result[0].keys() [u'skel_startc1', u'concept', u'frm_versionnumber', u'hrc_coord', u'refs_orig',.... and so on. Remember, the HEK query we made returns all the flares in the time-range stored in the HEK, regardless of the feature recognition method. The HEK parameter which stores the the feature recognition method is called “frm_name”. Using list comprehensions (which are very cool), it is easy to get a list of the feature recognition methods used to find each of the flares in the result object, for example: >>> [elem["frm_name"] for elem in result] [u'asainz', u'asainz', u'asainz', u'asainz', u'asainz', u'asainz', u'asainz', u'SSW Latest Events', u'SEC standard', u'Flare Detective - Trigger Module', u'Flare Detective - Trigger Module', u'SSW Latest Events', u'SEC standard', u'Flare Detective - Trigger Module', u'Flare Detective - Trigger Module', u'Flare Detective - Trigger Module', u'Flare Detective - Trigger Module', u'Flare Detective - Trigger Module'] It is likely each flare on the Sun was actually detected multiple times by many different methods. ## 4. More complex queries¶ The HEK client allows you to make more complex queries. There are two key features you need to know in order to make use of the full power of the HEK client. Firstly, the attribute module - hek.attrs - describes ALL the parameters stored by the HEK as listed in http://www.lmsal.com/hek/VOEvent_Spec.html, and the HEK client makes these parameters searchable. To explain this, let’s have a closer look at hek.attrs. The help command is your friend here; scroll down to section DATA you will see: >>> help(hek.attrs) AR = <sunpy.net.hek.attrs.AR object> Area = <sunpy.net.hek.attrs.Area object> Bound = <sunpy.net.hek.attrs.Bound object> BoundBox = <sunpy.net.hek.attrs.BoundBox object> CC = <sunpy.net.hek.attrs.CC object> CD = <sunpy.net.hek.attrs.CD object> CE = <sunpy.net.hek.attrs.CE object> CH = <sunpy.net.hek.attrs.EventType object> CJ = <sunpy.net.hek.attrs.EventType object> CR = <sunpy.net.hek.attrs.EventType object> CW = <sunpy.net.hek.attrs.EventType object> EF = <sunpy.net.hek.attrs.EF object> ER = <sunpy.net.hek.attrs.EventType object> Event = <sunpy.net.hek.attrs.Event object> FA = <sunpy.net.hek.attrs.EventType object> FE = <sunpy.net.hek.attrs.EventType object> FI = <sunpy.net.hek.attrs.FI object> FL = <sunpy.net.hek.attrs.FL object> FRM = <sunpy.net.hek.attrs.FRM object> etc etc... The object hek.attrs knows the attributes of the HEK. You’ll see that one of the attributes is a flare object: FL = <sunpy.net.hek.attrs.FL object> We can replace hek.attrs.EventType(‘FL’) with hek.attrs.FL - they do the same thing, setting the query to look for flare events. Both methods of setting the event type are provided as a convenience Let’s look further at the FRM attribute: >>> help(hek.attrs.FRM) Help on FRM in module sunpy.net.hek.attrs object: class FRM(__builtin__.object) \| Data descriptors defined here: \| \| __dict__ \| dictionary for instance variables (if defined) \| \| __weakref__ \| list of weak references to the object (if defined) \| \| ---------------------------------------------------------------------- \| Data and other attributes defined here: \| \| Contact = <sunpy.net.hek.attrs._StringParamAttrWrapper object> \| \| HumanFlag = <sunpy.net.hek.attrs._StringParamAttrWrapper object> \| \| Identifier = <sunpy.net.hek.attrs._StringParamAttrWrapper object> \| \| Institute = <sunpy.net.hek.attrs._StringParamAttrWrapper object> \| \| Name = <sunpy.net.hek.attrs._StringParamAttrWrapper object> \| \| ParamSet = <sunpy.net.hek.attrs._StringParamAttrWrapper object> \| \| SpecificID = <sunpy.net.hek.attrs._StringParamAttrWrapper object> \| \| URL = <sunpy.net.hek.attrs._StringParamAttrWrapper object> \| \| VersionNumber = <sunpy.net.hek.attrs._StringParamAttrWrapper object> Let’s say I am only interested in those flares identified by the SSW Latest Events tool. I can retrieve those entries only from the HEK with the following command: >>> result = client.search( hek.attrs.Time(tstart,tend), hek.attrs.EventType(event_type), hek.attrs.FRM.Name == 'SSW Latest Events') >>> len(result) 2 We can also retrieve all the entries in the time range which were not made by SSW Latest Events with the following command: >>> result = client.search( hek.attrs.Time(tstart,tend), hek.attrs.EventType(event_type),hek.attrs.FRM.Name != 'SSW Latest Events') >>> len(result) 17 We are using Python’s comparison operators to filter the returns from the HEK client. Other comparisons are possible. For example, let’s say I want all the flares that have a peak flux of over 4000.0: >>> result = client.search(hek.attrs.Time(tstart,tend), hek.attrs.EventType(event_type), hek.attrs.FL.PeakFlux > 4000.0) >>> len(result) 1 Multiple comparisons can be included. For example, let’s say I want all the flares with a peak flux above 1000 AND west of 800 arcseconds from disk center of the Sun >>> result = client.search(hek.attrs.Time(tstart,tend), hek.attrs.EventType(event_type), hek.attrs.Event.Coord1 > 800, hek.attrs.FL.PeakFlux > 1000.0) Multiple comparison operators can be used to filter the results back from the HEK. The second important feature about the HEK client is that the comparisons we’ve made above can be combined using Python’s logical operators. This makes complex queries easy to create. However, some caution is advisable. Let’s say I want all the flares west of 50 arcseconds OR have a peak flux over 1000.0: >>> result = client.search(hek.attrs.Time(tstart,tend), hek.attrs.EventType(event_type), (hek.attrs.Event.Coord1 > 50) or (hek.attrs.FL.PeakFlux > 1000.0) ) and as a check >>> [elem["fl_peakflux"] for elem in result] [None, None, None, None, None, None, None, 2326.86, 1698.83, None, None, 2360.49, 3242.64, 1375.93, 6275.98, 923.984, 1019.83] >>> [elem["event_coord1"] for elem in result] [51, 51, 51, 924, 924, 924, 69, 883.2, 883.2, 69, 69, 883.2, 883.2, 883.2, 883.2, 883.2, 883.2] Note that some of the fluxes are returned as “None”. This is because some feature recognition methods for flares do not report the peak flux. However, because the location of event_coord1 is greater than 50, the entry from the HEK for that flare detection is returned. Let’s say we want all the flares west of 50 arcseconds AND have a peak flux over 1000.0: >>> result = client.search(hek.attrs.Time(tstart,tend), hek.attrs.EventType(event_type), (hek.attrs.Event.Coord1 > 50) and (hek.attrs.FL.PeakFlux > 1000.0) ) >>> [elem["fl_peakflux"] for elem in result] [2326.86, 1698.83, 2360.49, 3242.64, 1375.93, 6275.98, 1019.83] >>> [elem["event_coord1"] for elem in result] [883.2, 883.2, 883.2, 883.2, 883.2, 883.2, 883.2] In this case none of the peak fluxes are returned with the value None. Since we are using an and logical operator we need a result from the (hek.attrs.FL.PeakFlux > 1000.0) filter. Flares that have None for a peak flux cannot provide this, and so are excluded. The None type in this context effectively means “Don’t know”; in such cases the client returns only those results from the HEK that definitely satisfy the criteria passed to it. ## 5. Getting data for your event¶ The ‘hek2vso’ module allows you to take an HEK event and acquire VSO records specific to that event. >>> from sunpy.net import hek2vso >>> h2v = hek2vso.H2VClient() There are several ways to use this capability. For example, you can pass in a list of HEK results and get out the corresponding VSO records. Here are the VSO records returned via the tenth result from the HEK query in Section 2 above: >>> result = client.search(hek.attrs.Time(tstart,tend),hek.attrs.EventType(event_type)) >>> vso_records = h2v.translate_and_query(result[10]) >>> len(vso_records[0]) 31 Result 10 is an HEK entry generated by the “Flare Detective” automated flare detection algorithm running on the AIA 193 angstrom waveband. The VSO records are for full disk AIA 193 images between the start and end times of this event. The ‘translate_and_query’ function uses exactly that information supplied by the HEK in order to find the relevant data for that event. Note that the VSO does not generate records for all solar data, so it is possible that an HEK entry corresponds to data that is not accessible via the VSO. You can also go one step further back, passing in a list of HEK attribute objects to define your search, the results of which are then used to generate their corresponding VSO records: >>> from sunpy.net import hek >>> q = h2v.full_query((hek.attrs.Time('2011/08/09 07:23:56', '2011/08/09 12:40:29'), hek.attrs.EventType('FL'))) The full capabilities of the HEK query module can be used in this function (see above). Finally, for greater flexibility, it is possible to pass in a list of HEK results and create the corresponding VSO query attributes. >>> vso_query = hek2vso.translate_results_to_query(result[10:11]) >>> vso_query[0] [<Time(datetime.datetime(2011, 8, 9, 7, 22, 44), datetime.datetime(2011, 8, 9, 7, 28, 56), None)>, <Source('SDO')>, <Instrument('AIA')>, <Wavelength(193.0, 193.0, 'Angstrom')>] This function allows users finer-grained control of VSO queries generated from HEK results. The ‘hek2vso’ module was developed with support from the 2013 Google Summer of Code.	2018-04-20 16:34:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2565571963787079, "perplexity": 6291.753808480108}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125944479.27/warc/CC-MAIN-20180420155332-20180420175332-00349.warc.gz"}
https://imomath.com/index.php?options=320&lmm=0	# Number Theory 1. (2 p.) Let $$n$$ be the largest positive integer for which there exists a positive integer $$k$$ such that $k\cdot n! = \frac{(((3!)!)!}{3!}.$ Determine $$n$$. 2. (11 p.) Find the sum of all positive integers of the form $$n = 2^a3^b$$ $$(a, b \geq 0)$$ such that $$n^6$$ does not divide $$6^n$$. 3. (59 p.) Let $$f$$ be a function defined along the rational numbers such that $$f(\tfrac mn)=\tfrac1n$$ for all relatively prime positive integers $$m$$ and $$n$$. The product of all rational numbers $$0< x< 1$$ such that $f\left(\dfrac{x-f(x)}{1-f(x)}\right)=f(x)+\dfrac9{52}$ can be written in the form $$\tfrac pq$$ for positive relatively prime integers $$p$$ and $$q$$. Find $$p+q$$. 4. (23 p.) Find the largest possible integer $$n$$ such that $$\sqrt n + \sqrt{n+60} = \sqrt m$$ for some non-square integer $$m$$. 5. (2 p.) $$n$$ is an integer between 100 and 999 inclusive, and $$n^{\prime}$$ is the integer formed by reversing the digits of $$n$$. How many possible values are for $$\|n-n^{\prime}\|$$? 2005-2020 IMOmath.com \| imomath"at"gmail.com \| Math rendered by MathJax	2020-09-29 09:52:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6372674107551575, "perplexity": 125.43072445950686}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401641638.83/warc/CC-MAIN-20200929091913-20200929121913-00143.warc.gz"}
https://www.nature.com/articles/s41467-017-00755-6?error=cookies_not_supported&code=da2a56a7-706d-401d-b787-dea2dfc59708	Article \| Open # Tropical explosive volcanic eruptions can trigger El Niño by cooling tropical Africa • Nature Communicationsvolume 8, Article number: 778 (2017) • doi:10.1038/s41467-017-00755-6 Accepted: Published online: ## Abstract Stratospheric aerosols from large tropical explosive volcanic eruptions backscatter shortwave radiation and reduce the global mean surface temperature. Observations suggest that they also favour an El Niño within 2 years following the eruption. Modelling studies have, however, so far reached no consensus on either the sign or physical mechanism of El Niño response to volcanism. Here we show that an El Niño tends to peak during the year following large eruptions in simulations of the Fifth Coupled Model Intercomparison Project (CMIP5). Targeted climate model simulations further emphasize that Pinatubo-like eruptions tend to shorten La Niñas, lengthen El Niños and induce anomalous warming when occurring during neutral states. Volcanically induced cooling in tropical Africa weakens the West African monsoon, and the resulting atmospheric Kelvin wave drives equatorial westerly wind anomalies over the western Pacific. This wind anomaly is further amplified by air–sea interactions in the Pacific, favouring an El Niño-like response. ## Introduction The El Niño Southern Oscillation (ENSO) is the leading mode of natural interannual climate variability1. It is associated with large-scale sea surface temperature (SST) anomalies in the central and eastern Pacific. Those anomalies grow as the result of the Bjerknes positive feedback2 between the ocean and atmosphere, whereby a central Pacific warm SST anomaly induces westerly wind anomalies that strengthen the initial SST anomaly. ENSO has two opposite phases with warm events being referred to as El Niño, and cold ones as La Niña. ENSO events occur spontaneously every couple of years, and usually peak at the end of the calendar year with worldwide impacts on climate through the alteration of the global atmospheric circulation1. Because of its global impacts and strong societal relevance, improving the understanding and forecasting of ENSO is a key objective in climate science. The understanding of ENSO intrinsic dynamics has improved over recent decades, in particular with the identification of useful predictors3. Equatorial Pacific heat content, measured as the volume of water above 20 °C equatorward of 5° in the Pacific (warm water volume (WWV)), is the most commonly used ENSO predictor4. An unusually high early-year WWV tends to precede an El Niño peaking 1 year later. The understanding of ENSO response to external forcing is however primitive compared to our understanding of its internal dynamics. There is, for instance, neither a clear consensus on how ENSO is influenced by anthropogenic forcing5 nor, as we will see below, how it is influenced by volcanism. Some explosive tropical volcanic eruptions are sufficiently strong to inject aerosols into the stratosphere. Those aerosols backscatter incoming solar radiation, and reduce the global surface temperature by a few tenths of a degree Celsius for a couple of years6. While this systematic, global effect of large tropical eruptions, such as that of Pinatubo in 1991, is relatively well understood, their impact on regional seasonal phenomena such as the Asian monsoon7 or on natural modes of climate variability such as ENSO has not yet settled to a consensus8. Observational in situ SST data sets dating back to 1882 indicate a large positive ENSO-like pattern following four out of five big eruptions during the historical period (Santa María in October 1902, Mt Agung in March 1963, El Chichón in April 1982 and Pinatubo in June 1991)9,10,11. While this is a small sample, and a warm event was already underway during two of these eruptions, longer ENSO-proxy records also suggest a more probable equatorial Pacific warming within 2 years after large tropical eruptions12,13,14,15,16. Modelling studies have however reported contrasting responses to volcanism, ranging from SST anomalies typical of La Niña conditions10, 17 to no clear signal18 or preferred El Niño events peaking about a year after the eruption16, 19,20,21,22,23,24,25. Part of this inconsistency has been attributed to the global volcanically induced surface cooling, which could hide a potential El Niño signal suggested by concomitant positive sea surface height anomalies in the central-eastern Pacific20. Modelling studies also do not agree on the mechanisms by which volcanism influences ENSO. Early studies using simple coupled ocean–atmosphere models26 proposed that following volcano-induced surface cooling, upwelling in the eastern equatorial Pacific acting on a reduced vertical temperature contrast between the ocean surface and interior leads to anomalous warming in this region, thereby favouring El Niño development the following year12, 27. In contrast, more complex coupled general circulation models (CGCMs) invoke the combined effect of regional changes in oceanic mixed layer depth, latent heat cooling and cloud feedbacks10, 19. Other studies have proposed that the southward shift of the Pacific Intertropical Convergence Zone in response to the Northern Hemisphere cooling21,22,23,24 or the increased land–sea thermal contrast between the maritime continent and western equatorial Pacific Ocean19, 25 could drive a weakening of the Pacific trades and lead to an El Niño. There is hence a need to reconcile in situ observations and paleo-proxies, which suggest an El Niño during the 2 years that follow the eruption, and modelling studies that have reached no consensus on either the sign or the mechanism of the ENSO response to volcanism. Here, we show that using SST anomalies relative to the tropical average (relative SST in the following) allows us to reconcile models and observations. This metric reveals that an El Niño event tends to follow large tropical eruptions in observed SST data sets and in the Coupled Model Intercomparison Project phase 5 (CMIP5) historical experiments. We further use the Institut Pierre Simon Laplace (IPSL)-CM5B28 climate model and complementary atmospheric and oceanic ensemble simulations, with and without a Pinatubo-like forcing, to identify the main mechanism at play. We choose the Pinatubo eruption in June 1991, as it is the largest and best-documented tropical stratospheric eruption of the instrumental period. We first show that the response to volcanism superposes linearly with the natural ENSO cycle. We then demonstrate that the cooling of tropical Africa (the largest tropical landmass) reduces precipitation, and forces an atmospheric Kelvin wave response that triggers westerly wind anomalies and an El Niño over the Pacific. We also show that, while modulated by the seasonal cycle of convection, the effect of volcanism on wind forcing over the Pacific persists during the second year after the eruption, implying that the Pacific El Niño-like response involves more external forcing than traditional, internally generated events. ## Results ### Relative SST isolates consistent ENSO response to volcanism An El Niño event followed four out of five large volcanic eruptions since the beginning of the observational record in the late nineteenth century (Supplementary Fig. 1). This means that the global cooling effect of volcanism is partly masked by the global warming effect of concomitant El Niños over the observing period29, which led some to conclude that the global post-eruption surface cooling is overestimated in the CMIP5 historical database8. In the equatorial Pacific, this aliasing may also explain the physical inconsistency between negative SST anomalies (indicative of La Niña) and concomitant positive sea surface height anomalies in the central-eastern Pacific indicative instead of El Niño20 after large eruptions. Here we isolate the intrinsic ENSO signal from the volcanically-induced surface cooling by using relative SST (designated RSST) anomalies rather than raw SST anomalies. RSST, which was initially proposed to better explain the response of tropical cyclones to climate fluctuations30, is defined as the residual signal obtained after removing the mean tropical (20 °N–20 °S) SST anomalies. In observations, RSST (Fig. 1a, b) reveals an equatorial Pacific warming after the five largest tropical volcanic eruptions of the historical period. The resulting RSST composite anomaly in the Niño3.4 region (a common ENSO indicator) reaches 1.1 °C during early boreal winter following the eruptions (Fig. 1a). The probability of having such an average anomaly for 5 randomly picked years is 0.3% according to bootstrap statistics (see Methods and Supplementary Fig. 2). Such low probability is suggestive of a tendency for tropical explosive volcanism to trigger an El Niño event11. Using RSST removes the apparent discrepancies among previous modelling studies that have reported responses to volcanism ranging from favouring La Niña to El Niño conditions. For each of the 106 CMIP531 historical simulations (see Supplementary Table 1), we identify El Niño events from averaged Niño3.4 region monthly RSST standardized anomalies that exceed 0.5 for three consecutive months. This analysis indicates that the number of CMIP5 simulations displaying an El Niño response increases by 30–60% relative to the climatological El Niño probability within 2 years that follows the 5 major volcanic events of the historical period (Fig. 1c and Supplementary Fig. 3a). The low probability (p < 0.05; assuming a normal distribution) to have such a high rate of El Niño occurrence during years with no eruption in the CMIP5 historical database indicates a tendency for volcanic forcing to induce El Niños after each major eruption. The increase of El Niño probability following large tropical eruptions remains undetected when using SST instead of RSST, not only because of the global warming trend, but also because of the volcanically induced global cooling (see Methods and Supplementary Fig. 4). Precipitation anomalies also match RSST much better than SST in the composite response to the Pinatubo eruption built from the CMIP531 database (Supplementary Fig. 5), suggesting that RSST better characterizes the atmospheric convective response to SST gradients and their fluctuations32, 33. This underlines the usefulness of RSST for characterizing ENSO response in the context of the global surface cooling induced by volcanic eruptions, which resolves apparent inconsistencies between previous observational9,10,11,12,13,14,15,16 and modelling10, 17,18,19,20,21,22,23,24,25 studies. ### Interaction between volcanism and the natural ENSO cycle While there are enhanced chances of El Niño events after the five major historic volcanic eruptions in the CMIP5 database, some members still display neutral or La Niña conditions following these eruptions. This can be explained by the interaction between volcanic forcing and the natural ENSO cycle. As mentioned earlier, the equatorial Pacific Ocean heat content (or WWV) is a well-known ENSO precursor4. Using the IPSL-CM5B model we ran three pairs of 3-year-long ensemble experiments, starting from anomalously high, nearly neutral and low equatorial Pacific heat content shortly before the Pinatubo eruption (see Methods). Each pair consists of a 30-member ensemble with and without the Pinatubo eruption radiative forcing. Paired anomalies between these two ensembles combine the response to volcanic forcing (signal) and internal climate variability (noise), the latter being reduced by averaging across ensemble members. In the absence of volcanic forcing, anomalously high heat content in late spring favours El Niño, low heat content favours La Niña and neutral heat content favours a nearly neutral state towards the end of the year in our model (Fig. 2), as in observations4. Relative to this, the volcanic forcing tends to favour an anomalous central Pacific RSST warming that peaks during the year following the eruption irrespective of the ENSO preconditioning, in line with CMIP5 results (Supplementary Fig. 3alight blue curve) and dedicated studies with other climate models16, 24, 25 (Supplementary Fig. 6). For ENSO events already underway, volcanic forcing tends to shorten La Niñas (Fig. 2a, b) and lengthen El Niños (Fig. 2c, d). In the absence of preconditioning, volcanic eruptions favour El Niño-like warmings instead of neutral states, but with a warming that peaks in boreal summer and fall rather than boreal winter (Fig. 2e, f). Figure 3a–f highlights the chain of mechanisms giving rise to an El Niño-like event in the IPSL-CM5B ensemble simulations. Given the relative insensitivity of the ENSO response to the Pacific heat content preconditioning, we have chosen to illustrate these processes for the case of neutral initial conditions (see Supplementary Figs. 7 and 8 for other cases). The tropical Pacific net shortwave reduction at the surface reaches about 10–12 W m−2 5 months after the eruption (Fig. 3a) as in observations34, before a gradual recovery until the end of the following year. A large-scale westerly wind anomaly develops in the western Pacific from late August of the eruption year (dotted frame in Figs 3d and 4a, b). The resulting downwelling equatorial oceanic Kelvin waves (Fig. 3e) induce anomalous eastward currents and a deepening of the thermocline (Fig. 3f) in the central Pacific, both of which drive a surface warming35 (Fig. 3b) late in the eruption year (year 0). RSST and thermocline depth anomalies temporarily decrease at the beginning of the following year, consistent with the delayed negative feedbacks usually associated with ENSO demise1. Yet, from spring onwards, westerly wind and positive SST anomalies grow again to reach a peak during the year after the eruption. ### Tracking the drivers of the Niño-like response The westerly wind anomaly in fall of the eruption year plays an essential role in favouring the development of an El Niño. We have developed targeted ensemble experiments using the atmospheric component of the IPSL-CM5B model forced by the outputs of the coupled model ensemble (see Methods and Supplementary Table 2) to investigate the respective role of the following three processes in its generation. First, we examine the direct effect of volcanic radiative forcing on clouds and atmospheric vapour content36, potentially altering the atmospheric vertical structure and inducing an atmospheric dynamical response (hereafter referred to as ATM). Volcanic aerosol forcing is included in ATM, but the surface albedo is modified so that continental surfaces do not cool in response to volcanic forcing in this experiment (see Methods, Fig. 4e, f). We also examine the indirect effects of volcanic forcing through the induced horizontal SST gradients10 (hereafter OCEAN, Fig. 4g, h). The OCEAN experiment includes not only the effect of SST anomalies that develop in response to volcanic forcing10, but also the positive Bjerknes feedback between the ocean and atmosphere once El Niño-like SST anomalies have started developing. Continents cool faster than the ocean in response to volcanic forcing, due to their lower heat capacity6. The third and last experiment (hereafter referred to as LAND) tests the effect of this differential surface cooling between land and ocean. In LAND, there is no prescribed aerosol forcing as in OCEAN, but the land surface albedo modification enforces a land surface cooling that is consistent with that of the IPSL-CM5B coupled model experiment (Fig. 4i, j). This strategy allows land surface temperature variations in OCEAN experiments, in response to SST anomalies (Fig. 4h), as described in the context of global warming37, 38. It is hence more physically relevant than experiments that directly constrain land surface temperature19, which can result in an exaggerated land cooling and land–ocean gradients. A complementary set of sensitivity experiments (see Methods, Supplementary Table 2) explores the effect of the faster land cooling only over specific regions: tropics (LAND-T), extra-tropics (LAND-ET), Africa (LAND-Africa), Southeast Asia (LAND-SEA) and the maritime continent (LAND-MC). The equatorial Pacific SST response to surface wind anomalies resulting from each process experiment is then estimated separately by forcing a linear ocean model (see Methods, Supplementary Fig. 9). An additional experiment that includes all three LAND, OCEAN and ATM processes (ALL) reproduces precipitation and surface temperature anomalies simulated by the IPSL-CM5B ensemble simulation (Fig. 4a–d and Supplementary Fig. 10), confirming the relevance of this two-tier approach for understanding how volcanic eruptions induce equatorial Pacific westerlies and RSST anomalies. ### Linking tropical Africa cooling to El Niño onset In ATM, the equatorial Pacific wind stress and SST response is negative but hardly significant at the end of the eruption year (Fig. 5a, byellow curve), indicating that in isolation from the surface cooling, the direct effect of the radiative forcing on the atmosphere cannot explain the volcanically induced western Pacific initial westerly wind anomaly (Fig. 4e, f) and the subsequent El Niño. The initial westerly wind anomaly after the eruption is instead largely driven by land–sea temperature contrasts, accounted for in the LAND experiment (light blue curve in Fig. 5a). These land–sea temperature contrasts induce a strong cooling over tropical continental surfaces (Fig. 4j) in late summer and early fall after the eruption, which results in reduced convective activity and drying of tropical America, Africa and the maritime continent (Fig. 4i). The experiment (see Methods, Supplementary Table 2) including only cooling over tropical Africa indicates that it contributes to more than 80% of the initial westerly wind anomaly (Fig. 5a, green curve). Rather than the maritime continent, Southeast Asia or the extra-tropics, it is hence the cooling over the largest tropical landmass, namely Africa, that has the strongest effect on Pacific winds (Fig. 5c). Indeed, the land surface cooling is maximum in the tropics during the first boreal summer and fall (Fig. 4j) and leads to a reduction of the West African monsoon (Fig. 6a, b). The reduced precipitation and tropospheric heating in the equatorial latitudes drive a Matsuno–Gill response39, 40 where atmospheric equatorial Rossby and Kelvin waves induce easterly wind anomalies over the Atlantic and westerly wind anomalies over the Indian Ocean and western Pacific (Figs. 6a, b and 7 and Supplementary Fig. 11). This Kelvin wave suppresses convection along its path, with reduced convection over the western Pacific that further strengthens the westerly wind signal (Fig. 6a, b). The easterly wind anomalies over the Atlantic induce a shallow thermocline and cold RSST anomalies, reminiscent of an Atlantic Niña41, which peak in fall of the eruption year (Fig. 7b). This may also contribute to generating westerly wind anomalies over the Pacific, as noted by recent studies41, 42. This western Pacific westerly wind anomaly initiates oceanic downwelling Kelvin waves that warm the central Pacific in winter of the eruption year (Fig. 7b), creating favourable conditions for El Niño development. While the westerly wind and positive RSST anomalies in the Pacific are initially driven by the continental—mostly African—cooling (light blue and green curves in Fig. 5a, b), the Bjerknes feedback (deep blue curves in Fig. 5a, b) strengthens those anomalies during the year after the eruption (Fig. 7c). The Pacific westerly anomalies driven by cooling over the continents weaken during boreal winter of the year following the eruption (Fig. 5a, c), probably because convection moves away from the equator during boreal winter (Fig. 6d–f). During the following year, the cooling of tropical land masses also induces a boreal fall westerly wind anomaly over the tropical Pacific again. This westerly wind anomaly is generated by similar mechanisms to those during the previous year (Fig. 6g–i), although the tropical African cooling and drying play a lesser role, as the tropical volcanic radiative cooling has significantly decreased by this time (Figs. 5a, c and 7c and Supplementary Fig. 12). This volcanically forced westerly wind anomaly also contributes to the second-year Niño-like warming during summer and fall (dark and light blue curves, Figs. 5a, b and 7d). Overall, this suggests a stronger role of externally forced westerly wind anomalies for volcanically induced El Niño-like events than for internally generated El Niños, for which the Bjerknes positive feedback is responsible for most of the anomalous conditions. ## Discussion Previous work has offered no consensus on how ENSO responds to volcanic forcing because the observational record is relatively short and there were inconsistencies between modelling studies10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25. In contrast, our results clearly indicate a tendency for large tropical volcanic eruptions to induce an El Niño-like response peaking during the year after the eruption. Part of the confusion in previous modelling studies arose because they did not properly separate the direct cooling effect of volcanism from that associated with ENSO. Global climate models, for instance, do not overestimate the volcanically induced tropical cooling43, 44 when accounting for the warm global SST anomalies associated with El Niños29. Similarly, it is necessary to remove the volcanically induced tropical cooling signal (for instance through the use of relative SST) in order to clearly isolate the ENSO response to volcanism. This is particularly true for the former studies suggestive of a La Niña response10, 17, which stumble on the physical inconsistencies between negative SST anomalies (indicative of La Niña) and concomitant positive sea surface height anomalies in the central-eastern Pacific indicative instead of El Niño20. Other studies have suggested an alternative mechanism for the influence of volcanically induced land cooling on ENSO. The larger landmass in the Northern Hemisphere for instance may induce a large cooling over Asia and weakening of the Asian monsoon, inducing a southward Intertropical Convergence Zone (ITCZ) shift in boreal summer and fall, and westerly anomalies over the Pacific21,22,23,24. This mechanism however does not apply in our experiments: the tropical cooling is rather symmetrical with respect to the equator in boreal fall after the eruption (Fig. 4), because the Pinatubo aerosol cloud is still mostly confined within the tropics at that time. Another work has underlined the potential role of the maritime continent cooling in generating the Pacific westerly wind anomalies19, 25. Our experiments do not yield any significant summer–fall Asian monsoon weakening or southward ITCZ shift (Fig. 4). Experiments with volcanically induced land cooling only applied over Southeast Asia, the maritime continent or the extra-tropics do not display any summer–fall westerly wind anomaly in the western Pacific (Fig. 5c). Our modelling results support a different mechanism from those invoked previously, in which the cooling of tropical landmasses, especially Africa, induces boreal summer–fall Pacific westerly wind anomalies that favour a Pacific warming (Fig. 7). This mechanism will however have to be tested in a wider variety of climate models. Our results may allow further reconciling the CMIP5 models response to volcanism with that from the limited observational record. RSST anomalies peak at the end of the eruption year for most of the observed eruptions, as in our experiments with a recharged equatorial Pacific heat content at the time of the eruption (Fig. 2). Ocean reanalyses60, 61, only available for the 1991 Pinatubo, 1982 El Chichón and 1963 Agung eruptions, suggest high equatorial Pacific heat content before those eruptions (Supplementary Fig. 13), which would tend to favour maximum RSST anomalies at the end of the year. On the other hand, CMIP5 historical simulations sample the full ENSO cycle (mixture of discharged, neutral and recharged states), and hence yield a response that is more representative of neutral initial conditions. Our results suggest that volcanically induced El Niño-like events involve a more persistent external wind forcing than classical El Niños, whose growth is dominated by internally generated wind anomalies through the Bjerknes feedback. This may explain the atypical timing of the volcanically induced El Niño-like warming in our experiments that peak in boreal summer rather than boreal winter (Fig. 2). Further analyses are however required to understand how the timing of the eruption influences the El Niño-like response16 and the link with the West African monsoon and to ascertain the robustness of the above mechanisms in other models. CMIP5 models use very diverse volcanic forcing products and strategies45 though, and so a unified approach is needed for investigating the effects of volcanism on climate variability, such as proposed in the upcoming CMIP6 VolMIP exercise8. An improved understanding and representation of ENSO response to volcanism in climate models may in particular yield extra ENSO predictability after major eruptions, beyond the traditional 6-month lead time limit. ## Methods ### Statistical testing in observations We investigated the equatorial Pacific (Niño3.4 and Niño3 regions) SST response to major tropical volcanic eruptions of the instrumental period using composites in HadISST observations46. We restricted our analyses to 1870–2010 as there are too many gaps in spatial sampling before 1870. The earliest part of the record (late nineteenth century) is, however, less reliable due to lower spatial resolution. We build composites using the five tropical volcanic eruptions in 1870–2010 with significant injection of sulphuric aerosol in the tropical stratosphere: Pinatubo (June 1991), El Chichón (April 1982), Mt Agung (March 1963), Santa María (October 1902) and Krakatau (August 1883). SST anomalies are computed relative to the 5-year mean preceding each eruption (to remove multidecadal variability and long-term trends) and smoothed with a 3-month Hanning filter (to filter out intraseasonal variability). To evaluate the level of statistical significance, we first count the number of events with anomalies that have the same sign, as shown in Fig. 1a. If we assume that there is the same probability of having positive and negative signs, according to the binomial law, the likelihood for having 5 anomalies of the same sign is 0.03, i.e., corresponding to a 97% confidence level. Second, we use a Monte Carlo approach to compute the distribution for the composite value of November–January RSST average anomalies of 5 randomly picked years within the historical period, the result of which is given in Supplementary Fig. 2. The probability in the Niño3.4 composite of RSST anomaly equal or above the observed 1.1 °C value is very small (0.3%). The level of statistical significance is similar in other data sets such as HadiSSTv1.1, HadSST31, ERSSTv2b247, Kaplan SST348 and Niño34hist449. Various methods to compute the anomalies (15-year high-pass filter, removing the mean value for the previous 1, 3, 5 or 10 years) were tested and did not affect the results. ### CMIP5 historical simulations We considered 26 CMIP5 models and a total of 106 members from the CMIP5 historical simulation database31 (Supplementary Table 1). Each member is a coupled ocean–atmosphere simulation using the CMIP5 recommended transient forcing from 1850–2005. We did not include simulations using anomalies of the solar constant to mimic volcanic forcing, interactive chemistry, carbon cycle or prognostic global aerosol models in our analyses, to restrict the potential noise associated with those extra degrees of freedom. Despite this, there is still a relatively large diversity of specified external forcings, as shown in Supplementary Table 1. Anomalies were computed relative to the 1961–1990 climatology of each model simulation. We analysed SST evolution in these simulations by using both the raw and relative (obtained by removing the global tropical 20 °N–20 °S mean SST) SST anomalies. For each of the 106 CMIP5 historical simulations31, we identified El Niño events from averaged Niño3.4 region monthly raw and relative SST standardized anomalies that exceed half a standard deviation over the 1870–2010 simulated period. The percentage of CMIP5 simulations displaying an El Niño-like state and its statistical significance are shown in Supplementary Figs. 35. Using various methods to compute the SST anomalies (removing the mean climatology of the previous 5 years, using the 1961–1990 or the 1870–2010 baseline climatology) do not affect the results. The difference between raw SST (Supplementary Fig. 1a, b) and RSST (Fig. 1a, b) is weak in observations, but it is large in CMIP5 models (see Supplementary Fig. 5). The largest uncertainties in the estimates of radiative forcing from CMIP5 historical simulations occur during periods of volcanic activity43 and generally models tend to overestimate the observed post-eruption global surface cooling44. Relying on relative SST and precipitation anomalies partly corrects this bias, so that it reveals the dynamical ENSO response to volcanic forcing in CMIP5 models. ### Coupled model simulations for the 1991 Pinatubo eruption To help analyse the response of ENSO to volcanic forcing, a suite of short ensemble experiments was run over January 1991 to December 1993 for the Mt Pinatubo eruption (June 1991) using the IPSL-CM5B-LR28 ocean atmosphere CGCM. The stratospheric aerosol optical depth (hereafter AOD) forcing due to the Pinatubo eruption in each member uses the Gao et al.51 data set. The stratospheric volcanic cloud spreads from the tropics to the poles from the start of the eruption in June 1991 until late 1992, and then slowly disappears as the aerosols are totally washed out of the atmosphere in 1993. The total AOD reaches its maximum in boreal autumn–winter 1991–1992. Corresponding control ensembles starting from the same initial conditions but without Pinatubo volcanic forcing were performed. ### IPSL-CM5B-LR coupled model simulations ensemble We use the IPSL-CM5B-LR CGCM. The Laboratoire de Météorologie Dynamique (LMD) developed the LMDZ5B atmospheric component, which is described in Hourdin et al.52. This model uses hybrid σp vertical coordinates, with 39 pressure levels including 18 in the stratosphere. The LMDZ5B model uses ORCHIDEE as land surface component. We used the low-resolution grid with 1.87° in latitude and 3.75° longitude. The oceanic component, Nucleus for European Modeling of the Ocean (NEMO) version 3.2, is an oceanic GCM with 31 vertical levels (whose thickness varies from 10 m near the surface to 500 m towards the bottom), a 2-level representation of sea ice and a mean spatial horizontal resolution of about 2° (with a refinement of the latitudinal resolution to 0.5° near the equator; ORCA2 grid). Previous works53 found that this model reproduces ENSO variability with two spectral peaks around 3–3.5 years and beyond 4 years, which is in good qualitative agreement with observations. The observed ENSO seasonal phase locking with a peak occurring mostly in boreal winter (November to January) and feedbacks are also well captured in this model53. Initial conditions were chosen in the IPSL-CM5B-LR standard historical run. To disentangle the role of Pinatubo eruption on ENSO evolution and avoid model spin-up, we chose restart days in the historical run with greenhouses gases and tropospheric aerosols concentrations close to the 1990s levels and more than 5 years after any volcanic eruption. We selected three restart dates corresponding to 1 June (i.e., shortly before the Pinatubo eruption) with initial conditions that would be favourable to the development of an El Niño, neutral, or La Niña phase in the absence of volcanic forcing. WWV (the volume of water above the 20 °C isotherm within 5 °N–5 °N 120 °E–80 °W) is a good precursor of the upcoming ENSO phase in nature4. This is also the case in our model as demonstrated by a peak correlation of r = 0.7 when the Niño 3.4 SST index lags the WWV in May by 6 months in the IPSL-CM5B CMIP5 historical simulation54. We hence selected 3 model states representing discharged, near-neutral and recharged WWV based on normalized WWV values in May. For the three initial states, 30 members (forced and unforced) were generated by adding a small white noise on the initial SST. Consistent with observations, the discharged, neutral and recharged states ensemble mean respectively produce a La Niña, near-neutral and El Niño state in the absence of volcanic forcing (see Fig. 2 and Supplementary Fig. 7). ### CNRM-CM5 coupled model simulations ensemble To evaluate the robustness of the results obtained with the IPSL climate model, we also used the CNRM-CM5 model developed jointly by CNRM-GAME (Centre National de Recherches Météorologiques—Groupe d’études de l’Atmosphère Météorologique) and Cerfacs (Centre Européen de Recherche et de Formation Avancée). CNRM-CM5, described by Voldoire et al.50, couples the ARPEGE-Climat (v5.2) atmosphere, NEMO (v3.2) ocean, the ISBA land surface and GELATO (v5) sea ice models through the OASIS (v3) coupler. The horizontal resolution is 1.4° in the atmosphere and 1° in the ocean (with a refinement of the latitudinal resolution to 0.5° near the equator). The model reproduces ENSO typical variability well with an 2‒7-year period and a seasonal phase locking with a peak occurring mostly in boreal winter, in good qualitative agreement with observations53. The ensemble of the CNRM-CM5 model was designed to explore the role of Pinatubo eruption on ENSO response when starting from random initial conditions. The initial dates were chosen randomly in the control and historical runs performed with the same model version50. ### The two-tier sensitivity experiments approach To explore the role of various processes that drive the climate response to volcanism, we ran two-tier experiments using an atmospheric model simulations ensemble (AGCM) and a linear ocean model. ### The atmospheric model simulations ensemble We ran several 30-member AGCM ensembles using the atmospheric component of the IPSL-CM5B coupled model52. Daily fields from each of the 30 members of the coupled model ensembles starting from neutral initial conditions are used to derive a set of boundary conditions (described below) for the stand-alone atmospheric component (LMDZ AGCM) of the IPSL-CM5B model, which includes the ORCHIDEE land surface module. The stand-alone atmospheric model is run for 2 years from 1 January 1991 onward (i.e., year 0; 5.5 months before the Pinatubo eruption) to enable the atmosphere to dynamically adjust to the boundary conditions. We have first run a Control 30-member AGCM ensemble using surface boundary conditions (i.e., SST and sea ice cover) from the coupled model control ensemble without including the changes induced the volcanic aerosol. The ALL experiment follows the same protocol, but includes the stratospheric AOD evolution corresponding to the Pinatubo eruption as in the coupled model experiments and specified SST and sea ice cover from each member of the coupled model Pinatubo ensemble. The anomalies of ALL relative to the Control are almost identical when compared to the anomalies of the coupled model experiment (Supplementary Fig. 10). The small differences between the coupled and the atmospheric experiment are explained by the differences in the surface flux due to coupling55. To address the role of the atmospheric, land and ocean responses in triggering surface wind anomalies, five complementary sets of experiments have been run (see summary in Supplementary Table 2). The ATM, LAND and OCEAN experiments were respectively designed to represent the effect of aerosol forcing through its direct impact on the atmospheric structure (ATM), and indirect impacts through changes in land (LAND) or ocean (OCEAN) surface temperatures. The ATM and LAND scenarios were developed using the same protocol as in ALL but with prescribed daily SST from the Control members (i.e., with a SST that excludes the impact of volcanic aerosol forcing). Volcanic aerosol forcing is included in the ATM experiments, but the surface albedo in this experiment is modified so that continental surfaces do not cool in response to volcanic forcing (details below). In LAND, there is no prescribed aerosol forcing, but the albedo modification enforces a land surface cooling that is consistent with that of the Pinatubo coupled model experiment. This strategy has the advantage of allowing surface land temperature to vary in the presence of SST anomalies (in the OCEAN experiment), as is seen in other model experiments in the context of global warming scenario37, 38. This is hence more physically relevant than experiments that constrain land surface temperature19 and which can result in unphysical and exaggerated land cooling, and hence exaggerated land–ocean gradients. We modify the land surface albedo to restore the atmospheric temperatures over land to that of the Control coupled model ensemble in ATM and to that of the Pinatubo coupled model ensemble in LAND. The boundary layer temperature changes ΔT in response to a given change in the net shortwave radiation ΔSWnetSFC at the surface can be approximated assuming linear dependency such as: $λΔT=Δ S W d o w n SFC - S W u p SFC =ΔSWne t SFC$ with λ being the feedback parameter including that due to the Planck emission (i.e., the feedback associated with changes in surface temperature), changes in vertical temperature lapse rate, water vapour and clouds. Here, we specify SWnetSFC by prescribing the surface albedo α. The surface albedo is related to the net shortwave radiation by: $SWne t SFC =SWdow n SFC 1 - α$ (1) In particular, we have: $SWne t cSFC =SWdow n cSFC 1 - α c$ (2) $SWne t pSFC =SWdow n pSFC 1 - α p$ (3) Here, the subscript c denotes the values of the control coupled model experiment, while the subscript p designates the values of the Pinatubo coupled model experiments. Then, we calculate Δα, the relative change of the incoming SW radiation due to stratospheric sulphate aerosols: $Δα= S W d o w n pSFC - S W d o w n cSFC ∕SWdow n cSFC$ (4) In ATM, the downward shortwave radiation is modified due to the direct radiative effect of stratospheric sulphate aerosols (i.e., SWdownSFCATM= SWdownpSFC), but we modify the surface albedo to αATM so that the surface-absorbed net shortwave radiation remains equal to that of the control run (i.e., SWnetSFCATM=SWnetcSFC). Using previous relationships together with (2) and (4), we obtain: $α ATM = α c + Δ α ∕ 1 + Δ α$ (5) Conversely, in LAND, the surface albedo is set to the αLAND value, with downward shortwave radiation equal to the control experiments (SWdownSFCLAND= SWdowncSFC), but surface-absorbed shortwave equal to that of the Pinatubo experiments (i.e., SWnetSFCLAND=SWnetpSFC): $α LAND = α p -Δα+ α p Δα$ (6) We diagnosed Δα from the control and Pinatubo mean ensemble simulations and performed ATM and LAND simulations imposing the albedo in order to restore the surface temperature towards control and Pinatubo values respectively. This protocol involves a linearity assumption of assumed constant values for the feedback parameter λ. Surface temperature anomalies in the ATM, LAND and OCEAN experiments (Fig. 4 and Supplementary Fig. 10) however add up quite linearly to surface temperature anomalies in the ALL atmospheric or Pinatubo coupled model experiment, indicating the validity of our approach. The LAND experiment isolates the effect of volcanically induced continental cooling but does not narrow down the region that plays a key role for inducing wind anomalies over the tropical Pacific. We hence also developed a suite of additional 30-member LAND sensitivity experiments (with and without Pinatubo radiative forcing) that allows exploring the respective roles of land cooling in the tropics (LAND-T), the extra-tropics (LAND-ET), Africa (LAND-Africa), Southeast Asia (LAND-SEA) and the maritime continent (LAND-MC) only (See Supplementary Table 2). ### The linear Indo-Pacific Ocean model The atmospheric experiments above isolate the wind changes over the Pacific associated with three different processes: direct effect of radiative forcing on the atmosphere (ATM) and indirect effect through the cooling of continents (LAND) and induced oceanic SST gradients (OCEAN). To translate those wind signals into SST responses, we use a linear continuously stratified56 (LCS) ocean model of the Indo-Pacific region combined with a linear SST equation. The LCS model is forced by the atmospheric forcing produced by each of the 30-member experiments above to deduce the equatorial Pacific Ocean response (and in particular the SST response). The LCS model for the Indo-Pacific Ocean resolves vertical baroclinic modes. It is used to simulate thermocline depth and near-surface current anomalies. It is combined with a linear SST equation57 to realistically resolve the variations of SST due to wind stress anomalies: $d S S T ′ d t =-U′ d S S T ¯ d x -V′ d S S T ¯ d y +γH′-βSST′$ The horizontal advection term is approximated from the anomalous advection of the climatological SST ($S S T ¯$) gradient (here from the IPSL historical run climatology) by the LCS 0–100 m averaged anomalous currents U' and V'. In the Indo-Pacific region, zonal advection dominates over meridional advection (not shown, see for example Vialard et al.35). The γH' term (H' is the thermocline depth anomaly obtained as H'=150 SSH' from the sea surface height anomaly SSH') parameterizes the vertical physics, i.e., the interannual modulation of mixing and upwelling by thermocline depth anomalies. The coefficient γ is constant to the east of 140 °W (γ=0.028 °C/month, as in Burgers58), and linearly decreases until 1/5 of its maximum value at the western boundary of the Pacific, as in McGregor et al.59. This mimics the decreasing effect of the upwelling as the climatological thermocline deepens towards the west. Heat fluxes are parameterized as a Newtonian damping –β SST′, with the associated damping timescale β−1=2 months, close to values used by McGregor et al.59 and Burgers58. The 5 first baroclinic modes are used (even though mainly only the first two modes contribute)57. This model has the advantage of being relatively realistic (Validation: high correlations of 0.8–0.9 with observations (TAO, AVISO, OSCAR), notably for zonal current and SST in Niño4 (central Pacific), for which the first two baroclinic modes are important, and also for SST and SSH in Niño3 (eastern Pacific) when the model is forced by ERAI (see details in Izumo et al.57), as well as rather simple and cheap to use. It is used to simulate the SST response to wind stress anomalies diagnosed from the ATM, LAND and OCEAN AGCM ensemble members. The high correlation of the IPSL-CM5B coupled model Niño3.4 SST anomalies with those simulated by the LCS model forced by the coupled model experiments (r = 0.9; Supplementary Fig. 9) and ALL (r = 0.8, not shown) ensemble wind stress justifies our approach. Furthermore, the favourable comparison of the ALL ensemble mean with the coupled model experiment wind stress anomalies in the western equatorial Pacific region (Supplementary Fig. 10) lends credence to our method and suggests that our two-tier methodology to investigate various physical mechanisms in forced experiments derived from the coupled one is sound. ### Estimating signal-to-noise ratios For the analysis of these Pinatubo sensitivity experiments, we used two approaches to evaluate the statistical significance of anomalies associated with volcanic forcing, relative to purely random, internal coupled ocean–atmosphere variability. The anomalies associated with the response to volcanism are computed as paired differences between each of the members of the Pinatubo experiment and its unforced counterpart. The first statistical calculation estimates the statistical significance of the anomaly between the two ensemble means using a two-tailed Welch’s t-test, and taking into account the size and variance of each (control and Pinatubo) ensemble. For CMIP5 model analyses, the significance of anomalies relative to the climatology is measured using a two-tailed Student’s t-test, considering each model as an independent degree of freedom. For both our paired Pinatubo ensembles and CMIP5 models, we also estimated at each time step and grid point the percentage for which at least two thirds of the individual simulations have anomalies of consistent sign, which is a commonly used technique for analysing outputs of CMIP models. This allows us to verify that statistically significant anomalies in the ensemble mean are representative of the ensemble members and not due to a few extreme members. Figure captions specify the test for each figure. ### Code and data availability The code and data that support the findings of this study are available from the corresponding authors on request. Change history: A correction to this article has been published and is linked from the HTML version of this paper. Publisher's note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## References 1. 1. McPhaden, M. J., Zebiak, S. E. & Glantz, M. H. ENSO as an integrating concept in earth science. Science 314, 1740–1745 (2006). 2. 2. Bjerknes, J. Atmospheric teleconnections from the equatorial Pacific. Mon. Weather Rev. 97, 163–172 (1969). 3. 3. Barnston, A. G., Tippett, M. K., L'Heureux, M. L., Li, S. & DeWitt, D. G. Skill of real-time seasonal ENSO model predictions during 2002–2011: is our capability increasing? Bull. Am. Meteorol. Soc. 93, 631–651 (2012). 4. 4. Meinen, C. S. & McPhaden, M. J. 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ANR-13-SENV-0002-02) and by a grant from European Commissions Seventh Framework Research Program (FP7) under the grant agreement 308378 (SPECS). It also benefited from the IPSL CMIP data access PRODIGUER. A.R. is supported by the US National Science Foundation grant AGS-1430051 and M.J.M. is supported by NOAA. This is PMEL contribution 4492. ## Author information ### Affiliations 1. #### Laboratoire d’Océanographie et du Climat: Expérimentations et approches numériques, Sorbonne Universités, UPMC Université Paris 06, IPSL, UMR CNRS/IRD/MNHN, F-75005, Paris, France • Myriam Khodri • , Takeshi Izumo • , Jérôme Vialard • , Serge Janicot • , Matthieu Lengaigne • , Juliette Mignot • , Guillaume Gastineau • , Eric Guilyardi • & Nicolas Lebas 2. #### Indo-French Cell for Water Sciences, IISc-NIO-IITM-IRD Joint International Laboratory, NIO, Goa, 403002, India • Takeshi Izumo • & Matthieu Lengaigne 3. #### CECI, CNRS, Cerfacs, Université de Toulouse, Toulouse, 31057, France • Christophe Cassou • Eric Guilyardi 5. #### Department of Environmental Sciences, Rutgers University, New Brunswick, NJ, 08901, USA • Alan Robock ### Contributions M.K. and J.V. designed the study with inputs from T.I., C.C., S.J., G.G., E.G., J.M., M.L., N.L., A.R. and M.J.M. M.K. developed the experimental strategy and ran the GCM models. T.I. performed the linear, continuously stratified, model simulations and the probability density function of observed SST anomalies. M.K. performed the observed and CMIP5 superimposed epoch composites over the historical period and analysed the model simulation outputs. All authors contributed to the interpretation and writing. ### Competing interests The authors declare no competing financial interests. ### Corresponding author Correspondence to Myriam Khodri.	2018-02-20 23:43:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6732090711593628, "perplexity": 6105.628334893343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813109.8/warc/CC-MAIN-20180220224819-20180221004819-00656.warc.gz"}
https://www.thejournal.club/c/paper/354353/	#### Rate-Distortion Theoretic Model Compression: Successive Refinement for Pruning ##### Berivan Isik, Albert No, Tsachy Weissman We study the neural network (NN) compression problem, viewing the tension between the compression ratio and NN performance through the lens of rate-distortion theory. We choose a distortion metric that reflects the effect of NN compression on the model output and then derive the tradeoff between rate (compression ratio) and distortion. In addition to characterizing theoretical limits of NN compression, this formulation shows that \emph{pruning}, implicitly or explicitly, must be a part of a good compression algorithm. This observation bridges a gap between parts of the literature pertaining to NN and data compression, respectively, providing insight into the empirical success of pruning for NN compression. Finally, we propose a novel pruning strategy derived from our information-theoretic formulation and show that it outperforms the relevant baselines on CIFAR-10 and ImageNet datasets. arrow_drop_up	2021-10-24 02:43:48	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8057653903961182, "perplexity": 1303.2710228145493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585837.82/warc/CC-MAIN-20211024015104-20211024045104-00393.warc.gz"}
https://www.vedantu.com/formula/barium-iodide-formula	Courses Courses for Kids Free study material Free LIVE classes More # Barium Iodide Formula ## The Chemical Formula of Barium Iodide Last updated date: 23rd Mar 2023 Total views: 195.6k Views today: 2.73k Barium iodide is an inorganic compound that is made up of barium and iodine. By definition, an iodide is any compound that consists of iodine having one negative charge due to the acceptance of an electron for achieving the octet valence electronic configuration. Hence, Barium Iodide is an ionic compound that consists of a barium atom donating one electron and iodine accepting one electron to form a metal halide. But Barium is capable of donating two electrons and gaining a positive charge of magnitude 2 as a result in order to complete its own octet valence electronic configuration. Hence, one barium atom forms an ionic compound with two iodine atoms. Thus, the resulting molecular formula of barium iodide can be given as BaI2 ### General Characteristics of Barium Iodide Barium Iodide, represented by the molecular formula as BaI2, is an ionic compound and is found in both anhydrous form and as a hydrate. The chemical formula of barium iodide in the hydrate form is given as BaI2(H2O)2. Both of the forms of the compound are solids and have white colour. The anhydrous form can be easily obtained by heating the hydrate. It is well established that the hydrated form of barium iodide is soluble in water, ethanol and acetone. The molecular weight of the anhydrous form as determined by the formula of barium iodide is 391.136 g/mol. Similarly, the molar mass of the hydrated form as can be calculated from the barium iodide formula is 427.167 g/mol. The properties of barium iodide and other halide compounds of barium are similar because they also are formed by the donation and subsequent acceptance of the electron. This is also visible from the commonality in their molecular formula. For example, the molecular formula of Barium Fluoride BaF2 is the same as the barium iodide formula with the change of iodine to fluorine. As already mentioned, Barium iodide exists as a solid in both forms - the anhydrous and the hydrated form. It exists as a crystal and has an orthorhombic crystalline structure. The crystal structure of the anhydrous form of barium iodide resembles that of Lead (II) Chloride where instead of the lead ion, barium ion is present in the centre of the crystal structure and is bound to nine iodide ligands. The similarity in the structure can be because of both lead and barium being heavy metals. The crystalline packing of barium iodide structure also shares similarity with the crystalline form of Barium Chloride (BaCl2). The image of the crystalline packaging of barium iodide with barium being in the centre surrounded by nine iodine atoms as ligands are shown below: [Image will be uploaded soon] Some of the Main Chemical Reactions of Barium Iodide are Summarized Below: • The usual method of preparation of anhydrous form of Barium Iodide is by the reaction of barium metal with 1,2-diiodoethane. This reaction takes place in a solution made up of ether. • Barium iodide is known to form organometallic compounds that are of particular research interest. Barium Iodide reacts with compounds of alkyl potassium in order to form the corresponding organobarium compounds. • Barium Iodide is known to undergo reduction with lithium biphenyl which produces a highly reactive form of Barium as a reaction product. ### Uses of Barium Iodide Barium iodide is toxic in nature just like other soluble salts containing barium. It has very limited uses as compared to many other halide forms. Some of the uses of barium iodide include the following: • It can be used for the preparation of Barium dioxide, but as such has no further currently known purposes. Also, there is no biological relevance of barium dioxide. • It is used for the production of other iodide compounds. • It can also be used for the identification of copper castings as has been pointed out by some researchers. ## FAQs on Barium Iodide Formula 1. What is the Name For BaI2? Ans: The given molecular formula represents an ionic compound made up of barium and iodine. In this compound, barium donates two electrons to achieve an octet electronic configuration in its outermost shell and these two electrons are accepted by one Iodine atom each for the same purpose. Also, any compound containing an iodine atom with -1 electrical charge is known as an iodide. Hence, the IUPAC name of the compound is Barium Iodide. 2. What is Barium Iodide Used For? Ans: There are a limited number of uses of barium iodide. It is mostly used for research purposes. Apart from that, it is used for the production of other important compounds of iodine. It can also be used for the preparation of barium dioxide. 3. What are the Ions in BaI2? Ans: The given molecular formula is of an ionic compound known as Barium Iodide. Now, any compound is named iodide when it consists of an iodine ion with a -1 electrical charge. In this case, there are two such iodine ions present. Also, Barium is a group 2 element in the periodic table with a capacity to lose two electrons and gain a +2 charge for getting a stable electronic configuration. Hence, the two ions that are present in the given compound are Ba2+ and I-.	2023-03-27 23:54:46	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8350933790206909, "perplexity": 2429.1117058556956}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00379.warc.gz"}
http://docs.thinkboxsoftware.com/products/sequoia/1.1/Documentation/manual/scripting/scripting_measure.html	# Scripting The Measurement Tool¶ ## Prerequisites¶ • The following methods are exposed by the MeasurementTool object implemented by the module Thinkbox.Sequoia.Tools • Before calling these methods, you must ensure the module is imported. import Thinkbox.Sequoia.Tools 1.0 ## Measurem Tool Methods¶ <List of Doubles> MeasurementTool.getStartPoint (); • Returns the start point position of the Measure gizmo. <void> MeasurementTool.setStartPoint ( <double>x, <double>y, <double>z ); • Sets the start point position of the Measure gizmo to the specified X, Y and Z values. <List of Doubles> MeasurementTool.getEndPoint (); • Returns the end point position of the Measure gizmo. <void> MeasurementTool.setEndPoint ( <double>x, <double>y, <double>z ); • Sets the end point position of the Measure gizmo to the specified X, Y and Z values.	2017-06-29 00:28:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28764939308166504, "perplexity": 12006.529048897826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323808.56/warc/CC-MAIN-20170629000723-20170629020723-00485.warc.gz"}
https://math.stackexchange.com/questions/2378348/show-that-if-b2-x-0-n-for-some-vector-x-neq-0-n-then-b-is-not-inverti/2378355	# Show that if $B^2 x = 0_n$ for some vector $x \neq 0_n$, then $B$ is not invertible If $B \in M_{n \times n}(\mathbb{R})$ and $B^2 x = 0_n$ for some vector $x \neq 0_n$, then $B$ is not invertible. I get that $$\mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated. A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel. If $$B^2 x = 0$$, then $$B(Bx) = 0$$, so $$Bx \in \ker B$$. Suppose $$Bx = 0$$, then this shows that $$B$$ has a non-trivial kernel, hence is not injective. Suppose $$Bx \neq 0$$ above, then $$Bx$$ is a non-trivial element of $$\ker B$$, so it is not injective for $$n \geq 3$$. Hence, either way it follows that $$B$$ is not injective. VIA CONTRAPOSITIVE : If $$B$$ is injective, then so is $$B^2$$, but then $$B^20 = 0$$, so $$B^2x$$ cannot be zero for any other $$x$$ by injectivity. Note that we can extend this to $$B^nx = 0$$ implies $$B$$ is not injective. If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $\det(B^2)=0$. But $\det(B^2)=\det(B)^2$, so $\det(B)=0$ and $B$ is not invertible. Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.	2020-09-22 04:06:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9871706962585449, "perplexity": 45.134085910007535}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400203096.42/warc/CC-MAIN-20200922031902-20200922061902-00227.warc.gz"}
http://mathhelpforum.com/math-topics/43586-really-easy-question.html	# Math Help - really easy question 1. ## really easy question You can only use the quadratic formula for equations that have the format "ax2 + bx + c = 0", right? They can't have degrees of more than two? 2. Correct. Hence, the name quadratic. 3. Originally Posted by dancingqueen9 You can only use the quadratic formula for equations that have the format "ax2 + bx + c = 0", right? They can't have degrees of more than two? You can also use it to solve a quartic equation of the form $ax^4+bx^2+c=0$. 4. I see. Thank you very much!	2014-10-22 12:31:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6662270426750183, "perplexity": 684.0089843179065}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507446943.4/warc/CC-MAIN-20141017005726-00037-ip-10-16-133-185.ec2.internal.warc.gz"}
https://eepower.com/news/ir-announces-strategic-collaboration-with-sanken/	News IR Announces Strategic Collaboration with Sanken May 17, 2001 by Jeff Shepard International Rectifier Corp. (IR, El Segundo, CA) announced a strategic collaboration with Sanken Electric Co. Ltd. (Japan) that includes joint design and development activity between the two power semiconductor companies for next-generation proprietary power-management products. The collaboration also includes a cross-shareholding agreement. IR has agreed to purchase approximately two percent of Sanken's shares, for about $13 million, and Sanken has agreed to purchase about 0.4 percent of IR's shares, for approximately$13 million. The IR shares to be purchased by Sanken will be newly issued pursuant to a private placement. The collaboration is planned to include the following: joint technology development of next-generation analog ICs; joint standards for systems architectures in certain automotive, ac/dc, dc/dc and motion control application areas; and collaboration on capacity management. In addition, the collaboration will include joint sales and marketing in certain regions. “Sanken and IR have worked very closely together for more than a decade," said Dr. Alex Lidow, CEO of IR. “We have been very successful with our joint product development activities, and both companies have benefited from greater access to the markets in Japan and the Americas. I see this new strategic effort as a natural progression of our relationship that can yield even grater benefits for both companies in the future."	2022-01-17 04:28:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4679439961910248, "perplexity": 9127.285845464248}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00619.warc.gz"}
https://www.physicsforums.com/threads/homework-help.42412/	# Homework Help? 1. Sep 9, 2004 ### tumbler19 I have been working on this problem for a while, and I still don't understand how to set it up for the first part of this problem: At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.2 m/s squared. At the same instant a truck, traveling with a constant speed of 9.5 m/s, overtakes and passes the automobile. (A) How far beyond the traffic signal will the automile overtake the truck? (B) How fast will the automobile be traveling at that instant? 2. Sep 9, 2004 ### Leong Use this formula, $$s=ut+\frac{1}{2}at^2$$ Initial velocity for the car is u=0. Constanst acceleration for the truck is a=0. $$s_{car}=1.1t^2$$ $$s_{truck}=9.5t$$ s is the position relative to the origin / traffic light. Part (a) When the automobile overtakes the truck, their positions relative to the traffic light is the same. so, we have $$s_{car}=s_{truck}$$. find the t when this happens and substitute back either of the equation for s above because they will be at the same position relative to traffic light. Part (b) Use this formula, $$v=u+at$$ File size: 1.2 KB Views: 30	2018-02-25 14:27:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5745185613632202, "perplexity": 847.3290094859649}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00696.warc.gz"}
https://physics.stackexchange.com/questions/215347/does-the-pressure-at-the-bottom-change-when-two-liquids-mix-and-dont-mix	# Does the pressure at the bottom change when two liquids mix and don't mix? My question is when these two liquids above mix and don't mix, does the pressure at the bottom change ? When they don't mix the pressure is 2hdg + 3hdg = 5hdg (d:density) But what happens when they mix ? The volume of the liquids is needed or not ? I saw someone says the pressure is equal either they mix or not. I'm confused, thanks in advance :) Knowing a specific density is rather handy, because it allows us to relate the volume and the mass of a substance. But to calculate the pressure, the density isn't necessary. What is needed is the mass. If the fluids mix, the volume (and therefore the density) may change, but the mass does not. The total mass remains constant. If prior to mixing, the two fluids have mass $m$, then the pressure at the bottom of the vessel due to the fluid, assuming a straight-sided vessel with area $A$: $$P = \frac{mg}A$$ Note that neither the volume nor the density appear here. As long as the mass and the shape of the container are constant, so is the pressure. • So the question was: "the pressure at the bottom", and not at any arbitrary depth ? Oct 29, 2015 at 16:19 • @FabriceNEYRET Yes yes, at the bottom Oct 30, 2015 at 20:35 • it would have been better to specify this in the question... (and it might be that the different opinions you heard was about pressure elsewhere, then). Oct 30, 2015 at 20:37 • @FabriceNEYRET I didn't see any necessities for that. Also I got my answer Nov 1, 2015 at 13:40 • It's like asking "what is the velocity of water in oceans accounting for currents" when you just want it on the ocean bottom. :-) And you have your answer, but your question reminds here forever for next students looking for answers, possibly misleading them. Nov 1, 2015 at 13:45 The pressure at a given depth below the surface is the one of the atmosphere above + increments of $\rho(z)~ dz$. So the pressure at a given depth, which corresponds to the weight of the pile above, depends of the $\rho$ values encoutered above: all at 1 then all at 3, or all at 3 then all at 1 (not very steady ;-) ), or mix of average 2, does not give the same curve $P(z)$. Besides, mixing sometime have strange effects, even if no chemical reaction occurs. E.g. alcool and water mixed occupy less volume than when separated. Which change the liquid level and the curve $P(z)$.	2022-08-20 05:11:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8365337252616882, "perplexity": 437.6994689684588}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573908.30/warc/CC-MAIN-20220820043108-20220820073108-00221.warc.gz"}
https://library.kiwix.org/datascience.stackexchange.com_en_all_2021-04/A/question/6735.html	## Can i use chi square test to remove a particular variable from the model? 3 1 I have 5 variables - $v_{1}$, $v_{2}$, $v_{3}$, $v_{4}$, and $v_{5}$. All are categorical variables. I conducted a $\chi^{2}$ (chi-square) test to understand the relationship between these variables. I could see that $v_{3}$ and $v_{4}$ have p-values of 0. I understand that these two variables are dependent on each other. Should I remove one of the variable from further analysis? What if the same $v_{3}$ and $v_{4}$ are independent of other variables, like $v_{1}$, $v_{2}$, and $v_{5}$? 1 – Aleksandr Blekh – 2015-08-10T04:50:12.297 What type of analysis do you intend to apply to the variables, this will affect the answer, as will the size of your data set. In some circumstance highly correlated variables may contain useful discriminative information. Ultimately run your analysis with the variable removed and with it left in and see how the results differ. – image_doctor – 2015-08-10T08:24:15.340 @image_doctor - Thank you very for your advice. My intention of doing chisquare test is to check for the collinearity between the variables. I got confused between the purposes of collinearity and chisquare test. Hence i raised the question. Thanks again. – Arun – 2015-08-10T10:36:31.123	2021-07-30 02:59:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6126275062561035, "perplexity": 500.3228791394407}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00465.warc.gz"}
https://askbot.fedoraproject.org/en/question/35761/how-can-i-disable-auto-gui-from-a-live-dvd-solved/?answer=35867	# How can I disable auto gui from a live dvd? [SOLVED] How can I disable automatic launching of kdm on my installed hard disk from a live session. Here's the scenario: Everything seems to be going well untill it's time for kdm to show up. Progress diamond fills up with a pretty f, The screen goes blank for a second (this is when kdm is supposed to show up) then the pretty f comes back. I can't open a terminal. Ctrl-Alt-Del reboots the machine. I try with the two previous kernals and even rescue entry in grub list, same stuff all around. I start a live session (with the same dvd I installed from months ago), mount root partition and check the boot log and everything seem fine up to the point KDE Login Manager starts. kdm log shows that it can't seem to find the nvidia screen. This I could most likely fix if get to console. So how can I keep kdm from tying up my machine? Perhaps I don't need to modify something from a live session. Is there a key sequence I can hit during bootup that will keep kdm from launching? edit retag close merge delete Sort by » oldest newest most voted Power on your computer, press esc when the diamond starts to fill and when all the text stops flying press ctrl+alt+F2 (or f3, f4, f5, f6). If the above does not work then you should try this: Power on your computer, and when it gives you a list of boot options, press e. Now use the arrow keys and find the bit that says rgba quiet and change it to text 3 then press F10. Now it will boot into text mode. more Thanks cjbayliss. I ended up adding a '3' just before 'ro' in the command line before hitting F10. That gave me a login prompt. Not sure if the position matters. Just following instructions. This all happened before I even saw your post. In case anybody is interested: It would seem that Apper or Yum or PackageKit twiddle with the grub config somehow so there is a 'blacklist=nouveau' in there even though it isn't in /etc/default/grub and it also makes the image names contain 'Schr?dingers Cat'. These both disappeared when I did my grub2-mkconfig so the nvidia drivers started conflicting with the nouveau drivers and BLAMMO! (I'm pretty sure that's the technical term for it...) I'm pretty sure the system was sitting there waiting for me to log in but had somehow lost track of the screen buffers so I couldn't see anything, even with a ctrl-alt-F2. So I added the 'blacklist=nouveau' to /etc/default/grub so that it was included when I mkconfig'd again. It would be interesting to know how the kernel update is modifying the grub config and see if there is anything else it is doing that I haven't noticed. Gordie more	2021-07-26 18:07:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17990639805793762, "perplexity": 2719.276851333078}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00140.warc.gz"}
https://nm.dev/courses/introduction-to-data-science/lessons/probability-distributions/topic/one-way-analysis-of-variance-anova/	### Introduction Analysis of Variance or ANOVA is a statistical procedure of inference of the acceptability of the null hypothesis when three or more population means are involved. ANOVA is a tool that splits the observed aggregate variability into systematic factors and random factors. ANOVA was developed by the famous statistician Ronald Fisher, and it is based on the Law of Total Variance. It uses the F-Test method to conclude the significant difference in the means of the population. ### One-Way ANOVA The one-way ANOVA determines the hypothesis acceptability of a single factor or variable only among the given number of populations. We use one-way ANOVA when we want to separately prove the significant difference of each factor or the main factor. One-way ANOVA follows a Completely Randomized Design where the treatment of the experimental units is random. The total variation is divided into two components, Treatment and Error. ### Algorithm The following are the steps to use one-way ANOVA to test the significant difference among various populations: Step 0: Initialize the null hypothesis and the alternative hypothesis. Step 1: Find the total number of observations, N. Step 2: Calculate the total of all observations, T. Step 3: Find the correction factor, T^2/N. Step 4: Calculate the total sum of squares, SST. Step 5: Calculate the column sum of squares, SSC. Step 6: Calculate the error sum of squares, SSE = SST – SSC. Step 7: Find the degrees of freedom between columns and for the error. Step 8: Calculate the mean sum of squares MSC (for column) and MSE (for error). Step 9: Calculate the F-ratio and compare it with the table value of F-test to infer whether the null hypothesis can be accepted or rejected. Note: In steps 4 and 5, we must subtract the sum of squares with the correction factor. ### Implementation Let us consider an example where various varieties of wheat are grown on various plots of land for a given duration. The values are given below: We can try to infer whether there is a significant difference between the 3 varieties of wheat using one-way ANOVA by following the algorithm mentioned above: We can initialize our null hypothesis to conclude the fact that the mean of all the wheat varieties is equal. The alternative hypothesis will conclude otherwise. The following table calculates the sum of elements as well as the sum of squares: Step 1: N = No. of rows(R) x No. of columns(C), i.e., 4 x 3 = 12. Step 2: T = 24 + 20 + 16 = 60. Step 3: Correction factor = 3600/12 = 300. Step 4: SST = 158 + 108 + 66 – 300 = 32. Step 5: SSC = (24)^2/4 + (20)^2/4 +(16)^2/4 – 300 = 8. Step 6: SSE = 32 – 8 = 24. Step 7: Degrees of freedom within the column = N – C = 12 – 3 = 9. Degrees of freedom between the columns = C – 1 = 2. We can tabulate the answers yielded to easily continue with step 8 and 9. From the table above: Step 8: MSC = 4 and MSE = 24/9. Step 9: F-ratio = 1.5. The table value of df = (9,2) at 5% significance is 4.2565. The F-test table values for various levels of significance can be found here. Since the F-ratio is less than the table value, we accept the null hypothesis. This means that there is no significant difference between the means of the yields of the variety of wheat crops among the various plots of land. Let us try to code the example above by creating a function that can return the final F-ratio value. fun one_anova(arr: Array): Double { val c = arr.size val r = arr[0].size val n = r * c var sum = 0.0 var sst = 0.0 var ssc = 0.0 var f = 0.0 val t = DoubleArray(c) val sqarr = DoubleArray(c) for (i in 0 until c) { for (j in 0 until r) { sum += arr[i][j] t[i] += arr[i][j] sqarr[i] += arr[i][j] * arr[i][j] } } val cfac = sum * sum / n for (i in sqarr.indices) { sst += sqarr[i] ssc += t[i] * t[i] } sst -= cfac ssc = ssc / r - cfac val sse = sst - ssc val msc = ssc / (c - 1) val mse = sse / (n - c) f = if (msc > mse) { msc / mse } else { mse / msc } return f } The table values can be taken as an input using 2-dimensional arrays wherein each 1-dimensional array represents the values of a population. Also, the table value of F for the given degrees of freedom can be found using the statistical tables. val arr = arrayOf( doubleArrayOf(6.0, 7.0, 3.0, 8.0), doubleArrayOf(5.0, 5.0, 3.0, 7.0), doubleArrayOf(5.0, 4.0, 3.0, 4.0)) val exp = 4.26 val ans = one_anova(arr) println("F value: "+ans) if (ans > exp) { println("Significant difference") } else { println("No significant difference") } Output: F value: 1.5 No significant difference	2022-08-15 12:26:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5398730039596558, "perplexity": 1511.855138144612}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572174.8/warc/CC-MAIN-20220815115129-20220815145129-00201.warc.gz"}