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https://mcwdn.org/Decimals/DivDec.html | 1,713,853,299,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00549.warc.gz | 362,615,293 | 2,417 | # Dividing Decimals
The division of decimals, like multiplying decimals, can be tricky. Again, less can be more and more less with dividing decimals. The other important precaution is that the placement of the decimal point's rules are different in different cases.
Here is a simple example:
# Two tenths
In this case, the quotient or answer has the same number of decimal places as the dividend or the number that is being divided up.
# Twenty groups of two hundredths.
In the example above, the four tenths is divided into groups of two hundredths. There are twenty groups of two hundredths in four tenths, which can also be called forty hundredths. In order to do this as a straight division problem, one needs to move the decimal point in the divisor to the right two places and then to the same to the dividend before dividing. Then the answer comes out correctly.
Dividing Decimals Videos
Place Value Adding Decimals Subtracting Decimals Multiplying Decimals Dividing Decimals Decimals & Money Metric Measurement Percents | 214 | 1,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-18 | latest | en | 0.893628 |
https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/8/lesson/8.1.6/problem/8-93 | 1,716,458,616,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00211.warc.gz | 259,518,032 | 17,718 | ### Home > A2C > Chapter 8 > Lesson 8.1.6 > Problem8-93
8-93.
You have seen that you can calculate values of the sine function using right triangles formed by a radius of the unit circle. Values of $θ$ that result in $30^\circ − 60^\circ − 90^\circ$ or $45^\circ − 45^\circ − 90^\circ$ triangles are used frequently on exercises and tests because their sines and cosines can be found exactly, without using a calculator. You should learn to recognize these values quickly and easily. The same is true for values of cosθ and sinθ that correspond to the $x$- and $y$-intercepts of the unit circle.
The central angles that correspond to these “special” values of $x$ are $30^\circ, 45^\circ, 60^\circ, 90^\circ, 120^\circ, 135^\circ, 150^\circ, 180^\circ, 210^\circ, 225^\circ, 240^\circ, 270^\circ, 300^\circ, 315^\circ, \text{and}\ 330^\circ$. What these angles have in common is that they are all multiples of $30^{\circ}\ \text{or}\ 45^{\circ}$, and some of them are also multiples of $60^\circ\ \text{or}\ 90^\circ$.
Copy and complete a table like the one below for all special angles between $0^{\circ}\ \text{and}\ 360^{\circ}$.
Degrees Radians $0$ $30$ $45$ $60$ $90$ $120$ $0$ $\frac { \pi } { 6 }$
Recall that $180^{^{\circ}}$ is $\pi$ radians. Since $45$ is $\frac{1}{4}$ of $180$, $45^{^{\circ}}$ is $\frac{1}{4}$ of $\pi$, or $\frac{\pi}{4}$.
All of the 'special' values of $x$ are multiples of either $30^{\circ}\ \text{or}\ 4^{\circ}$.
Knowing just these two values allows you to easily find the rest.
For Example, $225 \div 45 = 5. \text{ So } 225° = 5 \left( \frac{\pi}{4} \right) = \frac{5 \pi}{4} .$
Degrees Radians $0$ $30$ $45$ $60$ $90$ $120$ $180$ $270$ $0$ $\frac{\pi}{6}$ $\frac{\pi}{4}$ $\pi$ $\frac{3\pi}{2}$ | 607 | 1,745 | {"found_math": true, "script_math_tex": 43, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-22 | latest | en | 0.662138 |
https://www.pousadacaminhodascandeias.com.br/ac06p2yl/c16f0f-stirling-formula-in-physics | 1,618,424,161,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077843.17/warc/CC-MAIN-20210414155517-20210414185517-00553.warc.gz | 1,068,937,584 | 17,348 | ## stirling formula in physics
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 706.4 938.5 877 781.8 754 843.3 815.5 877 815.5 | δ n | 0 we have, by Lemmas 4 and 5 , Taking n= 10, log(10!) Stirling's formula synonyms, Stirling's formula pronunciation, Stirling's formula translation, English dictionary definition of Stirling's formula. David Mermin—one of my favorite writers among physicists—has much more to say about Stirling’s approximation in his American Journal of Physics article “Stirling’s Formula!” (leave it to Mermin to work an exclamation point into his title). /FontDescriptor 17 0 R endobj �L*���q@*�taV��S��j�����saR��h} ��H�������Z����1=�U�vD�W1������RR3f�� 444.4 611.1 777.8 777.8 777.8 777.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 523.8 585.3 585.3 462.3 462.3 339.3 585.3 585.3 708.3 585.3 339.3 938.5 859.1 954.4 endobj It is a good approximation, leading to accurate results even for small values of n. It is named after James Stirling, though it was first stated by Abraham de Moivre. << /LastChar 196 We will obtain an asymptotic expansion of γq(z) as |z| → ∞ in the right halfplane, which is uniform as q → 1, and when q → 1, the asymptotic expansion becomes Stirling's formula. /Resources<< /Type/Font Stirling’s formula can also be expressed as an estimate for log(n! << Stirling's approximation is also useful for approximating the log of a factorial, which finds application in evaluation of entropy in terms of multiplicity, as in the Einstein solid. ˇ15:104 and the logarithm of Stirling’s approxi-mation to 10! << Stirling’s Formula Steven R. Dunbar Supporting Formulas Stirling’s Formula Proof Methods Proofs using the Gamma Function ( t+ 1) = Z 1 0 xte x dx The Gamma Function is the continuous representation of the factorial, so estimating the integral is natural. The log of n! If the accuracy of ln( f(n) ) is in terms of abs( trueValue - estimatedValue ) and the desired accuracy is in terms of percentage, I think this should be possible. (/) = que l'on trouve souvent écrite ainsi : ! – Cheers and hth.- Alf Oct 15 '10 at 0:47 762.8 642 790.6 759.3 613.2 584.4 682.8 583.3 944.4 828.5 580.6 682.6 388.9 388.9 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 777.8 500 777.8 500 530.9 The version of the formula typically used in applications is ln n ! >> 0 0 0 0 0 0 0 0 0 0 777.8 277.8 777.8 500 777.8 500 777.8 777.8 777.8 777.8 0 0 777.8 1138.9 1138.9 892.9 329.4 1138.9 769.8 769.8 1015.9 1015.9 0 0 646.8 646.8 769.8 endobj At least two of these are named after James Stirling: the so-called Stirling approximation should probably be called the “first” Stirling approximation, since it can be seen as the first term in the Stirling series. /FirstChar 33 \approx (n+\frac{1}{2})\ln{n} – n + \frac{1}{2}\ln{2\pi}$$. We begin by calculating the integral (where ) using integration by parts. >> For every operator T ∈ L (ℝ n ) with s | n / 2 | ( T ) ⩾ 1 and every random space Y n ∈ X n . >> 756 339.3] /Subtype/Type1 12 0 obj /FirstChar 33 639.7 565.6 517.7 444.4 405.9 437.5 496.5 469.4 353.9 576.2 583.3 602.5 494 437.5 500 500 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 625 833.3 /BaseFont/QUMFTV+CMSY10 µ. /BaseFont/BPNFEI+CMR10 and other estimates, some cruder, some more refined, are developed along surprisingly elementary lines. = n ln n − n + O {\displaystyle \ln n!=n\ln n-n+O}, or, by changing the base of the logarithm, log 2 n ! /LastChar 196 892.9 1138.9 892.9] Stirling's Factorial Formula: n! 750 758.5 714.7 827.9 738.2 643.1 786.2 831.3 439.6 554.5 849.3 680.6 970.1 803.5 /Subtype/Type1 /Matrix[1 0 0 1 -6 -11] Stirling’s formula is also used in applied mathematics. is important in computing binomial, hypergeometric, and other probabilities. ∼ où le nombre e désigne la base de l'exponentielle. 388.9 1000 1000 416.7 528.6 429.2 432.8 520.5 465.6 489.6 477 576.2 344.5 411.8 520.6 ?ҋ���O���:�=�r��� ���?�{�\��4�z��?>�?��*k�{��@�^�5�xW����^e�֕�������^���U1��B� /LastChar 196 for n < 0. is approximated by. Trouble with Stirling's formula Thread starter stepheckert; Start date Mar 23, 2013; Mar 23, 2013 #1 stepheckert . Advanced Physics Homework Help. is. In Abraham de Moivre. /Subtype/Form 892.9 892.9 892.9 892.9 892.9 892.9 892.9 892.9 892.9 892.9 892.9 1138.9 1138.9 892.9 /Type/Font 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 277.8 777.8 472.2 472.2 777.8 /Widths[622.5 466.3 591.4 828.1 517 362.8 654.2 1000 1000 1000 1000 277.8 277.8 500 Article copyright remains as specified within the article. For instance, Stirling computes the area under the Bell Curve: Z +∞ −∞ e−x 2/2 dx = √ 2π. Physics 2053 Laboratory The Stirling Engine: The Heat Engine Under no circumstances should you attempt to operate the engine without supervision: it may be damaged if mishandled. /FontDescriptor 26 0 R /Subtype/Type1 /FirstChar 33 323.4 354.2 600.2 323.4 938.5 631 569.4 631 600.2 446.4 452.6 446.4 631 600.2 815.5 /FirstChar 33 /FontDescriptor 8 0 R n! 863.9 786.1 863.9 862.5 638.9 800 884.7 869.4 1188.9 869.4 869.4 702.8 319.4 602.8 ����B��i��%����aUi��Si�Ō�M{�!�Ãg�瘟,�K��Ĥ�T,.qN>�����sq������f����Օ 600.2 600.2 507.9 569.4 1138.9 569.4 569.4 569.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 693.8 954.4 868.9 777.8 777.8 1000 1000 777.8 777.8 1000 777.8] 465 322.5 384 636.5 500 277.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 << 575 1041.7 1169.4 894.4 319.4 575] /LastChar 196 The factorial function n! It is designed such that the two pistons operate a quarter cycle out of phase with each other so that when the heated piston is all the way out, the cooled piston is moving in, and the same heated/cooled air is shared between the two pistons. n! /Widths[791.7 583.3 583.3 638.9 638.9 638.9 638.9 805.6 805.6 805.6 805.6 1277.8 Selecting this option will search all publications across the Scitation platform, Selecting this option will search all publications for the Publisher/Society in context, The Journal of the Acoustical Society of America, Laboratory of Atomic and Solid State Physics, Cornell University, Ithaca, New York 14853. The Stirling Engine uses cyclic compression and expansion of air at different temperatures to convert heat energy into mechanical work. Shroeder gives a numerical evaluation of the accuracy of the approximations . /Name/F6 La formule de Stirling, du nom du mathématicien écossais James Stirling, donne un équivalent de la factorielle d'un entier naturel n quand n tend vers l'infini : → + ∞! Download Stirling Formula along with the complete list of important formulas used in maths, physics & chemistry. 2 π n n + 1 2 e − n ≤ n! 1135.1 818.9 764.4 823.1 769.8 769.8 769.8 769.8 769.8 708.3 708.3 523.8 523.8 523.8 >> 833.3 1444.4 1277.8 555.6 1111.1 1111.1 1111.1 1111.1 1111.1 944.4 1277.8 555.6 1000 There are quite a few known formulas for approximating factorials and the logarithms of factorials. Therefore, by the Hadamard inequality and the Stirling formula (recall that vol B 1 K = 2 K / k! /Name/Im1 x��\��%�u��+N87����08�4��H�=��X����,VK�!�� �{5y�E���:�ϯ��9�.�����? 319.4 958.3 638.9 575 638.9 606.9 473.6 453.6 447.2 638.9 606.9 830.6 606.9 606.9 Derive the Stirling formula:$$\ln(n!) 493.6 769.8 769.8 892.9 892.9 523.8 523.8 523.8 708.3 892.9 892.9 892.9 892.9 0 0 /FirstChar 33 611.1 798.5 656.8 526.5 771.4 527.8 718.7 594.9 844.5 544.5 677.8 762 689.7 1200.9 1000 1000 1055.6 1055.6 1055.6 777.8 666.7 666.7 450 450 450 450 777.8 777.8 0 0 C'est Abraham de Moivre [1] qui a initialement démontré la formule suivante : ! ): (1.1) log(n!) ∼ 2 π n n {\displaystyle n\,!\sim {\sqrt {2\pi n}}\,\left^{n}} où le nombre e désigne la base de l'exponentielle. vol B ⩽ ∑ σ vol B σ ⩽ ( [ ( 1 + κ ) k ] k ) ( 2 K ) k k ! /Name/F4 638.9 638.9 958.3 958.3 319.4 351.4 575 575 575 575 575 869.4 511.1 597.2 830.6 894.4 endobj 339.3 585.3 585.3 585.3 585.3 585.3 585.3 585.3 585.3 585.3 585.3 585.3 585.3 339.3 The factorial function n! 18 0 obj >> /LastChar 196 500 500 611.1 500 277.8 833.3 750 833.3 416.7 666.7 666.7 777.8 777.8 444.4 444.4 /BaseFont/OLROSO+CMR7 Selecting this option will search the current publication in context. In mathematics, Stirling's approximation is an approximation for factorials. /Subtype/Type1 1444.4 555.6 1000 1444.4 472.2 472.2 527.8 527.8 527.8 527.8 666.7 666.7 1000 1000 /Type/Font 277.8 500] Here is a simple derivation using an analogy with the Gaussian distribution: The Formula. ≈ √(2π) × n (n+1/2) × e -n Where, n = Number of elements 323.4 877 538.7 538.7 877 843.3 798.6 815.5 860.1 767.9 737.1 883.9 843.3 412.7 583.3 /BaseFont/FLERPD+CMMI10 You can derive better Stirling-like approximations of the form $$n! /BBox[0 0 2384 3370] n! 0 0 0 0 0 0 0 615.3 833.3 762.8 694.4 742.4 831.3 779.9 583.3 666.7 612.2 0 0 772.4 /Filter/FlateDecode /Name/F1 << Example 1.3. Visit http://ilectureonline.com for more math and science lectures! /Type/Font Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It is used in probability and statistics, algorithm analysis and physics. /FirstChar 33 = n log 2 n − n … ( n / e) n √ (2π n ) Collins English Dictionary. 506.3 632 959.9 783.7 1089.4 904.9 868.9 727.3 899.7 860.6 701.5 674.8 778.2 674.6 Appendix to III.2: Stirling’s formula Statistical Physics Lecture J. Fabian The Stirling formula gives an approximation to the factorial of a large number,N À1. Stirlings Factorial formula. 820.5 796.1 695.6 816.7 847.5 605.6 544.6 625.8 612.8 987.8 713.3 668.3 724.7 666.7 Calculation using Stirling's formula gives an approximate value for the factorial function n! = nlogn n+ 1 2 logn+ 1 2 log(2ˇ) + "n; where "n!0 as n!1. Stirling's formula [in Japanese] version 0.1.1 (57.9 KB) by Yoshihiro Yamazaki. 777.8 694.4 666.7 750 722.2 777.8 722.2 777.8 0 0 722.2 583.3 555.6 555.6 833.3 833.3 /BaseFont/YYXGVV+CMEX10 n ( n / e ) n when he was studying the Gaussian distribution and the central limit theorem. /Type/Font /Widths[277.8 500 833.3 500 833.3 777.8 277.8 388.9 388.9 500 777.8 277.8 333.3 277.8 323.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 323.4 323.4 This can also be used for Gamma function. stream Stirling Formula. Then, use Stirling's formula to find \lim_{n\to\infty} \frac{a_{n}}{\left(\frac{n}{e}\right)... Stack Exchange Network. 575 575 575 575 575 575 575 575 575 575 575 319.4 319.4 350 894.4 543.1 543.1 894.4 n a formula giving the approximate value of the factorial of a large number n, as n ! 24 0 obj This option allows users to search by Publication, Volume and Page. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 458.3 458.3 416.7 416.7 /FontDescriptor 23 0 R Stirling’s approximation to n!! \le e\ n^{n+{\small\frac12}}e^{-n}. In this thesis, we shall give a new probabilistic derivation of Stirling's formula. is approximately 15.096, so log(10!) In its simple form it is, N!…. /FontDescriptor 11 0 R ⩽ ( c 2 K k ) k . /Type/XObject /FirstChar 33 ∼ 2 π n (e n ) n. Furthermore, for any positive integer n n n, we have the bounds. can be computed directly, multiplying the integers from 1 to n, or person can look up factorials in some tables. /Type/Font /Widths[323.4 569.4 938.5 569.4 938.5 877 323.4 446.4 446.4 569.4 877 323.4 384.9 and its Stirling approximation di er by roughly .008. 869.4 818.1 830.6 881.9 755.6 723.6 904.2 900 436.1 594.4 901.4 691.7 1091.7 900 /Name/F3 >> /Type/Font = √(2 π n) (n/e) n. /Widths[719.7 539.7 689.9 950 592.7 439.2 751.4 1138.9 1138.9 1138.9 1138.9 339.3 /Name/F5 339.3 892.9 585.3 892.9 585.3 610.1 859.1 863.2 819.4 934.1 838.7 724.5 889.4 935.6 endobj In James Stirling …of what is known as Stirling’s formula, n! ∼ 2 π n (n e) n. n! /Type/Font 9 0 obj Note that xte x has its maximum value at x= t. That is, most of the value of the Gamma Function comes from values 1074.4 936.9 671.5 778.4 462.3 462.3 462.3 1138.9 1138.9 478.2 619.7 502.4 510.5 ≤ e n n + 1 2 e − n. \sqrt{2\pi}\ n^{n+{\small\frac12}}e^{-n} \le n! a formula giving the approximate value of the factorial of a large number n, as n! but the last term may usually be neglected so that a working approximation is. fq[����4ۻ!X69 �F�����9#�S4d�w�b^��s��7Nj��)�sK���7�%,/q���0 1277.8 811.1 811.1 875 875 666.7 666.7 666.7 666.7 666.7 666.7 888.9 888.9 888.9 /BaseFont/JRVYUL+CMMI7 To sign up for alerts, please log in first. /Subtype/Type1 << \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n. 0 0 0 0 0 0 691.7 958.3 894.4 805.6 766.7 900 830.6 894.4 830.6 894.4 0 0 830.6 670.8 680.6 777.8 736.1 555.6 722.2 750 750 1027.8 750 750 611.1 277.8 500 277.8 500 277.8 277.8 500 555.6 444.4 555.6 444.4 305.6 500 555.6 277.8 305.6 527.8 277.8 833.3 555.6 500 555.6 527.8 391.7 394.4 388.9 555.6 527.8 722.2 527.8 527.8 444.4 500 1000 500 /ProcSet[/PDF/Text] If n is not too large, then n! /FontDescriptor 14 0 R << 31 0 obj Copyright © HarperCollins Publishers. /BaseFont/ARTVRV+CMSY7 endobj 277.8 305.6 500 500 500 500 500 750 444.4 500 722.2 777.8 500 902.8 1013.9 777.8 888.9 888.9 888.9 888.9 666.7 875 875 875 875 611.1 611.1 833.3 1111.1 472.2 555.6 588.6 544.1 422.8 668.8 677.6 694.6 572.8 519.8 668 592.7 662 526.8 632.9 686.9 713.8 Stirling's formula definition is - a formula ... that approximates the value of the factorial of a very large number n. 646.5 782.1 871.7 791.7 1342.7 935.6 905.8 809.2 935.9 981 702.2 647.8 717.8 719.9 /FontDescriptor 29 0 R 511.1 575 1150 575 575 575 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /Name/F2 874 706.4 1027.8 843.3 877 767.9 877 829.4 631 815.5 843.3 843.3 1150.8 843.3 843.3 15 0 obj 21 0 obj Stirling Formula is provided here by our subject experts. /Subtype/Type1 noun. Stirling's Formula. 777.8 777.8 1000 500 500 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 Stirling’s formula was discovered by Abraham de Moivre and published in “Miscellenea Analytica” in 1730. 843.3 507.9 569.4 815.5 877 569.4 1013.9 1136.9 877 323.4 569.4] 877 0 0 815.5 677.6 646.8 646.8 970.2 970.2 323.4 354.2 569.4 569.4 569.4 569.4 569.4 It generally does not, and Stirling's formula is a perfect example of that. 597.2 736.1 736.1 527.8 527.8 583.3 583.3 583.3 583.3 750 750 750 750 1044.4 1044.4 791.7 777.8] /Widths[1000 500 500 1000 1000 1000 777.8 1000 1000 611.1 611.1 1000 1000 1000 777.8 /Font 32 0 R n! << 585.3 831.4 831.4 892.9 892.9 708.3 917.6 753.4 620.2 889.5 616.1 818.4 688.5 978.6 %PDF-1.2 594.7 542 557.1 557.3 668.8 404.2 472.7 607.3 361.3 1013.7 706.2 563.9 588.9 523.6 It makes finding out the factorial of larger numbers easy. Visit Stack Exchange. The Stirling formula or Stirling’s approximation formula is used to give the approximate value for a factorial function (n!). Basic Algebra formulas list online. La formule de Stirling, du nom du mathématicien écossais James Stirling, donne un équivalent de la factorielle d'un entier naturel n quand n tend vers l'infini: lim n → + ∞ n ! Stirling's formula in British English. The aim is to shed some light on why these approximations work so well, for students using them to study entropy and irreversibility in such simple statistical models as might be examined in a general education physics course. >> /Length 7348 Stirling's formula is one of the most frequently used results from asymptotics. 275 1000 666.7 666.7 888.9 888.9 0 0 555.6 555.6 666.7 500 722.2 722.2 777.8 777.8 /Subtype/Type1 >> 298.4 878 600.2 484.7 503.1 446.4 451.2 468.8 361.1 572.5 484.7 715.9 571.5 490.3 /Widths[1138.9 585.3 585.3 1138.9 1138.9 1138.9 892.9 1138.9 1138.9 708.3 708.3 1138.9 /Widths[350 602.8 958.3 575 958.3 894.4 319.4 447.2 447.2 575 894.4 319.4 383.3 319.4 1111.1 1511.1 1111.1 1511.1 1111.1 1511.1 1055.6 944.4 472.2 833.3 833.3 833.3 833.3 /LastChar 196 << 27 0 obj /BaseFont/SHNKOC+CMBX10 ��=8�^�\I�����Njx���U��!\�iV���X'&. Our motivation comes from sampling randomly with replacement from a group of n distinct alternatives. /Name/F8 797.6 844.5 935.6 886.3 677.6 769.8 716.9 0 0 880 742.7 647.8 600.1 519.2 476.1 519.8 Please show the declarations of exp and num.Especially exp.Without having checked Stirling's formula, there is also the possibility that you've exchanegd exp and num in the first call to pow-- perhaps you could also provide the formula? 2 π n n = 1 {\displaystyle \lim _{n\to +\infty }{n\,! 692.5 323.4 569.4 323.4 569.4 323.4 323.4 569.4 631 507.9 631 507.9 354.2 569.4 631 570 517 571.4 437.2 540.3 595.8 625.7 651.4 277.8] 666.7 666.7 666.7 666.7 611.1 611.1 444.4 444.4 444.4 444.4 500 500 388.9 388.9 277.8 >> >> = \sqrt{2 \pi n} \left(\dfrac{n}{e} \right)^n \left(1 + \dfrac{a_1}n + \dfrac{a_2}{n^2} + \dfrac{a_3}{n^3} + \cdots \right)$$ using Abel summation technique (For instance, see here), where $$a_1 = \dfrac1{12}, a_2 = \dfrac1{288}, a_3 = -\dfrac{139}{51740}, a_4 = - \dfrac{571}{2488320}, \ldots$$ The hard part in Stirling's formula is … \over {\sqrt {2\pi n}}\;\left^{n}}=1} que l'on trouve souvent écrite ainsi: n ! /FirstChar 33 /FormType 1 /Subtype/Type1 endobj In this video I will explain and calculate the Stirling's approximation. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 892.9 339.3 892.9 585.3 892.9 585.3 892.9 892.9 892.9 892.9 0 0 892.9 892.9 892.9 1138.9 585.3 585.3 892.9 d�=�-���U�3�2 l �Û �d"#�4�:u}�����U�{ ≅ (n / e) n Square root of √ 2πn, although the French mathematician Abraham de Moivre produced corresponding results contemporaneously. endobj Website © 2020 AIP Publishing LLC. 530.4 539.2 431.6 675.4 571.4 826.4 647.8 579.4 545.8 398.6 442 730.1 585.3 339.3 /Name/F7 If you need an account, please register here. /FontDescriptor 20 0 R 472.2 472.2 472.2 472.2 583.3 583.3 0 0 472.2 472.2 333.3 555.6 577.8 577.8 597.2 /LastChar 196 1 Stirling’s Approximation(s) for Factorials. 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Too large, then n! ) '10 at 0:47 Learn about this topic these. # XA0 ; & # XA0 ; Stirling & # stirling formula in physics ; s approximation formula is also in... | 10,027 | 27,216 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-17 | latest | en | 0.803837 |
https://sarunw.com/posts/getting-number-of-days-between-two-dates/?utm_campaign=AppCoda%20Weekly&utm_medium=email&utm_source=Revue%20newsletter | 1,607,053,830,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00379.warc.gz | 450,529,205 | 33,058 | There are a few variations when dealing with counting days. So, you need to ask yourself the definition of a day beforehand.
## What is your definition of a day? #
A day can be interpreted in two ways.
1. Day as time pass midnight
This is the kind of day that we use in our daily life. We consider it another day once the time passes midnight. For example, October 29, 23:59 and October 30, 00:01 consider one day different even it is only two minutes apart.
We don't care about time in this scenario.
2. Day as 24 hours time unit
This is a kind of day in a term of a unit of time. One day is 24 hours. For the same example, October 29, 23:59 and October 30, 00:01 consider two minutes different, not a day different.
In this kind of implementation, we care about time. You can use this when you want to count an elapsed time between two dates. That is the only scenario when I think this kind of counting makes sense.
## Do you include a start date? #
This is also an important question to ask before coding. There are scenarios that make sense for both of them.
1. Not include a start date
You usually have a start date in the counting when the number of days between two dates represents a remaining time between today until destination date. For example, if today is October 29 and your payment due date is October 30, people expected the number of days between two dates to be one (due date is tomorrow). So, in this kind of scenario, you don't include the start date in the counting.
2. Including a start date
One scenario that you need to include a start date is when you book a hotel room. If you book a hotel between October 29 and October 30, you would expect something that says two days and one night. You need to include a start date in this case.
After you know all the variation for counting the number of days between dates, you can pick the one that suits you. And the following are how you implement all of them.
## Number of days pass midnight, not including a start date #
``extension Calendar { func numberOfDaysBetween(_ from: Date, and to: Date) -> Int { let fromDate = startOfDay(for: from) // <1> let toDate = startOfDay(for: to) // <2> let numberOfDays = dateComponents([.day], from: fromDate, to: toDate) // <3> return numberOfDays.day! }}``
<1>, <2> As we mentioned before, counting day as a time pass midnight doesn't consider time components, so we use `startOfDay(for:)` to convert that date to the beginning of that day (00:00:00).
<3> Then we get a day difference between those two dates.
We create this method as an extension of `Calendar` because day always needs a time zone and calendar to interpret. We already discussed this in the previous post, Understanding Date and DateComponents. I encourage you to check it out if you haven't read it yet.
The midnight, in this case, will be based on a calendar that calls this function.
## Number of days pass midnight, including a start date #
This implementation is the same as the previous one. The only difference is we add one more day to the result.
``extension Calendar { func numberOfDaysBetween(_ from: Date, and to: Date) -> Int { let fromDate = startOfDay(for: from) let toDate = startOfDay(for: to) let numberOfDays = dateComponents([.day], from: fromDate, to: toDate) return numberOfDays.day! + 1 // <1> }}``
<1> We add another day to the result.
## Number of 24 hours days, not including a start date #
The implementation is similar to our previous example, but we get a day component from dates without stripping out time component.
``func numberOfDaysBetween(_ from: Date, and to: Date) -> Int { let numberOfDays = dateComponents([.day], from: from, to: to) return numberOfDays.day!}``
## Number of 24 hours days, including a start date #
I can't think of any scenario where this implementation makes sense, but I will put it here for completeness.
``func numberOf24DaysBetween(_ from: Date, and to: Date) -> Int { let numberOfDays = dateComponents([.day], from: from, to: to) return numberOfDays.day! + 1}`` | 966 | 4,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-50 | latest | en | 0.971861 |
https://gemseo.readthedocs.io/en/develop/_modules/gemseo.uncertainty.distributions.scipy.joint.html | 1,716,901,479,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00430.warc.gz | 228,070,399 | 9,270 | gemseo / uncertainty / distributions / scipy
# joint module¶
The SciPy-based joint probability distribution.
SPJointDistribution is a BaseJointDistribution based on the SciPy library.
Warning
For the moment, there is no copula that can be used with SPJointDistribution; if you want to introduce dependency between random variables, please consider OTJointDistribution.
class gemseo.uncertainty.distributions.scipy.joint.SPJointDistribution(distributions, copula=None, variable='')[source]
The SciPy-based joint probability distribution.
Parameters:
• distributions (Sequence[SPDistribution]) – The marginal distributions.
• copula (None) – A copula distribution defining the dependency structure between random variables; if None, consider an independent copula.
• variable (str) –
The name of the variable, if any; otherwise, concatenate the names of the random variables defined by distributions.
By default it is set to “”.
Raises:
NotImplementedError – When the copula is not None.
compute_cdf(vector)[source]
Evaluate the cumulative density function (CDF).
Evaluate the CDF of the components of the random variable for a given realization of this random variable.
Parameters:
vector (Iterable[float]) – A realization of the random variable.
Returns:
The CDF values of the components of the random variable.
Return type:
ndarray
compute_inverse_cdf(vector)[source]
Evaluate the inverse of the cumulative density function (ICDF).
Parameters:
vector (Iterable[float]) – A vector of values comprised between 0 and 1 whose length is equal to the dimension of the random variable.
Returns:
The ICDF values of the components of the random variable.
Return type:
ndarray
compute_samples(n_samples=1)
Sample the random variable.
Parameters:
n_samples (int) –
The number of samples.
By default it is set to 1.
Returns:
The samples of the random variable,
The number of columns is equal to the dimension of the variable and the number of lines is equal to the number of samples.
Return type:
ndarray
plot(index=0, show=True, save=False, file_path='', directory_path='', file_name='', file_extension='')
Plot both probability and cumulative density functions for a given component.
Parameters:
• index (int) –
The index of a component of the random variable.
By default it is set to 0.
• save (bool) –
If True, save the figure.
By default it is set to False.
• show (bool) –
If True, display the figure.
By default it is set to True.
• file_path (str | Path) –
The path of the file to save the figures. If the extension is missing, use file_extension. If empty, create a file path from directory_path, file_name and file_extension.
By default it is set to “”.
• directory_path (str | Path) –
The path of the directory to save the figures. If empty, use the current working directory.
By default it is set to “”.
• file_name (str) –
The name of the file to save the figures. If empty, use a default one generated by the post-processing.
By default it is set to “”.
• file_extension (str) –
A file extension, e.g. 'png', 'pdf', 'svg', … If empty, use a default file extension.
By default it is set to “”.
Returns:
The figure.
Return type:
Figure
plot_all(show=True, save=False, file_path='', directory_path='', file_name='', file_extension='')
Plot both probability and cumulative density functions for all components.
Parameters:
• save (bool) –
If True, save the figure.
By default it is set to False.
• show (bool) –
If True, display the figure.
By default it is set to True.
• file_path (str | Path) –
The path of the file to save the figures. If the extension is missing, use file_extension. If empty, create a file path from directory_path, file_name and file_extension.
By default it is set to “”.
• directory_path (str | Path) –
The path of the directory to save the figures. If empty, use the current working directory.
By default it is set to “”.
• file_name (str) –
The name of the file to save the figures. If empty, use a default one generated by the post-processing.
By default it is set to “”.
• file_extension (str) –
A file extension, e.g. 'png', 'pdf', 'svg', … If empty, use a default file extension.
By default it is set to “”.
Returns:
The figures.
Return type:
list[Figure]
DEFAULT_VARIABLE_NAME: Final[str] = 'x'
The default name of the variable.
JOINT_DISTRIBUTION_CLASS: ClassVar[type[BaseJointDistribution] | None] = None
The class of the joint distribution associated with this distribution, if any.
dimension: int
The number of dimensions of the random variable.
distribution: type
The probability distribution of the random variable.
distribution_name: str
The name of the probability distribution.
marginals: list[type]
The marginal distributions of the components of the random variable.
math_lower_bound: ndarray
The mathematical lower bound of the random variable.
math_upper_bound: ndarray
The mathematical upper bound of the random variable.
property mean: ndarray
The analytical mean of the random variable.
num_lower_bound: ndarray
The numerical lower bound of the random variable.
num_upper_bound: ndarray
The numerical upper bound of the random variable.
parameters: tuple[Any] | dict[str, Any]
The parameters of the probability distribution.
property range: list[ndarray]
The numerical range.
The numerical range is the interval defined by the lower and upper bounds numerically reachable by the random variable.
Here, the numerical range of the random variable is defined by one array for each component of the random variable, whose first element is the lower bound of this component while the second one is its upper bound.
property standard_deviation: ndarray
The analytical standard deviation of the random variable.
standard_parameters: dict[str, str] | None
The standard representation of the parameters of the distribution, used for its string representation.
property support: list[ndarray]
The mathematical support.
The mathematical support is the interval defined by the theoretical lower and upper bounds of the random variable.
Here, the mathematical range of the random variable is defined by one array for each component of the random variable, whose first element is the lower bound of this component while the second one is its upper bound.
transformation: str
The transformation applied to the random variable, e.g. ‘sin(x)’.
variable_name: str
The name of the random variable. | 1,379 | 6,459 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.698823 |
https://docs.gomafem.com/material_file/mechanical_and_constitutive/adaptive_viscosity_scaling.html | 1,726,088,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00403.warc.gz | 196,617,064 | 13,421 | ```Adaptive Viscosity Scaling = CONSTANT <float>
```
## Description / Usage#
This optional card is used to specify the adaptive viscosity scaling and the ε parameter associated with its usage (see theory section below). It requires one floating point number that scales the term. Definitions of the input parameters are as follows:
CONSTANT Name of the model for the adaptive viscosity scaling. - value of ε scaling parameter.
## Examples#
The following is a sample card that sets the adaptive viscosity scaling to 0.5:
```Adaptive Viscosity Scaling = CONSTANT 0.5
```
## Technical Discussion#
The momentum equation is modified with the addition of a numerical adaptive viscosity to help maintain the elliptic character of the equation set as stress and velocity gradient increase
where ηs is the solvent viscosity and ηp is the polymer viscosity. If we set the adaptive viscosity to zero (ηa= 0), we obtain the Standard EVSS Formulation of Guenette and Fortin (1995). For adaptive viscosity, we use the following definition
with 0<ε<1.
The equations are unchanged in the limit of h, the element size, going to zero.
Please see the viscoelastic tutorial for a discussion of usage for the adaptive viscosity scaling. The papers by Sun, et. al. (1996) and Sun, et. al (1999) provide a good discussion of the theory behind its usage. CRMPC presentations by R.R. Rao demonstrates its usefulness for Goma calculations.
## References#
GT-014.1: Tutorial for Running Viscoelastic Flow Problems with GOMA, June 21, 2000, R. R. Rao
Guenette, R. and M. Fortin, “A New Mixed Finite Element Method for Computing Viscoelastic Flows,” J. Non-Newtonian Fluid Mech., 60, 27-52 (1995).
Sun, J., N. Phan-Thien, R. I. Tanner, “An Adaptive Viscoelastic Stress Splitting Scheme and Its Applications: AVSS/SI and AVSS/SUPG,” J. Non-Newtonian Fluid Mech., 65, 75-91 (1996).
Sun, J., M. D. Smith, R. C. Armstrong, R. A. Brown, “Finite Element Method for Viscoelastic Flows Bases on the Discrete Adaptive Viscoelastic Stress Splitting and the Discontinuous Galerkin Method: DAVSS-G/DG,” J. Non-Newtonian Fluid Mech., 86, 281-307 (1999). | 534 | 2,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-38 | latest | en | 0.802094 |
http://mathforum.org/library/drmath/view/64538.html | 1,519,193,442,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813431.5/warc/CC-MAIN-20180221044156-20180221064156-00627.warc.gz | 218,787,493 | 3,837 | Associated Topics || Dr. Math Home || Search Dr. Math
### Integer Iteration Function
```Date: 12/24/2003 at 12:38:49
From: Miltos
Subject: The magic 123!!
Let X be a positive integer, A be the number of even digits in that
integer, B be the number of odd digits and C be the number of total
digits. We create the new integer ABC and then we apply that process
repeatedly. We will eventually get the number 123!
For example, suppose we start with 4673985. A = 3 evens, B = 4 odds
and X = 7 total digits. So now we have ABC = 347. Using the same
rule, there is 1 even, 2 odds and 3 total so we have 123. This always
works!
But how can we prove that?
```
```
Date: 12/24/2003 at 18:24:04
From: Doctor Vogler
Subject: Re: The magic 123!!
Hi Miltos,
This is an example I hadn't run into before of a function whose
iterations spiral into a small set, in this case a single 3-digit
number. (Other examples are to go from a number to the sum of its
digits, or to the sum of the squares of its digits.) They are all
analyzed in the same manner:
Let's say that f(n) is the function that goes from the number X to the
number ABC. That is, f(X) = ABC. If you start with a number X which
has more than three digits, then f(X) = ABC will have *fewer* digits
than X. But if you start with a number X which has three digits or
fewer, then f(X) = ABC will still have three digits or fewer. This is
the important point that shows that iterating (or repeating) this
function will take all large values closer and closer and finally into
the finite set of 3-digit numbers (there are only 999 of them), and
then it will stay there.
From there, we just have to look at what happens to those 999 numbers.
Fortunately, for your function, that's not very hard.
If X is a one-digit number, then f(X) is either 101 or 011. If X is a
two-digit number, then f(X) is one of 202, 112, or 022. If X is a
three-digit number, then f(X) is one of 303, 213, 123, or 033. All of
these converge to 123 like so:
one-digit-number -> one of these two:
101 -> 123
011 -> 123
two-digit-number -> one of these three:
202 -> 303 -> 123
112 -> 123
022 -> 303 -> 123
three-digit-number -> one of these four:
303 -> 123
213 -> 123
123 -> 123
033 -> 123
Does that answer your question? Write back if you have any more
questions.
- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 12/25/2003 at 08:21:02
From: Miltos
Subject: The magic 123!!
Thanks for the answer Dr.Vogler. I had the same thoughts but I still
haven't found how I can prove that this function will take X to f(X)
with fewer digits. I know that is obvious but we have a mathematical
problem here and I must.
```
```
Date: 12/25/2003 at 20:39:19
From: Doctor Vogler
Subject: Re: The magic 123!!
Miltos,
Well, okay. Let's suppose that X has n digits and n > 3.
Case 1:
If n < 10, then A, B, and C are one digit each, so f(X) has 3 digits.
Case 2:
If n >= 10, then A, B, and C are each less than or equal to n. So how
many digits can they each have? Well, no more than 1 + log n (to base
10). So f(X) can have no more than 3(1 + log n) digits. Then you
just need to show that 3(1 + log n) < n.
Show (for example by induction on n) that, for n > 9,
n < 2^(n-3).
Conclude that
n^3 < 8^(n-3) < 10^(n-3).
Since log (base 10) is an increasing function,
log (n^3) < log 10^(n-3)
3 log n < n - 3
3 + 3 log n < n
3(1 + log n) < n
Will that do it? Write back if you need more help.
- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 12/28/2003 at 13:22:47
From: Miltos
Subject: Thank you (The magic 123!!)
Thank you, Dr. Vogler. That's the complete answer to the problem.
Numbers are mysterious--have fun with your explorations of them!
```
Associated Topics:
College Number Theory
High School Functions
High School Number Theory
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 1,272 | 4,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2018-09 | latest | en | 0.907808 |
https://www.ademcetinkaya.com/2022/11/raymond-limited-stock-forecast-analysis.html | 1,695,747,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510214.81/warc/CC-MAIN-20230926143354-20230926173354-00371.warc.gz | 711,092,576 | 58,008 | Predicting the future price of financial assets has always been an important research topic in the field of quantitative finance. This paper attempts to use the latest artificial intelligence technologies to design and implement a framework for financial asset price prediction. We evaluate Raymond Limited prediction models with Ensemble Learning (ML) and Pearson Correlation1,2,3,4 and conclude that the NSE RAYMOND stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period: The dominant strategy among neural network is to Hold NSE RAYMOND stock.
Keywords: NSE RAYMOND, Raymond Limited, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.
## Key Points
1. Understanding Buy, Sell, and Hold Ratings
2. Is now good time to invest?
3. How do predictive algorithms actually work?
## NSE RAYMOND Target Price Prediction Modeling Methodology
Accurate prediction of stock price movements is highly challenging and significant topic for investors. Investors need to understand that stock price data is the most essential information which is highly volatile, non-linear, and non-parametric and are affected by many uncertainties and interrelated economic and political factors across the globe. Artificial Neural Networks (ANN) have been found to be an efficient tool in modeling stock prices and quite a large number of studies have been done on it. We consider Raymond Limited Stock Decision Process with Pearson Correlation where A is the set of discrete actions of NSE RAYMOND stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Pearson Correlation)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Ensemble Learning (ML)) X S(n):→ (n+8 weeks) $∑ i = 1 n a i$
n:Time series to forecast
p:Price signals of NSE RAYMOND stock
j:Nash equilibria
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## NSE RAYMOND Stock Forecast (Buy or Sell) for (n+8 weeks)
Sample Set: Neural Network
Stock/Index: NSE RAYMOND Raymond Limited
Time series to forecast n: 07 Nov 2022 for (n+8 weeks)
According to price forecasts for (n+8 weeks) period: The dominant strategy among neural network is to Hold NSE RAYMOND stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Adjusted IFRS* Prediction Methods for Raymond Limited
1. An entity is not required to restate prior periods to reflect the application of these amendments. The entity may restate prior periods if, and only if, it is possible without the use of hindsight and the restated financial statements reflect all the requirements in this Standard. If an entity does not restate prior periods, the entity shall recognise any difference between the previous carrying amount and the carrying amount at the beginning of the annual reporting period that includes the date of initial application of these amendments in the opening retained earnings (or other component of equity, as appropriate) of the annual reporting period that includes the date of initial application of these amendments.
2. Credit risk analysis is a multifactor and holistic analysis; whether a specific factor is relevant, and its weight compared to other factors, will depend on the type of product, characteristics of the financial instruments and the borrower as well as the geographical region. An entity shall consider reasonable and supportable information that is available without undue cost or effort and that is relevant for the particular financial instrument being assessed. However, some factors or indicators may not be identifiable on an individual financial instrument level. In such a case, the factors or indicators should be assessed for appropriate portfolios, groups of portfolios or portions of a portfolio of financial instruments to determine whether the requirement in paragraph 5.5.3 for the recognition of lifetime expected credit losses has been met.
3. The requirements in paragraphs 6.8.4–6.8.8 may cease to apply at different times. Therefore, in applying paragraph 6.9.1, an entity may be required to amend the formal designation of its hedging relationships at different times, or may be required to amend the formal designation of a hedging relationship more than once. When, and only when, such a change is made to the hedge designation, an entity shall apply paragraphs 6.9.7–6.9.12 as applicable. An entity also shall apply paragraph 6.5.8 (for a fair value hedge) or paragraph 6.5.11 (for a cash flow hedge) to account for any changes in the fair value of the hedged item or the hedging instrument.
4. For a financial guarantee contract, the entity is required to make payments only in the event of a default by the debtor in accordance with the terms of the instrument that is guaranteed. Accordingly, cash shortfalls are the expected payments to reimburse the holder for a credit loss that it incurs less any amounts that the entity expects to receive from the holder, the debtor or any other party. If the asset is fully guaranteed, the estimation of cash shortfalls for a financial guarantee contract would be consistent with the estimations of cash shortfalls for the asset subject to the guarantee
*International Financial Reporting Standards (IFRS) are a set of accounting rules for the financial statements of public companies that are intended to make them consistent, transparent, and easily comparable around the world.
## Conclusions
Raymond Limited assigned short-term Baa2 & long-term B3 forecasted stock rating. We evaluate the prediction models Ensemble Learning (ML) with Pearson Correlation1,2,3,4 and conclude that the NSE RAYMOND stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period: The dominant strategy among neural network is to Hold NSE RAYMOND stock.
### Financial State Forecast for NSE RAYMOND Raymond Limited Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*Baa2B3
Operational Risk 8941
Market Risk9048
Technical Analysis3453
Fundamental Analysis6541
Risk Unsystematic8532
### Prediction Confidence Score
Trust metric by Neural Network: 78 out of 100 with 858 signals.
## References
1. Akgiray, V. (1989), "Conditional heteroscedasticity in time series of stock returns: Evidence and forecasts," Journal of Business, 62, 55–80.
2. Bamler R, Mandt S. 2017. Dynamic word embeddings via skip-gram filtering. In Proceedings of the 34th Inter- national Conference on Machine Learning, pp. 380–89. La Jolla, CA: Int. Mach. Learn. Soc.
3. K. Boda, J. Filar, Y. Lin, and L. Spanjers. Stochastic target hitting time and the problem of early retirement. Automatic Control, IEEE Transactions on, 49(3):409–419, 2004
4. M. Puterman. Markov Decision Processes: Discrete Stochastic Dynamic Programming. Wiley, New York, 1994.
5. Bastani H, Bayati M. 2015. Online decision-making with high-dimensional covariates. Work. Pap., Univ. Penn./ Stanford Grad. School Bus., Philadelphia/Stanford, CA
6. Li L, Chen S, Kleban J, Gupta A. 2014. Counterfactual estimation and optimization of click metrics for search engines: a case study. In Proceedings of the 24th International Conference on the World Wide Web, pp. 929–34. New York: ACM
7. Hastie T, Tibshirani R, Friedman J. 2009. The Elements of Statistical Learning. Berlin: Springer
Frequently Asked QuestionsQ: What is the prediction methodology for NSE RAYMOND stock?
A: NSE RAYMOND stock prediction methodology: We evaluate the prediction models Ensemble Learning (ML) and Pearson Correlation
Q: Is NSE RAYMOND stock a buy or sell?
A: The dominant strategy among neural network is to Hold NSE RAYMOND Stock.
Q: Is Raymond Limited stock a good investment?
A: The consensus rating for Raymond Limited is Hold and assigned short-term Baa2 & long-term B3 forecasted stock rating.
Q: What is the consensus rating of NSE RAYMOND stock?
A: The consensus rating for NSE RAYMOND is Hold.
Q: What is the prediction period for NSE RAYMOND stock?
A: The prediction period for NSE RAYMOND is (n+8 weeks) | 2,049 | 8,728 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-40 | latest | en | 0.897235 |
https://www.statsdirect.com/help/data_preparation/dummy_variables.htm | 1,718,795,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00461.warc.gz | 905,111,110 | 2,884 | # Dummy Variables
This function creates dummy (or design) variables from one categorical variable.
The reference cell coding model is used (Kleinbaum et al., 1998):
- the source data may be numerical or text, representing categories. The coding scheme shown above is applied to your data in reverse alphanumeric order for the k categories found, so for three categories, say race equal to black, white or other, white (being the last in an alphabetical sorting) is coded 1,0,0 which reduces to dummy variables X (3) = 1, X (2) = 0.
In order to represent a categorical variable with more than two levels in a regression model you may wish to convert it to a series of dummy variables using this function.
Say a linear regression model is specified with three predictors; the first and third predictors are continuous data, and the second predictor is a classifier (categorical data) with three levels. The second predictor should be converted to two dummy dichotomous variables (e.g. the example below) and put into a multiple linear regression as two predictors.
The naming scheme for dummy variables is the original variable name suffixed with (1) if there are only two categories, or suffixed with (j+1) where there are j+1 categories giving rise to j dummy variables.
In general form, a regression model where the jth predictor variable is a classifier with k levels can be interpreted as follows, provided the jth variable is converted to dummy variables:
- where Y is the outcome variable, b is a regression coefficient, D is a dummy variable for a classifier variable of k levels and x is a non-classifier predictor variable.
Example
Group ID ---> Group ID (2) Group ID (3) 1 0 0 1 0 0 1 0 0 1 0 0 2 1 0 2 1 0 2 1 0 2 1 0 3 0 1 3 0 1 3 0 1 | 425 | 1,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.814791 |
https://www.coursehero.com/file/18102469/Midterm-1-Answer-Key/ | 1,544,520,029,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823614.22/warc/CC-MAIN-20181211083052-20181211104552-00089.warc.gz | 850,286,101 | 95,469 | # Midterm 1 -- Answer Key - Evans School of Public Affairs...
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Evans School of Public Affairs PBAF 516, Autumn 2012 Midterm 1 – Answer Key October 18, 2012 1. The Arroyo family only consumes heating oil and beef. Heating oil is a Giffen good for the Arroyo family. Beef is a normal good for the Arroyo family. Suppose the price of heating oil doubles and the price of beef also doubles. a. Fill-in the following table with one of the following signs: + or – or ? (3 points). Heating Oil Beef Income Effect + - Substitution Effect 0 0 Total Effect + - b. Explain the signs you gave for the substitution effects (1 point). Since the price ratio is unchanged, neither good becomes relatively more or less expensive. Thus, there is no reason to substitute one for the other. c. Explain the signs you gave for the income effects (1 point). With higher prices, the Arroyo’s cannot afford their original bundle. They will buy less of all normal goods (beef) and more of all inferior goods (heating oil – note: all Giffen goods are inferior goods). d. Explain the signs you gave for the total effects (1 point). Since there is no substitution effect, the total effects are equivalent to the income effects. 2. Presidential candidate Spano makes a speech where she argues that credit card companies should be limited to charging no more than a 15% interest rate for borrowers (banks currently charge 18% interest to borrowers). Spano claims that by lowering the borrowing rate of interest, it will increase the current consumption of citizens who are “borrowers” (i.e., those who consume more now than their current endowment). In response, banks complain that such a policy would lower the interest rate that they could
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# What is the least common multiple of 14 and49?
Updated: 8/20/2019
Wiki User
11y ago
Least Common Multiple (LCM) for 14 49 is 98.
Wiki User
11y ago
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Q: What is the least common multiple of 14 and49?
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Related questions
### What is the least common factors of 7 14 and49?
The LCF is 1. The GCF is 7. The LCM is 98.
### What is the least common multiple of 14 and 27?
The least common multiple of 14 and 27 is 378
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The least common multiple of 14 , 7 = 14
The GCF is 7.
### What is the least common multiple for 14 and 7?
The Least Common Multiple (LCM) for 14 7 is 14.
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The least common multiple of 4928 , 14 = 4,928
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### What is the least common multiple of 14 and 29?
The Least Common Multiple of 14 and 29 is 406.
### What is the least common multiple of the numbers 14 and 28?
56 is the least common multiple of 14 and 28 | 413 | 1,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-50 | latest | en | 0.941558 |
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# Lottery Bible ~ Large Print
1310 replies. Last post 6 days ago by helpmewin.
Page 16 of 88
United States
Member #76986
July 10, 2009
8001 Posts
Offline
Posted: January 25, 2013, 7:08 pm - IP Logged
Take a look at last week in ILLINOIS with the repeat 550 combo hits.
Notice that 238 follows. We saw that 238 as a part of the LB workout
for California's 550 in the simple 2378 string. Those 4 digits seem to mill
around 550 / 005. Check your state for this pattern. \$\$\$ MONEY \$\$\$
If we analyze all the hits with doubles like we did with California's
550/005, we could find the 4 basic digits that are the TRUE relatives.
Yes, this would apply for both Pick 3 & Pick 4. Save time and keep a log.
Keep an eye on the mirrors of the 4 digits. Are YOU taking notes?
Yes, I accept tips. \$\$\$ ... WINBIG!
Illinois, January 2013
13 030916 14 055751 15 655550 16 572532 17 328397 18 238102 19 505766
Look at this, Ladies & Gentlemen!
Striking evidence in the State of ILLINOIS Pick 4 Midday.
These events occured, back to back, on 1-24-2013 and 1-25-2013.
Please inspect, take notice and discuss amongst yourselves!
ILLINOIS Pick 4 Midday
Fri, Jan 25, 2013 0-0-5-9 Thu, Jan 24, 2013 3-8-5-2
I rest my case!
.
United States
Member #76986
July 10, 2009
8001 Posts
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Posted: January 26, 2013, 2:46 pm - IP Logged
Homework:
Find the TRUE relatives for the doubles in Pick 3 and/or Pick 4.
Once you find them, they will never change. Just add them to
your log and save them. Your log will be a very handy reference
tool to use, when those doubles appear again. Do this homework
now and watch your log, WINS and MONEY... GROW!
~~~
January 26, 2013 Results
Arizona 1-4-7 Arkansas 3-0-1 0-7-1 2-1-3-8 6-3-6-8 California 9-3-7 8-9-0 8-2-6-1 Connecticut 6-0-7 9-1-6 4-1-8-3 5-3-0-3 Delaware 5-2-6 0-1-4 6-5-5-9 7-7-0-9 Florida 7-9-9 7-2-1 9-3-9-8 4-0-9-0 Georgia 9-7-6 0-3-3 8-3-9-0 3-8-6-4 Idaho 0-2-4 6-1-1 Illinois 1-7-0 6-0-5 0-0-5-9 7-2-4-2 Illinois My3 7-0-9 5-5-9 Indiana 1-5-1 8-7-9 2-9-2-7 9-9-5-8 Iowa 1-7-0 6-0-5 0-0-5-9 7-2-4-2 Kansas 3-7-1 Kentucky 0-6-5 0-3-4 3-0-6-7 5-4-9-2 Louisiana 9-4-2 3-1-3-6 Maryland 3-6-9 4-3-0 4-8-9-1 8-9-6-7 Massachusetts 6-9-4-3 6-4-6-0 Michigan 6-1-4 8-5-3 4-0-2-5 7-6-1-3 Minnesota 8-6-0 Missouri 2-1-4 1-7-2 5-1-4-0 2-5-3-9 Nebraska 6-0-3 New Jersey 3-2-0 8-7-4 7-0-8-8 5-8-1-4 New Mexico 7-4-1 New York 4-6-6 6-1-8 7-0-4-9 8-5-7-6 North Carolina 0-5-9 6-3-8 9-5-6-0 1-6-5-6 Ohio 5-1-5 5-4-6 3-9-0-8 1-1-0-0 Oklahoma 9-3-1 6-0-8-1 Ontario 7-5-7 5-1-3 8-0-9-9 9-8-0-3 Oregon (1pm & 7pm) 0-6-7-2 2-2-5-3 Oregon (4pm & 10pm) 7-2-2-1 4-8-3-6 Pennsylvania 8-6-3 9-7-8 3-4-3-1 6-5-1-3 Puerto Rico 9-4-8 7-5-6-0 Québec 3-4-9 1-0-5-1 Rhode Island 9-3-6-2 South Carolina 9-1-5 0-0-2 9-7-9-7 0-1-9-7 Tennessee 7-5-7 5-1-4 2-0-2-7 1-2-1-1 Texas 4-1-2 1-1-9 2-4-0-5 7-9-4-2 Tri-State 5-7-2 3-9-9 8-5-7-7 9-1-2-9 Virginia 3-4-7 6-4-5 8-0-2-2 1-6-1-3 Washington 5-8-5 Washington, D.C. 9-4-8 7-9-0 9-1-7-5 2-9-4-4 West Virginia 9-7-3 5-6-3-5 Western Canada 2-0-1 Wisconsin 7-3-4 7-0-6-9
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Posted: January 26, 2013, 6:46 pm - IP Logged
Doubles Chart for your TRUE relatives homework assignment(s).
Match the Sisters. Example: 005 / 055
Use the Lottery Bible. Work at your own pace.
001, 002, 003, 004, 005, 006, 007, 008, 009, 011, 022, 033, 044, 055, 066, 077, 088, 099, 112, 113,
114, 115, 116, 117, 118, 119, 122, 133, 144, 155, 166, 177, 188, 199, 223, 224, 225, 226, 227, 228,
229, 233, 244, 255, 266, 277, 288, 299, 334, 335, 336, 337, 338, 339, 344, 355, 366, 377, 388, 399,
445, 446, 447, 448, 449, 455, 466, 477, 488, 499, 556, 557, 558, 559, 566, 577, 588, 599, 667, 668,
669, 677, 688, 699, 778, 779, 788, 799, 889, 899
.
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Posted: January 27, 2013, 10:01 pm - IP Logged
"Yes, absolutely! I'm doing the homework assignment, too. I'm in this for the long haul."
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Posted: January 31, 2013, 6:30 pm - IP Logged
Homework Help
How to break down a doubles Pick 4 to find the sister sets.
Take 3431
First identify the double digits... that would be ... 33
Next make two sets.... 33 and 33
Next fill in the 4 and the 1 with each 33
Like this ... 433 and 133
Now you've got one half of the sister set for each.
Finally, find the other half of the sisters for each of them.
433 / 344
133 / 311
Got it?
This goes for each Pick 4 doubles.
Next, run the lines found in the Lottery Bible.
Find the digits that show up the most.
The Top 4 digits ONLY for each sister set ... are the "TRUE relatives."
When you find the digits, wheel them for your game.
Wheel triads / triplets for Pick 3 & 4. Or play the top 4 digits together in Pick 4.
Once you do the homework, the info stays the same for sister sets.
You don't have to repeat the work, just keep a log, word doc. or excel record.
" TRUE Relatives " is a coined phrase by Harve\$t Moon / Harvest Moon.
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Posted: February 3, 2013, 4:04 pm - IP Logged
Louisiana's results from last week!
A few more opportunities to find
the TRUE relatives of the sisters.
Sat, Feb 2, 2013 Fri, Feb 1, 2013 641 5100 691 5546 327 7459 888 3257 519 8176 522 1737 971 5828
Both teams are already Winners! Congratulations! AND may your team WIN! Good Luck!
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Posted: February 3, 2013, 11:18 pm - IP Logged
Homework Help
How to break down a doubles Pick 4 to find the sister sets.
Take 3431
First identify the double digits... that would be ... 33
Next make two sets.... 33 and 33
Next fill in the 4 and the 1 with each 33
Like this ... 433 and 133
Now you've got one half of the sister set for each.
Finally, find the other half of the sisters for each of them.
433 / 344
133 / 311
Got it?
This goes for each Pick 4 doubles.
Next, run the lines found in the Lottery Bible.
Find the digits that show up the most.
The Top 4 digits ONLY for each sister set ... are the "TRUE relatives."
When you find the digits, wheel them for your game.
Wheel triads / triplets for Pick 3 & 4. Or play the top 4 digits together in Pick 4.
Once you do the homework, the info stays the same for sister sets.
You don't have to repeat the work, just keep a log, word doc. or excel record.
" TRUE Relatives " is a coined phrase by Harve\$t Moon / Harvest Moon.
OMG !!!
I called the final score in the homework assignment!
SUPER BOWL XLVll final score! 34-31 EXACTLY
BTW ... I was born in Baltimore, Maryland! RAVENS!
Michigan
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Posted: February 4, 2013, 7:51 am - IP Logged
OMG !!!
I called the final score in the homework assignment!
SUPER BOWL XLVll final score! 34-31 EXACTLY
BTW ... I was born in Baltimore, Maryland! RAVENS!
Awesome call,
Did you exchange a walk on part in the war ?
For a lead role in a cage?
From Pink Floyd's " Wish you were here"
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Posted: February 4, 2013, 3:01 pm - IP Logged
Awesome call,
Thanks, Sully16!
That number was plucked from the air
for the homework assignment.
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Posted: February 4, 2013, 3:20 pm - IP Logged
SUPER BOWL XLVll Score 34 - 31
All States Lottery History
Mon, Feb 4, 2013 South Carolina Pick 4 Midday 4-3-1-3 Fri, Jan 25, 2013 Pennsylvania Big 4 Midday 3-4-3-1 STRAIGHT Mon, Jan 21, 2013 Puerto Rico Pega 4 3-4-1-3 Thu, Dec 27, 2012 Ontario Pick 4 Evening 3-4-1-3 Mon, Dec 24, 2012 Kentucky Pick 4 Evening 1-3-3-4 Tue, Nov 27, 2012 Indiana Daily 4 Evening 1-3-3-4 Sat, Nov 17, 2012 Arkansas Cash 4 Midday 3-4-3-1 Thu, Sep 27, 2012 Texas Daily 4 Day 3-3-4-1, Sum It Up: 11 Sat, Sep 22, 2012 Oregon Pick 4 1pm 1-3-3-4 Sat, Sep 15, 2012 Oregon Pick 4 1pm 3-4-3-1 Thu, Aug 30, 2012 Missouri Pick 4 Midday 3-1-3-4 Sun, Aug 5, 2012 Oregon Pick 4 7pm 3-3-4-1 Mon, Jul 16, 2012 Virginia Pick 4 Day 4-1-3-3 Sat, Jun 23, 2012 Massachusetts Numbers Game Evening 3-4-1-3 Wed, Jun 13, 2012 South Carolina Pick 4 Midday 3-4-3-1 Tue, May 8, 2012 Louisiana Pick 4 3-3-1-4 Thu, Apr 19, 2012 Multi-State Tri-State Pick 4 Day 1-3-4-3 Thu, Mar 29, 2012 Tennessee Cash 4 Midday 3-4-1-3, Lucky Sum: 11 Thu, Mar 15, 2012 Illinois Pick 4 Midday 4-3-3-1 Thu, Mar 15, 2012 Indiana Daily 4 Evening 4-1-3-3 Thu, Mar 15, 2012 Iowa Pick 4 Midday 4-3-3-1 Thu, Mar 8, 2012 West Virginia Daily 4 4-3-3-1 Sat, Mar 3, 2012 Oregon Pick 4 1pm 3-4-3-1 Wed, Feb 29, 2012 Maryland Pick 4 Midday 3-4-3-1 Thu, Feb 9, 2012 Oklahoma Pick 4 3-3-1-4 Sat, Feb 4, 2012 Virginia Pick 4 Night 1-3-4-3 Tue, Jan 31, 2012 North Carolina Pick 4 Daytime 3-1-4-3 Fri, Dec 30, 2011 Delaware Play 4 Day 3-3-4-1 Sat, Dec 10, 2011 Tennessee Cash 4 Midday 4-1-3-3, Lucky Sum: 11 Fri, Dec 2, 2011 Massachusetts Numbers Game Midday 4-3-3-1 Mon, Nov 21, 2011 Illinois Pick 4 Evening 3-1-3-4 Mon, Nov 21, 2011 Iowa Pick 4 Evening 3-1-3-4 Mon, Nov 21, 2011 Tennessee Cash 4 Evening 3-4-1-3, Lucky Sum: 11 Wed, Nov 16, 2011 Washington, D.C. DC-4 Evening 4-3-3-1 Sat, Oct 8, 2011 Washington, D.C. DC-4 Evening 3-4-3-1 Sat, Aug 13, 2011 Louisiana Pick 4 3-4-3-1 Sun, Jul 24, 2011 North Carolina Pick 4 Evening 3-1-3-4 Sat, Jul 23, 2011 North Carolina Pick 4 Daytime 3-4-3-1 Sat, Jul 9, 2011 Texas Daily 4 Day 3-1-4-3, Sum It Up: 11 Mon, Jul 4, 2011 Oregon Pick 4 4pm 3-3-4-1 Sun, Jun 12, 2011 Oregon Pick 4 7pm 1-4-3-3 Sun, Apr 24, 2011 California Daily 4 4-1-3-3 Sat, Apr 16, 2011 Indiana Daily 4 Midday 1-3-3-4 Fri, Mar 18, 2011 New York Win 4 Evening 4-3-3-1, Lucky Sum: 11 Sat, Mar 12, 2011 Missouri Pick 4 Midday 3-4-1-3 Sat, Feb 19, 2011 Texas Daily 4 Day 1-4-3-3, Sum It Up: 11 Sat, Feb 5, 2011 South Carolina Pick 4 Evening 4-3-1-3 Wed, Feb 2, 2011 South Carolina Pick 4 Midday 4-1-3-3 Fri, Jan 21, 2011 Oregon Pick 4 10pm 3-1-4-3 Mon, Jan 3, 2011 New Jersey Pick 4 Evening 3-4-3-1 Sat, Nov 6, 2010 South Carolina Pick 4 Evening 4-3-1-3 Tue, Nov 2, 2010 Louisiana Pick 4 3-1-3-4 Mon, Oct 25, 2010 New Jersey Pick 4 Midday 3-3-4-1 Mon, Oct 25, 2010 Puerto Rico Pega 4 3-4-1-3 Mon, Oct 18, 2010 Puerto Rico Pega 4 3-1-3-4 Wed, Sep 29, 2010 Pennsylvania Big 4 Midday 3-1-4-3 Sun, Sep 12, 2010 Louisiana Pick 4 1-3-3-4 Wed, Sep 8, 2010 Virginia Pick 4 Night 3-3-1-4 Wed, Aug 18, 2010 Louisiana Pick 4 4-1-3-3 Tue, Jun 8, 2010 Florida Play 4 Evening 1-3-3-4 Fri, Jun 4, 2010 Washington, D.C. DC-4 Midday 3-4-3-1 Sat, May 29, 2010 Georgia Cash 4 Midday 1-3-4-3 Tue, May 11, 2010 Washington, D.C. DC-4 Midday 3-4-3-1 Tue, Mar 30, 2010 Texas Daily 4 Day 1-3-3-4, Sum It Up: 11 Fri, Mar 26, 2010 Pennsylvania Big 4 Evening 3-3-4-1 Thu, Mar 25, 2010 Virginia Pick 4 Day 3-3-1-4 Tue, Jan 12, 2010 Texas Daily 4 Night 1-3-3-4, Sum It Up: 11 Wed, Nov 18, 2009 Oregon Pick 4 4pm 3-1-4-3 Fri, Sep 25, 2009 Florida Play 4 Midday 1-3-4-3 Tue, Aug 11, 2009 California Daily 4 4-1-3-3 Mon, Jul 6, 2009 Georgia Cash 4 Midday 3-4-1-3 Tue, Jun 23, 2009 Texas Daily 4 Night 3-1-4-3, Sum It Up: 11 Tue, Jun 16, 2009 South Carolina Pick 4 Midday 3-1-3-4 Sun, Jun 14, 2009 Washington, D.C. DC-4 Evening 4-1-3-3 Wed, May 20, 2009 West Virginia Daily 4 3-1-4-3 Thu, May 14, 2009 Washington, D.C. DC-4 Evening 1-3-4-3 Mon, May 11, 2009 New Jersey Pick 4 Midday 3-3-1-4 Sat, May 9, 2009 Florida Play 4 Evening 1-3-4-3 Thu, May 7, 2009 North Carolina Pick 4 Evening 4-1-3-3 Sat, Mar 14, 2009 Indiana Daily 4 Evening 3-4-3-1 Sun, Feb 15, 2009 Kentucky Pick 4 Evening 3-3-1-4 Fri, Jan 30, 2009 Delaware Play 4 Night 3-3-1-4 Sun, Dec 7, 2008 Ohio Pick 4 Midday 1-3-4-3 Tue, Nov 25, 2008 Florida Play 4 Midday 4-1-3-3 Sun, Nov 23, 2008 Maryland Pick 4 Evening 4-3-3-1 Thu, Nov 13, 2008 Indiana Daily 4 Midday 4-3-1-3 Sun, Oct 26, 2008 Washington, D.C. DC-4 Midday 1-3-3-4 Sat, Oct 11, 2008 Tennessee Cash 4 Evening 3-3-4-1 Fri, Sep 12, 2008 Texas Daily 4 Night 1-3-4-3, Sum It Up: 11 Sat, Aug 30, 2008 Louisiana Pick 4 3-1-3-4 Wed, Jul 30, 2008 New York Win 4 Midday 4-3-1-3, Lucky Sum: 11 Mon, Jul 28, 2008 Connecticut Play 4 Day 1-3-3-4 Sat, Jul 12, 2008 Ohio Pick 4 Midday 3-4-1-3 Tue, Jul 8, 2008 South Carolina Pick 4 Evening 3-3-1-4 Thu, May 29, 2008 Pennsylvania Big 4 Evening 1-3-3-4 Tue, Apr 15, 2008 Ontario Pick 4 Evening 3-1-4-3 Thu, Apr 10, 2008 Delaware Play 4 Day 4-3-3-1 Tue, Mar 11, 2008 Pennsylvania Big 4 Evening 3-1-3-4 Tue, Feb 5, 2008 Tennessee Cash 4 Evening 3-3-4-1 Tue, Jan 29, 2008 South Carolina Pick 4 Evening 4-1-3-3 Fri, Jan 11, 2008 Pennsylvania Big 4 Midday 4-3-1-3 Wed, Dec 5, 2007 Delaware Play 4 Night 4-1-3-3 Tue, Dec 4, 2007 Tennessee Cash 4 Midday 3-3-4-1 Fri, Nov 23, 2007 West Virginia Daily 4 4-3-1-3 Tue, Nov 20, 2007 Missouri Pick 4 Midday 4-3-3-1 Tue, Nov 13, 2007 New York Win 4 Midday 3-3-4-1, Lucky Sum: 11 Tue, Nov 13, 2007 South Carolina Pick 4 Midday 4-1-3-3 Wed, Nov 7, 2007 Tennessee Cash 4 Evening 4-3-3-1 Wed, Oct 31, 2007 Oregon Pick 4 1pm 3-1-3-4 Tue, Oct 23, 2007 Indiana Daily 4 Midday 4-1-3-3 Mon, Oct 22, 2007 Oregon Pick 4 10pm 1-4-3-3 Sat, Oct 6, 2007 Illinois Pick 4 Evening 4-1-3-3 Sat, Oct 6, 2007 Iowa Pick 4 Evening 4-1-3-3 Sun, Aug 26, 2007 Michigan Daily 4 Evening 4-3-1-3 Fri, Aug 10, 2007 Massachusetts Numbers Game Evening 1-3-3-4 Mon, Jul 30, 2007 Maryland Pick 4 Midday 3-1-4-3 Thu, May 24, 2007 New Jersey Pick 4 Evening 1-3-3-4 Wed, May 16, 2007 Washington, D.C. DC-4 Evening 4-3-1-3 Fri, May 4, 2007 Pennsylvania Big 4 Evening 3-1-3-4 Wed, Mar 28, 2007 Kentucky Pick 4 Midday 3-1-3-4 Tue, Mar 27, 2007 Georgia Cash 4 Midday 3-1-3-4 Wed, Feb 7, 2007 Delaware Play 4 Night 1-3-3-4 Wed, Feb 7, 2007 Québec La Quotidienne 4 3-4-3-1 Sun, Feb 4, 2007 Oregon Pick 4 4pm 1-3-4-3 Fri, Dec 29, 2006 Kentucky Pick 4 Evening 4-3-1-3 Sun, Nov 19, 2006 Connecticut Play 4 Night 3-1-4-3 Mon, Nov 6, 2006 Connecticut Play 4 Day 3-4-3-1 Sun, Oct 15, 2006 Oregon Pick 4 7pm 3-3-1-4 Thu, Sep 28, 2006 Georgia Cash 4 Midday 3-3-4-1 Thu, Sep 21, 2006 Tennessee Cash 4 Midday 1-3-4-3 Tue, Sep 5, 2006 Kentucky Pick 4 Evening 3-4-1-3 Wed, Aug 30, 2006 New York Win 4 Midday 1-3-4-3 Sun, Jul 16, 2006 Florida Play 4 Evening 3-3-1-4 Sat, May 27, 2006 Ohio Pick 4 Midday 4-3-1-3 Mon, May 8, 2006 Missouri Pick 4 Evening 1-4-3-3 Thu, May 4, 2006 Washington, D.C. DC-4 Evening 1-4-3-3 Mon, Apr 10, 2006 Pennsylvania Big 4 Midday 3-3-1-4 Fri, Mar 31, 2006 Missouri Pick 4 Evening 3-4-1-3 Sat, Mar 25, 2006 New Mexico 4 This Way 3-4-1-3 Mon, Mar 6, 2006 Pennsylvania Big 4 Midday 4-3-3-1 Thu, Mar 2, 2006 Washington, D.C. DC-4 Evening 3-4-1-3 Fri, Feb 24, 2006 Michigan Daily 4 Evening 1-4-3-3 Sat, Feb 11, 2006 Oregon Pick 4 7pm 3-4-3-1 Mon, Feb 6, 2006 Ohio Pick 4 Evening 3-3-1-4 Mon, Jan 30, 2006 Illinois Pick 4 Evening 4-3-1-3 Mon, Jan 30, 2006 Iowa Pick 4 Evening 4-3-1-3 Wed, Jan 4, 2006 South Carolina Pick 4 Midday 3-1-4-3 Wed, Dec 28, 2005 Illinois Pick 4 Evening 3-4-3-1 Wed, Dec 28, 2005 Iowa Pick 4 Evening 3-4-3-1 Sat, Nov 5, 2005 Ohio Pick 4 Midday 1-4-3-3 Tue, Oct 25, 2005 Wisconsin Pick 4 4-3-3-1 Mon, Aug 15, 2005 Virginia Pick 4 Night 3-3-1-4 Fri, Jul 22, 2005 Wisconsin Pick 4 4-1-3-3 Wed, Jul 20, 2005 Kentucky Pick 4 Midday 4-3-3-1 Wed, Jul 13, 2005 Ohio Pick 4 Midday 1-3-4-3 Tue, Jun 14, 2005 West Virginia Daily 4 4-3-1-3 Sat, Jun 4, 2005 Maryland Pick 4 Midday 4-1-3-3 Mon, May 9, 2005 Illinois Pick 4 Midday 3-4-3-1 Mon, May 9, 2005 Iowa Pick 4 Midday 3-4-3-1 Wed, May 4, 2005 Connecticut Play 4 Day 1-3-3-4 Wed, May 4, 2005 Tennessee Cash 4 Evening 3-3-1-4 Sun, Apr 24, 2005 Oregon Pick 4 10pm 1-3-4-3 Thu, Apr 21, 2005 Massachusetts Numbers Game Evening 3-1-3-4 Fri, Mar 18, 2005 Virginia Pick 4 Night 3-1-4-3 Sun, Mar 6, 2005 Oregon Pick 4 1pm 4-1-3-3 Wed, Mar 2, 2005 New Mexico 4 This Way 3-4-1-3 Thu, Feb 10, 2005 West Virginia Daily 4 1-3-3-4 Fri, Jan 28, 2005 Washington, D.C. DC-4 Evening 4-3-3-1 Fri, Dec 10, 2004 Wisconsin Pick 4 3-3-4-1 Sat, Dec 4, 2004 Massachusetts Numbers Game Evening 1-4-3-3 Thu, Oct 28, 2004 Multi-State Tri-State Pick 4 Day 4-1-3-3 Tue, Oct 26, 2004 Multi-State Tri-State Pick 4 Evening 4-3-3-1 Sat, Oct 23, 2004 Washington, D.C. DC-4 Evening 1-3-4-3 Sat, Aug 28, 2004 Illinois Pick 4 Midday 1-4-3-3 Sat, Aug 28, 2004 Iowa Pick 4 Midday 1-4-3-3 Thu, Aug 12, 2004 Multi-State Tri-State Pick 4 Evening 3-4-3-1 Thu, Aug 5, 2004 Connecticut Play 4 Day 3-3-1-4 Thu, Aug 5, 2004 Multi-State Tri-State Pick 4 Evening 3-4-1-3 Fri, Jul 23, 2004 Florida Play 4 Evening 4-1-3-3 Fri, Jul 16, 2004 Connecticut Play 4 Day 3-4-1-3 Mon, Jun 21, 2004 Virginia Pick 4 Night 4-1-3-3 Thu, Jun 17, 2004 South Carolina Pick 4 Midday 1-3-3-4 Sun, Jun 13, 2004 Maryland Pick 4 Midday 3-3-1-4 Tue, May 25, 2004 Oregon Pick 4 7pm 1-3-4-3 Wed, May 19, 2004 New York Win 4 Midday 4-3-3-1 Thu, Apr 22, 2004 Michigan Daily 4 Midday 1-3-3-4 Mon, Mar 29, 2004 Oregon Pick 4 4pm 3-4-3-1 Mon, Mar 29, 2004 Oregon Pick 4 7pm 3-1-4-3 Sat, Mar 6, 2004 Indiana Daily 4 Midday 3-3-4-1 Wed, Feb 18, 2004 Puerto Rico Pega 4 3-3-4-1 Sat, Jan 17, 2004 Puerto Rico Pega 4 3-4-1-3 Tue, Dec 30, 2003 New Jersey Pick 4 Evening 1-3-4-3 Fri, Dec 26, 2003 Georgia Cash 4 Midday 4-3-3-1 Mon, Dec 15, 2003 Louisiana Pick 4 3-1-3-4 Thu, Nov 6, 2003 Illinois Pick 4 Evening 1-4-3-3 Thu, Nov 6, 2003 Iowa Pick 4 Evening 1-4-3-3 Fri, Oct 17, 2003 Connecticut Play 4 Night 1-3-4-3 Wed, Oct 8, 2003 Virginia Pick 4 Day 3-1-4-3 Fri, Oct 3, 2003 Wisconsin Pick 4 3-1-4-3 Wed, Oct 1, 2003 South Carolina Pick 4 Midday 1-4-3-3 Tue, Sep 9, 2003 Maryland Pick 4 Midday 4-3-1-3 Sun, Aug 24, 2003 New York Win 4 Midday 1-4-3-3 Fri, Aug 22, 2003 Québec La Quotidienne 4 3-3-4-1 Sun, Jul 13, 2003 Québec La Quotidienne 4 3-4-1-3 Fri, Jul 4, 2003 New York Win 4 Evening 3-4-1-3 Mon, Jun 30, 2003 Virginia Pick 4 Day 3-4-1-3 Sun, May 25, 2003 Multi-State Tri-State Pick 4 Evening 4-1-3-3 Fri, Apr 4, 2003 Georgia Cash 4 Evening 3-3-4-1 Thu, Mar 20, 2003 Illinois Pick 4 Evening 3-3-1-4 Fri, Mar 14, 2003 Puerto Rico Pega 4 4-3-3-1 Wed, Feb 5, 2003 Oregon Pick 4 1pm 3-4-1-3 Wed, Jan 29, 2003 Washington, D.C. DC-4 Midday 1-3-4-3 Mon, Jan 6, 2003 Virginia Pick 4 Night 4-3-3-1 Wed, Dec 25, 2002 Multi-State Tri-State Pick 4 Evening 1-3-4-3 Wed, Dec 18, 2002 Kentucky Pick 4 Midday 3-3-1-4 Thu, Dec 12, 2002 Georgia Cash 4 Midday 3-3-4-1 Sat, Nov 2, 2002 New Jersey Pick 4 Evening 3-3-1-4 Thu, Sep 19, 2002 Indiana Daily 4 Midday 3-1-3-4 Wed, Sep 4, 2002 Louisiana Pick 4 4-3-3-1 Sat, Jul 13, 2002 Indiana Daily 4 Evening 3-3-4-1 Sat, Jun 22, 2002 Québec La Quotidienne 4 3-3-4-1 Fri, Jun 21, 2002 Georgia Cash 4 Evening 1-3-4-3 Fri, Jun 21, 2002 Maryland Pick 4 Evening 1-4-3-3 Wed, May 22, 2002 Oregon Pick 4 1pm 3-4-1-3 Mon, Apr 1, 2002 Oregon Pick 4 4pm 1-4-3-3 Mon, Mar 4, 2002 Louisiana Pick 4 3-1-4-3 Sat, Feb 23, 2002 Delaware Play 4 Day 3-1-3-4 Fri, Jan 25, 2002 Rhode Island Numbers Game 3-1-3-4 Thu, Jan 24, 2002 Washington, D.C. DC-4 Evening 1-3-3-4 Sat, Jan 19, 2002 Washington, D.C. DC-4 Evening 3-4-3-1 Wed, Nov 21, 2001 West Virginia Daily 4 4-3-1-3 Tue, Nov 20, 2001 Virginia Pick 4 Night 3-1-3-4 Fri, Nov 9, 2001 Virginia Pick 4 Night 3-4-1-3 Wed, Oct 24, 2001 New York Win 4 Evening 4-1-3-3 Thu, Oct 4, 2001 Rhode Island Numbers Game 3-3-4-1 Wed, Oct 3, 2001 Virginia Pick 4 Night 3-1-3-4 Thu, Sep 27, 2001 Maryland Pick 4 Midday 3-4-3-1 Thu, Sep 13, 2001 Georgia Cash 4 Midday 3-1-3-4 Tue, Sep 4, 2001 Virginia Pick 4 Day 1-3-3-4 Sat, Sep 1, 2001 Virginia Pick 4 Night 1-3-3-4 Thu, Aug 9, 2001 Virginia Pick 4 Day 3-1-3-4 Mon, Aug 6, 2001 Delaware Play 4 Night 4-1-3-3 Sat, Jul 28, 2001 Washington, D.C. DC-4 Evening 1-3-3-4 Sat, Jun 30, 2001 Ohio Pick 4 Midday 1-3-3-4 Sat, Jun 23, 2001 Virginia Pick 4 Day 3-4-3-1 Fri, Jun 22, 2001 Georgia Cash 4 Evening 3-4-3-1 Fri, Jun 8, 2001 Michigan Daily 4 Evening 3-1-3-4 Mon, May 14, 2001 Wisconsin Pick 4 3-4-1-3 Sat, May 5, 2001 Massachusetts Numbers Game Evening 3-4-1-3 Tue, Apr 24, 2001 Oregon Pick 4 7pm 3-1-4-3 Tue, Apr 3, 2001 Georgia Cash 4 Evening 3-3-4-1 Fri, Jan 12, 2001 Florida Play 4 Evening 4-3-3-1 Sun, Dec 17, 2000 Louisiana Pick 4 3-3-4-1 Tue, Oct 31, 2000 Connecticut Play 4 Night 1-4-3-3 Sun, Oct 22, 2000 Washington, D.C. DC-4 Evening 1-3-3-4 Fri, Oct 20, 2000 New Jersey Pick 4 Evening 3-4-3-1 Wed, Oct 18, 2000 Maryland Pick 4 Evening 4-3-3-1 Tue, Sep 26, 2000 New Jersey Pick 4 Evening 4-1-3-3 Thu, Sep 7, 2000 Ohio Pick 4 Evening 4-3-1-3 Thu, Aug 17, 2000 Louisiana Pick 4 1-3-3-4 Mon, Jul 3, 2000 Pennsylvania Big 4 Evening 3-4-1-3 Mon, Jun 5, 2000 Indiana Daily 4 Evening 3-1-3-4 Fri, May 26, 2000 Ohio Pick 4 Midday 3-1-4-3 Sat, Apr 29, 2000 Rhode Island Numbers Game 1-3-3-4 Sat, Feb 19, 2000 New York Win 4 Evening 3-1-3-4 Thu, Feb 17, 2000 Georgia Cash 4 Evening 3-3-1-4 Wed, Dec 29, 1999 Multi-State Tri-State Pick 4 Evening 4-3-1-3 Wed, Dec 1, 1999 Louisiana Pick 4 4-3-1-3 Fri, Nov 26, 1999 Maryland Pick 4 Evening 3-4-3-1 Mon, Oct 18, 1999 Kentucky Pick 4 Midday 3-4-1-3 Mon, Oct 18, 1999 Virginia Pick 4 Night 4-1-3-3 Fri, Oct 15, 1999 Maryland Pick 4 Evening 1-3-4-3 Mon, Oct 4, 1999 Delaware Play 4 Day 1-3-3-4 Thu, Sep 23, 1999 Indiana Daily 4 Evening 4-1-3-3 Tue, Sep 14, 1999 Illinois Pick 4 Evening 1-4-3-3 Sun, Aug 29, 1999 Wisconsin Pick 4 3-3-1-4 Thu, Jul 22, 1999 Virginia Pick 4 Day 3-1-4-3 Sun, Jun 20, 1999 Québec La Quotidienne 4 3-4-3-1 Sat, May 29, 1999 Washington, D.C. DC-4 Evening 3-4-1-3 Mon, Apr 19, 1999 West Virginia Daily 4 4-3-3-1 Sat, Apr 17, 1999 New Jersey Pick 4 Evening 3-4-3-1 Sat, Apr 10, 1999 Washington, D.C. DC-4 Midday 4-1-3-3 Wed, Mar 10, 1999 Pennsylvania Big 4 Evening 3-4-3-1 Thu, Jan 28, 1999 Indiana Daily 4 Evening 4-3-3-1 Sat, Nov 28, 1998 Illinois Pick 4 Evening 3-3-4-1 Wed, Nov 25, 1998 Missouri Pick 4 Evening 3-3-4-1 Sun, Nov 22, 1998 Georgia Cash 4 Evening 4-1-3-3 Fri, Nov 20, 1998 Indiana Daily 4 Evening 1-3-4-3 Tue, Sep 29, 1998 Massachusetts Numbers Game Evening 1-3-3-4 Mon, Mar 9, 1998 Georgia Cash 4 Evening 4-3-3-1 Fri, Feb 20, 1998 Florida Play 4 Evening 4-3-3-1 Thu, Feb 12, 1998 Michigan Daily 4 Midday 4-3-1-3 Thu, Dec 25, 1997 Florida Play 4 Evening 3-3-1-4 Fri, Aug 29, 1997 Washington, D.C. DC-4 Midday 4-3-1-3 Thu, Aug 21, 1997 Washington, D.C. DC-4 Midday 3-4-3-1 Sat, Jul 12, 1997 Massachusetts Numbers Game Evening 1-4-3-3 Fri, Jun 6, 1997 Pennsylvania Big 4 Evening 1-3-3-4 Fri, Jan 3, 1997 Indiana Daily 4 Evening 3-1-3-4 Sun, Dec 29, 1996 Pennsylvania Big 4 Evening 1-3-3-4 Wed, Aug 14, 1996 West Virginia Daily 4 3-3-1-4 Fri, Jul 12, 1996 Virginia Pick 4 Night 3-3-4-1 Thu, Jul 4, 1996 Washington, D.C. DC-4 Midday 3-1-3-4 Tue, Jun 18, 1996 Delaware Play 4 Day 1-3-4-3 Sat, May 4, 1996 Maryland Pick 4 Midday 3-4-3-1 Sat, Apr 20, 1996 Virginia Pick 4 Night 3-3-1-4 Thu, Feb 15, 1996 Michigan Daily 4 Evening 3-1-4-3 Sat, Feb 10, 1996 Ohio Pick 4 Evening 1-3-4-3 Sun, Nov 12, 1995 Indiana Daily 4 Evening 3-4-3-1 Thu, Oct 26, 1995 Pennsylvania Big 4 Evening 3-4-1-3 Tue, Sep 26, 1995 Massachusetts Numbers Game Evening 1-4-3-3 Thu, Sep 21, 1995 Michigan Daily 4 Evening 3-1-4-3 Mon, Aug 28, 1995 Florida Play 4 Evening 1-3-3-4 Sat, Aug 19, 1995 Delaware Play 4 Day 4-1-3-3 Sat, Aug 5, 1995 Maryland Pick 4 Evening 4-1-3-3 Sat, Jul 8, 1995 Ohio Pick 4 Evening 1-3-4-3 Sat, Jun 10, 1995 Indiana Daily 4 Evening 1-3-3-4 Tue, May 16, 1995 New Jersey Pick 4 Evening 1-4-3-3 Sun, Feb 12, 1995 Kentucky Pick 4 Evening 4-1-3-3 Wed, Jan 18, 1995 Kentucky Pick 4 Evening 3-3-1-4 Fri, Nov 4, 1994 Florida Play 4 Evening 3-4-3-1 Fri, Sep 9, 1994 Delaware Play 4 Night 4-1-3-3 Wed, Sep 7, 1994 Illinois Pick 4 Evening 3-1-3-4 Fri, Sep 2, 1994 Ohio Pick 4 Evening 4-3-1-3 Tue, Jul 26, 1994 Pennsylvania Big 4 Evening 3-1-4-3 Fri, Jul 1, 1994 Delaware Play 4 Night 3-4-3-1 Fri, Apr 22, 1994 Multi-State Tri-State Pick 4 Evening 1-3-3-4 Tue, Feb 22, 1994 New York Win 4 Evening 3-4-3-1 Tue, Feb 15, 1994 Washington, D.C. DC-4 Midday 3-1-4-3 Sat, Jan 29, 1994 New York Win 4 Evening 3-1-4-3 Tue, Sep 28, 1993 Florida Play 4 Evening 3-1-4-3 Mon, Aug 30, 1993 Illinois Pick 4 Evening 4-1-3-3 Sun, Jun 20, 1993 Illinois Pick 4 Evening 3-4-3-1 Mon, Feb 15, 1993 Massachusetts Numbers Game Evening 3-4-1-3 Sat, Nov 14, 1992 Virginia Pick 4 Night 4-3-1-3 Wed, Sep 30, 1992 Ohio Pick 4 Evening 3-4-3-1 Mon, Aug 31, 1992 Delaware Play 4 Night 3-3-1-4 Mon, Mar 2, 1992 Maryland Pick 4 Evening 3-1-4-3 Fri, Dec 6, 1991 Washington, D.C. DC-4 Evening 3-1-3-4 Thu, Dec 5, 1991 Indiana Daily 4 Evening 3-3-4-1 Wed, Oct 2, 1991 Multi-State Tri-State Pick 4 Evening 3-3-4-1 Wed, Mar 13, 1991 New York Win 4 Evening 3-3-4-1 Sat, Oct 13, 1990 Ohio Pick 4 Evening 3-1-3-4 Fri, Sep 28, 1990 West Virginia Daily 4 3-1-3-4 Fri, Aug 24, 1990 Multi-State Tri-State Pick 4 Evening 1-4-3-3 Fri, Mar 30, 1990 Washington, D.C. DC-4 Evening 3-4-1-3 Mon, Mar 19, 1990 New Jersey Pick 4 Evening 1-4-3-3 Fri, Jan 26, 1990 Maryland Pick 4 Evening 4-3-3-1 Sun, Jan 21, 1990 Pennsylvania Big 4 Evening 4-3-1-3 Thu, Jan 11, 1990 Pennsylvania Big 4 Evening 3-4-3-1 Tue, Dec 26, 1989 Illinois Pick 4 Evening 3-3-1-4 Sat, Aug 5, 1989 Washington, D.C. DC-4 Evening 3-4-1-3 Fri, Aug 4, 1989 New Jersey Pick 4 Evening 3-1-3-4 Thu, Jul 13, 1989 Ohio Pick 4 Evening 1-3-3-4 Sat, Jun 10, 1989 Maryland Pick 4 Evening 1-3-3-4 Sat, Jun 3, 1989 Maryland Pick 4 Evening 3-3-1-4 Tue, Mar 14, 1989 Washington, D.C. DC-4 Evening 1-3-4-3 Tue, Oct 11, 1988 New Jersey Pick 4 Evening 3-3-1-4 Tue, Aug 2, 1988 New Jersey Pick 4 Evening 4-1-3-3 Mon, Jul 18, 1988 Illinois Pick 4 Evening 4-3-1-3 Tue, Jan 19, 1988 Multi-State Tri-State Pick 4 Evening 3-1-4-3 Mon, Dec 21, 1987 Michigan Daily 4 Evening 4-1-3-3 Sat, Dec 5, 1987 Ohio Pick 4 Evening 4-3-1-3 Mon, Nov 16, 1987 Multi-State Tri-State Pick 4 Evening 3-1-3-4 Mon, Jul 27, 1987 Ohio Pick 4 Evening 4-1-3-3 Sat, Jun 6, 1987 Delaware Play 4 Night 4-1-3-3 Tue, May 5, 1987 Multi-State Tri-State Pick 4 Evening 3-3-4-1 Wed, Feb 4, 1987 Delaware Play 4 Night 3-1-4-3 Tue, Dec 30, 1986 Illinois Pick 4 Evening 1-4-3-3 Sat, Dec 20, 1986 Michigan Daily 4 Evening 1-4-3-3 Wed, Oct 8, 1986 Multi-State Tri-State Pick 4 Evening 4-3-1-3 Sat, Sep 6, 1986 Washington, D.C. DC-4 Evening 1-3-4-3 Fri, Jun 6, 1986 Maryland Pick 4 Evening 1-4-3-3 Tue, Jun 3, 1986 Multi-State Tri-State Pick 4 Evening 1-3-4-3 Wed, Apr 9, 1986 Multi-State Tri-State Pick 4 Evening 3-1-4-3 Thu, Feb 20, 1986 Massachusetts Numbers Game Evening 1-3-3-4 Tue, Aug 27, 1985 New York Win 4 Evening 3-1-4-3 Tue, Aug 20, 1985 Maryland Pick 4 Evening 4-3-3-1 Wed, Jan 16, 1985 Illinois Pick 4 Evening 3-4-3-1 Sat, Jan 12, 1985 New Jersey Pick 4 Evening 3-4-1-3 Sat, Oct 6, 1984 New Jersey Pick 4 Evening 3-3-1-4 Mon, May 28, 1984 Maryland Pick 4 Evening 3-3-4-1 Fri, Mar 4, 1983 New Jersey Pick 4 Evening 3-4-3-1 Sat, Oct 30, 1982 Ohio Pick 4 Evening 4-3-3-1 Tue, Jun 29, 1982 Ohio Pick 4 Evening 1-4-3-3 Mon, Dec 7, 1981 Delaware Play 4 Night 1-4-3-3 Fri, Oct 31, 1980 New Jersey Pick 4 Evening 3-4-1-3 Fri, Mar 31, 1978 Massachusetts Numbers Game Evening 3-4-1-3
Michigan
United States
Member #81740
October 28, 2009
38973 Posts
Offline
Posted: February 4, 2013, 3:52 pm - IP Logged
Thanks, Sully16!
That number was plucked from the air
for the homework assignment.
Good Pluck, pluck some powerball numbers, and life will be easy street.
Did you exchange a walk on part in the war ?
For a lead role in a cage?
From Pink Floyd's " Wish you were here"
NYC
United States
Member #119024
November 13, 2011
825 Posts
Offline
Posted: February 4, 2013, 5:21 pm - IP Logged
I'm sure alot of people in South Carolina are happy today...
Mon, Feb 4, 2013 South Carolina Pick 4 Midday 4-3-1-3
United States
Member #76986
July 10, 2009
8001 Posts
Offline
Posted: February 6, 2013, 11:47 pm - IP Logged
Single digit combinations have TRUE relatives, also!
Find the 6 combinations (I'll call them, "the 6 brothers").
Find the 6 corresponding lines and line them up.
Count the digits.
Take the Top 4 numbers (sometimes there are only 3)
Use these numbers in Pick 3 and as triads / triplets for Pick 4.
Example: 317
317-812-058-034-270
371-494-630-838-074
137-496-632-274-410
173-814-050-470-616
713-072-218-850-096
731-236-472-458-694
Let's count the digits / numbers.
Numbers Count 8 8 2 8 6 8 5 4 9 4 7 12 0 12 4 12 3 11 1 11
Thanks, Manual, for counting these.
Use the Inspector 3 tool with your membership.
The TRUE relatives of the (six)6 brothers
of combination 317 are the digits 7, 0, 4.
Harve\$t Moon, this is so much FUN!
Look at some evidence and DISCUSS amongst yourselves.
Pick 3 New York, Sept 2012
2nd704 3rd137
Pick 4 New York, Sept 2012
3rd
7440
I rest my case, again!
Florida
United States
Member #135615
November 27, 2012
390 Posts
Offline
Posted: February 6, 2013, 11:49 pm - IP Logged
So what's the deal with the lottery bible? I did a little research on it awhile back (sadly, I have a very poor memory). I was going to look into ingetrating into my site somehow but wasn't sure if I'd get in trouble doing so...
United States
Member #76986
July 10, 2009
8001 Posts
Offline
Posted: February 7, 2013, 1:36 am - IP Logged
So what's the deal with the lottery bible? I did a little research on it awhile back (sadly, I have a very poor memory). I was going to look into ingetrating into my site somehow but wasn't sure if I'd get in trouble doing so...
This is the correct link for the Lottery Bible numbers.
To understand this work, you'll find the best place to start is on page 2.
The 1st page has errors in the Lottery Bible log.
The Lottery Bible log of numbers has no known author and is over 80 years old.
Get a big mug of coffee and some donuts!
There are no shortcuts and there are many, many proven winning workouts!
Try them all and find your favorites! Some can be automated and some can not.
A BIG thanks to those who shared their methods or responded in the thread.
Most of the creative ways to use this material, originated in my own creative head.
I backed this up with tons of research, discoveries, strained eyes and lack of sleep.
I can safely say that this thread is the most comprehensive work on the Lottery Bible
anywhere. I still have more ideas to create, explore, research and possibly ... share.
The Lottery Bible does not change, but the ways to use it are endless. I am creating
simple and true ways to use it to win the numbers games. That's "the deal" and my
reason for hosting this topic and thread.
Page 16 of 88 | 12,956 | 30,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-44 | latest | en | 0.772088 |
https://oeis.org/A045703 | 1,591,468,382,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348517506.81/warc/CC-MAIN-20200606155701-20200606185701-00481.warc.gz | 442,473,339 | 3,675 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A045703 Primes of the form F(i)^2 + F(j)^2, where F() are Fibonacci numbers. 1
2, 5, 13, 29, 73, 89, 173, 233, 1181, 1597, 3089, 23761, 28657, 54293, 142193, 372269, 426389, 514229, 2922509, 6680081, 119822837, 320356577, 433494437, 2971215073, 5748565541, 38580030893, 38580085013, 38582581133, 44208781349 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS T. D. Noe, Table of n, a(n) for n=1..300 FORMULA 5 = 1^2 + 2^2. MATHEMATICA f=Table[Fibonacci[i], {i, 1, 100}]; g=Union[Flatten[Table[f[[i]]^2+f[[j]]^2, {i, 1, 80}, {j, 1, 80}]]]; Table[g[[Select[Range[400], PrimeQ[g[[#]]]&]]]] (* Vincenzo Librandi, Jul 13 2012 *) CROSSREFS Cf. A000045. Sequence in context: A214633 A282831 A259762 * A289843 A242080 A178444 Adjacent sequences: A045700 A045701 A045702 * A045704 A045705 A045706 KEYWORD nonn AUTHOR STATUS approved
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Last modified June 6 14:04 EDT 2020. Contains 334827 sequences. (Running on oeis4.) | 499 | 1,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-24 | latest | en | 0.557381 |
https://www.coursehero.com/file/1654465/21-InstSolManual-PDF-Part13/ | 1,548,017,600,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583739170.35/warc/CC-MAIN-20190120204649-20190120230649-00295.warc.gz | 763,988,156 | 48,190 | 21_InstSolManual_PDF_Part13
# 21_InstSolManual_PDF_Part13 - Electromagnetic Induction...
• Notes
• 1
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21.56. Set Up: Problem 21.55 shows that initially is zero and rises to 25.0 V, that starts at 25.0 V and decreases to zero, and that i starts at zero and rises to 1.67 A. so is proportional to i. A reads the current i. Solve: The graphs are sketched in Figure 21.56. Figure 21.56 21.57. Set Up: Current decay in an R - L circuit is described by Eq. (21.28). Solve: and and Reflect: As R is decreased the time constant increases and it takes longer for the current to decay. For this circuit the time constant is After one time constant the current has decayed to about 37% of its original value. It is reasonable for it to take a little over two time constants for the current to decay to 10% of its original value. 21.58. Set Up: The current as a function of time is given by The energy stored in the inductor is U reaches its maximum value when i is times its maximum value. Solve: (a) The maximum current is gives and (b) and 21.59. Set Up: The resonant angular frequency of an L - C circuit is Solve: 21.60.
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• Spring '07
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http://www.tsusiatsoftware.net/jts/javadoc/com/vividsolutions/jts/algorithm/HCoordinate.html | 1,708,674,646,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00391.warc.gz | 71,513,802 | 3,771 | JTS Topology Suite version 1.12
## com.vividsolutions.jts.algorithm Class HCoordinate
```java.lang.Object
com.vividsolutions.jts.algorithm.HCoordinate
```
`public class HCoordinateextends java.lang.Object`
Represents a homogeneous coordinate in a 2-D coordinate space. In JTS `HCoordinate`s are used as a clean way of computing intersections between line segments.
Version:
1.7
Author:
David Skea
Field Summary
` double` `w`
` double` `x`
` double` `y`
Constructor Summary
`HCoordinate()`
`HCoordinate(Coordinate p)`
```HCoordinate(Coordinate p1, Coordinate p2)```
Constructs a homogeneous coordinate which is the intersection of the lines define by the homogenous coordinates represented by two `Coordinate`s.
```HCoordinate(Coordinate p1, Coordinate p2, Coordinate q1, Coordinate q2)```
```HCoordinate(double _x, double _y)```
```HCoordinate(double _x, double _y, double _w)```
```HCoordinate(HCoordinate p1, HCoordinate p2)```
Method Summary
` Coordinate` `getCoordinate()`
` double` `getX()`
` double` `getY()`
`static Coordinate` ```intersection(Coordinate p1, Coordinate p2, Coordinate q1, Coordinate q2)```
Computes the (approximate) intersection point between two line segments using homogeneous coordinates.
Methods inherited from class java.lang.Object
`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`
Field Detail
### x
`public double x`
### y
`public double y`
### w
`public double w`
Constructor Detail
### HCoordinate
`public HCoordinate()`
### HCoordinate
```public HCoordinate(double _x,
double _y,
double _w)```
### HCoordinate
```public HCoordinate(double _x,
double _y)```
### HCoordinate
`public HCoordinate(Coordinate p)`
### HCoordinate
```public HCoordinate(HCoordinate p1,
HCoordinate p2)```
### HCoordinate
```public HCoordinate(Coordinate p1,
Coordinate p2)```
Constructs a homogeneous coordinate which is the intersection of the lines define by the homogenous coordinates represented by two `Coordinate`s.
Parameters:
`p1` -
`p2` -
### HCoordinate
```public HCoordinate(Coordinate p1,
Coordinate p2,
Coordinate q1,
Coordinate q2)```
Method Detail
### intersection
```public static Coordinate intersection(Coordinate p1,
Coordinate p2,
Coordinate q1,
Coordinate q2)
throws NotRepresentableException```
Computes the (approximate) intersection point between two line segments using homogeneous coordinates.
Note that this algorithm is not numerically stable; i.e. it can produce intersection points which lie outside the envelope of the line segments themselves. In order to increase the precision of the calculation input points should be normalized before passing them to this routine.
Throws:
`NotRepresentableException`
### getX
```public double getX()
throws NotRepresentableException```
Throws:
`NotRepresentableException`
### getY
```public double getY()
throws NotRepresentableException```
Throws:
`NotRepresentableException`
### getCoordinate
```public Coordinate getCoordinate()
throws NotRepresentableException```
Throws:
`NotRepresentableException`
JTS Topology Suite version 1.12 | 750 | 3,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.47548 |
http://list.seqfan.eu/pipermail/seqfan/2014-October/028512.html | 1,701,397,316,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100264.9/warc/CC-MAIN-20231201021234-20231201051234-00305.warc.gz | 31,818,268 | 2,019 | # [seqfan] Two neighbors sum -- and odd ranks in S
Eric Angelini Eric.Angelini at kntv.be
Tue Oct 21 21:58:28 CEST 2014
```Hello SeqFans,
S is the lexically first permutation
of the Natural numbers such that
a(n)+a(n+1)=a(2n-1)
S = 1,0,2,3,5,4,8,6,9,7,12,10,14,11,15,13,16,17,19,18,22,20,24,21,25,23,26,27,28,30,29,31,33,32,36,34,37,35,40,38,42,39,44,41,45,43,46,47,48,50,49,51,53,52,55,54,...
Take n=1, then:
a(n)=a(1)=1
a(n+1)=a(2)=0
a(2n-1)=a(1)=1 and 1+0 is indeed 1
Take n=6, then:
a(n)=a(6)=4
a(n+1)=a(7)=8
a(2n-1)=a(11)=12 and 4+8 is indeed 12
Take n=21, then:
a(n)=a(21)=22
a(n+1)=a(22)=20
a(2n-1)=a(41)=42 and 22+20 is indeed 42
Best,
É.
```
More information about the SeqFan mailing list | 357 | 706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-50 | latest | en | 0.723293 |
https://www.coursehero.com/file/6582624/M334Sp06E1/ | 1,513,391,132,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948581033.57/warc/CC-MAIN-20171216010725-20171216032725-00022.warc.gz | 738,712,031 | 23,444 | M334Sp06E1
# M334Sp06E1 - Name Student ID Section Instructor Steven...
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Unformatted text preview: Name: Student ID: Section: Instructor: Steven McKay Math 334 (Ordinary Differential Equations) Exam 1 May 12,13,15 2006 Instructions: • For questions which require a written answer, show all your work. Full credit will be given only if the necessary work is shown justifying your answer. • Simplify your answers. • Calculators are not allowed. • Should you have need for more space than is allocated to answer a question, use the back of the page the problem is on and indicate this fact. • Please do not talk about the test with other students until after the last day to take the exam. Part I: Multiple Choice Mark the correct answer on the bubble sheet provided. 1. If y ( x ) is the solution of y = sin x cos y , y (0) = π , then the value of y ( π/ 2) is a) b) π/ 2 c) π d)- π/ 2 e)- π f) 1. 2. Which of the following equations are nonlinear and second order? a) y 00 + 3 y + ty = sin( t 4 ) b) y + y 2 = 0 c) y 00 + (sin t ) y = 0 d) yy 00 + y = 1 e) ( y ) 2 + 2 y = tan t f) y 000...
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Ask a homework question - tutors are online | 441 | 1,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-51 | latest | en | 0.797878 |
https://eleanorrigby-movie.com/what-is-the-height-of-a-badminton-net/ | 1,680,313,916,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00283.warc.gz | 271,900,154 | 17,769 | # What is the height of a badminton net?
## What is the height of a badminton net?
5 feet 1in
A standard badminton court is marked for both the singles and the doubles game. The court has two halves measuring 6.7m (22 feet) each and separated by a badminton net that stands at a height of 1.55m (5 feet 1in) at the ends and dips to 1.52m (5 feet) in the middle.
### How high is the net on a badminton court in inches?
How high is a badminton net? Badminton nets are regulated at a height of 5’1” (1.55 m) at the edges of the court and 5′ (1.52 m) at the center of the court.
#### What is the size of a badminton court in meters?
The overall length of a badminton court should be 13.4 metres long (44 feet) by 6.3 metres wide (20 feet). Diagonally, the full length of the badminton ground area is 14.366 metres.
What is the length and width of a badminton net?
In short, a badminton player should measure 5 foot 1 inch in height on either side of the court, with a dip down to 5 feet in the center of the court. This dip is achievable because of the width of the badminton net, which is 6.1 meters (20.01 feet).
What is Net play in badminton?
Badminton net play is mainly used to FORCE your opponent to lift the shuttle high in the air, so that you can execute a strong attacking shot. The competition is based on who gives up playing along the net and lifts the shuttle.
## What is the height of the net from the floor?
While the standard regulation volleyball net height for men is 7 feet, 11 ⅝ inches or 2.43 meters (this is the same for standing disabled men’s volleyball teams), men’s teams that fall into older age brackets (55-70+) are allowed to lower their nets to certain heights.
### How wide is a badminton net in feet?
#### What is set up in badminton?
Scoring: In badminton, a match is played best 2 of 3 games, with each game played up to 21 points. In tennis a match is played best of 3 or 5 sets, each set consisting of 6 games and each game ends when one player wins 4 points or wins two consecutive points at deuce points.
Rules
• A match consists of the best of three games of 21 points.
• The player/pair winning a rally adds a point to its score.
• At 20-all, the player/pair which first gains a 2-point lead wins that game.
• At 29-all, the side scoring the 30th point wins that game.
• The player/pair winning a game serves first in the next game.
What is a net drop?
It’s a very useful shot if your opponent is all the way in the back and he drops it to you towards the net and you want to drop it back to him while he’s all the way at the back of the court.
## What size is badminton court?
Badminton Courts have a length of 44′ (13.4 m), but double courts are 20′ (6.1 m) wide while single courts are reduced to 17′ (5.18 m); shrinking by 1.5′ (.
### What is the width of the net?
width n — the width of a net n. The width of a net is the length of its longest row, if any, else 0. For a string, the width and the height are the same.
#### What is 115 cm rule badminton?
According to the new rule, “the whole of the shuttle shall be below 1.15 metres from the surface of the court at the instant of being hit by the server’s racket”. Basically, the point of contact at the start of the serve cannot be more than 1.15 meters from the court.
What are the 10 badminton lets?
Here’s the full list of Badminton lets:
• Server and receiver both faulted.
• Shuttle getting caught on or in the net.
• Shuttle falls apart.
• Outside disruption, multiple shuttles on court.
• In or out? Too close to call.
• An accident or the unexpected occurs.
What is net lift in badminton?
A net lift is a stroke that is made from under the arm so to say. It is when the shuttle is caught closer to the ground as opposed to a toss which is caught when the shuttle is high up. It is generally played from the forecourt and is a return stroke to a drop or a net keep.
## What is the size of badminton ground?
Court dimensions The badminton court is 13.4m long and 6.1m wide. For singles the court is marked 5.18m wide. | 1,074 | 4,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-14 | longest | en | 0.953409 |
https://www.airmilescalculator.com/distance/ric-to-avp/ | 1,607,044,974,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00106.warc.gz | 548,802,501 | 20,702 | # Distance between Richmond, VA (RIC) and Wilkes-Barre, PA (AVP)
Flight distance from Richmond to Wilkes-Barre (Richmond International Airport – Wilkes-Barre/Scranton International Airport) is 278 miles / 447 kilometers / 241 nautical miles. Estimated flight time is 1 hour 1 minutes.
Driving distance from Richmond (RIC) to Wilkes-Barre (AVP) is 341 miles / 549 kilometers and travel time by car is about 6 hours 35 minutes.
## Map of flight path and driving directions from Richmond to Wilkes-Barre.
Shortest flight path between Richmond International Airport (RIC) and Wilkes-Barre/Scranton International Airport (AVP).
## How far is Wilkes-Barre from Richmond?
There are several ways to calculate distances between Richmond and Wilkes-Barre. Here are two common methods:
Vincenty's formula (applied above)
• 277.884 miles
• 447.210 kilometers
• 241.474 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 278.208 miles
• 447.732 kilometers
• 241.756 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Richmond International Airport
City: Richmond, VA
Country: United States
IATA Code: RIC
ICAO Code: KRIC
Coordinates: 37°30′18″N, 77°19′10″W
B Wilkes-Barre/Scranton International Airport
City: Wilkes-Barre, PA
Country: United States
IATA Code: AVP
ICAO Code: KAVP
Coordinates: 41°20′18″N, 75°43′24″W
## Time difference and current local times
There is no time difference between Richmond and Wilkes-Barre.
EST
EST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 66 kg (145 pounds).
## Frequent Flyer Miles Calculator
Richmond (RIC) → Wilkes-Barre (AVP).
Distance:
278
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
278
Round trip? | 510 | 2,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-50 | latest | en | 0.788276 |
https://unriskinsight.blogspot.com/2013/08/diffusion-reverted-what-was-temperature.html | 1,718,402,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00278.warc.gz | 540,870,241 | 16,980 | ## Pages
### Diffusion reverted - what was the temperature of Lake Traunsee?
In my recent post What is the temperature of LakeTraunsee, we started with a cold and a warm half of the lake
and calculated the lake temperature after while during which diffusion transported heat from the warm side to the cold one.
Now we ask: Can we mathematically revert this process? This means, we would measure the final temperature (with arbitrary precsion) and would like to know the (in our case) jumpy initial temperature.
To do some analysis, we observe that on the space-time domain (0,1) times (0,T)
builds a system of eigenfunctions for the heat equation
when Dirichlet boundary conditions u= 0 are to be satisfied. (In our case, the constant 20, which satisfies the heat equation just has to be added).
This means that if we can develop the initial condition into a Fourier sine series, then the solution can be calculated at any future time (always assuming that boundary conditions do not change. The "exp" in the eigenfunction damps high-frequent oscillations in the initial conditon.
Reverting?
With the same analysis, we could develop the final condition into a Fourier series. Reverting time means that the damping factor becomes an amplifying factor now. An error in the n-th Fourier coefficient is amplified by exp(n^2 pi ^2 t). We actually can observe this:
Backwards heat equation: 20 Fourier coefficients used.
However, so called regularization techniques help to overcome these oscillatory results. With a regularized singular value decomposition and a regularization parameter of 0.001 (top) and of 0.1 (bottom) we obtain:
This reproduction is reasonable, albeit not perfect. The jumps in the intial condition are the reason for slow convergence (in the appropriate Sobolev space).
Chapter 15 of the Workout deals with inverse problems arising in computational finance. | 410 | 1,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.907868 |
https://techcommunity.microsoft.com/t5/excel/time-out-of-100-vs-60/td-p/1823854 | 1,621,217,394,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991921.61/warc/CC-MAIN-20210516232554-20210517022554-00600.warc.gz | 578,607,349 | 76,098 | Time Out of 100 vs 60
Occasional Visitor
Time Out of 100 vs 60
Hello, I am looking for a very specific formula to solve all my problems. The program my work uses for time measures time in 60 minute ours (ex. 5:30pm closing time). Whenever an employee clocks in and out, the program automatically calculates their hours worked for the day. At the end of the pay period, a spreadsheet is auto-made that has everyone's worked hours for the week with time clocked in and out, as well as their hours worked in 60 minute hour intervals. (ex. 10 hours and 3 minutes worked shows as 10:03). I need help in coming up with a formula that will convert that to decimals for our payroll program. Payroll works in time out of 100, so 5 hours and 30 minutes of work is 5.5 hours in the payroll program. I am hoping there's some crazy formula that I can put into excel to turn my 5:30 hours worked to 5.5 hours worked for payroll. Any help would be a life saver and a huge time saver as well.
2 Replies
Betreff: Time Out of 100 vs 60
Here is an old work of mine, so you can calculate the time in decimal, just enter the start and end time of the work,
Where U was intended for vacation, K for sick, FB for further training.
You can rearrange the shape as you like or use the inserted file right away, if you like.
=IF(OR(I11="U",I11="K",I11="FB"),\$J\$2,IF(J11<I11,((J11+1)-I11)*24,(J11-I11)*24))
I would be happy to know if I could help.
Nikolino
I know I don't know anything (Socrates)
* Kindly Mark and Vote this reply if it helps please, as it will be beneficial to more Community members reading here.
Re: Time Out of 100 vs 60
Irrespective of whether you use the workbook offered, the conversion form a 'time' duration (measured as fractional part of a day) to hours (as a decimal number) is
= 24 * time | 480 | 1,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-21 | latest | en | 0.958252 |
http://www.slideserve.com/johana/part3 | 1,508,354,974,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823067.51/warc/CC-MAIN-20171018180631-20171018200631-00648.warc.gz | 558,868,376 | 19,694 | Born-Oppenheimer Approximation T N =0, V NN =ct
1 / 46
# part3 - PowerPoint PPT Presentation
Hartree-Fock equations. Born-Oppenheimer Approximation T N =0, V NN =ct. - Core integral (one-electron integral. - Coulombian integral (two-electron integral). - Exchange integral (two-electron integral).
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## PowerPoint Slideshow about 'part3' - johana
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Presentation Transcript
Hartree-Fock equations
Born-Oppenheimer Approximation TN=0, VNN=ct
- Core integral (one-electron integral
- Coulombian integral (two-electron integral)
- Exchange integral (two-electron integral)
Looking for the spin-orbitals which give the best EHF under the orthonormalization conditions and using the method of the Lagrange multipliers one obtains (see Jensen pg.62-63)
Hartree-Fock equations
Lagrangean multipliers are eigenvalues of the Fock operator:
- one-electron operator
• Hartree-Fock potential
• The average repulsive potential experienced by the i-th electron due to the remaining N-1 electrons
• Replaces the complicated 1/r12 two-electron operator (electron-electron repulsion is taken into account only in an average way)
With:
- represents the average local potential at position x1 arising from an electron located in j
Operating with Jj(x1) we obtain:
depends only on the value of i(x1) Jj is a local operator
The second term of vHF(x1) is the exchange contribution to the Hartree-Fock potential:
Kj(x1) exchanges the variables of the two spin-orbitals
depends on the value of i on all points in space because it depends on the integrating variable Kj(x1) is a non-local operator
JijKij0 Jii=Kii no self interaction in Hartree-Fock approximation
Conv.?
Since fi depends on i HF equations must be solved iteratively (SCF procedure)
No
Initial guess
vHF, f
{i}
Yes
SD, EHF
Molecular orbital energies:
Total energy:
For a closed-shell system in RHF formalism, the total energy and molecular orbital energies are given by (see Szabo and Ostlund, pag.83):
• Each occupied spin-orbital i contributes a term Hi to the energy
• Each unique pair of electrons (irrespective of their spins) in spatial orbitals i and j contributes the term Jij to the energy
• Each unique pair of electrons with parallel spins in spatial orbitals ψi and ψj contributes the term –Kij to the energy
Examples:
3
2
1
a)
b)
c)
d)
a) E=2H1+J12
b) E=H1+H2+J12-K12
c) E=H1+H2+J12
d) E=H1+2H2+H3+2J12+J22+J13+2J23-K12-K13-K23
Koopmans’ Theorem
What is the physical meaning of orbital energies?
Total electronic energy:
Summing all εi:
why?
because εi includes the Coulombian and exchange interactions with all the other electrons (also with εj). Similarly, εj will include also the interactions with εi, so that the interaction between the i-th and j-th electrons will be counted twice in the sum over orbital energies
So, what εi is?
Consider the ionization of the molecule (one electron removed from orbital number k) and suppose that no change of MO’s occurs during the ionization process
Koopmans’ theorem
vertical ionization potential (IP) photoelectron spectroscopy
(obtained without optimizing the cation geometry)
adiabatic ionization potential is obtained when the geometry of the cation is optimized
physical meaning of εi: the i-th orbital ionization energy!
Hartree-Fock-Roothaan Equations
LCAO-MO
i=1,2,...,K
{μ} – a set of known functions
The more complete set {μ}, the more accurate representation of the exact MO, the more exact the eigenfunctions of the Fock operator
The problem of calculating HF MO the problem of calculating the set cμi LCAO coefficients
matrix equation for the cμi coefficients
Multiplying by μ*(r1) on the left and integrating we get:
- Fock matrix (KxK Hermitian matrix)
- overlap matrix (KxK Hermitian matrix)
- Roothaan equations
More compactly:
FC=SC
where
• the matrix of the expansion coefficients
• (its columns describes the molecular orbitals)
The requirement that the molecular orbitals be orthonormal in the LCAO
approximation demands that:
The problem of finding the molecular orbitals {i} and orbital energies i involves solving the matrix equation FC=SC.
For this, we need an explicit expression for the Fock matrix
Charge density
For a closed shell molecule, described by a single determinant wave-function
The integral of this charge density is just the total number of electrons:
Inserting the molecular orbital expansion
into the expression for the charge density we get:
- elements of the density matrix
• the electronic population of the atomic overlap distribution
• give an indication of contributions to chemical binding when and
• centered on different atoms
- the net electronic charges residing in orbital
Where:
The integral of (r) is
By means of the last equation, the electronic charge distribution may be decomposed into contributions associated with the various basis functions of the LCAO expansion.
Off-diagonal elements
Diagonal elements
Population analysis ≡ allocate the electrons in the molecule in a fractional manner, among the various parts of the molecule (atoms, bonds, basis functions)
- Mulliken population analysis
Substituting the basis set expansion we get:
Basis set functions are normalized Sμμ=1
Pμμ - number of electrons associated with a particular BF
- net population of φμ
Qμ = 2PμSμ (μ≠)overlap population
associated with two basis functions which may be on the same atom or on two different atoms
Total electronic charge in a molecule consists of two parts:
first term is associated with individual BF
second term is associated with pairs of BF
- gross population for φμ
the net charge associated with the atom A
total overlap population between atoms A and B
Formaldehyde
Mulliken population analysis
#P RHF/STO-3G scf(conventional) Iop(3/33=6) Extralinks=L316 Noraff Symm=Noint Iop(3/33=1) pop(full)
Basis functions:
= sum over the line (or column) corresponding to the O(2px) basis function
Atomic populations (AP)
1 O 8.186789
2 C 5.926642
3 H 0.943285
4 H 0.943285
Total atomic charges (Q=Z-AP)
1 O -0.186789
2 C 0.073358
3 H 0.056715
4 H 0.056715
Basis Sets
with:
{μ} – a set of known functions
for UHF wave-functions two sets of coefficients are needed:
if μ AO LCAO-MO
if μ AO LCBF-MO
Basis functions
• mathematical functions designed to give the maximum flexibility to the molecular orbitals
• must have physical significance
• their coefficients are obtained variationally
Slater Type Orbitals (STO)
- similar to atomic orbitals of the hydrogen atom
- more convenient (from the numerical calculation point of view) than AO, especially when n-l≥2 (radial part is simply r2, r3, ... and not a polinom)
STO – are labeled like hydrogen atomic orbitals and their normalized form is:
• STO
• provide reasonable representations of atomic orbitals
• however they are not well suited to numerical (fast) calculations of especially two-electron integrals
• their use in practical molecular orbital calculations has been limited
STO
• Physically, the exponential dependence on distance from the nucleus is very close to the exact hydrogenic orbitals.
• Ensures fairly rapid convergence with increasing number of functions.
• Three and four center integrals cannot be performed analytically.
• No radial nodes. These can be introduced by making linear combinations of STOs.
• Practical Use:
• Calculations of very high accuracy, atomic and diatomic systems.
• Semi-empirical methods where 3- and 4-center integrals are neglected.
Gaussian Type Orbitals (GTO)
• introduced by Boys (1950)
• powers of x, y, z multiplied by
• α is a constant (called exponent) that determines the size (radial extent) of the function
or:
• N - normalization constant
• f - scaling factor
• scale all exponents in the related gaussians in molecular calculations
l, m, n are not quantum numbers
L=l+m+n - used analogously to the angular momentum quantum number for atoms to mark functions as s-type (L=0), p-type (L=1), d-type (L=2), etc (shells)
The absence of rn-1 pre-exponential factor restricts single gaussian primitives to approximate only 1s, 2p, 3d, 4f, ... orbitals.
However, combinations of gaussians are able to approximate correct nodal properties of atomic orbitals
GTO – uncontracted gaussian function (gaussian primitive)
- contracted gaussian function (gaussian contraction)
STO=
GTOs are inferior to STOs in three ways:
GTO’s behavior near the nucleus is poorly represented.
At the nucleus, the GTO has zero slope; the STO has a cusp. GTOs diminish too rapidly with distance. The ‘tail’ behavior is poorly represented.
Extra d-, f-, g-, etc. functions (from Cart. rep.)may lead to linear dependence of the basis set. They are usually dropped when large basis sets are used.
GTOs have analytical solutions. Use a linear combination of GTOs to overcome these deficiencies.
There are 6 possible d-type cartesian gaussians while there are only 5 linearly independent and orthogonal d orbitals
The gs, gx, gy and gz primitives have the angular symmetries of the four corresponding AO.
The 6 d-type gaussian primitives may be combined to obtain a set of 5 d-type functions:
gxy dxy
gxz dxz
gyz dyz
The 6-th linear combination gives an s-type function:
In a similar manner, the 10 f-type gaussian primitives may be combined to obtain a set of 7 f-type functions
GTOs are less satisfactory than STOs in describing the AOs close to the nucleus. The two type functions substantially differ for r=0 and also, for very large values of r.
cusp condition:
for STO: [d/dr e-ξr]r ≠ 0
for GTO:
With GTO the two-electron integrals are more easily evaluated. The reason is that the product of two gaussians, each on different centers, is another gaussian centered between the two centers:
where:
KAB=(2αβ/[(α+β)π])3/4exp(-αβ/(α+β)|RA-RB|2]
The exponent of the new gaussian centered at Rp is: p=α+β
and the third center P is on line joining the centers A and B (see the Figure below)
RP=(αRA+βRB)/(α+β)
The product of two 1s gaussian is a third 1s gaussian
allow a more rapidly and efficiently calculation of the two-electron integrals
GTO
have different functional behavior with respect to known functional behavior of AOs.
contractions (CGF or CGTO)
L – the length of the contraction
dpμ – contraction coefficients
How the gaussian primitives are derived?
• by fitting the CGF to an STO using a least square method
• varying the exponents in quantum calculations on atoms in order to minimize the energy
Example
STO-3G basis set for H2 molecule
Each BF is approximated by a STO, which in turn, is fitted to a CGF of 3 primitives
hydrogen 1s orbital in STO-3G basis set
For molecular calculations, first we need a BF to describe the H 1s atomic orbital
then: MO(H2) = LCBF
3 gaussian primitives:
exponent coefficient
0.222766 0.154329
0.405771 0.535328
0.109818 0.444636
If we use a scaling factor:
βi=αif2
! Using normalized primitives we do not need a normalization factor for the whole contraction
If the primitives are not normalized, we have to obtain a normalization factor. For this we use the condition:
S=F2[I1+I2+I3+2I4+2I5+2I6]
But:
so that:
Analogously:
and thus:
Now,
Imposing that S=1 we obtain:
In the general case of a contraction of dimension n, the above expression become:
Summary
The 1s hydrogen orbital in STO-3G basis set will be:
with:
- normalization factors for primitives
- normalization factor for the whole contraction (when un-normalized primitives or segmented contractions are used)
N=1.0000002
Explicitly:
If the exponents are not scaled:
STO-3G basis set example
http://www.chem.utas.edu.au/staff/yatesb/honours/modules/mod5/c_sto3g.html
This is an example of the STO-3G basis set for methane in the format produced by the "gfinput" command in the Gaussian computer program. The first atom is carbon. The other four are hydrogens.
Standard basis: STO-3G (5D, 7F) Basis set in the form of general basis input:
1 0
S 3 1.00
.7161683735D+02 .1543289673D+00
.1304509632D+02 .5353281423D+00
.3530512160D+01 .4446345422D+00
SP 3 1.00
.2941249355D+01 -.9996722919D-01 .1559162750D+00
.6834830964D+00 .3995128261D+00 .6076837186D+00
.2222899159D+00 .7001154689D+00 .3919573931D+00
****
2 0
S 3 1.00
.3425250914D+01 .1543289673D+00
.6239137298D+00 .5353281423D+00
.1688554040D+00 .4446345422D+00
****
3 0
S 3 1.00
.3425250914D+01 .1543289673D+00
.6239137298D+00 .5353281423D+00
.1688554040D+00 .4446345422D+00
****
4 0
S 3 1.00
.3425250914D+01 .1543289673D+00
.6239137298D+00 .5353281423D+00
.1688554040D+00 .4446345422D+00
****
5 0
S 3 1.00
.3425250914D+01 .1543289673D+00
.6239137298D+00 .5353281423D+00
.1688554040D+00 .4446345422D+00
****
Split valence basis sets
http://www.chem.utas.edu.au/staff/yatesb/honours/modules/mod5/split_bas.html
Valence orbitals are represented by more than one basis function, (each of which can in turn be composed of a fixed linear combination of primitive Gaussian functions). Depending on the number of basis functions used for the reprezentation of valence orbitals, the basis sets are called valence double, triple, or quadruple-zeta basis sets. Since the different orbitals of the split have different spatial extents, the combination allows the electron density to adjust its spatial extent appropriate to the particular molecular environment.
Split is often made for valence orbitals only, which are chemically important.
3-21G basis set
The valence functions are split into one basis function with two GTOs, and one with only one GTO. (This is the "two one" part of the nomenclature.) The core consists of three primitive GTOs contracted into one basis function, as in the STO-3G basis set.
1 0 //C atom
S 3 1.00
.1722560000D+03 .6176690000D-01
.2591090000D+02 .3587940000D+00
.5533350000D+01 .7007130000D+00
SP 2 1.00
.3664980000D+01 -.3958970000D+00 .2364600000D+00
.7705450000D+00 .1215840000D+01 .8606190000D+00
SP 1 1.00
.1958570000D+00 .1000000000D+01 .1000000000D+01
****
2 0 //H atom
S 2 1.00
.5447178000D+01 .1562850000D+00
.8245472400D+00 .9046910000D+00
S 1 1.00
.1831915800D+00 .1000000000D+01
****
6-311G basis set
Extended basis sets
The most important additions to basis sets are polarization functions and diffuse basis functions.
Polarization basis functions
The influence of the neighboring nuclei will distort or polarize the electron density near a given nucleus. In order to take this effect into account, orbitals that have more flexible shapes in a molecule than the s, p, d, etc shapes in the free atoms are used.
An s orbital is polarized by using a p-type orbital
A p orbital is polarized by mixing in a d-type orbital
6-31G(d)
a set of d orbitals is used as polarization functions on heavy atoms
6-31G(d,p)
a set of d orbitals are used as polarization functions on heavy atoms
and a set of porbitals are used as polarization functions on hydrogen atoms
Diffuse basis functions
For excited states and anions where the electronic density is more spread out over the molecule, some basis functions which themselves are more spread out are needed (i.e. GTOs with small exponents).
These additional basis functions are called diffuse functions. They are normally added as single GTOs.
6-31+G - adds a set of diffuse sp orbitals to the atoms in the first and second rows (Li - Cl).
6-31++G - adds a set of diffuse sp orbitals to the atoms in the first and second rows (Li- Cl) and a set of diffuse s functions to hydrogen.
Diffuse functions can also be added along with polarization functions.
This leads, for example, to the 6-31+G(d), 6-31++G(d), 6-31+G(d,p) and 6-31++G(d,p) basis sets.
Standard basis: 6-31+G (6D, 7F) Basis set in the form of general basis input:
1 0
S 6 1.00
.3047524880D+04 .1834737130D-02
.4573695180D+03 .1403732280D-01
.1039486850D+03 .6884262220D-01
.2921015530D+02 .2321844430D+00
.9286662960D+01 .4679413480D+00
.3163926960D+01 .3623119850D+00
SP 3 1.00
.7868272350D+01 -.1193324200D+00 .6899906660D-01
.1881288540D+01 -.1608541520D+00 .3164239610D+00
.5442492580D+00 .1143456440D+01 .7443082910D+00
SP 1 1.00
.1687144782D+00 .1000000000D+01 .1000000000D+01
SP 1 1.00
.4380000000D-01 .1000000000D+01 .1000000000D+01
****
2 0
S 3 1.00
.1873113696D+02 .3349460434D-01
.2825394365D+01 .2347269535D+00
.6401216923D+00 .8137573262D+00
S 1 1.00
.1612777588D+00 .1000000000D+01
****
Number of primitives and basis functions for 1,2-Benzosemiquinone free radical with the STO-3G basis set
Primitives:
atom C: nr.primitives = 15 x nr. atoms = 6 → 90
atom H: nr.primitives = 3 x nr. atoms = 4 → 12
atom O: nr.primitives = 15 x nr. atoms = 2 → 30
TOTAL: 132 GTO primitives
Basis functions:
atom C: nr. BF = 5 x nr.atoms = 6 → 30
atom H: nr. BF = 1 x nr.atoms = 4 → 4
atom O: nr. BF = 5 x nr.atoms = 2 → 10
TOTAL: 44BF
Number of primitives and basis functions for 1,2-Benzosemiquinone free radical with the 6-31+G* (6-31+G(d)) basis set
Primitives:
atom C: nr.primitives = 32 x nr. atoms = 6 → 192
atom H: nr.primitives = 4 x nr. atoms = 4 → 16
atom O: nr.primitives = 32 x nr. atoms = 2 → 64
TOTAL: 272 GTO primitives
Basis functions:
atom C: nr. BF = 19 x nr.atoms = 6 → 114
atom H: nr. BF = 2 x nr.atoms = 4 → 8
atom O: nr. BF = 19 x nr.atoms = 2 → 38
TOTAL: 160BF
Effective core potentials (ECPs)
• Core electrons, which are not chemically very important, require a large number of basis functions for an accurate description of their orbitals.
• An effective core potential is a linear combination of specially designed Gaussian functions that model the core electrons, i.e., the core electrons are represented by a effective potential and one treats only the valence electrons explicitly.
• ECP reduces the size of the basis set needed to represent the atom (but introduces additional approximations)
• for heavy elements, pseudopotentials can also include of relativistic effects that otherwise would be costly to treat
ECP potentials are specified as parameters of the following equation:
where p is the dimension of the expansion di are the coefficients for the expansion terms, r0 is the distance from nucleus and ξi represents the exponents for each term.
• Saving computational effort
• Taking care of relativistic effects partly
• Important for heavy atoms, e.g., transition metal atoms
ECP example
complexPd1.chk
complex Pd v1
0 2
C 8.89318310 9.90388210 6.72569337
C 9.52931379 8.77525770 6.27102032
H 9.29586123 7.93893890 6.60431879
C 10.52592748 8.89096200 5.30965653
H 10.95942133 8.13380930 4.98695425
C 10.85850598 10.13123090 4.84438728
H 11.51852449 10.22866610 4.19609286
C 10.20972534 11.23549650 5.34144511
etc.
etc.
H 4.15752044 17.83312399 10.48668123
H 5.63848578 17.14049639 11.10318367
N C O H 0
6-31G(d)
****
Pd 0
CEP-121G
****
Pd 0
CEP-121G
• Recomendations for basis set selection
• Always a compromise between accuracy and computational cost!
• With the increase of basis set size, calculated energy will converge (complete basis set (CBS) limit).
• Special cases (anion, transition metal, transtion state)
• Use smaller basis sets for preliminary calculations and for heavy duties (e.g., geometry optimizations), and use larger basis sets to refine calculations.
• Use larger basis sets for critical atoms (e.g., atoms directly involved in bond-breaking/forming), and use smaller basis sets for unimportant atoms (e.g., atoms distant away from active site).
• Use popular and recommended basis sets. They have been tested a lot and shown to be good for certain types of calculations. | 5,933 | 20,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-43 | latest | en | 0.838042 |
https://www.hackmath.net/en/word-math-problems/units?page_num=95 | 1,607,116,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00704.warc.gz | 702,452,820 | 10,470 | # Units - math word problems
#### Number of problems found: 3477
• Truncated cone
A truncated cone has a bases radiuses 40 cm and 10 cm and a height of 25 cm. Calculate its surface area and volume.
• AL wire
What is the weight of an aluminum wire 250 m long with a diameter of 2 mm, if the density of aluminum is p = 2700 kg/m cubic. Determine to the nearest gram.
• Storm and roof
The roof on the building is a cone with a height of 3 meters and a radius equal to half the height of the roof. How many m2 of roof need to be repaired if 20% were damaged in a storm?
• Cube into cylinder
If we dip a wooden cube into a barrel with a 40cm radius, the water will rise 10 cm. What is the size of the cube edge?
• Train speed
The train speed is decreased during 50 sec from 72 km/h to 36 km/h. Assuming that the train movement is equally slowing, find the the acceleration and the distance that it travels at.
• Supermarket cashiers
When at the supermarket are opened only 2 cash people waiting in the front approximately 12 minutes. How many will shorten the average waiting time in a front where supermarket open another three cashiers?
• Mukls
The working group, in which 6 workers would ordered job completed within 21 working days. How many workers must be accept yet that the contract will be completed in 14 working days?
• 3rd dimension
The block has a surface of 42 dm2 and its dimensions are 3 dm and 2 dm. What is the third dimension?
• Diamond ABCD
In the diamond ABCD is the diagonal e = 24 cm and size of angle SAB is 28 degrees, where S is the intersection of the diagonals. Calculate the circumference of the diamond.
• Cuboid and ratio
Find the dimensions of a cuboid having a volume of 810 cm3 if the lengths of its edges coming from the same vertex are in ratio 2: 3: 5
• Sisters
Sisters Hanka and Vera goes mostly tograndmother by bicycle. Journey took Hanka 30 minutes and Vera 20 minutes. How long will Vera catch up with Hanka when started from home five minutes later than Hanka?
• Thermometer
The thermometer showed -12 degrees Celsius in the morning then the temperature rises by 4 degrees and later again increased by 2 degrees at the evening has fallen by 5 degrees and then falls 3 degrees. What end temperature does the thermometer show?
• Dig water well
Mr. Zeman digging a well. Its diameter is 120 cm, and plans to 3.5 meters deep. How long (at least) must be a ladder, after which Mr. Zeman would have eventually come out?
• Aquarium height
How high does the water in the aquarium reach, if there are 36 liters of water in it? The length of the aquarium is 60 cm and the width is 4 dm.
• The car
The car has traveled the distance between A and B for four hour. If we increased the average by 17 km/h the car travel this distance an hour earlier. Determine the initial speed of the car and the distance between A and B.
• Average speed
The average speed of a pedestrian who walked 10 km was 5km/h, the average speed of a cyclist on the same track was 20km/h. In how many minutes did the route take more than a cyclist? Q
• Tin with oil
Tin with oil has the shape of a rotating cylinder whose height is equal to the diameter of its base. Canned surface is 1884 cm2. Calculate how many liters of oil is in the tin.
• Water depth
How many square meters of the inside of the pool wet the water if the pool is 25 meters long and 10 meters wide and the water depth is 1.2 meters everywhere.
• The sides
The sides of a rectangle are in a ratio of 2:3, and its perimeter is 1 1/4 inches. What are the lengths of its side? Draw it.
• Hydrostatic force
What hydrostatic force is applied to an area of 30 cm2 in water at a depth of 20 m? (Water density is 1000 kg/m3)
Do you have an interesting mathematical word problem that you can't solve it? Submit a math problem, and we can try to solve it.
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Please do not submit problems from current active competitions such as Mathematical Olympiad, correspondence seminars etc... | 1,002 | 4,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2020-50 | latest | en | 0.939659 |
http://heycere.com/statistics/central-limit-theorem/?lang=en | 1,560,801,290,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998558.51/warc/CC-MAIN-20190617183209-20190617205209-00294.warc.gz | 80,807,805 | 10,758 | Central limit theorem
Pocket
The Central Limit theorem is, together with the Normal Distribution, an essential point to understand statistical inference.
Central Limit Theorem
Let’s consider a sample with size $n$ from a large population which has average $\mu$ and variance $\sigma^2$. If the sample size $n$ is enough large, the distribution of the average of the samples $\bar{X}$ obeys a Normal distribution with average $\mu$, and variance $\sigma^2/n$.
This theorem has two important points.
1. Although population does not follow a Normal Distribution, the average of the sample follows a Normal Distribution. However, if the distribution tail of the population is fat (e.g. power-law distribution) and the average or/and standard deviation in the population become infinite, then theorem does not hold.
2. If we consider about standard deviation, the standard deviation of sample average $\bar{X}$ is $1/\sqrt{n}$ of population standard deviation $\sigma$. This means that by increasing the sample size $n$, the deviation between the sample average $\bar{X}$ and population average $\mu$ decreases. (By increasing the sample size $n$, the accuracy increases.)
If the population has infinite average or variance, the central limit theorem does not hold. In this case, we can use another theorem that involves the stable distribution. Did you know the case that the population distribution has infinite average or variance? A typical example is when the tail of the distribution follows a power-law, which is also called a fat-tailed ditribution.
Let us consider the case when the tail of the population distribution follows a power-law $f(k)\propto k^{-\gamma}$ with exponent $\gamma$.
(1) When the exponent $\gamma > 3$, the population average and variance are finite and the central limite theorem holds.
(2) When the exponent $3> \gamma > 2$, the population average is finite, but variance is infinite. In this case, the central limit theorem can not be applied.
(3) When the exponent $2> \gamma > 1$, both the population average and variance are infinite and the central limite theorem can not be applied either. | 452 | 2,135 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-26 | latest | en | 0.885116 |
http://stackoverflow.com/questions/14663071/translating-arrays-from-matlab-to-numpy/14663198 | 1,435,967,199,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096290.39/warc/CC-MAIN-20150627031816-00200-ip-10-179-60-89.ec2.internal.warc.gz | 242,578,629 | 17,817 | # Translating arrays from MATLAB to numpy
I am defining an array of two's, with one's on either end. In MATLAB this can be acheived by
``````x = [1 2*ones(1,3) 1]
``````
In Python, however, numpy gives something quite different:
``````import numpy
numpy.array([[1],2*numpy.ones(3),[1]])
``````
What is the most efficient way to perform this MATLAB command in Python?
-
``````In [33]: import numpy as np
In [34]: np.r_[1, 2*np.ones(3), 1]
Out[34]: array([ 1., 2., 2., 2., 1.])
``````
Alternatively, you could use `hstack`:
``````In [42]: np.hstack(([1], 2*np.ones(3), [1]))
Out[42]: array([ 1., 2., 2., 2., 1.])
``````
``````In [45]: %timeit np.r_[1, 2*np.ones(300), 1]
10000 loops, best of 3: 27.5 us per loop
In [46]: %timeit np.hstack(([1], 2*np.ones(300), [1]))
10000 loops, best of 3: 26.4 us per loop
In [48]: %timeit np.append([1],np.append(2*np.ones(300)[:],[1]))
10000 loops, best of 3: 28.2 us per loop
``````
Thanks to DSM for pointing out that pre-allocating the right-sized array from the very beginning, can be much much faster than appending, using `r_` or `hstack` on smaller arrays:
``````In [49]: %timeit a = 2*np.ones(300+2); a[0] = 1; a[-1] = 1
100000 loops, best of 3: 6.79 us per loop
In [50]: %timeit a = np.empty(300+2); a.fill(2); a[0] = 1; a[-1] = 1
1000000 loops, best of 3: 1.73 us per loop
``````
-
Thanks unutbu. In the future, how should I go about trying to find routines like this? – Doubt Feb 2 '13 at 15:07
It should be noted though that `r_` can be 4+ times slower than simply doing `2*ones(n+2)` and then patching the edges. This won't be an issue unless it's in an inner loop, of course. – DSM Feb 2 '13 at 15:08
@DSM: Could you elaborate? Perhaps it depends on the version of numpy? I compared `r_` with `hstack` and got roughly comparable results. – HappyLeapSecond Feb 2 '13 at 15:18
@unutbu: I mean versus `a = 2*np.ones(3+2); a[0] = 1; a[-1] = 1;`, which for me (1.6.2) is consistently about 4 times faster. – DSM Feb 2 '13 at 15:24
@Doubt: Experience helps. There is no shortcut. I learned a lot of numpy functions by just going through tutorials (like the numpy book), taking notes and writing example code snippets. If memory serves, I think I first saw `r_` used here. – HappyLeapSecond Feb 2 '13 at 15:27
Use numpy.ones instead of just ones:
``````numpy.array([[1],2*numpy.ones(3),[1]])
``````
-
That will give a very different object, not `array([1, 2, 2, 2, 1])` -- that's the OP's problem. (If the OP's code worked, probably a `from numpy import *` happened, or the OP simply copied the code incorrectly.) – DSM Feb 2 '13 at 15:01
Thanks, I have fixed this. Sorry for the confusion. – Doubt Feb 2 '13 at 15:05 | 937 | 2,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2015-27 | latest | en | 0.756576 |
http://radio.radiotrician.org/2019/07/intro-to-electronics-part-1-terms.html | 1,726,288,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00742.warc.gz | 24,290,420 | 10,696 | ## Saturday, July 27, 2019
### intro to electronics - part 1 - the terms
I plan to introduce the subject as if it was a day one student being taught. This may to basic for some but a good review has never hurt me. So we begin.
When the world was evolving towards our modern electronic wonder man kind had no knowledge of basic sciences. The alchemist was trying to turn lead to gold and doctors were bleeding their patients in an effort to cure them. Then the battery was invented. The battery was used to shock frog legs and make them jump! Electroplating was developed ...............
As the periodic table was developed man kind became aware of the characteristics of the elements and a new term was coined. The gram molecular weight which in a strange way becomes the basic for our world of electricity and electronics.................
To electroplate a gram molecular weight of a material requires a specific numbers of electrons. If we define this number of electrons as a coulomb it could be stated "when one gram molecular weight of XXXXX is deposited by the electroplating process one coulomb of electrons flowed through the solution."..........
A quick search would reveal the number but the actual number is not important. We use the unit to define our other terms, that is why it helps to know what it is.
Ohm's law says when we apply 1 volt of potential across a 1 ohm resistor we have 1 ampere of current flow. One coulomb of electrons flowing through a circuit in 1 second is 1 ampere. One ampere (amp) of current produces 1 watt of power in a 1 ohm resistor. So the units were developed based on current flow and the potential to produce it through a specific value of resistance.
So we can use Ohm's law to find voltage (E), current (I) or resistance (R) in a circuit.
E=I*R
R=E/I
I=E/R | 406 | 1,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-38 | latest | en | 0.963112 |
https://www.teacherspayteachers.com/Product/4OAA1-4OAA2-4OAA3-TASK-CARDS-2423278 | 1,529,390,670,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00443.warc.gz | 895,234,597 | 16,051 | Subject
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This set has 16 task cards covering three different standards;
CCSS.Math.Content.4.OA.A.1
Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
CCSS.Math.Content.4.OA.A.2
Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.1
CCSS.Math.Content.4.OA.A.3
Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
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Task Cards 4.OA.B.4 FACTORS AND MULTIPLES
OA.B.4 Assessment/ Factors and Multiples
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https://billsschedulechallenge.com/q73um6m8/ | 1,606,341,749,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184870.26/warc/CC-MAIN-20201125213038-20201126003038-00055.warc.gz | 207,872,974 | 9,031 | # Common Core Standards Math Grade 7 Worksheets
Chianna Flora November 17, 2020 Worksheet
1st grade math worksheets and my Mom’s math teaching style. Math won’t be as terrible as it seems if parents take interest in preparing their little ones for math before school age. I grew up not understanding how it is that people talk about math as difficult as they do, it was my best subject at school. It was easy because of my upbringing that ensured that math and I got acquainted long before school. My mother who was a primary grade teacher told me how she began teaching me math in different guises at home before I got to school age.
it does need is a change in attitude, and a solid foundation of basic skills on which to build. Mathematics worksheets can help you provide your preschooler with a solid grounding that will help them conquer math.
My students are engaged in the activity because they are always eager to find out what the next scene will be, and how the math problems will be nestled within. They also like how within each handout I inscribe the title in a way that fits with the theme of that particular scene – another attention catching technique. And since this review activity only takes about fifteen minutes of class time, it is quick yet extremely beneficial. Point is, whatever it takes to get students actively involved with the reviewing process where they are not bored and effectively reviewing grade level material in order to prepare them for state or quarterly assessments. Hopefully this has inspired you to develop exciting and engaging review worksheets for your class when needed and your students achieve as much as they can when it comes time to test.
As a parent, I’m very aware of what my own children are learning in school. For the most part, I’ve been happy with their progress, but as they rise in grade level, I’m starting to see more emphasis on a loose understanding of the concepts and less emphasis on skills–particularly skills with arithmetic of fractions. The main problem with what I see with my students and my own children is that kids are taught ”concepts” and are not taught skills–unless they’re lucky enough to have a teacher who knows better. Most particularly, children are not taught mastery of arithmetic with fractions. Unfortunately, virtually all of their future math education depends on being able to do fractional arithmetic.
The math worksheets are specially designed for kids and adults. They are very helpful in improving mathematical aptitude and skills. They can be easily used by school students as well as college goers. They are available from elementary to advanced level. You can also buy customized worksheets. Customized sheets can be planned according to the level of your school going child. You can find several types of sheets online and offline. You can choose among multiplication, Addition, Subtraction, Division, Geometry, Decimal, Shapes and Space worksheets. These sheets help the users to practice mathematical problems. Solving these problems become much easier with the help of mathematical worksheets.
Teachers are actually doing their best to educate children. The problems with education aren’t so much on the level of teachers as with the institution as a whole. It’s kind of like the state of communications in our country before the deregulation of the telephone companies. Before deregulation, one and only one advancement–the touch tone phone. After deregulation, well you have cell phones, the Internet, instant messaging, you name it! What dedicated teachers and parents need to do is to supplement public school instruction with strategies that work, that have always worked, to get kids to really master the fundamental skills of elementary math.
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Latest News | 951 | 4,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-50 | latest | en | 0.97321 |
https://eagle-cmr.com/symphony-environmental-dqkto/euler-circuit-calculator-cba689 | 1,632,884,337,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780061350.42/warc/CC-MAIN-20210929004757-20210929034757-00222.warc.gz | 254,309,004 | 5,521 | An Euler circuit is a circuit that uses every edge in a graph with no repeats. Title: Microsoft Word - eulerpath.doc Author: direland Created Date: 9/24/2003 5:29:07 AM Power, Voltage, Current & Resistance (P,V,I,R) Calculator. We can use the same vertices for multiple times. Inductance of Straight Wire & Electrode Calculator. Euler's Formula is used in many scientific and engineering fields. Required Value of Resistor for LED’s Circuit Calculator. Example. Buried in that proof is a description of an algorithm for nding such a circuit. (a) First, pick a vertex to the the \start vertex." Advance Voltage Drop Calculator and Voltage Drop Formula. (b) Find at random a cycle that begins and ends at the start vertex. The second is shown in … The graph below has several possible Euler circuits. euler path calculator, Euler’s circuit theorem The Euler characteristic for connected planar graphs is also V – E +F, where F is the number of faces in the graph, including the exterior face. Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student The problem is same as following question. 3, 4, 5 and 6 Band Resistor Color Code Calculators. vertex has even degree, then there is an Euler circuit in the graph. Euler Formula and Euler Identity interactive graph. An Euler path, in a graph or multigraph, is a walk through the graph which uses every edge exactly once. Our goal is to find a quick way to check whether a graph (or multigraph) has an Euler path or circuit. If a graph is connected and every vertex has an even number of edges, then it has at least one Euler circuit, a path that starts and ends at the same vertex and uses Below is an interactive graph that allows you to explore the concepts behind Euler's famous - and extraordinary - formula: e iθ = cos(θ) + i sin(θ) When we set θ = π, we get the classic Euler's Identity: e iπ + 1 = 0. How to find whether a given graph is Eulerian or not? “Is it possible to draw a given graph without lifting pencil from the paper and without tracing any of the edges more than once”. Fortunately, we can find whether a given graph has a Eulerian Path … To improve this 'Euler's method(1st-derivative) Calculator', please fill in questionnaire. A graph is called Eulerian if it has an Eulerian Cycle and called Semi-Eulerian if it has an Eulerian Path. Male or Female ? An Euler circuit is an Euler path which starts and stops at the same vertex. Get the free "Euler critical buckling load" widget for your website, blog, Wordpress, Blogger, or iGoogle. Here’s a couple, starting and ending at vertex A: ADEACEFCBA and AECABCFEDA. Zener Diode & Zener Voltage Regulator Calculator. Section 4.5 Euler Paths and Circuits Investigate! The problem seems similar to Hamiltonian Path which is NP complete problem for a general graph. Being a circuit, it must start and end at the same vertex. The Euler path is a path, by which we can visit every edge exactly once. Find more Engineering widgets in Wolfram|Alpha. When the starting vertex of the Euler path is also connected with the ending vertex of that path, then it is called the Euler Circuit. The Euler Circuit is a special type of Euler path. Eulerian Circuit is an Eulerian Path which starts and ends on the same vertex. | 773 | 3,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-39 | latest | en | 0.902152 |
https://docs.cupy.dev/en/v12.0.0a2/reference/generated/cupy.linalg.lstsq.html | 1,696,273,388,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511002.91/warc/CC-MAIN-20231002164819-20231002194819-00129.warc.gz | 236,855,091 | 17,498 | # cupy.linalg.lstsq#
cupy.linalg.lstsq(a, b, rcond='warn')[source]#
Return the least-squares solution to a linear matrix equation.
Solves the equation a x = b by computing a vector x that minimizes the Euclidean 2-norm || b - a x ||^2. The equation may be under-, well-, or over- determined (i.e., the number of linearly independent rows of a can be less than, equal to, or greater than its number of linearly independent columns). If a is square and of full rank, then x (but for round-off error) is the “exact” solution of the equation.
Parameters
• a (cupy.ndarray) – “Coefficient” matrix with dimension `(M, N)`
• b (cupy.ndarray) – “Dependent variable” values with dimension `(M,)` or `(M, K)`
• rcond (float) – Cutoff parameter for small singular values. For stability it computes the largest singular value denoted by `s`, and sets all singular values smaller than `s` to zero.
Returns
A tuple of `(x, residuals, rank, s)`. Note `x` is the least-squares solution with shape `(N,)` or `(N, K)` depending if `b` was two-dimensional. The sums of `residuals` is the squared Euclidean 2-norm for each column in b - a*x. The `residuals` is an empty array if the rank of a is < N or M <= N, but iff b is 1-dimensional, this is a (1,) shape array, Otherwise the shape is (K,). The `rank` of matrix `a` is an integer. The singular values of `a` are `s`.
Return type
tuple
Warning
This function calls one or more cuSOLVER routine(s) which may yield invalid results if input conditions are not met. To detect these invalid results, you can set the linalg configuration to a value that is not ignore in `cupyx.errstate()` or `cupyx.seterr()`. | 443 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.782975 |
http://catalog.uab.edu/coursedescriptions/qm/ | 1,516,737,039,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892238.78/warc/CC-MAIN-20180123191341-20180123211341-00589.warc.gz | 58,668,906 | 7,158 | # QM-Quantitative Methods Courses
### Courses
QM 214. Quantitative Analysis I. 3 Hours.
This course is an introductory course to statistics and data analyses for business students. Students will be exposed to basic statistical concepts and procedures to handle the data. Students are expected to recognize the nature of their data, select appropriate procedures, perform requisite calculations, demonstrate computer proficiency, and explain the results to layperson.
Prerequisites: (MA 105 [Min Grade: C] or MA 109 [Min Grade: C] or MA 125 [Min Grade: C]) and BUS 110 [Min Grade: C]
QM 215. Quantitative Analysis II. 3 Hours.
QM215 is the sequel of QM214 and a continuation of introductory statistics and data analyses. Built upon students' understanding of descriptive statistics and probability, this course exposes students to inferential statistics. Upon finishing the course, students are expected to know how to formulate hypotheses, collect data, select appropriate model(s), conduct statistical analyses, and present statistical findings with proper support of statistical graphs.
Prerequisites: QM 214 [Min Grade: C]
QM 350. Quantitative Methods for Finance. 3 Hours.
Development of the mathematical foundations of undergraduate level financial modeling and analysis, including applications of calculus, probability theory, linear algebra and Monte Carlo simulation to the measurement of asset returns and the assessment of risk, to the pricing of options and other financial derivatives, and to the solution of important financial optimization problems.
Prerequisites: (QM 215 [Min Grade: C] and GPAT and GPAO 2.00) and (QM 215 [Min Grade: C] and GPAU 2.00 and GPAO 2.00)
QM 400. Survey of Management Science. 3 Hours.
Application of mathematical and statistical techniques to management problems. Network planning techniques, linear programming, inventory systems, queuing theory, simulation, and decision analysis.
Prerequisites: (GPAT and GPAO 2.00 and AC 201 [Min Grade: C] and EC 210 [Min Grade: C] and EC 211 [Min Grade: C] and LS 246 [Min Grade: C] and QM 215 [Min Grade: C]) or (GPAU 2.00 and GPAO 2.00 and AC 201 [Min Grade: C] and EC 210 [Min Grade: C] and EC 211 [Min Grade: C] and LS 246 [Min Grade: C] and QM 215 [Min Grade: C])
QM 410. Nonparametric Statistics. 3 Hours.
Nonparametric methods applied to business decisions. Nonparametric tests for medians using one and two samples; tests for randomness and independence. Contingency tables and goodness of fit tests. All applications in area of behavioral sciences, particularly marketing and management. Completion of all pre-business requirements required.
Prerequisites: (EC 210 [Min Grade: C] and GPAT and GPAO 2.00) or (EC 210 [Min Grade: C] and GPAU 2.00 and GPAO 2.00) or (EC 211 [Min Grade: C] and GPAT and GPAO 2.00) or (EC 211 [Min Grade: C] and GPAU 2.00 and GPAO 2.00) or (QM 215 [Min Grade: C] and GPAT and GPAO 2.00) or (QM 215 [Min Grade: C] and GPAU 2.00 and GPAO 2.00)
QM 416. Sampling Techniques. 3 Hours.
Sampling procedures and application to estimation problems in business. Simple random, stratified, and cluster sampling reviewed and applied to simple and ratio estimators. Completion of all pre-business requirements required.
Prerequisites: (QM 215 [Min Grade: C] and EC 210 [Min Grade: C] and EC 211 [Min Grade: C] and GPAT and GPAO 2.00) or (QM 215 [Min Grade: C] and EC 210 [Min Grade: C] and EC 211 [Min Grade: C] and GPAU 2.00 and GPAO 2.00)
QM 420. Applied Forecasting. 3 Hours.
Practical use of various forecasting techniques on business and economic data. Topics include dynamic regression models, exponential smoothing, forecast criteria, moving averages, seasonality, and univariate Box Jenkins ARIMA modeling. Completion of all pre-business requirements required.
Prerequisites: (GPAT and GPAO 2.00 and AC 201 [Min Grade: C] and EC 210 [Min Grade: C] and EC 211 [Min Grade: C] and LS 246 [Min Grade: C] and QM 215 [Min Grade: C]) or (GPAU 2.00 and GPAO 2.00 and AC 201 [Min Grade: C] and EC 210 [Min Grade: C] and EC 211 [Min Grade: C] and LS 246 [Min Grade: C] and QM 215 [Min Grade: C])
QM 425. Applied Regression Analysis. 3 Hours.
Simple, multilinear, and polynomial regression analysis. Model selection, inferential procedures, and application with computer. Completion of all pre-business requirements required.
Prerequisites: (GPAT and GPAO 2.00 and EC 210 [Min Grade: C] and EC 211 [Min Grade: C] and QM 215 [Min Grade: C]) or (GPAU 2.00 and GPAO 2.00 and EC 210 [Min Grade: C] and EC 211 [Min Grade: C] and QM 215 [Min Grade: C])
QM 442. Statistics for Quality and Productivity. 3 Hours.
Application of statistics to improve quality and productivity throughout organization. Process analysis and improvement via numerical and graphical procedures illustrated with construction and interpretation of control charts. Tolerances, specifications, process capability studies, and elements of total quality program as espoused by Deming and Ishikawa.
Prerequisites: (MG 403 [Min Grade: C] and GPAT and GPAO 2.00) or (MG 403 [Min Grade: C] and GPAU 2.00 and GPAO 2.00)
QM 490. Advanced Topics in Statistics/Management Science. 3 Hours.
Statistics/management science application to problems in business and economics.
Prerequisites: (GPAT and GPAO 2.00) or (GPAU 2.00 and GPAO 2.00)
QM 499. Directed Readings in Quantitative Methods. 1-3 Hour.
Readings and independent study in selected areas.
Prerequisites: (GPAT and GPAO 2.00) or (GPAU 2.00 and GPAO 2.00) and EC 211 [Min Grade: C] and QM 215 [Min Grade: C] and EC 210 [Min Grade: C] | 1,479 | 5,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-05 | longest | en | 0.847447 |
https://byjus.com/question-answer/myoglobin-stores-oxygen-for-metabolic-processes-in-muscles-chemical-analysis-shows-that-it-contains-0-4/ | 1,709,503,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00619.warc.gz | 144,304,042 | 24,259 | 0
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Question
# Myoglobin stores oxygen for metabolic processes in muscles. Chemical analysis shows that it contains 0.35% Fe by mass. If there is one Fe atom and 6 N atoms per molecule of myoglobin, which of the following option is not correct about myoglobin? (Molar Mass of Fe = 56 g/mol)
A
Molar mass of myoglobin is 16000 g/mol
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B
One molecule of myoglobin contains 84 gm nitrogen atoms.
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C
% content of nitrogen in myoglobin is 0.525%
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D
If one molecule of myoglobin can absorb one molecule of oxygen gas then after absorption of oxygen there will 0.2% increase in mass.
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Solution
## The correct option is B One molecule of myoglobin contains 84 gm nitrogen atoms.Let, molar mass of myoglobin be x. As x contains 0.35% of Fe. So, x×0.35100=56 x=56×1000.35=16000 g/mol (a) Molar mass of Myoglobin is 16000 g/mol (b) As one molecule of myoglobin contains 6 nitrogen i.e. 84 amu nitrogen atom. (c) Fraction of nitrogen atoms present is given by =84×10016000 =0.525% (d) When one molecule of myoglobin absorbs one O2 (molar mass=32 g/mol) % increase of oxygen =32×10016000=0.2%
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Join BYJU'S Learning Program | 469 | 1,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-10 | latest | en | 0.826736 |
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1d kinematics worksheet ap physics pdf | 2,434 | 10,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-31 | latest | en | 0.872318 |
https://algorithm-wiki.csail.mit.edu/wiki/Diameter_3_vs_7 | 1,716,457,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00129.warc.gz | 63,951,563 | 5,857 | # Diameter 3 vs 7 (Graph Metrics)
## Description
Given a graph $G = (V, E)$, distinguish between diameter 3 and diameter 7. In other words, approximate diameter within a factor of $9/4-\epsilon$.
## Related Problems
Generalizations: Approximate Diameter
## Parameters
$n$: number of nodes
$m$: number of edges
## Table of Algorithms
Currently no algorithms in our database for the given problem.
## Reductions FROM Problem
Problem Implication Year Citation Reduction
3-OV If: to-time: $O(N^{({3}/{2}-\epsilon)}$ where $N=n^{2} d^{2}$ and $\epsilon > {0}$
Then: from-time: $n^{3-{2}\epsilon} poly(d)$ | 167 | 610 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.697797 |
https://www.dataunitconverter.com/kilobit-per-second-to-byte-per-minute | 1,713,140,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816904.18/warc/CC-MAIN-20240414223349-20240415013349-00726.warc.gz | 668,732,005 | 16,498 | # kbps to Byte/Min → CONVERT Kilobits per Second to Bytes per Minute
expand_more
info 1 kbps is equal to 7,500 Byte/Min
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
Input Kilobits per Second (kbps) - and press Enter.
## Kilobits per Second (kbps) Versus Bytes per Minute (Byte/Min) - Comparison
Kilobits per Second and Bytes per Minute are units of digital information used to measure storage capacity and data transfer rate.
Kilobits per Second is a "decimal" unit where as Bytes per Minute is a "basic" unit. One Kilobit is equal to 1000 bits. One Byte is equal to 8 bits. There are 0.008 Kilobit in one Byte. Find more details on below table.
Kilobits per Second (kbps) Bytes per Minute (Byte/Min)
Kilobits per Second (kbps) is a unit of measurement for data transfer bandwidth. It measures the number of Kilobits that can be transferred in one Second. Bytes per Minute (Byte/Min) is a unit of measurement for data transfer bandwidth. It measures the number of Bytes that can be transferred in one Minute.
## Kilobits per Second (kbps) to Bytes per Minute (Byte/Min) Conversion - Formula & Steps
The kbps to Byte/Min Calculator Tool provides a convenient solution for effortlessly converting data rates from Kilobits per Second (kbps) to Bytes per Minute (Byte/Min). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Kilobit) and target (Byte) data units.
Source Data Unit Target Data Unit
Equal to 1000 bits
(Decimal Unit)
Equal to 8 bits
(Basic Unit)
The conversion from Data per Second to Minute can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Kilobits per Second (kbps) to Bytes per Minute (Byte/Min) can be expressed as follows:
diamond CONVERSION FORMULA Byte/Min = kbps x 1000 ÷ 8 x 60
Now, let's apply the aforementioned formula and explore the manual conversion process from Kilobits per Second (kbps) to Bytes per Minute (Byte/Min). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Bytes per Minute = Kilobits per Second x 1000 ÷ 8 x 60
STEP 1
Bytes per Minute = Kilobits per Second x 125 x 60
STEP 2
Bytes per Minute = Kilobits per Second x 7500
Example : By applying the previously mentioned formula and steps, the conversion from 1 Kilobits per Second (kbps) to Bytes per Minute (Byte/Min) can be processed as outlined below.
1. = 1 x 1000 ÷ 8 x 60
2. = 1 x 125 x 60
3. = 1 x 7500
4. = 7,500
5. i.e. 1 kbps is equal to 7,500 Byte/Min.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Kilobits per Second to Bytes per Minute using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Kilobit ?
A Kilobit (kb or kbit) is a decimal unit of digital information that is equal to 1000 bits. It is commonly used to express data transfer speeds, such as the speed of an internet connection and to measure the size of a file. In the context of data storage and memory, the binary-based unit of Kibibit (Kibit) is used instead.
arrow_downward
#### What is Byte ?
A Byte is a unit of digital information that typically consists of 8 bits and can represent a wide range of values such as characters, binary data and it is widely used in the digital world to measure the data size and data transfer speed.
## Excel Formula to convert from Kilobits per Second (kbps) to Bytes per Minute (Byte/Min)
Apply the formula as shown below to convert from 1 Kilobits per Second (kbps) to Bytes per Minute (Byte/Min).
A B C
1 Kilobits per Second (kbps) Bytes per Minute (Byte/Min)
2 1 =A2 * 125 * 60
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kilobits per Second (kbps) to Bytes per Minute (Byte/Min) Conversion
You can use below code to convert any value in Kilobits per Second (kbps) to Kilobits per Second (kbps) in Python.
kilobitsperSecond = int(input("Enter Kilobits per Second: "))
bytesperMinute = kilobitsperSecond * 125 * 60
print("{} Kilobits per Second = {} Bytes per Minute".format(kilobitsperSecond,bytesperMinute))
The first line of code will prompt the user to enter the Kilobits per Second (kbps) as an input. The value of Bytes per Minute (Byte/Min) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Kilobits(kbit) are there in a Byte?expand_more
There are 0.008 Kilobits in a Byte.
#### What is the formula to convert Byte to Kilobit(kbit)?expand_more
Use the formula kbit = Byte x 8 / 1000 to convert Byte to Kilobit.
#### How many Bytes are there in a Kilobit(kbit)?expand_more
There are 125 Bytes in a Kilobit.
#### What is the formula to convert Kilobit(kbit) to Byte?expand_more
Use the formula Byte = kbit x 1000 / 8 to convert Kilobit to Byte.
#### Which is bigger, Kilobit(kbit) or Byte?expand_more
Kilobit is bigger than Byte. One Kilobit contains 125 Bytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,427 | 5,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.76117 |
https://www.studypool.com/discuss/544651/elementary-algebra-question-15-1?free | 1,508,673,572,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825227.80/warc/CC-MAIN-20171022113105-20171022133105-00169.warc.gz | 1,000,136,749 | 14,467 | ##### Elementary Algebra Question(15)
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Can I get help on the 15th question from this link?
May 18th, 2015
hello there welcome to study pool!
2x-9y=15----------------(1)
3x-2y=11---------------(2)
(1)*3-(2)*2
6x-27y-6x+4y=45-22
-23y=23
y=-1
from (1)
2x-9*(-1)=15
2x=15-9
x=6/2
x=3
so y=-1 and x=3
May 18th, 2015
I don't understand how you've got the first part.
May 18th, 2015
sorry i'll do it again for you
May 18th, 2015
i multifly the first equation by 3
then we got
3*2x-9y*3=15*3
simplify
6x-27y=45------------(3)
and also i have multiflied 2nd equation by 2
then we got
3x*2-2y*2=11*2
simplify
6x-4y=22-----------------(4)
then i have substract (4) by (3)
then we got
6x-27y-6x-4y=45-22
-23y=23
y=-1
got it now?
May 18th, 2015
still have any doubt ?
May 18th, 2015
...
May 18th, 2015
...
May 18th, 2015
Oct 22nd, 2017
check_circle | 391 | 958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-43 | latest | en | 0.871204 |
https://discuss.leetcode.com/topic/18522/one-line-java-solution-using-bitcount | 1,513,347,661,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948572676.65/warc/CC-MAIN-20171215133912-20171215155912-00636.warc.gz | 555,067,593 | 12,136 | # One line java solution using bitCount
• This is kind of cheating, but the idea is that a power of two in binary form has and only has one "1".
``````public class Solution {
public boolean isPowerOfTwo(int n) {
return n>0 && Integer.bitCount(n) == 1;
}
}``````
• This post is deleted!
• ``````public boolean isPowerOfTwo(int n) {
if(n<=0)
return false;
int log_base_2= (int)(Math.log(n)/Math.log(2));
return (((n & (1<< log_base_2)) ==n)?true:false);
}
``````
I used shift operator !
• Same idea here. Count the number of ones (should be 1) by right shifting n by 32 times (be sure to kick out 0x80000000).
``````public boolean isPowerOfTwo(int n) {
if (n == 0x80000000) return false;
int ones = 0;
for (int i = 1; i <= 32; ++i) {
ones += n & 1;
n = n >> 1;
}
return ones == 1;
}
``````
• @qxx I have the same, but this is actually not efficient =(
• ``````public static boolean isPowerOfTwo(int n) {
return n > 0 && Integer.highestOneBit(n) == Integer.lowestOneBit(n) ? true : false;
}
``````
This is my solution. so easy to understand.
• What is the runtime complexity of Integer.bitCount()? Is it O(n) where n is the number of bits of the integer?
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 379 | 1,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-51 | latest | en | 0.748794 |
https://fractalforums.org/fractal-mathematics-and-new-theories/28/m-brot-distance-estimation-versus-claudes-fake-de/382/0;wap2 | 1,631,848,097,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780054023.35/warc/CC-MAIN-20210917024943-20210917054943-00493.warc.gz | 320,063,127 | 3,925 | Fractal Related Discussion > Fractal Mathematics And New Theories
M-brot distance estimation versus Claude's "fake" DE
(1/2) > >>
gerrit:
Looking at the formulas for the true distance estimation (DE) and the normalized (smoothed) iteration count (NC), it seems that the norm of the gradient of the NC (g), proposed by Claude as "fake DE" (FDE) is in fact identical to the "true" DE, if we take FDE=1/(g*log(2)), in the limit of infinite escape radius and inf X inf grid. Except it is not calculated analytically during the iterations, but numerically later on using finite differences.
Numerically however we get g=0 at some locations, so I tried a "regularization": FDE = 1/((g+ep)*log(2)) where I guess ep such that when g=0 FDE equals dx/4 (maximum distance) where dx is the size of the region. This gives ep = 1/(log(2)*dx/4).
g is here the norm of the finite difference (viz. f(x+1)-f(x)) divided by h=pixel spacing, i.e., the actual numerical estimate.
With this regularization the true DE and "fake" DE look almost identical.
Examples below comparing the "fake-DE" (titled Claude DE) and the "true" DE.
gerrit:
What I wrote was not correct. Simply FDE=1/(g*log(2)) when g>0 and undefined when g=0 (i.e., in interior).
With this I computed FDE and true DE for some examples. Error as measured by mean(FDE-DE)/max(DE) for various gradient finite difference formulas was typically of the order of 0.05% for the Diag2X2 gradient method, but about 100 times higher for the central and forward formulas, which I didn't expect (or I could have a bug).
gerrit:
--- Quote from: gerrit on October 01, 2017, 03:32:59 AM ---Simply FDE=1/(g*log(2)) when g>0 and undefined when g=0 (i.e., in interior).
--- End quote ---
I investigated some more. Theoretically analytical distance estimate (DE) d is related to numerical gradient norm g of smoothed iteration map through
d = 1/(g*log(2)) when g>0 and undefined when g=0 (i.e., in interior). (Natural log).
On an example I computed numerical gradient norm g(x) (and analytical DE) with four methods:
1) diag: finite difference (FD) over diagonals of 2X2 stencil with x bottom left.
2) forward: FD from x to all its 8 neighbours (left/right and diagonal)
3) central: FD across x (viz (f(x+1)-f(x-1))/2) in all 4 directions (left/right and diagonal)
4) xy: just FD in positive x and y direction
3) has order h^2 error, others are O(h).
Following table lists difference between analytic and numeric DE on an example for four methods, and antialiasing factor (AA ).
Metric is 100% *||num DE - an. DE||/||an DE|| over the image.
--- Code: --- AA diag forward central xy
1 5.5900 3.6700 3.390 7.0500
2 2.8600 1.6500 1.520 2.5100
3 2.0500 2.0500 0.949 1.7700
4 1.4400 0.7340 0.6770 1.2800
5 1.1400 0.5640 0.5210 1.0800
--- End code ---
claude:
--- Quote from: gerrit on October 04, 2017, 05:24:05 AM ---Following table lists difference between analytic and numeric DE on an example for four methods, and antialiasing factor (AA ).
Metric is 100% *||num DE - an. DE||/||an DE|| over the image.
--- End quote ---
I guess that is the mean of the realtive error over the image. I think it would be interesting to see more statistics on the error, like maybe RMS average, or min/max values, maybe even histogram - perhaps a few very bad points are spoiling it for the rest of them?
gerrit:
--- Quote from: claude on October 04, 2017, 10:20:59 PM ---I guess that is the mean of the realtive error over the image. I think it would be interesting to see more statistics on the error, like maybe RMS average, or min/max values, maybe even histogram - perhaps a few very bad points are spoiling it for the rest of them?
--- End quote ---
||.|| stands for the 2-norm, i.e., RMS. 1-norm is about the same, infiniti norm (=max) is large and all over the place (the few bad points you mention). I didn't save those numbers.
I plan to look at another example when I have some time. | 1,155 | 4,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-39 | latest | en | 0.895139 |
https://trickledown.wordpress.com/2008/06/23/when-to-use-the-composite-functionchain-rule-for-derivatives/ | 1,521,941,399,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651481.98/warc/CC-MAIN-20180325005509-20180325025509-00498.warc.gz | 753,213,053 | 23,357 | ## When to Use the Composite Function/Chain Rule for Derivatives
Ah, the composite function rule/chain rule. (Wikipedia) (Mathematics for Economists).
When would you want to use the composite function/chain rule? (Note: I’m no math expert, so take this all with a grain of salt). Well, if you have a function that’s a function of another function, i.e. a composite function, sometimes the easiest way to find the derivative of the composite function is to use the composite function rule/chain rule.
From Wikipedia:
In mathematics, a composite function, formed by the composition of one function on another, represents the application of the former to the result of the application of the latter to the argument of the composite. The functions f: XY and g: YZ can be composed by first applying f to an argument x and then applying g to the result. Thus one obtains a function g o f: XZ defined by (g o f)(x) = g(f(x)) for all x in X. The notation g o f is read as “g circle f“, or “g composed with f“, “g following f“, or just “g of f“.
It may be helfpul to think about what a function is. Functions are generally formulas that you apply to some input, and which “map” the input to some output. The “value” of the output, the dependent variable, is usually named some variable like y, and the name of the function is usually something like f or g; the input, the independent variable, is usually named some variable like x, inside parentheses next to the name of the function, like f(x). From the http://www.math.csusb.edu/ website:
A function (or map) is a rule or correspondence that associates each element of a set X called the domain with a unique element of another set Y called the codomain. We typically give the rule a name such as a letter like f or g (or any letter of your choice) or a name agreed upon by convention like sine or log or square root.
Now, functions can be very simple, such as y=f(x)=x, in which case the function basically doesn’t do anything but map x back to itself. You can have more complicated functions such as $y=f(x)=x^3 + 2x + 5$, a polynomial, which does quite a few things to the input x before outputting the output value y.
Functions are interesting because basically anything in a mathematical expression can be called a function. Take $y=x^3 + 2x + 5$ for example. You could say $x^3$ is a function which maps x to some variable z, and you could name the function g(x). You could say 2x is a function which maps x to some variable u, and you could name the function h(x). You could even say 5 is a function which maps x to the constant 5 each time and name the function i(x) and the name the constant c. You can write 5 as a function of x here if you want to, $c=5 * x^0=5*1=5$. So pretty much anything in a math expression can be called a function, even constants.
So what about composition of functions? This is another area where I think you can basically find a function to be a composition of functions whenever you want–but there are only certain circumstances in which it matters enough for you to think about using the composite function rule.
One example of a situation in which you have a noticeable composite function is when instead of a lone x or some other independent variable within the parentheses of the function notation, you have other things going on, such as y=f(5x) instead of just y=f(x). In this case, the 5x within the parentheses is a whole other function, you could name the function g for example, and name the output of the function g(x) a dependent variable such as u, and then you would have u=g(x)=5x. Then since y=f(5x), and 5x=u, y is a function of u, a function of the function g(x), and also a function of x, since u is a function of x. In this case we have the composite function y=f(u)=f(g(x))=f(5x).
Now, here’s how to use the composite function rule/chain rule (see Wikipedia and Mathematics for Economists). To find dy/dx, you can first find dy/du then multiply that times du/dx. What if $y=f(x)=u^2$ and $u=g(x)=5x$? Then y=f(u)=f(g(x). By the power rule, dy/du would be 2u. Then, where u=g(x)=5x, du/dx would equal 5. By the composite function rule, the derivative dy/dx = dy/du * du/dx = 2u * 5 = 2*5x *5 =50x. This is why I said that there are some cases in which you want to use the composite function rule and in other cases you won’t need to think about it: in this case it might have been simpler to distribute the power of 2 in the beginning, so if we had y= (5x)^2, then y=25x^2, and using the power rule then dy/dx=50x, which is what we got by using the composite function rule above. So sometimes you can simplify first or figure the problem out without explicitly using the composite function/chain rule, and other times it’s easier to start out by using the composite function/chain rule.
When you are multiplying or dividing terms with the variable you are differentiating with respect to, when you are multiplying different functions (see the above about how just about anything can be called a function), in order to differentiate the resulting function, a function which is a product or quotient of two other functions, you can use the product and quotient rules. Once again, you only need to use these rules when it would be easier than multiplying out or dividing out the functions, or when the functions can’t be simplified any further. For example, if you had y=$5x^2 * 3x^3$ you might as well just multiply this out and then take the derivative of the result. You could have $5x^2 * 3x^3=15x^5$, then use the power rule to get dy/dx=75x^4. or you could use the product rule to get the same result, but it would take more effort. You could even use the product rule on y=f(x)=5x, since $5=5*x^0$, and here $y =5*x^0 * x$, in case you were wondering; there are many functions where there’s no point in using the product rule. But if the functions you start with are complicated enough, it can be simpler and easier to use the product rule to begin with instead of multiplying out the functions then taking the derivative of the product. (See product and quotient rules, Wikipedia and Mathematics for Economists). And to sum up, when the output of one function is the input into another function, then you use the composite function rule/chain rule to find the derivative. (See composite function rule/chain rule, Wikipedia and Mathematics for Economists).
### 4 Responses
1. Yep! Sounds pretty right on to me!
2. very interesting,but would have been more interesting and better if yohad used less wordings.
Maths about simple steps and English essay,hope you don’t mind.No hard feelings at all.
Good work.
Cheers
• This might not have hit the mark entirely and is very rough, but I believe the sparse, terse explanations given in most math textbooks only fit with the learning preferences of a small subset of students. Hence the popularity of books like the “Dummies” series, etc. There are a ton of non-math/non-science, humanities oriented people who could benefit from more context and explanation, rather than typical, terse “here’s a theorem, here’s what obviously follows from it” math textbook fare. Those math books basically are tailored to a small subset of people whose minds think that way to begin with, not for people who may need a little more explanation of how/what/why to ease their minds into hwo the math actually works.
3. …so i wa meant to say and ‘not English essay’ | 1,772 | 7,416 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-13 | latest | en | 0.902284 |
https://id.scribd.com/document/217001229/CBSE-Class-12-Engineering-Drawing-Solved-Sample-Paper-02-for-2011 | 1,569,144,174,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575402.81/warc/CC-MAIN-20190922073800-20190922095800-00330.warc.gz | 529,386,546 | 78,113 | Anda di halaman 1dari 9
# Model Test Paper No: 2
ENGINEERING DRAWING
Time allowed: 3 hours. Note: (i) Attempt all the questions. (ii) Use both sides of the drawing sheet, if necessary. (iii) All dimensions are in millimeters. (iv) Missing and mismatching dimensions, if any, may be suitably assumed. (v) Follow the SP: 46 -1988 codes, (with First Angle method of projection). (vi) In no view of question 1, are hidden edges / lines required. (vii) In question 3, hidden edges/ lines are to be shown in all full views. Q.1. (a) Construct an isometric scale. 4 (b) Construct an isometric projection to isometric scale, of the frustum of a regular hexagonal pyramid, having its axis vertical and two base edges parallel to V.P. The top edge = 20 mm, base edge = 40 mm and height = 80 mm. Give all the dimensions. Draw the axis and indicate the direction of viewing. 7 (c) A sphere of diameter 80 mm is resting centrally on its curved surface on top of a vertical pentagonal prism, having its base edge = 40 mm and height = 80 mm, keeping one of its rectangular face, in front, parallel to the V.P. Draw an isometric projection of the combination of solids. Draw the common axis and indicate the direction of viewing. Give all the dimensions. 14 Q.2. (a) Draw to scale 1 : 1, the standard profile of a METRIC SCREW THREAD (external), taking an enlarged pitch = 50 mm. Give all the standard dimensions. 9 OR Draw to scale 1 : I, the front view and top view of a SQUARE NUT, for a nominal diameter d = 20 mm, keeping its axis perpendicular to H.P. and two sides of the square parallel to V.P. Give all the standard dimensions. (b) Sketch freehand, the front view and the top view of a COUNTER SUNK HEAD(60) RIVET of diameter = 20 mm, keeping its axis vertical. Give all the standard dimensions. 6 OR Sketch freehand, three views of a WOODRUFF KEY for using on a shaft of diameter = 60 mm. Give all the standard dimensions. Maximum marks: 70
13
Q.3.
Fig.1. shows the parts of a KNUCKLE JOINT. Assemble the parts correctly and then draw the front view, showing upper half in section, using the scale 1 : 1. 24 Print title and scale used. Give 6 important dimensions. 6
14
OR Fig. 2. Shows the assembly of an UNPROTECTED FLANGE COUPLING. Disassemble the parts and draw the following views to scale 1 :1. (a) FLANGE A (i) Front view, upper half in section. (ii) Side view, viewing from the left hand side. (b) SHAFT A (i) Front view., (ii) Side view, viewing from the right hand side. 18
Print titles, symbol of projection and scale used. Draw the projection symbol. Give 6 important dimensions. 6
15
## Model Test Paper No: 2
Value Points
Q.1 (a) Isometric Scale 4
(i) Drawing 45 inclined lines showing true length. 1 (ii) Projections on 30 inclined line showing isometric length with 1 mm sub divisions. 2 (iii) Writing titles, subtitles and angles. 1
(b) Isometric Projection of Frustum of a Hexagonal Pyramid. (i) (ii) (iii) (iv) Helping view of a hexagon. Drawing hexagons. Drawing slant edges. Axis, Dimensioning, Direction of Viewing.
16
7 1 2 2 2
(c) Isometric Projection of Combination of solids. (i) (ii) (iii) (iv) Helping view of a pentagon. Drawing isometric pentagons with vertical lines. Drawing Sphere with true radius from isometric center. Common axis, Dimensioning, Direction of Viewing.
14 1 5 4 4
17
Q.2
(a) Metric Screw Thread Profile (External) (i) Drawing two crusts and two roots. (ii) Drawing conventional break with sectioning. (iii) Writing title and standard values.
9 4 3 2
OR (a) Square Nut (i) Drawing Front View across flat with details. (ii) Drawing Top View with details. (iii) Writing title and standard values. 9 4 3 2
18
Q.2
(b) Counter Sunk Head Rivet (i) Sketching head with shank. (ii) Sketching conventional top view. (iii) Writing title, standard values and axis.
6 3 1 2
OR (b) Woodruff Key (i) Sketching Front View, Top View and Side View. (ii) Writing title and standard values. 6 4 2
19
Q.3
## Assembly of Knuckle joint
30
(a) Front View, Upper Half in section. 24 (i) Drawing eye end, fork end and sectioning lines as per convention for upper half. 12 (ii) Drawing eye end and fork end for lower half. 6 (iii) Bolt Pin, Collar in section with Taper Pin in position. 6 (b) Others (i) Important Dimensions. (ii) Titles, Symbol of Projection, scale and line work. 6 2 4
20
## OR Q.3 A Disassembly of Flange Coupling FLANGE-A 30 18 11 7 4 7 5 2 6 4 2 6 2 4
(a) Front View, Upper half in section. (i) Drawing sectional upper half with hole and key way. (ii) Drawing lower half. (b) Side View, viewing from the left hand side. (i) Drawing four circles with key way. (ii) Drawing four PCD holes. B SHAFT-A
(i) Drawing conventional Front view with key way. (ii) Drawing Side View, viewing from the right hand side. C Others
(i) Important Dimensions. (ii) Titles, Symbol of Projection, scale and line work.
21 | 1,368 | 4,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-39 | latest | en | 0.873288 |
https://math.stackexchange.com/questions/1736192/interesting-property-of-primes-involving-modulo | 1,627,720,437,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154085.58/warc/CC-MAIN-20210731074335-20210731104335-00136.warc.gz | 395,496,671 | 38,298 | # Interesting Property of Primes involving Modulo?
## Primes & Modulo
What I have observed is that for the following expression, choose a positive integer $m$, and if it is prime then for positive integers $n=1,2,3,\ldots$ the results will be $0,1,1,1,\ldots$ such that the sequence is made of $0$'s and $1$'s which repeat in a way that there is a $0$, then $(m-1)$ $\times$ $1$'s and then again $0$ and $1$'s...
$$\sum^{m-1}_{k=0} n^k\mod{m}$$
For $m=2$ we have for $n=1,2,3,\ldots$ following results: $0,1,0,1,0,1,0,1,0,1,\ldots$
For $m=3$ we have: $0,1,1,0,1,1,0,1,1,0,1,1,\ldots$
For $m=5$ we have: $0,1,1,1,1,0,1,1,1,1,0,1,\ldots$
And for non-prime numbers we get a specific mix of repeating numbers that repeat every $m$ numbers, for example:
$m=6$, We get a repeating sets of $0,3,4,3,0,1$
$m=8$, We have: $0,7,0,5,0,3,0,1$
I have accidentally stumbled upon this property, in fact I don't know if its 100% true since I've checked a handful of numbers only.
## Questions
Is there an explanation for this property, why does it work for prime numbers like that?
Is there any significance or any kind of a pattern in numbers that appear for non-prime numbers?
Is there any significance to this expression, and does this expression or something similar appear somewhere in mathematics? In a specific field, equations, theorems... somewhere in the mathematical history... or simply mentioned somewhere else before?
• The $0$ terms are probably obvious, since $p^m\equiv 0\pmod p$... – abiessu Apr 10 '16 at 16:06
• That they repeat (at least) every $m$ numbers is a consequence of modular arithmetic: If $u \equiv v \mod m$, then $\sum_{k=0}^{m-1} u^k \equiv \sum_{k=0}^{m-1} v^k \mod m$. That they are $1$'s for prime $m$ and for $n \nmid m$ follows from the geometric series formula $\sum_{k=0}^{m-1} n^k = \dfrac{1-n^m}{1-n}$ and Fermat's little theorem (which says that $n^m \equiv n \mod m$). (You need to argue that $1-n$ is coprime to $m$ and thus can be cancelled.) – darij grinberg Apr 10 '16 at 16:08
The sum of the power series can be written as:
$$\sum_{k=0}^{m-1} n^k = \frac{n^m-1}{n-1}$$
You then need to know the result:
When $m$ prime and $n$ an integer, $n^m-n$ is divisible by $m$.
So if $n-1$ is not divisible by $m$, then $\frac{n^m-1}{n-1} = \frac{n^m-n}{n-1} + 1\equiv 1\pmod {m}$.
When $n-1$ is divisible by $m$, then $n^k-1$ is divisible by $m$ for all $k$, and thus $$\sum_{k=0}^{m-1} n^k\equiv \sum_{k=0}^{m-1} (n^k-1) \equiv 0\pmod {m}$$
• $n=1$ is covered by the case where $n-1$ is divisible by $m$. @user254665 – Thomas Andrews Apr 10 '16 at 17:25
• I mean for the sum of the power series. I'm just nit-picking. – DanielWainfleet Apr 10 '16 at 17:58 | 974 | 2,700 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-31 | latest | en | 0.866074 |
http://www.docstoc.com/docs/79312810/L07-Photometry | 1,430,045,434,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246654264.98/warc/CC-MAIN-20150417045734-00146-ip-10-235-10-82.ec2.internal.warc.gz | 456,241,212 | 41,450 | # L07 Photometry by ert554898
VIEWS: 12 PAGES: 6
• pg 1
``` Vision Science II - Monocular Sensory Aspects of Vision
Lecture 7 - Radiometry, Photometry, the V(λ ) function
A.
Q. What is the difference between radiometry and photometry?
A.
Energy and power
The basic unit of energy is the joule. Power is defined as energy per unit time. The basic unit for power is
the watt.
1 watt = 1 joule / 1 second
A light bulb rated at 60 watts of radiant power produces 60 joules of energy each second. If it’s left on for 10
seconds it produces 600 joules of energy. This describes the total energy output of the light source.
A point source emits radiant energy in all directions. If located at the center of a sphere, its energy or power
would be distributed across the inside surface of the sphere. In some cases you need to know how
concentrated the power is. That is, how much power is contained within a defined volume. The amount of
power contained within a defined cone-shaped volume is termed the radiant intensity. The more power in
the cone, the greater its radiant intensity. The unit for solid angle (cone size) is the steradian (Schwartz Fig.
4-5).
2
Steradians = Area at cone opening/(cone length)
To get a feel for the size of one steradian, imagine a fat ice cream cone with an opening that is 6.75” in
diameter and has a length of 6”. Such a cone has an angular volume of about 1 steradian. Radiant
intensity quantifies the concentration of light coming from a point source only and is expressed in
An extended source can be thought of as a collection of points. The amount of radiant power emitted from
2
2
The amount of radiant power falling on a surface is the irradiance. Irradiance is expressed in watts/m . Be
careful not to confuse radiance and irradiance. Radiance refers to the energy emitted from a surface (energy
off). Irradiance refers to energy falling onto a surface (energy on).
PHOTOMETRY
With this general background on radiometry, we will next study the topic of photometry. Radiometry
describes the physical properties of light, but photometry describes light from a perceptual frame of
reference. Radiometry measures energy; photometry quantifies light as it would be perceived by a standard
human observer.
In optometry we are usually more interested in photometry than radiometry, but there are certain cases in
which radiometry is more relevant. For example, when studying the effect of lasers on ocular tissues, we are
Lecture 7 – Photometry, V(λ)
not so concerned about how bright it looks, but rather, how much energy is transferred to the tissues. In that
case, you would be more interested in radiometric than photometric data.
Photometry is closely related to radiometry, but be careful not to confuse them. Obviously for visible light,
more energy or more power will make it appear brighter. That is why a 100-watt light bulb appears brighter
than a 60-watt bulb.
When comparing the apparent brightness of lights of different wavelength, you must take into account the
sensitivity of the eye for different wavelengths. For example a 5-mW laser green laser will look brighter than
a 5-mW red laser, because the eye is more sensitive to green than red light.
THE CIE LUMINOSITY FUNCTION
For simplicity, let’s first consider a monochromatic light source that is visible to the human eye. In order to
compute its brightness for a standard observer, that is, the photometric intensity of the light, you must
know its:
• wavelength
• the eye’s sensitivity at that wavelength
The eye’s sensitivity to different wavelengths, for a standard observer, was established by the CIE
(Commission Internationale de l’Eclairage or International Commission on Illumination) in 1924. This
standardized data set is fundamental to the field of photometry and is known as the
• luminous efficiency function of the human eye
• spectral luminosity function
• photopic luminosity function or
• 1924 CIE V(λ) function or V lambda curve for cone (photopic) vision
This function describes the normal relative sensitivity of the eye for different wavelengths under light-adapted
conditions. That is, during daylight rather than night viewing conditions. In this case, the cone
photoreceptors are working.
During dark adaptation, the eye’s maximum sensitivity shifts toward shorter wavelengths, when the rods are
working. The corresponding sensitivity function under dark adapted conditions is the scotopic luminosity
function or the 1951 CIE V’(λ) function (V-prime lambda function). We will usually work with the photopic
V(λ) function, but you should be aware that there is a different spectral luminosity function for scotopic (dark
Figure 1 shows the V(λ) and V’(λ) functions on the same graph. Note that both curves are bell-shaped. The
V(λ) curve peaks at about 555 nm (peak cone sensitivity), therefore the luminous efficiency of the human eye
at this wavelength is given a value of 1.0 V(λ555) = 1.0.
The V’(λ) (scotopic) curve is shifted toward shorter wavelengths and peaks at about 507 nm (peak rod
sensitivity). The luminous efficiency under scotopic conditions, at this wavelength, is given a value of 1.0.
V’(λ507) = 1.0. (Schwartz Fig. 4-8)
2
Lecture 7 – Photometry, V(λ)
1
V (l)
0.8
V' (l)
0.6
Luminous
efficiency
0.4
0.2 Figure 1. The CIE V(λ) (right,
red) and V’(λ) (left, blue) curves
0
400 450 500 550 600 650 700
Wavelength
Derivation of the photopic luminosity function
How can you experimentally determine the photopic luminosity function? That is, how bright do different
wavelengths appear to the human eye? One way (Schwartz Fig. 4-9) would be to compare a reference light
with fixed wavelength and radiance, with another light of different wavelength. The subject would adjust the
radiance of the test light until it appears the same brightness as the reference light (Fig. 2, below).
Unfortunately this procedure gives variable results because it is very difficult to match the brightness of
stimuli that have different wavelengths.
Other λ
λ
Figure 2. Experiment to compare relative luminous efficiency for different wavelengths.
Another clever method, known as heterochromatic flicker photometry (HFP), was developed to overcome
this problem. A single illuminated stimulus is designed so that it alternately flickers between two
wavelengths (Fig. 3 A, and Schwartz Fig. 4-10) at a rate of about 15 cycles/sec (~ 15 Hz). For example, one
wavelength may be 555 nm with a fixed radiance (Fig. 3, B), while the other wavelength is variable, and its
radiance can be adjusted (Fig. 3, C). The alternating colors will appear to fuse into another in-between color,
but if their perceived brightnesses are not equal the light will still flicker. The radiance of the test wavelength
is adjusted until the flicker disappears or is minimized. At that point, the luminances (perceived brightness
for a standard human observer) are equal. The procedure is repeated for many test wavelengths. The CIE
1924 data is based on experiments using this method.
3
Lecture 7 – Photometry, V(λ)
555 nm green,
Alternates B
10 watts
A
C
650 nm red
100 watts
Figure 3. Heterochromic flicker photometry. The yellow spot, whose color is seen as a
fusion of the two alternating colors, appears to flicker until the luminances of the two colors
become equal (equal perceived brightnesses). In this example, red must be set to ten times
the radiant power of the reference green light, therefore the eye must be 1/10th as sensitive
at 650 nm; hence the V(λ) value at 650 nm is 0.1.
Minimally Distinct Border method is shown in Schwartz Fig. 4-11. The standard wavelength and test
wavelength stimuli occupy two halves of square patch (similar to the stimulus shown in Fig. 2) and share a
the brightnesses appear to be equal, so luminances are equal. The results agree well with HFP results.
BASIC QUANTITIES MEASURED IN PHOTOMETRY
Luminous power
Whereas radiant power is simply a function of how much energy is present, luminous power indicates
perceived brightness (for a standard human observer). 10 watts at 555 nm is much brighter than 10 watts at
400 nm. Even though the radiant power is the same, the luminous power at 400 nm is different.. Luminous
power at one particular wavelength is expressed in lumens, where one lumen is defined as:
lumens = (radiant power in watts)Vλ(680)
This equation appears in Schwartz and uses a constant that has been rounded to 680. Other references
may use 683 or 685. Note that this formula is for photopic lumens. See the examples in Schwartz Fig. 4-2.
The V(λ) curve refers to the photopic luminosity function. Since rods are more sensitive to light, that is, they
can see dim lights better than cones, the perceived brightness of a light in the scotopic system is different,
and there is a different formula for scotopic lumens.
scotopic lumens = (radiant power in watts)V’λ1700)
It turns out that at 555 nm, the scotopic luminous efficiency is 0.4, so for a 1-watt light source at that
wavelength, there are 680 scotopic lumens. We won’t use scotopic lumens much in this course. In the
photopic system, luminous efficiency at 555 nm is 1.0, so for a 1-watt source at that wavelength, there are
680 photopic lumens.
4
Lecture 7 – Photometry, V(λ)
1600 Lumens/watt
Scotopic lumens/watt
1400
1200
Lumens
per
1000
watt
800
600
400
200 Figure 4 Photopic & scotopic
0 lumens compared
400 450 500 550 600 650 700
Wavelength
Figure 4 shows the number of photopic and scotopic lumens at each wavelength, assuming a radiant power
of 1 watt. Schwartz mentions that at 555 nm, both the number of photopic and scotopic lumens is equal to
680. At 507 nm there are 1699 scotopic lumens per watt. Schwartz rounds it to 1700.
If the light source is polychromatic (that is it has multiple wavelengths), the total luminous power is equal to
the sum of the luminous power computed separately for each wavelength. The additivity of luminous power
at each wavelength is called Abney’s law of additivity. (Schwartz Fig. 4-3)
Luminous intensity
This photometric term is similar to radiant intensity (watts/steradian). It is used for a point source only and
the unit for luminous intensity is the candela.
1 candela = 1 lumen / 1 steradian
Luminance
The perceived brightness of an extended source (for the standard observer) is referred to as the luminance
and is similar to radiance in that it quantifies light given off by an extended surface area. The basic metric
unit for luminance is the nit.
2
1 nit = 1 candela / 1 m
Another metric unit for luminance is the apostilb and the similar English unit is the footlambert. These are
specifically used with Lambertian surfaces, which we will discuss in the next Lecture. They are defined as
follows:
2
1 apostilb = (candela / m ) / π) = (1/π)nits
2
1 footlambert = (candela / ft ) / π
Illuminance
This term is similar to irradiance in that it quantifies light falling onto a surface. The metric unit for
illuminance is the lux. The English unit for illuminance is the foot-candle.
2
1 lux = 1 lumen / m
2
1 foot-candle = 1 lumen / ft
5
Lecture 7 – Photometry, V(λ)
Schwartz Table 4-1 lists recommended illuminance values for various sites or activities, and may be a useful
reference for you someday when you are in practice. Other units for luminance and illuminance are in
Schwartz Table 4-2.
Be careful not to confuse luminance and illuminance. Luminance refers to the brightness of light coming off
a surface. Illuminance refers to the brightness of the light falling on a surface.
Table 1, below, shows that there is a parallel between radiometric and photometric units. Photometry is
concerned with how bright a light looks and that depends both on the radiant power and the V(λ) value for
each particular wavelength considered.
Table 1 Comparison of radiometric and photometric units
energy energy joule
joules/sec
energy/time (power) radiant power luminous power lumen
(watt)
point source intensity intensity (candela)
2
lum/str/m
energy emitted from 2 2
an extended source
(nit)
2
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http://oeis.org/A192544 | 1,544,989,870,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827992.73/warc/CC-MAIN-20181216191351-20181216213351-00260.warc.gz | 212,304,502 | 4,339 | This site is supported by donations to The OEIS Foundation.
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A192544 Number bases n such that all integers m having the commuting property r(m)^2=r(m^2), where r is cyclic replacement of digits d->(d+1) mod n, are of the form m=A^kB, where B=n/2, A=B-1, and 0<=k<=n-3. 1
8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The number bases n form the arithmetic sequence n=8+4*k, k>=0, so B is necessarily even. The number bases n=2 and n=4 have B as the only number with the commuting property. No odd base n has the commuting property. LINKS EXAMPLE In base 8, B=4, A=3, and the numbers with the commuting property are 4, 34, 334, 3334, 33334, 333334. CROSSREFS Cf. A059558, A124354, A192544, A117755, A127856, A127857, A127859, A127860, A127861. Sequence in context: A253296 A081925 A049199 * A302139 A160392 A242272 Adjacent sequences: A192541 A192542 A192543 * A192545 A192546 A192547 KEYWORD nonn,base AUTHOR Walter Kehowski, Jul 04 2011 STATUS approved
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Last modified December 16 14:50 EST 2018. Contains 318167 sequences. (Running on oeis4.) | 582 | 1,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-51 | longest | en | 0.757756 |
https://en.oxforddictionaries.com/definition/diophantine_equation | 1,540,301,303,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516194.98/warc/CC-MAIN-20181023132213-20181023153713-00077.warc.gz | 636,085,805 | 24,679 | # Definition of Diophantine equation in English:
## Diophantine equation
### noun
Mathematics
• A polynomial equation with integral coefficients for which integral solutions are required.
• ‘Wolfram displays a table of some of the simplest possible Diophantine equations, distinguishing between those known to have integer solutions, those known to have no integer solutions, and those for which the question is still open.’
• ‘Turán goes on to say that Carl Siegel and Klaus Roth generalised the classes of Diophantine equations for which these conclusions would hold and even bounded the number of solutions.’
• ‘In addition Poinsot worked on number theory and on this topic he studied Diophantine equations, how to express numbers as the difference of two squares and primitive roots.’
• ‘That conjecture offers a new way of expressing Diophantine problems, in effect translating an infinite number of Diophantine equations (including the equation of Fermat's last theorem) into a single mathematical statement.’
### Origin
Early 18th century: named after Diophantus.
## Diophantine equation
/ˌdʌɪəˈfantɪn//-tʌɪn/ | 249 | 1,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-43 | latest | en | 0.918184 |
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Homework03-sol
# Homework03-sol - 4:23 PM Homework#03(Solution Binary...
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2/7/03 4:23 PM ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2003 Page 1 of 3 Homework #03 (Solution) Binary arithmetic and floating-point representation 1a. (4 points) What is the smallest and largest unsigned numbers that can be represented using 16-bits? 0 and 65,535 Grading: 2pts for each correct number, 1pt for trying. 1b. (4 points) What is the largest unsigned number that can be represented using n bits? 2 n -1 Grading: 4pts for correct, 2pts for trying. 1c. (4 points) How many unsigned numbers can you represent using n bits? 2 n Grading: 4pts for correct, 2pts for trying. 2a. (4 points) What is the smallest 16-bit signed integer in hexadecimal form? 0x8000 Grading: 4pts for correct, 1pt for trying. 2b. (4 points) What is the largest 16-bit signed integer in hexadecimal form? 0x7FFF Grading: 4pts for correct, 1pt for trying. 2c. (4 points) Negate the signed 8-bit number represented by 0x8E. Flip all bits and add 1 0x8e= 1000 1110 flip = 0111 0001 add 1 = 0111 0010 =0x72 Grading: 4pts for correct, 1pt for trying.
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https://pytorch.org/docs/1.1.0/_modules/torch/nn/modules/padding.html | 1,638,023,015,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358189.36/warc/CC-MAIN-20211127133237-20211127163237-00636.warc.gz | 559,027,783 | 7,636 | Shortcuts
from .module import Module
from .utils import _pair, _quadruple, _ntuple
from .. import functional as F
from ..._jit_internal import weak_module, weak_script_method
# TODO: grad_output size asserts in THNN
@weak_module
def __init__(self, value):
self.value = value
@weak_script_method
def forward(self, input):
def extra_repr(self):
[docs]@weak_module
r"""Pads the input tensor boundaries with a constant value.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in both boundaries. If a 2-tuple, uses
(:math:\text{padding\_left}, :math:\text{padding\_right})
Shape:
- Input: :math:(N, C, W_{in})
- Output: :math:(N, C, W_{out}) where
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.randn(1, 2, 4)
>>> input
tensor([[[-1.0491, -0.7152, -0.0749, 0.8530],
[-1.3287, 1.8966, 0.1466, -0.2771]]])
>>> m(input)
tensor([[[ 3.5000, 3.5000, -1.0491, -0.7152, -0.0749, 0.8530, 3.5000,
3.5000],
[ 3.5000, 3.5000, -1.3287, 1.8966, 0.1466, -0.2771, 3.5000,
3.5000]]])
>>> input = torch.randn(1, 2, 3)
>>> input
tensor([[[ 1.6616, 1.4523, -1.1255],
[-3.6372, 0.1182, -1.8652]]])
>>> m(input)
tensor([[[ 3.5000, 3.5000, 1.6616, 1.4523, -1.1255, 3.5000, 3.5000],
[ 3.5000, 3.5000, -3.6372, 0.1182, -1.8652, 3.5000, 3.5000]]])
>>> # using different paddings for different sides
>>> m = nn.ConstantPad1d((3, 1), 3.5)
>>> m(input)
tensor([[[ 3.5000, 3.5000, 3.5000, 1.6616, 1.4523, -1.1255, 3.5000],
[ 3.5000, 3.5000, 3.5000, -3.6372, 0.1182, -1.8652, 3.5000]]])
"""
[docs]@weak_module
r"""Pads the input tensor boundaries with a constant value.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in all boundaries. If a 4-tuple, uses (:math:\text{padding\_left},
:math:\text{padding\_right}, :math:\text{padding\_top}, :math:\text{padding\_bottom})
Shape:
- Input: :math:(N, C, H_{in}, W_{in})
- Output: :math:(N, C, H_{out}, W_{out}) where
:math:H_{out} = H_{in} + \text{padding\_top} + \text{padding\_bottom}
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.randn(1, 2, 2)
>>> input
tensor([[[ 1.6585, 0.4320],
[-0.8701, -0.4649]]])
>>> m(input)
tensor([[[ 3.5000, 3.5000, 3.5000, 3.5000, 3.5000, 3.5000],
[ 3.5000, 3.5000, 3.5000, 3.5000, 3.5000, 3.5000],
[ 3.5000, 3.5000, 1.6585, 0.4320, 3.5000, 3.5000],
[ 3.5000, 3.5000, -0.8701, -0.4649, 3.5000, 3.5000],
[ 3.5000, 3.5000, 3.5000, 3.5000, 3.5000, 3.5000],
[ 3.5000, 3.5000, 3.5000, 3.5000, 3.5000, 3.5000]]])
>>> # using different paddings for different sides
>>> m = nn.ConstantPad2d((3, 0, 2, 1), 3.5)
>>> m(input)
tensor([[[ 3.5000, 3.5000, 3.5000, 3.5000, 3.5000],
[ 3.5000, 3.5000, 3.5000, 3.5000, 3.5000],
[ 3.5000, 3.5000, 3.5000, 1.6585, 0.4320],
[ 3.5000, 3.5000, 3.5000, -0.8701, -0.4649],
[ 3.5000, 3.5000, 3.5000, 3.5000, 3.5000]]])
"""
[docs]@weak_module
r"""Pads the input tensor boundaries with a constant value.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in all boundaries. If a 6-tuple, uses
(:math:\text{padding\_left}, :math:\text{padding\_right},
:math:\text{padding\_top}, :math:\text{padding\_bottom},
:math:\text{padding\_front}, :math:\text{padding\_back})
Shape:
- Input: :math:(N, C, D_{in}, H_{in}, W_{in})
- Output: :math:(N, C, D_{out}, H_{out}, W_{out}) where
:math:D_{out} = D_{in} + \text{padding\_front} + \text{padding\_back}
:math:H_{out} = H_{in} + \text{padding\_top} + \text{padding\_bottom}
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.randn(16, 3, 10, 20, 30)
>>> output = m(input)
>>> # using different paddings for different sides
>>> m = nn.ConstantPad3d((3, 3, 6, 6, 0, 1), 3.5)
>>> output = m(input)
"""
@weak_module
@weak_script_method
def forward(self, input):
def extra_repr(self):
[docs]@weak_module
r"""Pads the input tensor using the reflection of the input boundary.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in all boundaries. If a 2-tuple, uses
(:math:\text{padding\_left}, :math:\text{padding\_right})
Shape:
- Input: :math:(N, C, W_{in})
- Output: :math:(N, C, W_{out}) where
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.arange(8, dtype=torch.float).reshape(1, 2, 4)
>>> input
tensor([[[0., 1., 2., 3.],
[4., 5., 6., 7.]]])
>>> m(input)
tensor([[[2., 1., 0., 1., 2., 3., 2., 1.],
[6., 5., 4., 5., 6., 7., 6., 5.]]])
>>> # using different paddings for different sides
>>> m(input)
tensor([[[3., 2., 1., 0., 1., 2., 3., 2.],
[7., 6., 5., 4., 5., 6., 7., 6.]]])
"""
[docs]@weak_module
r"""Pads the input tensor using the reflection of the input boundary.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in all boundaries. If a 4-tuple, uses (:math:\text{padding\_left},
:math:\text{padding\_right}, :math:\text{padding\_top}, :math:\text{padding\_bottom})
Shape:
- Input: :math:(N, C, H_{in}, W_{in})
- Output: :math:(N, C, H_{out}, W_{out}) where
:math:H_{out} = H_{in} + \text{padding\_top} + \text{padding\_bottom}
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.arange(9, dtype=torch.float).reshape(1, 1, 3, 3)
>>> input
tensor([[[[0., 1., 2.],
[3., 4., 5.],
[6., 7., 8.]]]])
>>> m(input)
tensor([[[[8., 7., 6., 7., 8., 7., 6.],
[5., 4., 3., 4., 5., 4., 3.],
[2., 1., 0., 1., 2., 1., 0.],
[5., 4., 3., 4., 5., 4., 3.],
[8., 7., 6., 7., 8., 7., 6.],
[5., 4., 3., 4., 5., 4., 3.],
[2., 1., 0., 1., 2., 1., 0.]]]])
>>> # using different paddings for different sides
>>> m = nn.ReflectionPad2d((1, 1, 2, 0))
>>> m(input)
tensor([[[[7., 6., 7., 8., 7.],
[4., 3., 4., 5., 4.],
[1., 0., 1., 2., 1.],
[4., 3., 4., 5., 4.],
[7., 6., 7., 8., 7.]]]])
"""
@weak_module
@weak_script_method
def forward(self, input):
def extra_repr(self):
[docs]@weak_module
r"""Pads the input tensor using replication of the input boundary.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in all boundaries. If a 2-tuple, uses
(:math:\text{padding\_left}, :math:\text{padding\_right})
Shape:
- Input: :math:(N, C, W_{in})
- Output: :math:(N, C, W_{out}) where
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.arange(8, dtype=torch.float).reshape(1, 2, 4)
>>> input
tensor([[[0., 1., 2., 3.],
[4., 5., 6., 7.]]])
>>> m(input)
tensor([[[0., 0., 0., 1., 2., 3., 3., 3.],
[4., 4., 4., 5., 6., 7., 7., 7.]]])
>>> # using different paddings for different sides
>>> m(input)
tensor([[[0., 0., 0., 0., 1., 2., 3., 3.],
[4., 4., 4., 4., 5., 6., 7., 7.]]])
"""
[docs]@weak_module
r"""Pads the input tensor using replication of the input boundary.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in all boundaries. If a 4-tuple, uses (:math:\text{padding\_left},
:math:\text{padding\_right}, :math:\text{padding\_top}, :math:\text{padding\_bottom})
Shape:
- Input: :math:(N, C, H_{in}, W_{in})
- Output: :math:(N, C, H_{out}, W_{out}) where
:math:H_{out} = H_{in} + \text{padding\_top} + \text{padding\_bottom}
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.arange(9, dtype=torch.float).reshape(1, 1, 3, 3)
>>> input
tensor([[[[0., 1., 2.],
[3., 4., 5.],
[6., 7., 8.]]]])
>>> m(input)
tensor([[[[0., 0., 0., 1., 2., 2., 2.],
[0., 0., 0., 1., 2., 2., 2.],
[0., 0., 0., 1., 2., 2., 2.],
[3., 3., 3., 4., 5., 5., 5.],
[6., 6., 6., 7., 8., 8., 8.],
[6., 6., 6., 7., 8., 8., 8.],
[6., 6., 6., 7., 8., 8., 8.]]]])
>>> # using different paddings for different sides
>>> m = nn.ReplicationPad2d((1, 1, 2, 0))
>>> m(input)
tensor([[[[0., 0., 1., 2., 2.],
[0., 0., 1., 2., 2.],
[0., 0., 1., 2., 2.],
[3., 3., 4., 5., 5.],
[6., 6., 7., 8., 8.]]]])
"""
[docs]@weak_module
r"""Pads the input tensor using replication of the input boundary.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in all boundaries. If a 6-tuple, uses
(:math:\text{padding\_left}, :math:\text{padding\_right},
:math:\text{padding\_top}, :math:\text{padding\_bottom},
:math:\text{padding\_front}, :math:\text{padding\_back})
Shape:
- Input: :math:(N, C, D_{in}, H_{in}, W_{in})
- Output: :math:(N, C, D_{out}, H_{out}, W_{out}) where
:math:D_{out} = D_{in} + \text{padding\_front} + \text{padding\_back}
:math:H_{out} = H_{in} + \text{padding\_top} + \text{padding\_bottom}
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.randn(16, 3, 8, 320, 480)
>>> output = m(input)
>>> # using different paddings for different sides
>>> m = nn.ReplicationPad3d((3, 3, 6, 6, 1, 1))
>>> output = m(input)
"""
[docs]@weak_module
r"""Pads the input tensor boundaries with zero.
For N-dimensional padding, use :func:torch.nn.functional.pad().
Args:
padding (int, tuple): the size of the padding. If is int, uses the same
padding in all boundaries. If a 4-tuple, uses (:math:\text{padding\_left},
:math:\text{padding\_right}, :math:\text{padding\_top}, :math:\text{padding\_bottom})
Shape:
- Input: :math:(N, C, H_{in}, W_{in})
- Output: :math:(N, C, H_{out}, W_{out}) where
:math:H_{out} = H_{in} + \text{padding\_top} + \text{padding\_bottom}
:math:W_{out} = W_{in} + \text{padding\_left} + \text{padding\_right}
Examples::
>>> input = torch.randn(1, 1, 3, 3)
>>> input
tensor([[[[-0.1678, -0.4418, 1.9466],
[ 0.9604, -0.4219, -0.5241],
[-0.9162, -0.5436, -0.6446]]]])
>>> m(input)
tensor([[[[ 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000],
[ 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000],
[ 0.0000, 0.0000, -0.1678, -0.4418, 1.9466, 0.0000, 0.0000],
[ 0.0000, 0.0000, 0.9604, -0.4219, -0.5241, 0.0000, 0.0000],
[ 0.0000, 0.0000, -0.9162, -0.5436, -0.6446, 0.0000, 0.0000],
[ 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000],
[ 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000]]]])
>>> # using different paddings for different sides
>>> m = nn.ZeroPad2d((1, 1, 2, 0))
>>> m(input)
tensor([[[[ 0.0000, 0.0000, 0.0000, 0.0000, 0.0000],
[ 0.0000, 0.0000, 0.0000, 0.0000, 0.0000],
[ 0.0000, -0.1678, -0.4418, 1.9466, 0.0000],
[ 0.0000, 0.9604, -0.4219, -0.5241, 0.0000],
[ 0.0000, -0.9162, -0.5436, -0.6446, 0.0000]]]])
"""
## Docs
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## Tutorials
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View Tutorials | 4,760 | 11,186 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-49 | latest | en | 0.345261 |
http://www.talkstats.com/showthread.php/1800-Clueless | 1,369,322,215,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703489876/warc/CC-MAIN-20130516112449-00079-ip-10-60-113-184.ec2.internal.warc.gz | 735,251,440 | 10,681 | 1. ## Clueless
I am in a really big fix. I have a midterm today and I just came back from a two week vacation to the Carribean...and I don't get the missed classes at all. Can anyone help a bit? What is the formula to find a probability of something? Also I know there is a ^ in a lot of problems, what does that mean? Thanks for any help!
2. ## ?
Ex: p(1-p)^(x-1) Does this mean that p(1-p) is to the power of (x-1)?
3. The formula for finding the probability depends on the random experiment you have.
If the set of possible outcomes (sample space) is a finite set and the outcomes are equally likely, you can use the classical definition of probability which is the ratio of favourable no. of outcomes to the total no. of outcomes in the sample space.
But if the sample space cotains infine no. of elements, you have to use other methods to find the probability.
p(1-p)^(x-1) means (1-p)*(1-p)*(1-p)....(x-1) times * p.
#### Posting Permissions
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• You may not edit your posts | 279 | 1,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2013-20 | latest | en | 0.934376 |
https://forum.allaboutcircuits.com/threads/16-bit-latch.119893/ | 1,631,930,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00579.warc.gz | 311,272,108 | 20,658 | # 16 bit latch
#### The_Cook
Joined May 29, 2014
48
Hi guys, I'm new to logic design so I need some help getting started. Can anyone help me design a 16-bit latch with a load-enable bit and 16-bit data input/output. I want to create it in a logic simulator program, doesn't really matter which as long as it is correct.
#### sailorjoe
Joined Jun 4, 2013
363
Look at the datasheet for a 74HC374, or 74HC259. Use two 8 bit latches to make a 16 bit latch.
Also, Google "8 bit latch" and you'll see a variety of options.
#### Papabravo
Joined Feb 24, 2006
16,775
There is a trick to building a good transparent latch which is easy to see if you use a Karnaugh map. The basic circuit is combinatorial involving the data input D, the enable input G, and the output Q. If the enable G, is HIGH(LOW), then Q = D. If the enable G, is LOW(HIGH), then Q = Q. This plan gives you two alternatives, one with the enable HIGH and one with the enable LOW. Because you are feeding the output back into the decision making process there is a chance under certain conditions for the output to have a momentary moment of indecision. Can you find the solution for a 1-bit latch? If so you can replicate it 16 times.
Hint: The solution contains the sum of three product terms.
Last edited:
#### The_Cook
Joined May 29, 2014
48
Okay I will take a look. However, I am only allowed to use 'and' 'or' and 'not' gates
#### Papabravo
Joined Feb 24, 2006
16,775
Okay I will take a look. However, I am only allowed to use 'and' 'or' and 'not' gates
That is all you need. | 429 | 1,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-39 | longest | en | 0.903033 |
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# ps7 - (b Are X and Y statistically independent Justify your...
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ECE 804, Random Signal Analysis Nov. 16, 2009 OSU, Autumn 2009 Due: Nov. 23, 2009 Problem Set 7 Problem 1 Random variables X and Y have a joint density, f X,Y ( x, y ) = k y inside the triangular region shown below and 0 elsewhere. -2 -1 0 1 2 -1 -0.5 0 0.5 1 1.5 2 X Y In what follows you may or may not need the following: R 1 0 ln x dx = - 1. (a) Find k . (b) Find and plot the marginal densities f X ( x ) and f Y ( y ). (c) Which of the following statements is correct? Explain. (i) X and Y are independent. (ii) X and Y are uncorrelated. (d) Let Z = XY . Find f Y,Z ( y, z ) Problem 2 Let Θ and R be independent random variables, where Θ is uniformly distributed on [0 , π 2 ], and f R ( r ) = λe - λr u ( r ). Let X = R cos(Θ) and Y = R sin(Θ). (a) Find the joint pdf of X and Y . Be sure to clearly state the region in the ( x, y ) plane in which this joint pdf is nonzero.
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Unformatted text preview: (b) Are X and Y statistically independent? Justify your answer. 1 Problem 3 At a party n people put their hats in the center of a room, where the hats are mixed together. Each person then randomly selects one with equal probability. Let us define ~ X = [ X 1 X 2 ··· X n ], where X i = ( 1 , if the i th person selects his or her own hat , otherwise. (a) Find E [ X i ] and var( X i ). (b) Find the covariance matrix, C ~ X . For n = 3, find its eigenvalues and eigenvectors. Write it in the form C ~ X = V Λ V T , where Λ is a diagonal matrix. (c) Let S = X 1 + X 2 + ··· + X n . Find E [ S ] and σ 2 S . 2...
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ps7 - (b Are X and Y statistically independent Justify your...
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Ask a homework question - tutors are online | 646 | 2,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-13 | latest | en | 0.838231 |
https://convert-dates.com/days-before/4126/2024/04/04 | 1,713,222,847,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00199.warc.gz | 157,439,403 | 4,347 | ## 4126 Days Before April 4, 2024
Want to figure out the date that is exactly four thousand one hundred twenty six days before Apr 4, 2024 without counting?
Your starting date is April 4, 2024 so that means that 4126 days earlier would be December 17, 2012.
You can check this by using the date difference calculator to measure the number of days before Dec 17, 2012 to Apr 4, 2024.
December 2012
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1. 30
2. 31
December 17, 2012 is a Monday. It is the 352nd day of the year, and in the 51st week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 31 days in this month. 2012 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 12/17/2012, and almost everywhere else in the world it's 17/12/2012.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 4126 weekdays before Apr 4, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Apr 4, 2024, which falls on a Thursday. Counting forward, the next day would be a Friday.
To get exactly four thousand one hundred twenty six weekdays before Apr 4, 2024, you actually need to count 5776 total days (including weekend days). That means that 4126 weekdays before Apr 4, 2024 would be June 11, 2008.
If you're counting business days, don't forget to adjust this date for any holidays.
June 2008
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
1. 1
2. 2
3. 3
4. 4
5. 5
6. 6
7. 7
1. 8
2. 9
3. 10
4. 11
5. 12
6. 13
7. 14
1. 15
2. 16
3. 17
4. 18
5. 19
6. 20
7. 21
1. 22
2. 23
3. 24
4. 25
5. 26
6. 27
7. 28
1. 29
2. 30
June 11, 2008 is a Wednesday. It is the 163rd day of the year, and in the 163rd week of the year (assuming each week starts on a Sunday), or the 2nd quarter of the year. There are 30 days in this month. 2008 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 06/11/2008, and almost everywhere else in the world it's 11/06/2008.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 934 | 2,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-18 | latest | en | 0.918875 |
https://www.unitconverters.net/radiation-absorbed-dose/teragray-to-joule-gram.htm | 1,632,689,832,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057973.90/warc/CC-MAIN-20210926205414-20210926235414-00629.warc.gz | 1,043,949,789 | 2,800 | Home / Radiation-Absorbed Dose Conversion / Convert Teragray to Joule/gram
# Convert Teragray to Joule/gram
Please provide values below to convert teragray [TGy] to joule/gram [J/g], or vice versa.
From: teragray To: joule/gram
### Teragray to Joule/gram Conversion Table
Teragray [TGy]Joule/gram [J/g]
0.01 TGy10000000 J/g
0.1 TGy100000000 J/g
1 TGy1000000000 J/g
2 TGy2000000000 J/g
3 TGy3000000000 J/g
5 TGy5000000000 J/g
10 TGy10000000000 J/g
20 TGy20000000000 J/g
50 TGy50000000000 J/g
100 TGy100000000000 J/g
1000 TGy1000000000000 J/g
### How to Convert Teragray to Joule/gram
1 TGy = 1000000000 J/g
1 J/g = 1.0E-9 TGy
Example: convert 15 TGy to J/g:
15 TGy = 15 × 1000000000 J/g = 15000000000 J/g | 285 | 712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-39 | latest | en | 0.434909 |
http://www.mathworks.com/matlabcentral/fileexchange/39949-using-analytical-tools-to-gain-insight-and-speed-up-num-analysis-in-matlab---symbolic-math-toolbox/content/AnalyticalInsightsWebinar/inverseKinematics/beginREADME_InverseKinematics.m | 1,427,907,183,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131305143.64/warc/CC-MAIN-20150323172145-00265-ip-10-168-14-71.ec2.internal.warc.gz | 606,384,215 | 8,533 | Code covered by the BSD License
# Using Analytical Tools to Gain Insight and Speed-up Num. Analysis in MATLAB & Symbolic Math Toolbox
### Deepak Ramaswamy (view profile)
28 Jan 2013 (Updated )
files from the webinar
```%% Begin README : Inverse Kinematics of a Two-link Manipulator
% The <http://www.mathworks.com/products/symbolic/description5.html Symbolic Math Notebook app>
% and the <http://www.mathworks.com/products/symbolic/index.html Symbolic Math Toolbox> are
% used to model the inverse kinematics of a two-link manipulator. The
% equations for the manipulator are then brought into Simulink.
%
% <<image/image.PNG>>
%
%
% Copyright 2012 The MathWorks, Inc.
%% References
%
% # Find the related webinar: *Using Analytical Tools to Gain Insight and Speed Up
% Numerical Analysis in MATLAB* <http://www.mathworks.com/wbnr73465 here>
% # This example is a stripped down version of a larger example that
% <http://www.mathworks.com/company/events/webinars/wbnr66719.html?id=66719&p1=961661812&p2=961661830
% How a Differential Equation Becomes a Robot: Expanding the Power of MATLAB
% with Simulink and the Symbolic Math Toolbox>.
%
%% Solving Inverse Kinematics in the Symbolic Math Notebook
% The inverse kinematics problem is <http://www.mathworks.com/help/symbolic/mupad_ref/solve.html solved> for a two-link manipulator in
% the <http://www.mathworks.com/products/symbolic/description5.html Notebook>. A simple animation is also shown.
%
% <matlab:open('MUInverseKinematics.mn') Open notebook>
%% Solving Inverse Kinematics using Symbolic in MATLAB
% The inverse kinematics problem is solved in MATLAB. A very simple
% animation is performed.
%
% <matlab:open('MLInverseKinematics.m') Open script>
%
% <matlab:MLInverseKinematics Run script>
%% Bring in Symbolic expression from Notebook and convert to a block in Simulink
% The symbolic expression is accessed from the <http://www.mathworks.com/products/symbolic/description5.html Notebook>, and then converted
% into a block in a Simulink project using the
% <http://www.mathworks.com/help/symbolic/matlabfunctionblock.html matlabFunctionBlock>
% function of the <http://www.mathworks.com/products/symbolic/index.html Symbolic Math Toolbox>. It is informative to run the
% script in cell mode
% | 565 | 2,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2015-14 | longest | en | 0.652082 |
https://www.physicsforums.com/threads/dc-dc-converters-kcl.836293/ | 1,603,983,601,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904287.88/warc/CC-MAIN-20201029124628-20201029154628-00426.warc.gz | 845,835,230 | 24,812 | # DC-DC Converters & KCL?
I'm confused, how is it that DC-converters work with respect to KCL in terms of conservation of charge?
When a boost/buck converter would output low/high current output than input?
I_in =/ I_out
I'm teaching myself the basics of circuits, understood KCL and KVL, but this point is confusing me when relating power converters(of all classes).
## Answers and Replies
Related Electrical Engineering News on Phys.org
berkeman
Mentor
I'm confused, how is it that DC-converters work with respect to KCL in terms of conservation of charge?
When a boost/buck converter would output low/high current output than input?
I_in =/ I_out
I'm teaching myself the basics of circuits, understood KCL and KVL, but this point is confusing me when relating power converters(of all classes).
No, most DC-DC switching converters are constant power converters, so Iin does not equal Iout (unless Vin = Vout).
The input current in a Buck converter only flows when the high-side switch transistor is conducting. When it snaps off, the flywheel diode at the output keeps the current flowing. Since the buck circuit is driving into an inductor, the inductor current (Iout) ramps up while the switch is on, and ramps down when the switch is off and the flywheel diode is conducting.
Does that make sense? Use Google Images to find some current and voltage waveforms for Buck DC-DC converters...
PhiowPhi
berkeman
Mentor
Last edited by a moderator:
PhiowPhi
berkeman
One thing though, is it valid to have a "varying" power converters? Where feedback loops would exist to change the Vout(and Iout) with respect to the change of the input.
Acting similar to the behavior of constant voltage/current sources.
berkeman
Mentor
One thing though, is it valid to have a "varying" power converters? Where feedback loops would exist to change the Vout(and Iout) with respect to the change of the input.
Acting similar to the behavior of constant voltage/current sources.
I'm not sure I understand the question. Normally you will use voltage feedback from the output in order to adjust the PWM circuit to maintain the output voltage at the desired value, independent of the input voltage.
Alternately (as is used in some LED driver circuits), you can use feedback from the output current to control the PWM circuit to maintain a constant output current at some set value, independent of the input voltage.
Does that help?
PhiowPhi
Yes it does thanks, I confused myself.
It should be a change input(V or I) to maintain a constant (V or I) depending on the setup.
berkeman
Mentor
BTW, one way to use a DC-DC buck converter as a constant current source is to have a low-side sensing resistor (small value) to give the converter a small voltage that represents the current flowing through the load. The DC-DC converter uses PWM to maintain that constant average current value.
Like this Maxim circuit for driving LEDs with a constant current:
https://www.maximintegrated.com/en/images/appnotes/3668/3668Fig01.gif
PhiowPhi
meBigGuy
Gold Member
Assuming a 100% efficient DC to DC converter, the power in ALWAYS equals the power out. Exactly. If you reduce the load, the converter draws less. If you reduce the input voltage, the converter draws more current (assuming constant output voltage and load).
Baluncore
Science Advisor
2019 Award
One simple way to view a buck converter is as an LC low-pass filter. The inductor input is switched rapidly between zero and the input voltage with a duty cycle that determines the output voltage. Since there are two distinct phases in each cycle, you can apply KL to only one phase at the time.
Fundamentally there are two independent circuits in a buck converter. The first circuit is from the supply, through the switch, the inductor, to the load, then back to the supply via the ground return. The second is from the ground, through the diode, inductor and load.
The polarity of the voltage across the inductor is reversed during each phase of the cycle. Since VL = L * di/dt, the inductor current alternatively rises and falls during the two phases of each cycle. di/dt = VL / L.
PhiowPhi and berkeman
I'm struggling with one aspect related to this topic, applying Ohm's law with the respect to the output. Let me give a simple example:
Assume a constant DC power supply, that is 30W connected to a boost-converter to output higher voltage(with lower current) to the load, since Pin = Pout(assume 100% efficiency). The wire's resistance is 0.5ohms(total), the load's resistance is 5 ohms(so net resistance is 5.5), I think that the applied voltage from the PS would be 12.84V and the current is 2.33A. Now the input range of this boost-converter is 5-20VDC,0.5-5A(example numbers,their all made up), and I set the output voltage to 30V, in my head I know it has to be 1A current since P = 30W, but if I apply ohms law... I always get higher current than 1A which can't be true, so what am I doing wrong here?
berkeman
Mentor
I'm struggling with one aspect related to this topic, applying Ohm's law with the respect to the output. Let me give a simple example:
Assume a constant DC power supply, that is 30W connected to a boost-converter to output higher voltage(with lower current) to the load, since Pin = Pout(assume 100% efficiency). The wire's resistance is 0.5ohms(total), the load's resistance is 5 ohms(so net resistance is 5.5), I think that the applied voltage from the PS would be 12.84V and the current is 2.33A. Now the input range of this boost-converter is 5-20VDC,0.5-5A(example numbers,their all made up), and I set the output voltage to 30V, in my head I know it has to be 1A current since P = 30W, but if I apply ohms law... I always get higher current than 1A which can't be true, so what am I doing wrong here?
I'm not following your example at all. First you set the output voltage at 12.84V, then to 30V. What do you set your boosted output voltage to? What is your load resistance? That gives you your output power. What is the input voltage set to? That determines the input current drawn to provide the output power.
I'm not following your example at all. First you set the output voltage at 12.84V, then to 30V.
That would be the input from 30W PS, since resistance is 5.5ohms, that 12.84VDC will be inputted to the boost-converter.
What do you set your boosted output voltage to? What is your load resistance? That gives you your output power. What is the input voltage set to? That determines the input current drawn to provide the output power.
Output would be set to 30V(from 12.84 input), load resistance is 5ohms(with 0.5ohms from the wire so I totaled the resistance to 5.5ohms).
Input voltage should be set to 12.84V, I assumed the input current would be 2.33A(not sure though).
berkeman
Mentor
Now you've got 5.5 Ohms at both the input and output?
Can you just draw a sketch? The input current is determined by the output power and the input voltage.
Baluncore
Science Advisor
2019 Award
Assume a constant DC power supply, that is 30W
You have an input power of 30W from your supply, then the output power from a 100% efficient converter will be 30W.
If the output load is 5.5 ohm total, then Vo/Io = 5.5R and Vo*Io = 30W. Then Io = 2.3355 amp and Vo = 12.845 volt.
in my head I know it has to be 1A current since P = 30W, but if I apply ohms law... I always get higher current than 1A which can't be true, so what am I doing wrong here?
You are assuming too many fixed parameters.
You cannot change the output voltage of the 30W power supply to 30 volt without increasing the load resistance from 5.5 ohm to 30 ohm. That is because that would require more than the 30 watt power available at the input.
You should ignore power as an input parameter in your computational games.
Energy is conserved in a 100% efficient converter. Power is the rate of flow of energy.
Power is the only “computational bridge” between input and output.
Follow this computational process.
Specify the input voltage and the output voltage. Vi, Vo.
Specify the output load resistance. Ro.
Compute the output current. Io = Vo / Ro.
Compute the output power. Wo = Io * Vo.
For a 100% efficient converter. Wi = Wo
Compute the input current. Ii = Wi / Vi.
Or just note that since Wi = Wo then Vi*Ii = Vo*Io and Ro = Vo / Io.
@berkeman , sorry for the confusion, but the load is 5ohm at output, the 0.5 is the wire's resistance.
@Baluncore you made me realize my flaw here, thank you, but I'm stuck on a few things(bear with me):
For the computational process what is the resistance prior to the converter and after? It should be 30ohms all around? Or 0.5ohms(for wires) before for input calculations, and 30ohms for output calculations?
Ri = 0.5ohms(wires) ##\therefore ## Ro = 29.5ohms(load) + 0.5ohms(wires).
meBigGuy
Gold Member
Where do you come up with this stuff? You say 30W power source, but do you even understand what that means?
First off, the concept of a dynamic 30W power supply is not realistic. That in its self is a complex system guaranteed to confuse. It is neither constant voltage nor constant current. Is it an instantaneous 30W supply? Or does the supply average 30W. (for example does it limit the charge cycles of the DC to DC converter then go to infinite voltage when a switch opens). Supplying "30W" to a dynamically changing load makes no practical sense.
When we speak of power-in to power-out in a dynamic converter, we are speaking average power averaged over the energy storage times of the converters energy storage components.
So, first define your converters output characteristics. Is is constant voltage? Is it constant current? Is is current limited? Is the current limit "fold-back"?
'
Maybe you want to define some sort of non-linear input to output relationship? Well, define it and write the equations. But, playing with that can get tricky when you start talking dynamic systems with response times, feedback, damping factor, etc. That's a whole new subject.
Then define your load. Is it constant, or changing? If it is changing, write the equations.
Now, if you want to apply some weird power source to all that, then again, define the power source mathematically.
Start with a constant voltage supply, define a converter, define a load, then look at what is happening. Then change 1 thing at a time and look at the effects back at the power source.
Start with a constant voltage supply, define a converter, define a load, then look at what is happening. Then change 1 thing at a time and look at the effects back at the power source.
lol, well I was in the process of doing that... and I think I've defined a lot about the circuit from the previous posts relative to the example(hopefully)?
But I'll give that approach a go, it's perfect way of analyzing the circuit(and other things).
I need a clarification on point #16 though because my calculations are based off that.
I have a constant voltage DC power supply(forgot to mention "voltage" on post #11), that's supplying 30W.
Ri (which is just the wire connecting the PS to other components) = 0.5ohms
Vi = 3.8V
Ii = 7.75A
Pi = Vi x Ii = 30W
That is the input to the boost-converter that is meant to output 30V to the 29.5ohm load:
Vo = 30V
Iout = 1A
Ro = 30ohm( 29.5ohm + 0.5ohm)
Pout = 30W
Pi= Po ##\checkmark##
Am I right?
Baluncore
Science Advisor
2019 Award
Ri (which is just the wire connecting the PS to other components) = 0.5ohms
No. Your 0.5 ohm must be modelled on the output side with the load if you add it to the load. Wire resistance on the input side will lower the supply voltage to the converter. It is unimportant to the computations.
A converter is the DC equivalent of an AC transformer. It transforms the V/I ratio from one side to the other.
Do not cross the converter with anything other than the power. Treat input and output as totally separate circuits.
Baluncore
Science Advisor
2019 Award
Ri (which is just the wire connecting the PS to other components) = 0.5ohms
By introducing a poorly specified series resistance, one that can jump around the circuit and across the transformer without being transformed, you are making it both unreal and more complex than it needs to be.
The output load is 30 ohms. Now forget the 0.5 ohm wire. All wires are now perfect conductors.
Now forget the 0.5 ohm wire. All wires are now perfect conductors.
That 0.5 ohm of the wire is what made me determine the input current & voltage, If I neglect it, I don't know what the input values would be from the PS...
Baluncore
Science Advisor
2019 Award
The output voltage is regulated to 30 volt. The output load is fixed at 30 ohm.
The output current must be 1 amp. The output power is then 30 watt.
For converter efficiency = 100%, the input power must be 30 watt / 100% = 30 watt.
The power supply has a regulated output voltage of Vs =3.8 volt.
So the input current must be 30 watt / 3.8 volt = 7.895 amp.
The load appears to be 30 ohms when viewed from the converter.
But when seen from the power supply, the converter transforms the load to look like a different value resistance.
That converted load resistance then appears to be 3.8 volt / 7.895 amp = 0.4813 ohms.
So the converter is transforming the load of 30 ohms to look like 0.4813 ohms to the lower voltage power supply.
Now, the idea that there might be Rs = 0.5 ohm of supply output resistance before the converter is clearly impossible.
3.8 volt / 0.5 ohm = 7.6 amp. There is clearly insufficient voltage to supply the 7.895 amp current needed by the converter.
The maximum Rs that could function would be when series resistance = apparent converter and load resistance.
In that situation the input voltage to the converter will be reduced to half of the 3.8 volt = 1.9 volt.
Because converter input voltage is halved, the converter input current would have to rise to 30 watt / 1.9 volt = 15.79 amp.
1.9 volt / 15.79 amp = 0.1203 ohm. Which is the maximum possible supply series resistance for the circuit to still function.
Overall efficiency would then have fallen to 50% simply due to resistance of the supply and cable to the converter.
The total power supplied would then need to be 60 watt, 30W to the supply series resistance and 30 watt to the load.
That is why you should use short thick cable between the power supply and the boost converter.
PhiowPhi
I had something that I'm not sure about at all, here is a diagram of a buck-converter(a simple one):
The out voltage(##V_o##) is it in series with the load? Any component from point a-b are considered in series with the output current and voltage or parallel?
I find it parallel, because of the diode and the capacitor are parallel, but not sure...
meBigGuy
Gold Member
That question tells me you have no concept of voltage, current, and circuits.
Baluncore
Science Advisor
2019 Award
The voltage appears across the load. Vo is measured at (a) relative to (b).
The current flows through the load.
Series and parallel refer to the connection relationship between components. | 3,706 | 15,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-45 | longest | en | 0.887607 |
https://www.puzzle-nurikabe.com/?size=13 | 1,725,929,894,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00228.warc.gz | 907,751,019 | 9,618 | # Nurikabe
Translate this site.
Nurikabe is a logic puzzle with a bit complicated rules and challenging solutions.
The rules are not so simple. :)
You have a grid of squares Some cells of the grid start containing numbers. The goal is to determine whether each of the cells of the grid is "black" or "white" (Islands in the Stream calls these "water" and "land" respectively). The black cells form "the nurikabe" (Islands in the Stream calls it "the stream"): they must all be orthogonally contiguous (form a single polyomino), number-free, and contain no 2x2 or larger solid rectangles (Islands in the Stream calls such illegal blocks "pools"). The white cells form "islands" (which is where Islands in the Stream got its name): each number n must be part of an n-omino composed only of white cells. All white cells must belong to exactly one island; islands must have exactly one numbered cell. Solvers will typically shade in cells they have deduced to be black and dot (non-numbered) cells deduced to be white.
Left click on a square to make it black. Right click to mark with dot. Click and drag to mark more than one square.
Video Tutorial
Hide the rules
Share
Special Monthly Nurikabe - Sep 01, 2024
ShareShare
Progress Permalink: Progress Screenshot:
2024-09-10 00:58:13
www.puzzle-nurikabe.com | 345 | 1,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.883544 |
https://www.coolstuffshub.com/speed/convert/inches-per-minute-to-yards-per-minute/ | 1,660,411,949,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00211.warc.gz | 638,962,045 | 6,179 | # Inches per minute to yards per minute conversion
1 in/min = 0.027779188891711 yd/min
## How to convert inches per minute to yards per minute?
To convert inches per minute to yards per minute, multiply the value in inches per minute by 0.02777918889171112141.
You can use the formula :
yards per minute = inches per minute × 0.02777918889171112141
1 inch per minute is equal to 0.027779188891711 yards per minute.
## Inches per minute to yards per minute conversion table
Inches per minute Yards per minute
1 in/min 0.027779188891711 yd/min
2 in/min 0.055558377783422 yd/min
3 in/min 0.083337566675133 yd/min
4 in/min 0.11111675556684 yd/min
5 in/min 0.13889594445856 yd/min
6 in/min 0.16667513335027 yd/min
7 in/min 0.19445432224198 yd/min
8 in/min 0.22223351113369 yd/min
9 in/min 0.2500127000254 yd/min
10 in/min 0.27779188891711 yd/min
20 in/min 0.55558377783422 yd/min
30 in/min 0.83337566675133 yd/min
40 in/min 1.1111675556684 yd/min
50 in/min 1.3889594445856 yd/min
60 in/min 1.6667513335027 yd/min
70 in/min 1.9445432224198 yd/min
80 in/min 2.2223351113369 yd/min
90 in/min 2.500127000254 yd/min
100 in/min 2.7779188891711 yd/min | 415 | 1,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.430938 |
https://fr.scribd.com/document/373263751/Report-Writing-State-Spelling-Challenge | 1,576,448,903,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541310866.82/warc/CC-MAIN-20191215201305-20191215225305-00199.warc.gz | 367,866,452 | 71,733 | Vous êtes sur la page 1sur 2
# Based on the tables below, write a report on the 2010 and 2011 State Spelling Challenge Prizes and
the
number of participants in the 2011 Competition. You are to link the information given. Your report must be
150 – 200 words.
## Prize State Spelling Challenge
2010 2011
Primary school Secondary school Primary school Secondary school
1st prize RM2000 RM3000 RM3000 RM5000
2nd prize RM1500 RM2500 RM2000 RM4000
3rd prize RM1000 RM2000 RM1000 RM3000
Overall Best
RM500 RM500 RM500 RM800
Individual
Overall Best
RM1300 RM18000 RM2600 RM3600
Group
## Participating States No. of participants in Secondary Schools
2010 2011
Melaka 20 36
Pulau Pinang 17 30
Pahang 11 22
Selangor 22 40
Federal Territory 29 54
Title State Spelling Challenge Prizes and the Number of Participants (2010-2011)
Introduction Table 1 shows the State Spelling Challenge Prizes (2010-2011) and Table 2
shows the number of participants in secondary schools (2010-2011).
Overview/Overall trend The increase in the prize money for secondary schools in the State Spelling
Challenge from 2010 to 2011 might have influenced the increase in the
number of participants in 2011.
In 2010, primary and secondary schools’ first prize money was the highest at
RM2000 and RM3000 respectively and a similar trend was recorded for both
schools in 2011.
However, the lowest prize money went to the overall best individual for both
schools at RM5000 each in 2010.
The third prize for primary school remained unchanged at RM1000 but there
was a rise of RM1000 from RM2000 to RM3000 for secondary school over
the two years.
Both primary and secondary overall best group prize money doubled from
RM1300 to RM2600 and from RM1800 to RM3600 respectively between
2010 and 2011.
Secondary school’s second prize money had a higher increase from RM2500
(2010) to RM4000 (2011) than primary school’s which increased from
RM1500 to RM2000 respectively in the two years.
Federal Territory recorded the most participants in 2010 and 2011 with 29
and 54 respectively
Meantime, Pahang charted the least with 11 and 22 participants each in 2010
and 2011.
## The almost doubled growth in the number of secondary school participants
in all states in 2011 was probably due to the increase in prize money from
2010 and 2011.
Conclusion In conclusion, the increase in prize money from 2010 to 2011 has an effect
on the number of participants in secondary school in 2011. | 684 | 2,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-51 | latest | en | 0.923275 |
http://phpcantho.com/a-gentle-introduction-to-applied-machine-learning-as-a-search-problem/ | 1,527,252,242,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867092.48/warc/CC-MAIN-20180525121739-20180525141739-00334.warc.gz | 231,571,430 | 34,128 | Applied machine learning is challenging because the designing of a perfect learning system for a given problem is intractable.
There is no best or best algorithm for your problem, only the best that you can discover.
The application of machine learning is best thought of as search problem for the best mapping of inputs to outputs given the knowledge and resources available to you for a given project.
In this post, you will discover the conceptualization of applied machine learning as a search problem.
After reading this post, you will know:
• That applied machine learning is the problem of approximating an unknown underlying mapping function from inputs to outputs.
• That design decisions such as the choice of data and choice of algorithm narrow the scope of possible mapping functions that you may ultimately choose.
• That the conceptualization of machine learning as a search helps to rationalize the use of ensembles, the spot checking of algorithms and the understanding of what is happening when algorithms learn.
Let’s get started.
A Gentle Introduction to Applied Machine Learning as a Search Problem
Photo by tonko43, some rights reserved.
## Overview
This post is divided into 5 parts; they are:
1. Problem of Function Approximation
2. Function Approximation as Search
3. Choice of Data
4. Choice of Algorithm
5. Implications of Machine Learning as Search
## Problem of Function Approximation
Applied machine learning is the development of a learning system to address a specific learning problem.
The learning problem is characterized by observations comprised of input data and output data and some unknown but coherent relationship between the two.
The goal of the learning system is to learn a generalized mapping between input and output data such that skillful predictions can be made for new instances drawn from the domain where the output variable is unknown.
In statistical learning, a statistical perspective on machine learning, the problem is framed as the learning of a mapping function (f) given input data (X) and associated output data (y).
We have a sample of X and y and do our best to come up with a function that approximates f, e.g. fprime, such that we can make predictions (yhat) given new examples (Xhat) in the future.
As such, applied machine learning can be thought of as the problem of function approximation.
Machine learning as the mapping from inputs to outputs
The learned mapping will be imperfect.
The problem of designing and developing a learning system is the problem of learning a useful approximate of the unknown underlying function that maps the input variables to the output variables.
We do not know the form of the function, because if we did, we would not need a learning system; we could specify the solution directly.
Because we do not know the true underlying function, we must approximate it, meaning we do not know and may never know how close of an approximation the learning system is to the true mapping.
## Function Approximation as Search
We must search for an approximation of the true underlying function that is good enough for our purposes.
There are many sources of noise that introduce error into the learning process that can make the process more challenging and in turn result in a less useful mapping. For example:
• The choice of the framing of the learning problem.
• The choice of the observations used to train the system.
• The choice of how the training data is prepared.
• The choice of the representational form for the predictive model.
• The choice of the learning algorithm to fit the model on the training data.
• The choice of the measure by which to evaluate predictive skill.
And so much more.
You can see that there are many decision points in the development of a learning system, and none of the answers are known beforehand.
You can think of all possible learning systems for a learning problem as a huge search space, where each decision point narrows the search.
Search space of all possible mapping functions from inputs to outputs
For example, if the learning problem was to predict the species of flowers, one of millions of possible learning systems could be narrowed down as follows:
• Choose to frame the problem as predicting a species class label, e.g. classification.
• Choose measurements of the flowers of a given species and their associated sub-species.
• Choose flowers in one specific nursery to measure in order to collect training data.
• Choose a decision tree model representation so that predictions can be explained to stakeholders.
• Choose the CART algorithm to fit the decision tree model.
• Choose classification accuracy to evaluate the skill of models.
And so on.
You can also see that there may be a natural hierarchy for many of the decisions involved in developing a learning system, each of which further narrows the space of possible learning systems that we could build.
This narrowing introduces a useful bias that intentionally selects one subset of possible learning systems over another with the goal of getting closer to a useful mapping that we can use in practice. This biasing applies both at the top level in the framing of the problem and at low levels, such as the choice of machine learning algorithm or algorithm configuration.
## Choice of Data
The chosen framing of the learning problem and the data used to train the system are a big point of leverage in the development of your learning system.
You do not have access to all data: that is all pairs of inputs and outputs. If you did, you would not need a predictive model in order to make output predictions for new input observations.
You do have some historical input-output pairs. If you didn’t, you would not have any data with which to train a predictive model.
But maybe you have a lot of data and you need to select only some of it for training. Or maybe you have the freedom to generate data at will and are challenged by what and how much data to generate or collect.
The data that you choose to model your learning system on must sufficiently capture the relationship between the input and output data for both the data that you have available and data that the model will be expected to make predictions on in the future.
Choice of training data from the universe of all data for a problem
## Choice of Algorithm
You must choose the representation of the model and the algorithm used to fit the model on the training data. This, again, is another big point of leverage on the development of your learning system.
Choice of algorithm from the universe of all algorithms for a problem
Often this decision is simplified to the selection of an algorithm, although it is common for the project stakeholders to impose constraints on the project, such as the model being able to explain predictions which in turn imposes constraints on the form of the final model representation and in turn on the scope of mappings that you can search.
Effect of choosing an approximate mapping from inputs to outputs
## Implications of Machine Learning as Search
This conceptualization of developing learning systems as a search problem helps to make clear many related concerns in applied machine learning.
This section looks at a few.
### Algorithms that Learn Iteratively
The algorithm used to learn the mapping will impose further constraints, and it, along with the chosen algorithm configuration, will control how the space of possible candidate mappings is navigated as the model is fit (e.g. for machine learning algorithms that learn iteratively).
Here, we can see that the act of learning from training data by a machine learning algorithm is in effect navigating the space of possible mappings for the learning system, hopefully moving from poor mappings to better mappings (e.g. hill climbing).
Effect of a learning algorithm iteratively training on data
This provides a conceptual rationale for the role of optimization algorithms in the heart of machine learning algorithms to get the most out of the model representation for the specific training data.
### Rationale for Ensembles
We can also see that different model representations will occupy quite different locations in the space of all possible function mappings, and in turn have quite different behavior when making predictions (e.g. uncorrelated prediction errors).
This provides a conceptual rationale for the role of ensemble methods that combine the predictions from different but skillful predictive models.
Interpretation of combining predictions from multiple final models
### Rationale for Spot Checking
Different algorithms with different representations may start in different positions in the space of possible function mappings, and will navigate the space differently.
If the constrained space that these algorithms are navigating is well specified by an appropriating framing and good data, then most algorithms will likely discover good and similar mapping functions.
We can also see how a good framing and careful selection of training data can open up a space of candidate mappings that may be found by a suite of modern machine learning algorithms.
This provides rationale for spot checking a suite of algorithms on a given machine learning problem and doubling down on the one that shows the most promise, or selecting the most parsimonious solution (e.g. Occam’s razor).
This section provides more resources on the topic if you are looking to go deeper.
## Summary
In this post, you discovered the conceptualization of applied machine learning as a search problem.
Specifically, you learned:
• That applied machine learning is the problem of approximating an unknown underlying mapping function from inputs to outputs.
• That design decisions such as the choice of data and choice of algorithm narrow the scope of possible mapping functions that you may ultimately choose.
• That the conceptualization of machine learning as search helps to rationalize the use of ensembles, the spot checking of algorithms and the understanding of what is happening when algorithms learn.
Do you have any questions?
Ask your questions in the comments below and I will do my best to answer.
No tags for this post. | 1,922 | 10,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-22 | longest | en | 0.917384 |
https://www.analystforum.com/t/heads-up20-fcff-fcfe-and-depreciation/74511 | 1,607,036,857,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141732835.81/warc/CC-MAIN-20201203220448-20201204010448-00246.warc.gz | 503,438,008 | 3,948 | # Heads-Up20: FCFF/FCFE and depreciation
Assuming 40% tax, if depreciation increases by \$100, then:
FCFF = NI + Depr + Int(1-t) - FCinv -WCinv,
should increase/decrease by how much?
How about FCFE, increase/decrease by how much?
What if If interest expense increases by \$100, what’s the impact on FCFF and FCFE?
NI will be lower by 100*(1-0.4)
Dep will be bigger by 100, so FCFF bigger by 40. same with FCFE
if int exp inc by 100, then you NI will be lower by 100*(1-0.4)
FCFF will add 100*(1-0.4) so no change for him
FCFE will be lower by 100*(1-.4) since you don’T add it to FCFE
• 40 for both, and -60 for FCFE. This is first problem from free cash flow valuation from EOC. It’s tricky. Strongly recommend to go through it at least one time.
thanks for the tip waldz. just did it and it was worth it.
FCFF:+40, 0
FCFE:+40, -60 | 271 | 846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-50 | latest | en | 0.885774 |
http://ftn.uns.ac.rs/n2044407211/discrete-mathematics | 1,679,852,973,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00195.warc.gz | 18,808,225 | 8,286 | #### Faculty of Technical Sciences
Subject: Discrete Mathematics (17 - ESI002)
Basic Information
Category Theoretical-methodological Scientific or art field: Teorijska i primenjena matematika Interdisciplinary yes ECTS 4
Native organizations units
Course specification
Course is active from 15.11.2012..
The main aim of the course is to train students abstract thinking and acquire basic knowledge in the field of classical combinatorial objects, non-classical combinatorial objects and graph theory. Students will learn to classify combinatorial problems and solve them using well-known combinatorial methods, through the acquisition of theoretical knowledge and solving practical examples. Through the learning of well-known concepts and theorems from graph theory, students will be able to set graphic formal models from other fields (e.g. computer science and transport engineering). Properties of graphs will be precisely mathematically proved, with the aim of further development of students skills for deriving proofs.
As outcome of the course, students will acquire basic knowledge in the field of classical combinatorial objects, non-classical combinatorial objects and graph theory, with their abstract thinking and the skills of proofing being greatly improved. Students will be able to recognize combinatorial objects and solve them by known methods, as well as to develop and analyse graph models in some other fields.
Classical combinatorial objects (permutations, variations and combinations with and without repetition), partition sets, Stirling numbers, recurrent formulas, generative functions, basic concepts of graph theory, graph representation, special classes of graphs, operations on graphs, isomorphism of graphs, connected graphs, trees, planar graphs, Euler and Hamiltonian graphs.
In lectures theoretical part of the course is presented accompanied by characteristic and representative examples in order to better understand the matter. In practice, which follows lectures, typical problems are solved and lectured theory is deepened.
AuthorsNameYearPublisherLanguage
Ratko TošićKombinatorika1999Univezitet u Novom SaduSerbian language
Robin J. WilsonIntroduction to Graph Theory1996Robin WilsonEnglish
I. Bošnjak, D. Mašulović, V. Petrović, R. TošićZbirka zadataka iz teorije grafova2006Prirodno-matematicki fakultet, Departman za matematiku i informatikuSerbian language
D. Mašulović, M.PechZbirka zadataka iz kombinatorike2015Prirodno-matematički fakultet, Departman za matematiku i informatikuSerbian language
Course activity Pre-examination ObligationsNumber of points
TestYesYes10.00
TestYesYes10.00
Written part of the exam - tasks and theoryNoYes30.00
Lecture attendanceYesYes5.00
Computer exercise attendanceYesYes5.00
Theoretical part of the examNoYes40.00
Name and surnameForm of classes
Lectures | 629 | 2,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.913493 |
https://www.jiskha.com/display.cgi?id=1271350878 | 1,516,291,625,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887423.43/warc/CC-MAIN-20180118151122-20180118171122-00445.warc.gz | 911,643,671 | 4,372 | # Math
posted by .
Identify which types of sampling is used:random,stratified, systematic,convenience.
1. a sample consists of every 20th student who leaves the bookstore.
2. a pollster uses a computer to generate 500 random numbers, then interviews the voters corrosponding to those numbers.
• Math -
Number 2 doesn't look like stratified to me: there's no stratification involved. Stratification is the process of grouping members of the population into relatively homogeneous subgroups before sampling, whereas here the pollster is simply picking voters at random, with each voter presumably ("presumably" because we don't know anything about the random number generator he's using) having an equal chance of being picked. That's simple random sampling.
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More Similar Questions | 661 | 3,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-05 | latest | en | 0.902231 |
http://mathhelpforum.com/peer-math-review/index2.html | 1,534,621,861,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213737.64/warc/CC-MAIN-20180818193409-20180818213409-00065.warc.gz | 274,552,134 | 14,642 | Page 2 of 3 First 123 Last
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Peer Math Review Forum: For peer math review of your original math work
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,
### my maths forum.com
Click on a term to search for related topics.
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted. | 1,074 | 2,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-34 | longest | en | 0.705476 |
https://www.jiskha.com/questions/653681/find-the-legth-of-the-third-side-of-the-right-triangle-if-a-23-cm-and-b-13-cm-i-thought | 1,611,753,505,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704824728.92/warc/CC-MAIN-20210127121330-20210127151330-00503.warc.gz | 861,639,671 | 4,736 | # math
Find the legth of the third side of the right triangle if a= 23 cm and b= 13 cm. I thought it was 144cm but i am not sure.
1. 👍
2. 👎
3. 👁
1. See your 12:06pm post.
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1. Find the length of the missing side of the right triangle (A triangle is shown to have a base of 15 cm and a height of 8 cm. The slope of it is unmarked A. 289 cm B. 17 cm *** C. 23 cm D. 4.79 cm 2. Find the length of the
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An art class is making a mural for their school which has a triangle drawn in the middle. The length of the bottom of the triangle is x. Another side is 10 more than four times the length of the bottom of the triangle. The last
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Use a proportion to find the length of side x for the pair of smilar figures below. They're both right triangles. Triangle 1 has 14cm on the the right side and 10cm on the bottom. Triangle 2 has x on the right side and 22cm on the
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Surveying A surveyor wishes to find the distance across a swamp. The bearing from A to B (Segment AB is opposite side of triangle) is N 32° W. The surveyor walks 50 meters from A to C, and at the point C the bearing to B is N
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A triangle has a perimeter of 32 inches. The medium side is 7 more than the short side, and the longest side is3 times the length of the shortest side. Find the shortest side. | 776 | 2,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-04 | latest | en | 0.89428 |
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``` Name: __________________________
How to Read a Topographic Map
One special kind of map is called a Topographic Map. It has contour lines to show the
shape and elevation of the land. They are sometimes called "level lines" because they
show points that are at the same level. Here's how contour lines work:
The top of this drawing is a contour map showing the hills that are illustrated at the
bottom.
On this map, the vertical distance between each contour line is 10 feet.
1. Which is higher, hill A or hill B?
_____________________________________________________
2. Which is steeper, hill A or hill B?
_____________________________________________________
3. How many feet of elevation are there between contour lines?
_____________________________________________________
4. How high is hill A? _______________ Hill B? _______________
5. Are there contour lines closer together on hill A or hill B?
_____________________________________________________
Name: __________________________
Advantages to Topographic Maps
Look at this picture. It shows a river valley and several nearby hills.
1. On the illustration, locate the following things:
A church
A bridge over the river
An oceanside cliff
A stream that flows into the main river
A hill that rises steeply on one side and more smoothly on the other.
Here is a topographic map of the same place.
2. Find the items you located on the illustration on the topographic map.
Circle the symbol for a church
Draw a church symbol here.
Put a square around the map symbol for a bridge.
Draw a bridge symbol here.
Put an X on the oceanside cliff.
3. What is the elevation of the contour line at the top of that cliff? ______________
4. Locate a stream that flows into the main river. Draw a pencil line down that stream.
Put an X where the stream joins the main river. On a real topographic map, streams
are shown in blue and contour lines are shown in brown.
5. Find the hill that rises steeply on one side and more smoothly on the other. On the
topographic map, draw a path up the hill to the highest point that would be easy to
climb. (Hint: remember that when contour lines are close together, the ground is
very steep). Draw another path (dotted line) showing a very steep way up the hill.
6. Tell how you might use a topographic map if you were selecting:
a) A route for a hike
__________________________________________________________
__________________________________________________________
b) The best location for an airport
__________________________________________________________
__________________________________________________________
c) A route for a new road
__________________________________________________________
__________________________________________________________
```
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https://sciencedocbox.com/Physics/103254172-10-noether-normalization-and-hilbert-s-nullstellensatz.html | 1,714,059,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00023.warc.gz | 449,690,629 | 27,302 | # 10. Noether Normalization and Hilbert s Nullstellensatz
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1 10. Noether Normalization and Hilbert s Nullstellensatz Noether Normalization and Hilbert s Nullstellensatz In the last chapter we have gained much understanding for integral and finite ring extensions. We now want to prove an elementary but powerful theorem stating that every finitely generated algebra R over a field K (so in particular every coordinate ring of a variety by Remark 1.31) is a finite extension ring of a polynomial ring K[z 1,...,z r ] and hence of a very simple K-algebra that is easy to deal with. Let us start by giving the geometric idea behind this so-called Noether Normalization theorem, which is in fact very simple. Example 10.1 (Idea of Noether Normalization). Let R = C[x 1,x 2 ]/(x 1 x 2 1) be the coordinate ring of the variety X = V (x 1 x 2 1) A 2 C as in Example 9.4 (b). We know already that R is not integral (and hence not finite) over C[x 1 ]; this is easily seen geometrically in the picture below on the left since this map does not satisfy the Lying Over property for the origin as in Example R coordinate change x 1 = y 2 + y 1 x 2 = y 2 y 1 R C[x 1 ] x 1 x 2 1 = 0 not finite C[y 1 ] y 2 2 y2 1 1 = 0 finite It is easy to change this however by a linear coordinate transformation: if we set e. g. x 1 = y 2 +y 1 and x 2 = y 2 y 1 then we can write R also as R = C[y 1,y 2 ]/(y 2 2 y2 1 1), and this is now finite over C[y 1] by Proposition 9.5 since the polynomial y 2 2 y2 1 1 is monic in y 2. Geometrically, the coordinate transformation has tilted the space X as in the picture above on the right so that e. g. the Lying Over property now obviously holds. Note that this is not special to the particular transformation that we have chosen; in fact, almost any linear coordinate change would have worked to achieve this goal. In terms of geometry, we are therefore looking for a change of coordinates so that a suitable coordinate projection to some affine space A r K then corresponds to a finite ring extension of a polynomial ring over K in r variables. Note that this number r can already be thought of as the dimension of X (a concept that we will introduce in Chapter 11) as finite ring extensions correspond to surjective geometric maps with finite fibers by Example 9.19, and thus should not change the dimension (we will prove this in Lemma 11.8). As we have seen above already, the strategy to achieve our goal is to find a suitable change of coordinates so that the given relations among the variables become monic. The first thing we have to do is therefore to prove that such a change of coordinates is always possible. It turns out that a linear change of coordinates works in general only for infinite fields, whereas for arbitrary fields one has to allow more general coordinate transformations. Lemma Let f K[x 1,...,x n ] be a non-zero polynomial over an infinite field K. Assume that f is homogeneous, i. e. every monomial of f has the same degree (in the sense of Exercise 0.16). Then there are a 1,...,a n 1 K such that f (a 1,...,a n 1,1) Proof. We will prove the lemma by induction on n. The case n = 1 is trivial, since a homogeneous polynomial in one variable is just a constant multiple of a monomial.
2 92 Andreas Gathmann So assume now that n > 1, and write f as f = d i=0 f i x1 i where the f i K[x 2,...,x n ] are homogeneous of degree d i. As f is non-zero, at least one f i has to be non-zero. By induction we can therefore choose a 2,...,a n 1 such that f i (a 2,...,a n 1,1) 0 for this i. But then f (,a 2,...,a n 1,1) K[x 1 ] is a non-zero polynomial, so it has only finitely many zeroes. As K is infinite, we can therefore find a 1 K such that f (a 1,...,a n 1,1) 0. Lemma Let f K[x 1,...,x n ] be a non-zero polynomial over an infinite field K. Then there are λ K and a 1,...,a n 1 K such that λ f (y 1 + a 1 y n,y 2 + a 2 y n,...,y n 1 + a n 1 y n,y n ) K[y 1,...,y n ] is monic in y n (i. e. as an element of R[y n ] with R = K[y 1,...,y n 1 ]). Proof. Let d be the degree of f in the sense of Exercise 0.16, and write f = k1,...,k n c k1,...,k n x k 1 1 xk n n with c k1,...,k n K. Then the leading term of λ f (y 1 + a 1 y n,y 2 + a 2 y n,...,y n 1 + a n 1 y n,y n ) = λ c k1,...,k n (y 1 + a 1 y n ) k1 (y n 1 + a n 1 y n ) k n 1 y k n n k 1,...,k n in y n is obtained by always taking the second summand in the brackets and only keeping the degree-d terms, i. e. it is equal to λ c k1,...,k n a k 1 1 ak n 1 n 1 yk 1+ +k n n = λ f d (a 1,...,a n 1,1)y d n, k 1,...,k n k 1 + +k n =d where f d is the (homogeneous) degree-d part of f. Now pick a 1,...,a n 1 by Lemma 10.2 such that f d (a 1,...,a n 1,1) 0, and set λ = f d (a 1,...,a n 1,1) 1. Exercise Let f K[x 1,...,x n ] be a non-zero polynomial over an arbitrary field K. Prove that there are λ K and a 1,...,a n 1 N such that is monic in y n. λ f (y 1 + y a 1 n,y 2 + y a 2 n,...,y n 1 + y a n 1 n,y n ) K[y 1,...,y n ] Proposition 10.5 (Noether Normalization). Let R be a finitely generated algebra over a field K, with generators x 1,...,x n R. Then there is an injective K-algebra homomorphism K[z 1,...,z r ] R from a polynomial ring over K to R that makes R into a finite extension ring of K[z 1,...,z r ]. Moreover, if K is an infinite field the images of z 1,...,z r in R can be chosen to be K-linear combinations of x 1,...,x n. Proof. We will prove the statement by induction on the number n of generators of R. The case n = 0 is trivial, as we can then choose r = 0 as well. So assume now that n > 0. We have to distinguish two cases: (a) There is no algebraic relation among the x 1,...,x n R, i. e. there is no non-zero polynomial f over K such that f (x 1,...,x n ) = 0 in R. Then we can choose r = n and the map K[z 1,...,z n ] R given by z i x i for all i, which is even an isomorphism in this case. (b) There is a non-zero polynomial f over K such that f (x 1,...,x n ) = 0 in R. Then we choose λ and a 1,...,a n 1 as in Lemma 10.3 (if K is infinite) or Exercise 10.4 (for any K) and set y 1 := x 1 a 1 x n,..., y n 1 := x n 1 a n 1 x n, y n := x n (so that x 1 = y 1 + a 1 y n,..., x n 1 = y n 1 + a n 1 y n, x n = y n ) or y 1 := x 1 x a 1 n,..., y n 1 := x n 1 x a n 1 n, y n := x n (so that x 1 = y 1 + y a 1 n,..., x n 1 = y n 1 + y a n 1 n, x n = y n ),
3 10. Noether Normalization and Hilbert s Nullstellensatz 93 respectively. Note that in both cases these relations show that the K-subalgebra K[y 1,...,y n ] of R generated by y 1,...,y n R is the same as that generated by x 1,...,x n, i. e. all of R. Moreover, y n is integral over the K-subalgebra K[y 1,...,y n 1 ] of R, since λ f (y 1 + a 1 y n,...,y n 1 + a n 1 y n,y n ) or λ f (y 1 + y a 1 n,...,y n 1 + y a n 1 n,y n ), respectively, is monic in y n and equal to λ f (x 1,...,x n ) = 0. Hence R = K[y 1,...,y n ] is finite over K[y 1,...,y n 1 ] by Proposition 9.5. In addition, the subalgebra K[y 1,...,y n 1 ] of R is finite over a polynomial ring K[z 1,...,z r ] by the induction hypothesis, and thus R is finite over K[z 1,...,z r ] by Lemma 9.6 (a). Moreover, if K is infinite we can always choose the coordinate transformation of Lemma 10.3, and thus y 1,...,y n (i. e. also the images of z 1,...,z r by induction) are linear combinations of x 1,...,x n. Remark Let R = A(X) be the coordinate ring of a variety X over a field K. (a) In the Noether Normalization of Proposition 10.5, the (images of) z 1,...,z r in R are algebraically independent functions on X in the sense that there is no polynomial relation among them with coefficients in K. On the other hand, every other element of R is algebraically dependent on z 1,...,z r, i. e. it satisfies a (monic) polynomial relation with coefficients in K[z 1,...,z r ]. We can therefore think of r as the number of parameters needed to describe X, i. e. as the dimension of X as already mentioned in Example In fact, we will see in Remark that the number r in Proposition 10.5 is uniquely determined to be the dimension of X in the sense of Chapter 11. (b) As one would have guessed already from the geometric picture in Example 10.1, the proof of Lemma 10.2 shows that most choices of linear coordinate transformations are suitable to obtain a Noether normalization: in each application of this lemma, only finitely many values of a 1 K have to be avoided. Hence we can translate Proposition 10.5 into geometry by saying that a sufficiently general projection to an r-dimensional linear subspace corresponds to a finite ring extension, and hence to a surjective map with finite fibers (where r is the dimension of X as in (a)). Exercise Find a Noether normalization of the C-algebra C[x,y,z]/(xy+z 2,x 2 y xy 3 +z 4 1). Exercise Let R R be an integral ring extension, and assume that R is a finitely generated algebra over some field K. Moreover, let P 1 P 3 be prime ideals in R and P 1 P 3 be prime ideals in R such that P 1 R = P 1 and P 3 R = P 3. (a) Prove: If there is a prime ideal P 2 in R with P 1 P 2 P 3, then there is also a prime ideal P 2 in R with P 1 P 2 P 3. R : P 1 P 2 P 3 (b) Can we always find P 2 in (a) such that in addition P 2 R = P 2 holds? R: P 1 P 2 P 3 As an important application of Noether Normalization we can now give rigorous proofs of some statements in our dictionary between algebra and geometry, namely of the correspondence between (maximal) ideals in the coordinate ring A(X) of a variety X over an algebraically closed field and subvarieties (resp. points) of X. There are various related statements along these lines, and they are all known in the literature by the German name Hilbert s Nullstellensatz ( theorem of the zeroes ). Let us start with the simplest instance of this family of propositions. Still very algebraic in nature, it is the statement most closely related to Noether Normalization, from which the geometric results will then follow easily. Corollary 10.9 (Hilbert s Nullstellensatz, version 1). Let K be a field, and let R be a finitely generated K-algebra which is also a field. Then K R is a finite field extension. In particular, if in addition K is algebraically closed then R = K.
4 94 Andreas Gathmann Proof. By Noether Normalization as in Proposition 10.5 we know that R is finite over a polynomial ring K[z 1,...,z r ], and thus also integral over K[z 1,...,z r ] by Proposition 9.5. But R is a field, hence K[z 1,...,z r ] must be a field as well by Corollary 9.21 (a). This is only the case for r = 0, and so R is finite over K. In particular, if K is algebraically closed then there are no algebraic extension fields of K since all zeroes of polynomials over K lie already in K. Hence by Proposition 9.5 there are no finite extensions either in this case, and we must have R = K. Corollary (Hilbert s Nullstellensatz, version 2). Let K be an algebraically closed field. Then all maximal ideals of the polynomial ring K[x 1,...,x n ] are of the form for some a = (a 1,...,a n ) A n K. I(a) = (x 1 a 1,...,x n a n ) 19 Proof. Let P K[x 1,...,x n ] be a maximal ideal. Then K[x 1,...,x n ]/P is a field by Lemma 2.3 (b), and in addition certainly a finitely generated K-algebra. Hence K[x 1,...,x n ]/P = K by Corollary 10.9, i. e. the natural map K K[x 1,...,x n ]/P, c c is an isomorphism. Choosing inverse images a 1,...,a n of x 1,...,x n we get x i = a i for all i, and thus (x 1 a 1,...,x n a n ) P. But (x 1 a 1,...,x n a n ) is a maximal ideal by Example 2.6 (c), and so we must already have P = (x 1 a 1,...,x n a n ) = I(a). Remark (Points of a variety correspond to maximal ideals). It is easy to extend Corollary to a statement about an arbitrary variety X A n K over an algebraically closed field K: if R = A(X) is the coordinate ring of X we claim that there is a one-to-one correspondence {points of X} 1:1 a I(a) V (I) I. {maximal ideals in A(X)} In fact, the maximal ideals of A(X) = K[x 1,...,x n ]/I(X) are in one-to-one correspondence with maximal ideals I K[x 1,...,x n ] such that I I(X) by Lemma 1.21 and Corollary 2.4. By Corollary this is the same as ideals of the form I(a) = (x 1 a 1,...,x n a n ) containing I(X). But I(a) I(X) is equivalent to a X by Lemma 0.9 (a) and (c), so the result follows. Remark (Zeroes of ideals in K[x 1,...,x n ]). Another common reformulation of Hilbert s Nullstellensatz is that every proper ideal I K[x 1,...,x n ] in the polynomial ring over an algebraically closed field K has a zero: by Corollary 2.17 we know that I is contained in a maximal ideal, which must be of the form I(a) by Corollary But I I(a) implies a V (I) by Lemma 0.9 (a) and (c), and hence V (I) /0. Note that this statement is clearly false over fields that are not algebraically closed, as e. g. (x 2 + 1) is a proper ideal in R[x] with empty zero locus in A 1 R. In order to extend the correspondence between points and maximal ideals to arbitrary subvarieties we need another algebraic preliminary result first: recall that in any ring R the radical of an ideal I equals the intersection of all prime ideals containing I by Lemma We will show now that it is in fact sufficient to intersect all maximal ideals containing I if R is a finitely generated algebra over a field. Corollary (Hilbert s Nullstellensatz, version 3). For every ideal I in a finitely generated algebra R over a field K we have I = P. P maximal P I Proof. The inclusion follows immediately from Lemma 2.21, since every maximal ideal is prime.
5 10. Noether Normalization and Hilbert s Nullstellensatz 95 For the opposite inclusion, let f R with f / I; we have to find a maximal ideal P I with f / P. Consider the multiplicatively closed set S = { f n : n N}. As f / I implies I S = /0, we get by Exercise 6.14 (a) a prime ideal P R with P I and P S = /0, in particular with f / P. Moreover, we can assume by Exercise 6.14 (a) that S 1 P is maximal. It only remains to show that P is maximal. To do this, consider the ring extension K R/P (R/P) f = R f /P f, where the subscript f denotes localization at S as in Example 6.5 (c). Note that the second map is in fact an inclusion since R/P is an integral domain, and the stated equality holds by Corollary 6.22 (b). Moreover, R f /P f is a field since P f is maximal, and finitely generated as a K-algebra (as generators we can choose the classes of generators for R together with 1 f ). So K R f /P f is a finite field extension by Corollary 10.9, and hence integral by Proposition 9.5. But then R/P R f /P f is integral as well, which means by Corollary 9.21 (a) that R/P is a field since R f /P f is. Hence P is maximal by Lemma 2.3 (b). Corollary (Hilbert s Nullstellensatz, version 4). Let X A n K be a variety over an algebraically closed field K. Then for every ideal I A(X) we have I(V (I)) = I. In particular, there is a one-to-one correspondence {subvarieties of X} 1:1 Y I(Y ) V (I) I. Proof. Let us first prove the equality I(V (I)) = I. {radical ideals in A(X)} : Assume that f / I. By Corollary there is then a maximal ideal P A(X) with P I and f / P. But by Remark this maximal ideal has to be of the form I(a) = (x 1 a 1,...,x n a n ) for some point a X. Now I(a) I implies a V (I) by Lemma 0.9 (a) and (c), and f / I(a) means f (a) 0. Hence f / I(V (I)). : Let f I, i. e. f n I for some n N. Then ( f (a)) n = 0, and hence f (a) = 0, for all a V (I). This means that f I(V (I)). The one-to-one correspondence now follows immediately from what we already know: the two maps are well-defined since I(Y ) is always radical by Remark 1.10, and they are inverse to each other by Lemma 0.9 (c) and the statement I(V (I)) = I proven above.
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### Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35
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### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE ABOUT VARIETIES AND REGULAR FUNCTIONS.
ALGERAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE AOUT VARIETIES AND REGULAR FUNCTIONS. ANDREW SALCH. More about some claims from the last lecture. Perhaps you have noticed by now that the Zariski topology | 12,115 | 42,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-18 | latest | en | 0.898418 |
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## Re: [Bug-gnubg] Gammon output setup
From: Øystein Schønning-Johansen Subject: Re: [Bug-gnubg] Gammon output setup Date: Sun, 11 Dec 2011 13:56:07 +0100
Hi Mark!
How's your rally driving going. ;-)
On Sun, Dec 11, 2011 at 4:45 AM, Mark Higgins wrote:
I notice in gnubg and other neural networks the probability of gammon gets its own output node, alongside the probability of (any kind of) win.
Doesn't this sometimes mean that the estimated probability of gammon could be larger than the probability of win, since both sigmoid outputs run from 0 to 1?
There is a sanity check function called after the neural net evaluation, that check that gammons don't exceed wins and backgammon does not exceed gammons.
I'm playing around with making the gammon node represent the probability of a gammon win conditioned on a win; then the unconditional probability of a gammon win = prob of win * conditional prob of gammon win. In that setup, both outputs are free to roam (0,1) without causing inconsistencies.
That's a possibility, but I go not believe it gains anything. (This is of course just a guess, since I've not tried. And you are of course free to try.) I guess you also need a similar scheme for backgammons?
Is there something I'm missing here about why this is suboptimal? Is there some other way people tend to ensure that prob of gammon win <= prob of any kind of win?
I guess you have to divide by the win prob in the training, which is still just an estimate. Hmmm.. I'm still thinking, maybe it can gain something, since they are kind of depending on each other.
However... what I would rather try is to have six outputs with a softmax activation function. Several neural net experts recommends softmax in their books and papers, and other parameter update rules (other than backpropagation) has been developed based on softmax outputs.
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# Newton’s First Law of Motion (Law of Inertia)
Definition:
- A body will continue its state of rest or uniform motion in a straight
line, unless it is acted upon by an external resultant force.
Newton’s First Law of Motion (Law of Inertia)
## - When there is an external force acting, there is a change of motion:
a) Rest ------> motion
b) Motion --------> Rest
c) Change of speed
d) Change in direction
e) Change in both speed and direction
## - The natural tendency of objects to resist changes in their state of
motion is described as INERTIA and inertia depends on mass.
Case A: The car and the wall
- What will happen to the driver if the car is halted by the collision with
the wall?
## Case B: The truck and the ladder
Newton’s Second Law of Motion
Definition:
- The rate of change of momentum of a
body is directly proportional to the
resulting force acting on it, and the
momentum change occurs in the
direction of the force.
## An object requires a force in order
for it to move with constant
velocity? NO
Newton’s Third Law of Motion
## Definition: If a body A exerts a force on body B, then body B exerts an
equal force but in the opposite direction on body A.
OR
To every action there is an equal but opposite reaction.
Newton’s Third Law of Motion
**Consider the flying motion of birds. A bird flies by use of its wings.
The wings of a bird push air downwards. Since forces result from
mutual interactions, the air must also be pushing the bird upwards.
## ** Bird flies at constant height.
FREE BODY DIAGRAM
To solve problem involving multiple forces, it is best to sketch each
and every force that is acting on the body
The steps involving are:
1. Sketch the force
2. Isolate the Object of Interest
3. Choose a convenient coordinate system
4. Resolve the forces into component
5. Apply Newton's Second Law to each coordinate direction
EXAMPLE
Moe, Larry and Curly push on a 752-kg boat that
floats next to a dock. They each exert an 80.5-N
force parallel to the dock.
a) What is the magnitude of the boat if all push in
the same direction? Give direction and
magnitude
b) What are the magnitude and direction of the
boat’s acceleration if Larry and Curly push in
the opposite direction to Moe’s push?
EXAMPLE
## Jack and Jill lift upward on 1.30-kg pail of water,
with jack exerting a force of F1 of magnitude 7.0
N and Jill exerting force of F2 of magnitude 11 N.
Jill’s force is exerted at an angle of 28° with the
vertical.
a) At what angle θ with respect to the vertical should
Jack exert his force if the pail is to accelerate straight
upward
b) Determine the acceleration of the pail of water given
the weight, W, has a magnitude of 12.8 N
TRY
A 4.60-kg sled is pulled across a smooth ice surface.
The force acting on the sled is of magnitude 6.20
N and points in a direction 35.0° above the
horizontal. If the sled starts at rest, how fast is it
going after being pulled for 1.15 s?
WEIGHT
Weight, W, of an object on the Earth’s surface is
the gravitational force exerted on it by the Earth
The formula is:
## The SI unit still as Newton ,N,.
The kg usually associate with weight is actually
mass, m,
APPARENT WEIGHT
The feeling of heavier and lighter when riding an
elevator is an apparent weight.
If the force is greater than our weight, it will feel
heavy.
If the force is lower than our weight, we will feel
lighter.
Sum of force:
EXAMPLE
The fire alarm goes off, and a 97-kg fireman slides
3.0 m down a pole to the ground floor. Suppose
the fireman starts from rest, slides with constant
acceleration, and reaches the ground floor in 1.2
s. What was upward force F exerted by the pole
on the fireman?
EXAMPLE
As part of an attempt to combine physics and
biology in the same class, an instructor asks
student to weigh a 5.0-kg salmon by hanging it
from a fish scale attached to the ceiling of an
elevator. What is the apparent weight of the
salmon Wa, if the elevator
a) Is at rest
## c) Moves with a downward acceleration of 3.2 m/s2
TRY
A newborn baby’s brain grows rapidly. In fact, it
has been found to increase in mass by about 1.6
mg per minute.
a) How much does the brain’s weight increase in
one day? [0.023 N]
b) How long does it take for the brain’s weight to
increases by 0.15 N? [6.6 days]
NORMAL FORCE
An object that is at rest on a surface has a total of
zero acceleration.
This mean that there is an upward force acted on
the object to counter the downward force acted on
the object in the form of weight.
This force is Normal force.
## It called Normal force is because it is
perpendicular to the surface.
EXAMPLE
A 6.0-kg block of ice is acted on by two force, F1 and
F2, as shown in the diagram. If the magnitude of
the force F1 = 13 N and F2 = 11 N, find
a) The acceleration of the ice
## b) The normal exerted on it by the table
TRY
Find the normal force exerted on a 2.9-kg book
resting on a surface inclined at 36° above the
horizontal.
[23 N]
Friction Force
Definition: Friction is the force that opposes the relative motion of two
surfaces in contact due to the roughness of the surfaces.
## - There are two types of frictional forces namely:
a) Static friction
b) Kinetic friction
motion occurs.
## Normal Force/Reaction R = Weight W
Applied Force F < Static Force Fr
** Object resists to move
Applied Force F > Static Force Fr
** Object starts to move
Static Friction Force
## • Coefficient of static friction,
• Static friction, Fs = µR
EXMPLE
A flatbed truck slowly tilts its bed upward to
dispose of a 95.0-kg crate. For small angles of tilt
the crate stays put, but when the tilt angle
exceeds 23.2°, the crate begins to slide. What is
the coefficient of static friction between the bed of
the truck and the crate?
Friction Force
motion occurs.
## • Coefficient of kinetic friction,
• Kinetic friction, Fk = µR
EXAMPLE
Someone at the other end of the table asks you to
pass the salt. Feeling quite dashing, you slide the
50.0-g salt shaker in their direction, giving it an
initial speed of 1.15 m/s
a) If the shaker comes to rest with constant
acceleration in 0.840 m, what is the coefficient
of kinetic friction between the shaker and the
table
b) How much time is required for the shaker to
come to rest if you side it with an initial speed
of 1.32 m/s
HOOKE’S LAW
It is the law that govern the elastic of spring
Its formula are:
## Where x is the length the spring strecth or
compress from its original
k is the force constant
EXAMPLE
An increasingly popular device for improving the
air flow through nasal passages is the nasal strip,
which consists of two flat, polyester springs
enclosed by an adhesive tape covering ,
Measurement shows that a nasal strip can exert
an outward force of 0.22 N on the ne, casing it to
expand by 3.5 mm.
a) Treating the nose as a ideal spring, find its force
constant in Newton per meter
b) How much force would be required to expand
the by 4.0 mm? | 1,853 | 6,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-45 | latest | en | 0.886641 |
https://www.unitconverters.net/length/link-to-x-unit.htm | 1,718,403,511,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00250.warc.gz | 945,421,069 | 4,092 | Home / Length Conversion / Convert Link to X-unit
Please provide values below to convert link [li] to X-unit [X], or vice versa.
### Link to X-unit Conversion Table
0.01 li20075043908.67 X
0.1 li200750439086.7 X
1 li2007504390867 X
2 li4015008781734 X
3 li6022513172601 X
5 li10037521954335 X
10 li20075043908670 X
20 li40150087817340 X
50 li1.0037521954335E+14 X
100 li2.007504390867E+14 X
1000 li2.007504390867E+15 X
### How to Convert Link to X-unit
1 li = 2007504390867 X
1 X = 4.9813091545375E-13 li
Example: convert 15 li to X:
15 li = 15 × 2007504390867 X = 30112565863005 X | 225 | 587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.424455 |
https://rdrr.io/cran/multiridge/man/predictIWLS.html | 1,638,982,932,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00103.warc.gz | 552,046,504 | 8,625 | # predictIWLS: Predictions from ridge fits In multiridge: Fast Cross-Validation for Multi-Penalty Ridge Regression
## Description
Produces predictions from ridge fits for new data.
## Usage
`1` ```predictIWLS(IWLSfit, X1new = NULL, Sigmanew) ```
## Arguments
`IWLSfit` List, containing fits from either `IWLSridge` (linear, logistic ridge) or `IWLSCoxridge` `X1new` Matrix. Dimension `nnew x p_0`, representing unpenalized covariates for new data. `Sigmanew` Matrix. Dimensions `nnew x n`. Sample cross-product from penalized variables, usually computed by first applying `createXXblocks` and then `SigmaFromBlocks`.
## Details
Predictions rely purely on the linear predictors, and do not require producing the parameter vector.
## Value
Numerical vector of linear predictor for the test samples.
`IWLSridge` (`IWLSCoxridge`) for fitting linear and logistic ridge (Cox ridge). `betasout` for obtaining parameter estimates. `Scoring` to evaluate the predictions. A full demo and data are available from:
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21``` ```#Example below shows how to create the input argument Sigmanew (for simulated data) #Simulate Xbl1 <- matrix(rnorm(1000),nrow=10) Xbl2 <- matrix(rnorm(2000),nrow=10) Xbl1new <- matrix(rnorm(200),nrow=2) Xbl2new <- matrix(rnorm(400),nrow=2) #check whether dimensions are correct nrow(Xbl1)==nrow(Xbl1new) nrow(Xbl2)==nrow(Xbl2new) ncol(Xbl1)==nrow(Xbl2) ncol(Xbl1new)==ncol(Xbl2new) #create cross-product XXbl <- createXXblocks(list(Xbl1,Xbl2),list(Xbl1new,Xbl2new)) #suppose penalties for two data types equal 5,10, respectively Sigmanew <- SigmaFromBlocks(XXbl,c(5,10)) #check dimensions (should be nnew x n) dim(Sigmanew) ``` | 518 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | latest | en | 0.604476 |
https://www.aqua-calc.com/page/density-table/substance/tin-op-ii-cp--blank-chloride | 1,709,198,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474795.48/warc/CC-MAIN-20240229071243-20240229101243-00523.warc.gz | 658,969,065 | 10,543 | # Density of Tin(II) chloride [SnCl2 or Cl2Sn]
## Tin(II) chloride weighs 3 950 kg/m³ (246.59044 lb/ft³)
• Density of Tin(II) chloride in a few select units of density measurement:
• Density of Tin(II) chloride g cm3 = 3.95 g/cm³
• Density of Tin(II) chloride g ml = 3.95 g/ml
• Density of Tin(II) chloride g mm3 = 0.004 g/mm³
• Density of Tin(II) chloride kg m3 = 3 950 kg/m³
• Density of Tin(II) chloride lb in3 = 0.14 lb/in³
• Density of Tin(II) chloride lb ft3 = 246.59 lb/ft³
• See density of Tin(II) chloride in hundreds of units of density measurement grouped by weight.
### Tin(II) chloride density values, grouped by weight and shown as value of density, unit of density
60.96 gr/cm³ 60 957.82 gr/dm³ 1 726 133.11 gr/ft³ 998.92 gr/in³ 60 957 815.68 gr/m³ 0.06 gr/mm³ 46 605 593.78 gr/yd³ 60 957.82 gr/l 15 239.45 gr/metric c 914.37 gr/metric tbsp 304.79 gr/metric tsp 60.96 gr/ml 14 421.9 gr/US c 1 802.52 gr/fl.oz 230 750.43 gr/US gal 28 843.8 gr/pt 57 687.61 gr/US qt 901.37 gr/US tbsp 300.46 gr/US tsp
3.95 g/cm³ 3 950 g/dm³ 111 851.54 g/ft³ 64.73 g/in³ 3 950 000 g/m³ 0 g/mm³ 3 019 991.69 g/yd³ 3 950 g/l 987.5 g/metric c 59.25 g/metric tbsp 19.75 g/metric tsp 3.95 g/ml 934.52 g/US c 116.8 g/fl.oz 14 952.38 g/US gal 1 869.05 g/pt 3 738.09 g/US qt 58.41 g/tbsp 19.47 g/tsp
0 kg/cm³ 3.95 kg/dm³ 111.85 kg/ft³ 0.06 kg/in³ 3 950 kg/m³ 3.95 × 10-6 kg/mm³ 3 019.99 kg/yd³ 3.95 kg/l 0.99 kg/metric c 0.06 kg/metric tbsp 0.02 kg/metric tsp 0 kg/ml 0.93 kg/US c 0.12 kg/fl.oz 14.95 kg/US gal 1.87 kg/pt 3.74 kg/US qt 0.06 kg/tbsp 0.02 kg/tsp
3.89 × 10-6 long tn/cm³ 0 long tn/dm³ 0.11 long tn/ft³ 6.37 × 10-5 long tn/in³ 3.89 long tn/m³ 3.89 × 10-9 long tn/mm³ 2.97 long tn/yd³ 0 long tn/l 0 long tn/metric c 5.83 × 10-5 long tn/metric tbsp 1.94 × 10-5 long tn/metric tsp 3.89 × 10-6 long tn/ml 0 long tn/US c 0 long tn/fl.oz 0.01 long tn/US gal 0 long tn/pt 0 long tn/US qt 5.75 × 10-5 long tn/US tbsp 1.92 × 10-5 long tn/US tsp
3 950 000 µg/cm³ 3 950 000 000 µg/dm³ 111 851 544 070 µg/ft³ 64 728 902.8 µg/in³ 3 950 000 000 000 µg/m³ 3 950 µg/mm³ 3 019 991 689 100 µg/yd³ 3 950 000 000 µg/l 987 500 000 µg/metric c 59 250 000 µg/metric tbsp 19 750 000 µg/metric tsp 3 950 000 µg/ml 934 523 536.15 µg/US c 116 815 441.92 µg/fl.oz 14 952 376 531 µg/US gal 1 869 047 068.35 µg/pt 3 738 094 136.7 µg/US qt 58 407 720.96 µg/tbsp 19 469 240.28 µg/tsp
3 950 mg/cm³ 3 950 000 mg/dm³ 111 851 544.07 mg/ft³ 64 728.9 mg/in³ 3 950 000 000 mg/m³ 3.95 mg/mm³ 3 019 991 689.1 mg/yd³ 3 950 000 mg/l 987 500 mg/metric c 59 250 mg/metric tbsp 19 750 mg/metric tsp 3 950 mg/ml 934 523.54 mg/US c 116 801.5 mg/fl.oz 14 952 376.57 mg/US gal 1 869 047.07 mg/pt 3 738 094.14 mg/US qt 58 407.72 mg/tbsp 19 469.24 mg/tsp
0.14 oz/cm³ 139.33 oz/dm³ 3 945.45 oz/ft³ 2.28 oz/in³ 139 332.15 oz/m³ 0 oz/mm³ 106 527.07 oz/yd³ 139.33 oz/l 34.83 oz/metric c 2.09 oz/metric tbsp 0.7 oz/metric tsp 0.14 oz/ml 32.96 oz/US c 4.42 oz/fl.oz 527.43 oz/US gal 65.93 oz/pt 131.86 oz/US qt 2.06 oz/tbsp 0.69 oz/tsp
2.54 dwt/cm³ 2 539.91 dwt/dm³ 71 922.21 dwt/ft³ 41.62 dwt/in³ 2 539 908.99 dwt/m³ 0 dwt/mm³ 1 941 899.74 dwt/yd³ 2 539.91 dwt/l 634.98 dwt/metric c 38.1 dwt/metric tbsp 12.7 dwt/metric tsp 2.54 dwt/ml 600.91 dwt/US c 75.11 dwt/fl.oz 9 614.6 dwt/US gal 1 201.83 dwt/pt 2 403.65 dwt/US qt 37.56 dwt/US tbsp 12.52 dwt/US tsp
0.01 lb/cm³ 8.71 lb/dm³ 246.59 lb/ft³ 0.14 lb/in³ 8 708.26 lb/m³ 8.71 × 10-6 lb/mm³ 6 657.94 lb/yd³ 8.71 lb/l 2.18 lb/metric c 0.13 lb/metric tbsp 0.04 lb/metric tsp 0.01 lb/ml 2.06 lb/US c 0.28 lb/fl.oz 32.96 lb/US gal 4.12 lb/pt 8.24 lb/US qt 0.13 lb/tbsp 0.04 lb/tsp
4.35 × 10-6 short tn/cm³ 0 short tn/dm³ 0.12 short tn/ft³ 7.14 × 10-5 short tn/in³ 4.35 short tn/m³ 4.35 × 10-9 short tn/mm³ 3.33 short tn/yd³ 0 short tn/l 0 short tn/metric c 6.53 × 10-5 short tn/metric tbsp 2.18 × 10-5 short tn/metric tsp 4.35 × 10-6 short tn/ml 0 short tn/US c 0 short tn/fl.oz 0.02 short tn/US gal 0 short tn/pt 0 short tn/US qt 6.44 × 10-5 short tn/US tbsp 2.15 × 10-5 short tn/US tsp
0 sl/cm³ 0.27 sl/dm³ 7.66 sl/ft³ 0 sl/in³ 270.66 sl/m³ 2.71 × 10-7 sl/mm³ 206.94 sl/yd³ 0.27 sl/l 0.07 sl/metric c 0 sl/metric tbsp 0 sl/metric tsp 0 sl/ml 0.06 sl/US c 0.01 sl/fl.oz 1.02 sl/US gal 0.13 sl/pt 0.26 sl/US qt 0 sl/tbsp 0 sl/tsp
0 st/cm³ 0.62 st/dm³ 17.61 st/ft³ 0.01 st/in³ 622.02 st/m³ 6.22 × 10-7 st/mm³ 475.57 st/yd³ 0.62 st/l 0.16 st/metric c 0.01 st/metric tbsp 0 st/metric tsp 0 st/ml 0.15 st/US c 0.02 st/fl.oz 2.35 st/US gal 0.29 st/pt 0.59 st/US qt 0.01 st/US tbsp 0 st/US tsp
3.95 × 10-6 t/cm³ 0 t/dm³ 0.11 t/ft³ 6.47 × 10-5 t/in³ 3.95 t/m³ 3.95 × 10-9 t/mm³ 3.02 t/yd³ 0 t/l 0 t/metric c 5.93 × 10-5 t/metric tbsp 1.98 × 10-5 t/metric tsp 3.95 × 10-6 t/ml 0 t/US c 0 t/fl.oz 0.01 t/US gal 0 t/pt 0 t/US qt 5.84 × 10-5 t/tbsp 1.95 × 10-5 t/tsp
0.13 oz t/cm³ 127 oz t/dm³ 3 596.11 oz t/ft³ 2.08 oz t/in³ 126 995.45 oz t/m³ 0 oz t/mm³ 97 094.99 oz t/yd³ 127 oz t/l 31.75 oz t/metric c 1.9 oz t/metric tbsp 0.63 oz t/metric tsp 0.13 oz t/ml 30.05 oz t/US c 3.76 oz t/fl.oz 480.73 oz t/US gal 60.09 oz t/pt 120.18 oz t/US qt 1.88 oz t/US tbsp 0.63 oz t/US tsp
0.01 troy/cm³ 10.58 troy/dm³ 299.68 troy/ft³ 0.17 troy/in³ 10 582.95 troy/m³ 1.06 × 10-5 troy/mm³ 8 091.25 troy/yd³ 10.58 troy/l 2.65 troy/metric c 0.16 troy/metric tbsp 0.05 troy/metric tsp 0.01 troy/ml 2.5 troy/US c 0.31 troy/fl.oz 40.06 troy/US gal 5.01 troy/pt 10.02 troy/US qt 0.16 troy/US tbsp 0.05 troy/US tsp
Tin(II) chloride density values in 285 units of density, in the form of a matrix
Density = weight ÷ volumemicrogram (µg)milligram (mg)gram (g)kilogram (kg)tonne (t)ounce (oz)pound (lb)volume unitgrain (gr)slug (sl)short ton (short tn)long ton (long tn)stone (st)troy ounce (oz t)troy pound (troy)pennyweight (dwt)
cubic millimeter3 9503.95<0.01<0.01<0.01<0.01<0.01cubic millimeter0.06<0.01<0.01<0.01<0.01<0.01<0.01<0.01
cubic centimeter3 950 0003 9503.95<0.01<0.010.140.01cubic centimeter60.96<0.01<0.01<0.01<0.010.130.012.54
cubic decimeter3 950 000 0003 950 0003 9503.95<0.01139.338.71cubic decimeter60 957.820.27<0.01<0.010.6212710.582 539.91
cubic meter3 950 000 000 0003 950 000 0003 950 0003 9503.95139 332.158 708.26cubic meter60 957 815.68270.664.353.89622.02126 995.4510 582.952 539 908.99
milliliter3 950 0003 9503.95<0.01<0.010.140.01milliliter60.96<0.01<0.01<0.01<0.010.130.012.54
liter3 950 000 0003 950 0003 9503.95<0.01139.338.71liter60 957.820.27<0.01<0.010.6212710.582 539.91
metric teaspoon19 750 00019 75019.750.02<0.010.70.04metric teaspoon304.79<0.01<0.01<0.01<0.010.630.0512.7
metric tablespoon59 250 00059 25059.250.06<0.012.090.13metric tablespoon914.37<0.01<0.01<0.010.011.90.1638.1
metric cup987 500 000987 500987.50.99<0.0134.832.18metric cup15 239.450.07<0.01<0.010.1631.752.65634.98
cubic inch64 728 902.864 728.964.730.06<0.012.280.14cubic inch998.92<0.01<0.01<0.010.012.080.1741.62
cubic foot111 851 544 070111 851 544.07111 851.54111.850.113 945.45246.59cubic foot1 726 133.117.660.120.1117.613 596.11299.6871 922.21
cubic yard3 019 991 689 1003 019 991 689.13 019 991.693 019.993.02106 527.076 657.94cubic yard46 605 593.78206.943.332.97475.5797 094.998 091.251 941 899.74
US teaspoon19 469 240.2819 469.2419.470.02<0.010.690.04US teaspoon300.46<0.01<0.01<0.01<0.010.630.0512.52
US tablespoon58 407 720.9658 407.7258.410.06<0.012.060.13US tablespoon901.37<0.01<0.01<0.010.011.880.1637.56
US fluid ounce116 815 441.92116 801.5116.80.12<0.014.420.28US fluid ounce1 802.520.01<0.01<0.010.023.760.3175.11
US cup934 523 536.15934 523.54934.520.93<0.0132.962.06US cup14 421.90.06<0.01<0.010.1530.052.5600.91
US pint1 869 047 068.351 869 047.071 869.051.87<0.0165.934.12US pint28 843.80.13<0.01<0.010.2960.095.011 201.83
US quart3 738 094 136.73 738 094.143 738.093.74<0.01131.868.24US quart57 687.610.26<0.01<0.010.59120.1810.022 403.65
US gallon14 952 376 53114 952 376.5714 952.3814.950.01527.4332.96US gallon230 750.431.020.020.012.35480.7340.069 614.6
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The troy pound per cubic meter density measurement unit is used to measure volume in cubic meters in order to estimate weight or mass in troy pounds
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sl/m³ to g/ft³ conversion table, sl/m³ to g/ft³ unit converter or convert between all units of density measurement.
#### Calculators
Area of a circle calculator, circumference and area formulas | 4,341 | 9,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.269034 |
http://jelv.is/talks/stlc.html | 1,524,448,641,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945669.54/warc/CC-MAIN-20180423011954-20180423031954-00140.warc.gz | 155,467,364 | 7,130 | Tikhon Jelvis (tikhon@jelv.is)
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# 1 Untyped Lambda Calculus
• simple model of functions
• few parts:
\begin{align} e ::&= x & \text{variable}\\ &|\quad \lambda x. e & \text{abstraction} \\ &|\quad e_1 e_2 & \text{application} \end{align}
# 2 Untyped Lambda Calculus
• simple evaluation
• just function application!
$$\frac{(\lambda x. e) e'}{[e'/x]e}$$
• replace $$x$$ with the argument in the body
# 3 The Song that Never Ends
• Lambda calculus is Turing-complete (Church-Turing thesis)
• infinite loops:
$$(\lambda x. x x) (\lambda y. y y) \Rightarrow (\lambda y. y y) (\lambda y. y y)$$
• good for programming, bad for logic
# 4 Preventing Self-Application
• problem: self-application
• $$xx$$ leads to infinite loops
• we need a rule to prevent self-application (and infinite loops in general)
• simple
• syntactic
• static
• conservative by necessity
$$\newcommand{\t}[1]{\mathbf{#1}}$$ $$\newcommand{\ite}[3]{\text{if }#1\text{ then }#2\text{ else }#3}$$ $$\newcommand{\case}[5]{\text{case }#1\text{ of }#2 \to #3 \quad|\ #4 \to #5}$$
# 5 Why?
• helps lambda calculus as a logic
• provides simple model of real type systems
• helps design new types and type systems
• usual advantages of static typing
# 6 Base Types
• ints, booleans… whatever
• even just the $$\t{unit}$$ type is fine
• base types have values:
• $$()$$ is of type $$\t{unit}$$
• $$1$$ is of type $$\t{int}$$
• ultimately, the exact base types don't matter
# 7 Function Types
• one type constructor: $$\to$$ (like axiom schema)
• represents function types
• $$\t{unit} \to \t{unit}$$
• $$\t{int} \to \t{unit} \to \t{int}$$
• values are functions
# 8 Assigning Types
• we need some way to give a type to an expression
• only depends on the static syntax
• typing judgement: $$x : \tau$$
# 9 Context
• depends on what's in scope (typing context): $$\Gamma \vdash x : \tau$$
• things in scope: "context", $$\Gamma$$
• set of typing judgements for free variables:$$\Gamma = \{x : \tau, y : \tau \to \tau, ... \}$$
# 10 New Syntax
\begin{align} \tau ::&= \t{unit} & \text{unit type}\\ &|\quad \t{int} & \text{int type}\\ &|\quad \tau_1 \to \tau_2 & \text{function types} \end{align}
\begin{align} e ::&= () & \text{unit value}\\ &|\quad n & \text{integer}\\ &|\quad e_1 + e_2 & \text{arithmetic}\\ &|\quad x & \text{variable}\\ &|\quad \lambda x:\tau. e & \text{abstraction}\\ &|\quad e_1 e_2 & \text{application} \end{align}
# 11 Typing Rules
• we can assign types following a few "typing rules"
• idea: if we see expression "x", we know "y"
• just like implication in logic $$\frac{\text{condition}}{\text{result}}$$
• remember the context matters: $$\Gamma$$
# 12 Base rules
• note: no prerequisites!
$$\frac{}{\Gamma \vdash n : \t{int}}$$ $$\frac{}{\Gamma \vdash () : \t{unit}}$$
• base cases for recursion
# 13 Main Rules
• contexts:
$$\frac{x : \tau \in \Gamma}{\Gamma \vdash x : \tau}$$
• function bodies:
$$\frac{\Gamma, x : \tau \vdash e : \tau'}{\Gamma \vdash (\lambda x:\tau. e) : \tau \to \tau'}$$
# 14 Main Rules
• application:
$$\frac{\Gamma \vdash e_1 : \tau \to \tau' \quad \Gamma \vdash e_2 : \tau}{\Gamma \vdash e_1 e_2 : \tau'}$$
• recursive cases in the type system
• think of a function over syntactic terms
• similar to evaluation!
# 15 Domain-Specific Rules
• we add rules for our "primitive" operations
$$\frac{\Gamma \vdash e_1 : \t{int} \quad \Gamma \vdash e_2 : \t{int}}{\Gamma \vdash e_1 + e_2 : \t{int}}$$
• imagine other base types like booleans
$$\frac{\Gamma \vdash c : \t{bool} \quad \Gamma \vdash e_1 : \tau \quad \Gamma \vdash e_2 : \tau}{\Gamma \vdash \ite{c}{e_1}{e_2} : \tau}$$
• easy to extend
# 16 No Polymorphsim
• we do not have any notion of polymorphism
• function arguments have to be annotated
• untyped: $$\lambda x. x$$
• typed:
• $$\lambda x:unit. x$$
• $$\lambda x:int. x$$
• $$\lambda x:int \to unit \to int. x$$
# 17 Numbers
• remember numbers as repeated application
• untyped:
• 0: $$\lambda f. \lambda x. x$$
• 1: $$\lambda f. \lambda x. f x$$
• 2: $$\lambda f. \lambda x. f (f x)$$
• 3: $$\lambda f. \lambda x. f (f (f x))$$
# 18 Typed Numbers
• 0: $$\lambda f : \t{unit} \to \t{unit}. \lambda x : \t{unit}. x$$
• 1: $$\lambda f : \t{unit} \to \t{unit}. \lambda x : \t{unit}. f x$$
• 2: $$\lambda f : \t{unit} \to \t{unit}. \lambda x : \t{unit}. f (f x)$$
• 3: $$\lambda f : \t{unit} \to \t{unit}. \lambda x : \t{unit}. f (f (f x))$$
• numbers: $$(\t{unit} \to \t{unit}) \to \t{unit} \to \t{unit}$$
# 19 Pairs
• remember pair encoding:
• cons: $$\lambda x. \lambda y. \lambda f. f x y$$
• first: $$\lambda x. \lambda y. x$$
• second: $$\lambda x. \lambda y. y$$
• lets us build up data types, like lisp
# 20 Typed Pairs
• cons: $$\lambda x : \tau. \lambda y : \tau . \lambda f : \tau \to \tau \to \tau. f x y$$
• but we want pairs of different types!
• we should add pairs ("product types") to our system
# 21 Product Types
• new type syntax: $$\tau_1 \times \tau_2$$
• like Haskell's (a, b) or OCaml's a * b
• constructor:
$$\frac{\Gamma \vdash e_1 : \tau_1 \quad \Gamma \vdash e_2 : \tau_2}{\Gamma \vdash (e_1, e_2) : \tau_1 \times \tau_2}$$
# 22 Product Types
• accessors ($$\text{first}$$ and $$\text{second}$$):
$$\frac{\Gamma \vdash e : \tau_1 \times \tau_2}{\Gamma \vdash \text{first } e : \tau_1}$$ $$\frac{\Gamma \vdash e : \tau_1 \times \tau_2}{\Gamma \vdash \text{second } e : \tau_2}$$
# 23 Sum Types
• sum types: disjoint/tagged unions, variants
• like Haskell's Either
• new type syntax: $$\tau_1 + \tau_2$$
• construction:
$$\frac{\Gamma \vdash e : \tau_1}{\Gamma \vdash \text{left } e : \tau_1 + \tau_2}$$ $$\frac{\Gamma \vdash e : \tau_2}{\Gamma \vdash \text{right } e : \tau_1 + \tau_2}$$
# 24 Sum Types
• matching
$$\frac{\Gamma\ \vdash\ e : \tau_1 + \tau_2 \atop \Gamma,\ x : \tau_1 \ \vdash\ e_1 : \tau' \quad \Gamma,\ y : \tau_2 \ \vdash\ e_2 : \tau'}{\Gamma \vdash (\case{e}{x}{e_1}{y}{e_2}) : \tau'}$$
# 25 Algebraic Data Types
• this basically gives us algebraic data types
• now we just need recursive types and polymorphism | 2,205 | 6,080 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-17 | longest | en | 0.56683 |
https://ethereum.stackexchange.com/questions/99763/whats-exactly-ether-unit-return-web3-eth-estimategas-wei-gwei/126027 | 1,716,817,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00266.warc.gz | 212,991,062 | 41,194 | # Whats exactly Ether unit return web3.eth.estimateGas, wei, gwei ...?
I am testing with function web3.eth.estimateGas and is not clear what Ether unit is returning.
Looks like it's gwei, but my final calculations don't match.
In my code, the EGAS returned is 84753. I think it is too little to be wei or gwei
How can I convert to wei the uint returned by estimateGas function?
I think I found the answer, but if you go to the source code "eth.py", function signature says it's returning "Wei":
``````def estimateGas(self, transaction: TxParams, block_identifier: BlockIdentifier=None) -> Wei:
``````
This is actually very misleading. I tested this out. It returned "239231", but in what unit? You should interpret the number as follows:
``````gas_estimate
239231
gas_price = w3.eth.gasPrice
gas_price
5000000000000
gas_estimate_wei = gas_estimate * gas_price
gas_estimate_wei
1196155000000000000
gas_estimate_in_coin = gas_estimate_wei / (1000000000 * 1000000000)
gas_estimate_in_coin
1.196155
``````
How you should interpret the numbers are as follows:
gas_estimate = 239231 units
gas_price = 5000,000,000,000 wei per unit
Or
gas_price = 5000 gwei per unit
So, "gas_estimate_in_coin" = 1.196155. Now, what "coin"? It depends on the chain you're interacting with. If you're interacting with CRONOS, that's 1.196155 CRO, which is about 49.9 cents in USD (CROUSDT = 0.4168 on 20220413)! On the other hand, on Ethereum mainnet, that's 1.196155 ETH, which changed everything!!!
References
(Wei, Gwei, ETH conversion)
https://blog.oasis.app/gas-fees-a-small-guide/
https://www.thebalance.com/gwei-5194614
https://norman-lm-fung.medium.com/interact-with-cronos-single-usdc-lp-with-web3-py-1e14c62a0d9c
Similar question: Smart Contract - approve function: wad
It returns the gas limit, not the gas price. Gas limit is how many units of gas to use at most, and a unit of gas is your gas price.
• I dont want to know te gasPrice. There is a web3js function for that. I need to know the real fee for a transaccion in Wei. May 27, 2021 at 9:56
• For that you need to use `web3.eth.getGasPrice`. May 27, 2021 at 10:06
• getGasPrice is a fixed value by the chain. I dont need that. May 27, 2021 at 19:43
• What? It literally returns the thing you want, the gas price in Wei. Also it is NOT a fixed value. Try multiplying `estimateGas` with `getGasPrice` for a transaction's total fee in wei. May 28, 2021 at 3:47
• this awnser is prerry correct to what was asked, why somebody downvoted it. estimateGas returns the gas limit, its unit is gas 'unit'. Sep 13, 2022 at 0:48 | 750 | 2,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.802323 |
https://exercism.io/tracks/ruby/exercises/wordy/solutions/7f28c3b3863941dc9d8efb9846b03614 | 1,604,030,390,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107907213.64/warc/CC-MAIN-20201030033658-20201030063658-00433.warc.gz | 308,032,356 | 7,506 | π Exercism Research is now launched. Help Exercism, help science and have some fun at research.exercism.io π
# apfinger's solution
## to Wordy in the Ruby Track
Published at Jul 13 2018 · 0 comments
Instructions
Test suite
Solution
#### Note:
This solution was written on an old version of Exercism. The tests below might not correspond to the solution code, and the exercise may have changed since this code was written.
Parse and evaluate simple math word problems returning the answer as an integer.
What is 5 plus 13?
Evaluates to 18.
Handle large numbers and negative numbers.
## Iteration 2 β Subtraction, Multiplication and Division
Now, perform the other three operations.
What is 7 minus 5?
2
What is 6 multiplied by 4?
24
What is 25 divided by 5?
5
## Iteration 3 β Multiple Operations
Handle a set of operations, in sequence.
Since these are verbal word problems, evaluate the expression from left-to-right, ignoring the typical order of operations.
What is 5 plus 13 plus 6?
24
What is 3 plus 2 multiplied by 3?
15 (i.e. not 9)
## Bonus β Exponentials
If you'd like, handle exponentials.
What is 2 raised to the 5th power?
32
For installation and learning resources, refer to the exercism help page.
For running the tests provided, you will need the Minitest gem. Open a terminal window and run the following command to install minitest:
``````gem install minitest
``````
If you would like color output, you can `require 'minitest/pride'` in the test file, or note the alternative instruction, below, for running the test file.
Run the tests from the exercise directory using the following command:
``````ruby wordy_test.rb
``````
To include color from the command line:
``````ruby -r minitest/pride wordy_test.rb
``````
## Source
Inspired by one of the generated questions in the Extreme Startup game. https://github.com/rchatley/extreme_startup
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### wordy_test.rb
``````require 'minitest/autorun'
require_relative 'wordy'
# Common test data version: 1.0.0 5b8ad58
class WordyTest < Minitest::Test
# skip
question = 'What is 1 plus 1?'
end
skip
question = 'What is 53 plus 2?'
end
skip
question = 'What is -1 plus -10?'
end
skip
question = 'What is 123 plus 45678?'
end
def test_subtraction
skip
question = 'What is 4 minus -12?'
end
def test_multiplication
skip
question = 'What is -3 multiplied by 25?'
end
def test_division
skip
question = 'What is 33 divided by -3?'
end
skip
question = 'What is 1 plus 1 plus 1?'
end
skip
question = 'What is 1 plus 5 minus -2?'
end
def test_multiple_subtraction
skip
question = 'What is 20 minus 4 minus 13?'
end
skip
question = 'What is 17 minus 6 plus 3?'
end
def test_multiple_multiplication
skip
question = 'What is 2 multiplied by -2 multiplied by 3?'
end
skip
question = 'What is -3 plus 7 multiplied by -2?'
message = "You should ignore order of precedence. -3 + 7 * -2 = -8, not #{answer}"
end
def test_multiple_division
skip
question = 'What is -12 divided by 2 divided by -3?'
end
def test_unknown_operation
skip
question = 'What is 52 cubed?'
assert_raises ArgumentError do
end
end
def test_non_math_question
skip
question = 'Who is the President of the United States?'
assert_raises ArgumentError do
end
end
# Problems in exercism evolve over time, as we find better ways to ask
# questions.
# The version number refers to the version of the problem you solved,
#
# Define a constant named VERSION inside of the top level BookKeeping
# module, which may be placed near the end of your file.
#
# In your file, it will look like this:
#
# module BookKeeping
# VERSION = 1 # Where the version number matches the one in the test.
# end
#
# http://ruby-doc.org/docs/ruby-doc-bundle/UsersGuide/rg/constants.html
def test_bookkeeping
skip
assert_equal 1, BookKeeping::VERSION
end
end``````
``````module BookKeeping
VERSION = 1
end
class WordProblem
BOOL_OPERATORS = {
'plus' => '+',
'minus' => '-',
'multiplied by' => '*',
'divided by' => '/',
'raised to the power of' => '**'
}
def initialize(q)
raise ArgumentError unless valid?(q)
@question = sanitize(q)
end
def valid?(q)
q.match(/What is -?\d+(?: (?:#{BOOL_OPERATORS.keys.join('|')}) -?\d+)+\?/)
end
def sanitize(s)
s.gsub!('What is ', '').gsub!('?', '')
end
a = String.new(@question)
while a.match(/(?:#{BOOL_OPERATORS.keys.join('|')})/) # While a boolean operation exists...
next_operation = a.match(/^-?\d+ (?:#{BOOL_OPERATORS.keys.join('|')}) -?\d+/).to_s # Get the first operation
eval_string = String.new(next_operation)
BOOL_OPERATORS.each { |words,symbol| eval_string.gsub!(words, symbol) }
a.gsub!(next_operation, eval(eval_string).to_s)
end
a.to_i
end
end`````` | 1,325 | 4,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-45 | latest | en | 0.911179 |
http://codeforces.com/problemset/problem/121/A | 1,632,774,350,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00140.warc.gz | 17,010,234 | 13,393 | A. Lucky Sum
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | 385 | 1,130 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-39 | latest | en | 0.714889 |
https://metallurgicheskiy.academic.ru/12640/strain_rate_compatibility_condition | 1,725,981,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00330.warc.gz | 361,572,916 | 11,994 | # strain rate compatibility condition
strain rate compatibility condition
Смотри условие совместности скоростей деформации.
Энциклопедический словарь по металлургии. — М.: Интермет Инжиниринг. . 2000.
### Смотреть что такое "strain rate compatibility condition" в других словарях:
• условие совместности скоростей деформации — [strain rate compatibility condition] имеет аналогичный с условием совместности деформации физический смысл и записывается в виде двух групп уравнений: ∂2ηxy/∂x∂y = ∂2ξx/∂y2 + ∂2ξy/∂x2 ∂2ηyx/∂y∂z = ∂2ξy/∂z2 + ∂2ξz/∂y2 ∂2ηzx/∂z∂x = ∂2ξz/∂x2 +… … Энциклопедический словарь по металлургии
• solids, mechanics of — ▪ physics Introduction science concerned with the stressing (stress), deformation (deformation and flow), and failure of solid materials and structures. What, then, is a solid? Any material, fluid or solid, can support normal forces.… … Universalium
• Failure theory (material) — v · d · e Materials failure modes Buckling · Corro … Wikipedia
• Fluid dynamics — Continuum mechanics … Wikipedia
• List of mathematics articles (S) — NOTOC S S duality S matrix S plane S transform S unit S.O.S. Mathematics SA subgroup Saccheri quadrilateral Sacks spiral Sacred geometry Saddle node bifurcation Saddle point Saddle surface Sadleirian Professor of Pure Mathematics Safe prime Safe… … Wikipedia
• Germany — /jerr meuh nee/, n. a republic in central Europe: after World War II divided into four zones, British, French, U.S., and Soviet, and in 1949 into East Germany and West Germany; East and West Germany were reunited in 1990. 84,068,216; 137,852 sq.… … Universalium
• Nobel Prizes — ▪ 2009 Introduction Prize for Peace The 2008 Nobel Prize for Peace was awarded to Martti Ahtisaari, former president (1994–2000) of Finland, for his work over more than 30 years in settling international disputes, many involving ethnic,… … Universalium
• HIV — Classification and external resources Diagram of HIV … Wikipedia
• Navier–Stokes equations — Continuum mechanics … Wikipedia
• radiation — radiational, adj. /ray dee ay sheuhn/, n. 1. Physics. a. the process in which energy is emitted as particles or waves. b. the complete process in which energy is emitted by one body, transmitted through an intervening medium or space, and… … Universalium
### Поделиться ссылкой на выделенное
##### Прямая ссылка:
Нажмите правой клавишей мыши и выберите «Копировать ссылку» | 682 | 2,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.403558 |
https://blogs.warwick.ac.uk/emmacooke/entry/our_qvc_script/ | 1,642,407,795,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300343.4/warc/CC-MAIN-20220117061125-20220117091125-00271.warc.gz | 185,268,275 | 16,988 | ## September 30, 2010
### Our QVC script
This is an approximate transcript of our group's QVC presentation on the square. We didn't have any script at all when we were doing it, so this is just based on my memory.
Emma: (In obnoxiously cheerful voice) Hello everybody! Today we’ve got some fabulous products for you which I’m really excited about. Our first product today is the Square, which is available at Today’s Special Value price of £17.99. Now I’m joined by Paul today who’s going to tell you a little bit more about the Square. So Paul, what’s so special about the Square?
Paul: Well Emma, I would say that the Square is really the Obama or the Ed Milliband of the shape world, because it’s all about equality. You see each side of a square is equal.
Emma: Are you saying each side is exactly the same length? That’s amazing!
Paul: Yes Emma, they are exactly equal. And that’s not all, all four angles in the Square are also exactly equal.
Emma: Unbelievable! And what angle is that?
Paul: Well I’ll tell you this Emma, (chuckles) none of them are wong!
Entire group laughs in that horribly fake way QVC presenters do
Emma: Well that is truly outstanding, and don’t forget today you can buy the Square for just £17.99. Now Charlie, can you tell me any of the other features of the square?
Charlie: Well Emma, you might not know this but the Square actually has a lot of symmetry (folds the square in several ways) it has not one, not two…
Emma: no, surely not three!
Charlie: …three, four lines of symmetry!
Emma: Four lines of symmetry for only £17.99!
Charlie: But wait, there’s more! Not only does it have reflectional symmetry, it also has rotational symmetry! (Demonstrates)
Paul: What a truly outstanding product the square is! I wonder, Naresh, do you know of any famous places that are named after the Square?
Naresh: Yes, in fact there are many places named after the Square! There’s Leicester Square, where people have some really good parties, none of which would be possible with out the Square. There’s Trafalgar Square (holds up picture of pigeon with Nelson’s column in the background) where there are giant pigeons and no people… There’s also St Peter’s square, which is where the Pope goes.
Emma: So are you saying that the square has religious benefits too? And that in fact by buying the square you are actually getting closer to God? And that by paying just £17.99 you can guarantee yourself a place in heaven?
Naresh: That’s exactly what I’m saying Emma. But if you’re not religious, there’s also Red Square…
Emma: Now tell me Lydia, are there any other uses for the Square?
Lydia: Yes Emma, if you purchase 64 Squares you can put them together as shown (holds up picture) and make yourself a chessboard!
Emma: So you can actually make yourself a game! And all for just £17.99 times 64! Which I can’t work out…
Paul: Emma, I don’t need to work it out to know it’s a bargain!
Lydia: And you can also use lots of the squares put together to tile your bathroom floor!
Paul: So there’ll be no more urinating on carpets, which means a much more hygienic bathroom!
Emma: In fact, the Square is so much more hygienic, you could probably throw away all your other cleaning products. I am giving you the chance to revolutionise your entire cleaning routine from just £17.99!
Lydia: Giulian has some other ideas for what you can do with Squares.
Giulian: (shaking with uncontrollable laughter, having still not recovered from Paul’s urinating comment) you can buy six squares and put them together as a cube, taking you into the fabulous third dimension, for when two dimensions are just not enough!
Emma: Call the number on the bottom of your screen if you’d like to buy the Square for today’s Special Value Price of just £17.99.
### 2 comments by 2 or more people
1. #### Lydia Clarke
i’ve just read this in the library. not a good plan, as i had to try so hard not to laugh out loud
02 Oct 2010, 13:06
2. This was the most hilarious thing I have seen in a long while. I’ve got high expectations of your peer tutoring now! :p
10 Oct 2010, 01:47
You are not allowed to comment on this entry as it has restricted commenting permissions. | 1,001 | 4,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.955478 |
https://news.mit.edu/2010/marathon-1022 | 1,680,116,507,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00766.warc.gz | 487,834,736 | 38,365 | # Fueling up for 26.2 miles
HST student develops model that can help marathon runners pace themselves to a strong finish.
Caption:
Benjamin Rapoport, an MD/PhD student in the Harvard-MIT Division of Health Sciences and Technology, runs in the 2009 Chicago marathon.
Credits:
Photo courtesy of Benjamin Rapoport
Caption:
Rapoport’s new mathematical model can help marathon runners pick the pace and fueling strategy that will get them to the finish line.
Credits:
Photo: Patrick Gillooly
Twenty miles into the 2005 New York Marathon, Benjamin Rapoport realized something was wrong. Up to that point, he had been on pace to finish the race in about three hours — his best marathon time yet. But as he entered Manhattan for the last several miles, his legs just didn’t want to keep up the pace.
Rapoport, an MD/PhD student in the Harvard-MIT Division of Health Sciences and Technology, was experiencing a common phenomenon known as “hitting the wall.” Essentially, the body runs out of fuel, forcing the runner to slow down dramatically.
“You feel like you’re not going anywhere,” says Rapoport. “It’s a big psychological letdown, because you feel powerless. You can’t will yourself to run any faster.”
Most marathon runners know they need to consume carbohydrates before and during a race to avoid hitting the wall, but many don’t have a good fueling strategy, says Rapoport. After his experience, he decided to take a rigorous approach to calculating just how much carbohydrate fuel a runner needs to get through 26.2 miles, and what pace that runner can reasonably expect to sustain. The result is a new model, described in the Oct. 21 issue of PLoS Computational Biology, that allows runners to calculate those targets using an estimate of their aerobic capacity.
On the run
Of the hundreds of thousands of people who run a marathon each year, more than 40 percent hit the figurative wall, and 1 to 2 percent drop out before finishing.
“People think hitting the wall is inevitable, but it’s not,” says Rapoport, who has run 18 marathons, including a personal best of two hours and 55 minutes at this year’s Boston Marathon. “In order to avoid it, you need to know what your capabilities are. You need to set a target pace that will get you to the finish without hitting the wall. Once you do that, you need to make sure you appropriately carbo-load.”
During strenuous exercise such as running, the body relies on carbohydrates for most of its energy, even though fat stores are usually much larger. Most of those carbohydrates come from glycogen stored in the liver and in the leg muscles. A small amount of glucose is also present in the blood.
Hitting the wall occurs when those stored carbohydrates are completely depleted, forcing the body to start burning fat. When that happens, the runner’s pace can drop about 30 percent, and ketones, the byproducts of fat metabolism, start building up the body, causing pain and fatigue.
To create his new model, Rapoport identified fundamental physiologic factors that limit performance in endurance runners — aerobic capacity and the ability of the leg muscles to store carbohydrates as glycogen. Aerobic capacity, also known as VO2max, is a measure of how efficiently the muscles can use oxygen during exercise. Oxygen is critical to muscle performance because glucose can only be broken down completely in the presence of oxygen.
The average untrained male has a VO2max of 45 ml/kg/min, but VO2max can be boosted with training, and elite marathoners often have VO2max in excess of 75 ml/kg/min. Measuring exact VO2max requires a treadmill stress test at maximum effort, but it can be estimated by measuring heart rate while running at a constant pace on a treadmill.
Avoiding the wall
Using Rapoport’s model, any runner training for a marathon who estimates his or her VO2max can figure out the pace he or she can sustain without hitting the wall. For example, a man with a VO2max of 60 ml/kg/min could run the race in three hours, 10 minutes, without consuming any carbs during the race.
A VO2max of 60 ml/kg/min is about the highest that most men can attain through training, and 3:10 happens to be a gold standard in marathoning: It’s the time that men ages 18 to 34 must achieve to qualify for the Boston Marathon. For women of the same age, the qualifying time is 3:40, which is also the time that Rapoport’s model predicts for a runner with a VO2max of 52 ml/kg/min, about the highest level than the average woman can attain through training.
The model’s predictions also depend on the runner’s leg muscle mass, because larger muscles can store more glycogen. In the examples above, those finishing times assume that the runner’s leg muscles make up at least 7.5 percent of body mass, which is true of most people. For men, the values range from 14 to 27.5 percent, and in women, they range from 18 to 22.5 percent.
Rapoport’s model also allows runners to calculate how much carbohydrate they need to consume during the race if they want to run a faster pace without hitting the wall. For example, a runner with a VO2max of 50 ml/kg/min who wanted to achieve the 3:10 Boston Marathon qualifying time would need to consume 30 calories of carbohydrate per kilogram of body weight (about 2,090 calories for a 154-pound runner), assuming that his legs make up at least 15 percent of his body mass.
Jake Emmett, professor of kinesiology and sports studies at Eastern Illinois University, says he believes the model could be useful to marathon runners and coaches. “There is a lot of guesswork out there about carbo-loading and about carb intake during a marathon,” he says. “This would at least provide a framework for people to work out a good approximation of how to adapt their diet in the days leading up to a marathon, and what they consume during the race.”
While physiological models like Rapoport’s can help runners plan for their races, Rapoport says that other factors such as mental toughness and racecourse terrain also play important roles in how a runner will perform. One of the most important things a runner should do during a marathon is stick to his or her target pace, Rapoport advises. When runners start out too fast, they burn a higher percentage of carbohydrates, increasing the risk of hitting the wall.
“Once you figure out your target pace, you have to stay at it,” he says. “People sometimes get too excited or change their game plan on the day of the race, and that’s a tactical mistake.”
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MIT engineers discover new carbonation pathways for creating more environmentally friendly concrete. | 1,640 | 7,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-14 | longest | en | 0.950708 |
http://www.torontokidscomputer.com/uncategorized/tuesday-python-21-0119-17-2/ | 1,675,280,356,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00309.warc.gz | 92,105,615 | 11,403 | TORONTO KIDS COMPUTER CLUB | Tuesday 17:30 Python Homework 21.01.19.
18960
# Tuesday 17:30 Python Homework 21.01.19.
## 20 Jan Tuesday 17:30 Python Homework 21.01.19.
Question 1
Make a variable and assign a number to it (any number you like). Then display your variable using print.
Question 2
Modify your variable, either by replacing the old value with a new value, or by adding something to the old value. Display the new value using print.
Question 3
Make another variable and assign a string (some text) to it.
Then display it using print.
Question 4
Make a variable for DaysPerWeek, HoursPerDay, and MinutesPerHour (or make up your own names), and then multiply them together.
Question 5
People are always saying there’s not enough time to get everything done. How many minutes would there be in a week if there were 26 hours in a day? (Hint: Change the HoursPerDay variable.)
Question 6
Write a program to solve the following question: Three people ate dinner at a restaurant and want to split the bill. The total is \$35.27, and they want to leave a 15 percent tip. How much should each person pay?
Question 7
Write a program to solve the following question: Calculate the area and perimeter of a rectangular room, 12.5 meters by 16.7 meters.
Question 8
Write a program to convert temperatures from Fahrenheit to Celsius. The formula for that is:
`C = 5 / 9 * (F - 32)` | 359 | 1,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-06 | longest | en | 0.845948 |
https://socratic.org/questions/the-solubility-of-scandium-iii-fluoride-scf-3-in-pure-water-is-2-0-x-10-5-moles- | 1,576,128,913,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00227.warc.gz | 553,176,755 | 6,327 | # The solubility of scandium (III) fluoride, ScF_3, in pure water is 2.0 x 10^-5 moles per liter. How do you calculate the value of K_(sp) for scandium (III) fluoride from this data?
Sep 12, 2017
${K}_{\text{sp}} \left(S c {F}_{3}\right) = 4.32 \times {10}^{-} 18$
We interrogate the reaction.....
#### Explanation:
We interrogate the reaction.....
$S c {F}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s S {c}^{3 +} + 3 {F}^{-}$
And as written, we can directly write the solubility expression, ${K}_{\text{sp}}$, where ${K}_{\text{sp}} = \left[S {c}^{3 +}\right] {\left[{F}^{-}\right]}^{3}$
And we are given the solubility of $S c {F}_{3}$ in water, $2.0 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$....
And thus $\left[S {c}^{3 +}\right] = 2.0 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$
And $\left[{F}^{-}\right] = 3 \times 2.0 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$
Capisce?
And then we input these numbers into the ${K}_{\text{sp}}$ expression.....
${K}_{\text{sp}} = \left(2.0 \times {10}^{-} 5\right) \times {\left(6.0 \times {10}^{-} 5\right)}^{3} = 4.32 \times {10}^{-} 18$
PS I have not got a good text at hand, but the values I read for ${K}_{\text{sp}}$ for scandium fluoride seem to vary widely. I would check the source of these data. | 489 | 1,284 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-51 | longest | en | 0.616789 |
http://www.danfletcherblog.ca/2015/09/2015-16-term-1-week-3-learning-journal-uopeople/ | 1,669,611,687,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00546.warc.gz | 62,642,873 | 21,665 | # 2015-16 Term 1- Week 3 Learning Journal – UoPeople
This weeks learning journal is more programming based then usual. I traditionally try to keep these posts appropriate to non-CS students, so I refrained from using any code. I tried to keep the technical wording to a minimal because I think the lesson I learned this week can still apply to students taking business administration as well. Always, always double check work before submitting it!
At the beginning of week 3, I caught my first major mistake at UoPeople. I submitted a programming assignment that had a fairly critical bug in it during week 2, and didn’t catch it until the first day of week 3. Of course it was too late to correct the mistake by then.
The way I caught the bug was by reading the answer to the assignment before I sat down to do the assessments. My program was supposed to roll a pair of dice until they hit Snake Eyes (two 1’s), and then repeat the process a thousand times, printing out the average number of rolls to get Snake Eyes.
I noticed that the answer code’s output said the average was around 30 rolls, and mine was much lower. I knew right away I had made a mistake but didn’t know for sure where it was. My first thought was that the dice must not actually be Snake Eyes, when the program thinks they are.
So my first reaction was to use a line of code to check the value of my dice at the time the program thinks it rolled Snake Eyes. It turned out I was right; the program thought Snake Eyes had been rolled, but only when one of the dice was a 1, and not both. So of course this must mean something is wrong with the logic that controls my loop.
It was a sloppy oversight, but the bug was that my condition for re-rolling the dice was that both dice could not be equal to 1. The condition should be that one, or the other dice can’t be equal to one. The former logic means that the dice will stop being rolled if only one dice is equal to 1, but of course we want both, which is why the latter is correct.
Once I corrected the problem, I thought it would be interesting to share the bug on the forum, and see if anyone else had trouble seeing it. Only one student bit unfortunately so I didn’t get much feedback.
I didn’t feel bad about the mistake, but I did feel like I was careless to have submitted work like that. In my confidence I wrote the code for both the assignment and the challenge in one go. I didn’t bother checking the value of the dice because in my head at the time, the code made perfect sense. On the surface it seemed to work correctly, and I actually suspect that a student or two might miss the error in my program when assessing it.
I felt careless because it’s a mistake I know how to prevent. I should have been printing the value of the dice to the console from the beginning to ensure my code was working as expected. Moving forward, I’ll have to make sure to force myself to be more attentive to small details like this.
From the experience in my learning journal this week, you can see how easy it is to make small, but critical errors. The prevention was easy, because I already knew the steps that I should have taken to ensure I had no errors in my assignment. However, probably due to too much confidence, I skipped important steps that I would normally do when programming.
Confidence is important, but it can also be a pitfall. For me, I was too caught up in the bigger picture of my assignment, that I had missed a small, but detrimental detail. The worst fact, is that it would have only taken a matter of seconds to prevent my mistake. This is why we should always be diligent in our work (performing all of the appropriate steps, and double checking) no matter how confident we become.
Cheers,
Dan
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 836 | 3,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-49 | latest | en | 0.978741 |
https://cstheory.stackexchange.com/questions/31395/courcelles-theorem-for-bounded-clique-width-graphs | 1,726,867,210,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00013.warc.gz | 160,114,538 | 39,957 | # Courcelle's theorem for bounded clique-width graphs
Courcelle's theorem states that "Every graph property which is expressible in monadic second order logic is decidable in linear time for bounded tree-width graphs". Later it was extended to bounded clique-width graphs [CMR99].
The theorem also holds with linear time replaced by logarithmic space [EJT10] for bounded tree-width graphs. My question is: Does this theorem still hold for bounded clique-width graphs if we replace linear time with logarithmic space? Let me know of the recent progress in this direction.
• You should be careful with what you mean by "monadic second order logic". MSO${}_1$ (vertex set) properties can be extended to bounded clique-width, but MSO${}_2$ (edge set) properties only work with bounded treewidth. Commented May 6, 2015 at 20:38
• I will assume that you are talking about $MSO_1$. I am not aware of any work, however as far as I know about the proof in Tantau et al.'s paper Section 4 can be probably translated to any graph grammar based on disjoint union, fusion and quantifier-free operations (which will include clique-width). Now, the problem will be how to construct a rank-decomposition (or clique-width expression) in logspace. Commented May 7, 2015 at 7:18 | 295 | 1,262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.941793 |
https://www.justintools.com/unit-conversion/length.php?k1=zeptometers&k2=zettameters | 1,582,633,075,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146066.89/warc/CC-MAIN-20200225110721-20200225140721-00471.warc.gz | 778,127,084 | 27,798 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# LENGTH Units Conversionzeptometers to zettameters
1 Zeptometers
= 1.0E-42 Zettameters
Category: length
Conversion: Zeptometers to Zettameters
The base unit for length is meters (SI Unit)
[Zeptometers] symbol/abbrevation: (zm)
[Zettameters] symbol/abbrevation: (Zm)
How to convert Zeptometers to Zettameters (zm to Zm)?
1 zm = 1.0E-42 Zm.
1 x 1.0E-42 Zm = 1.0E-42 Zettameters.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [length] => (meters), 1 Zeptometers (zm) is equal to 1.0E-21 meters, while 1 Zettameters (Zm) = 1.0E+21 meters.
1 Zeptometers to common length units
1 zm = 1.0E-21 meters (m)
1 zm = 1.0E-24 kilometers (km)
1 zm = 1.0E-19 centimeters (cm)
1 zm = 3.2808398950131E-21 feet (ft)
1 zm = 3.9370078740157E-20 inches (in)
1 zm = 1.0936132983377E-21 yards (yd)
1 zm = 6.2137119223733E-25 miles (mi)
1 zm = 1.056970721911E-37 light years (ly)
1 zm = 3.7795280352161E-18 pixels (PX)
1 zm = 62500000000000 planck length (pl)
Zeptometersto Zettameters (table conversion)
1 zm = 1.0E-42 Zm
2 zm = 2.0E-42 Zm
3 zm = 3.0E-42 Zm
4 zm = 4.0E-42 Zm
5 zm = 5.0E-42 Zm
6 zm = 6.0E-42 Zm
7 zm = 7.0E-42 Zm
8 zm = 8.0E-42 Zm
9 zm = 9.0E-42 Zm
10 zm = 1.0E-41 Zm
20 zm = 2.0E-41 Zm
30 zm = 3.0E-41 Zm
40 zm = 4.0E-41 Zm
50 zm = 5.0E-41 Zm
60 zm = 6.0E-41 Zm
70 zm = 7.0E-41 Zm
80 zm = 8.0E-41 Zm
90 zm = 9.0E-41 Zm
100 zm = 1.0E-40 Zm
200 zm = 2.0E-40 Zm
300 zm = 3.0E-40 Zm
400 zm = 4.0E-40 Zm
500 zm = 5.0E-40 Zm
600 zm = 6.0E-40 Zm
700 zm = 7.0E-40 Zm
800 zm = 8.0E-40 Zm
900 zm = 9.0E-40 Zm
1000 zm = 1.0E-39 Zm
2000 zm = 2.0E-39 Zm
4000 zm = 4.0E-39 Zm
5000 zm = 5.0E-39 Zm
7500 zm = 7.5E-39 Zm
10000 zm = 1.0E-38 Zm
25000 zm = 2.5E-38 Zm
50000 zm = 5.0E-38 Zm
100000 zm = 1.0E-37 Zm
1000000 zm = 1.0E-36 Zm
1000000000 zm = 1.0E-33 Zm
(Zeptometers) to (Zettameters) conversions | 995 | 2,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-10 | latest | en | 0.779213 |
http://physics.stackexchange.com/tags/coriolis-force/hot | 1,369,550,165,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706635063/warc/CC-MAIN-20130516121715-00079-ip-10-60-113-184.ec2.internal.warc.gz | 200,937,076 | 13,329 | # Tag Info
31
Because the rotation of the earth is very smooth and doesn't change, the centripetal acceleration we feel is very nearly constant. This means that the (small) centrifugal force from the rotation gets added to gravity to make up the "background force" we don't notice. Earthquakes are not at all smooth and the accelerations involved are large and change ...
21
Dan's answer is essentially good, but miss one effect : the Coriolis effect. You can imagine a planet spinning much more rapidly than the earth, but at a constant angular speed. On that quickly rotating planet, the explanation of Dan would still stand, but as soon as on moves, we would feel a lateral Coriolis force. The Coriolis acceleration is ...
10
The whirl is due to the net angular momentum the water has before it starts draining, which is pretty much random. If the circulation were due to Coriolis forces, the water would always drain in the same direction, but I did the experiment with my sink just now and observed the water to spin different directions on different trials. The Coriolis force is ...
9
Ok, here is my (hopefully rigorous) demonstration of the origin of these forces here, from first principles. I've tried to be pretty clear what's happening with the maths. Bear with me, it's a bit lengthy! Angular velocity vector Let us start with the principal equation defining angular velocity in three dimensions, \dot{\vec{r}} = \vec{\omega} \times ...
7
Yes, the ball would land in front of you. If you watch from outside the space station, the ball moves in a straight line at constant speed while you move in a circle at constant speed. That means the distance the ball takes to get from point A (where you release it) to point B (where it hits the floor) is shorter than the distance you take. Further, ...
5
The calculation of the Coriolis force is dependent on latitude: $F = m a$ where $a = 2 \Omega sin(lat)$, with $\Omega$ being the Earth's angular velocity $m$ is the mass of the object in question The Earth's angular velocity is (about) $7.29 \times 10^{-5}$ rad/sec So, for a sink with a couple gallons of water in it at 45 degrees north... the Coriolis ...
5
Of course that there would be forces that would try to bend the track but they would be tiny. Each segment of the track would be under the action of $-2m \Omega \times v$ Coriolis force. Note that the Coriolis force only depends on velocities, not accelerations as you stated! In other words, there is the Coriolis acceleration, $-2\Omega\times v$, and you ...
4
Since you want to explain it to your daughter, take a plastic bottle, cut the bottom open, turn it upside town, hold the top closed and fill it with water. Give her that bottle and have her release the top (which is on the bottom now, sorry for the bad phrasing). The water will whirl in different orientations whenever you repeat this (if it whirls at all) ...
3
The whirl happens in the draining tube, whose optimal solution to drain the bathtub is a laminar flow allowing for some rotation in the tube. What you see in the surface is the match between the solution of flow in the tube and the solution of flow in the surface. Angular momentum of the flow gets modified a lot as the tube twists and twists, sometimes even ...
3
It does have an effect. Also see this paper about modelling tsunami propagation. As noted in the paper, the Coriolis force only becomes important over large distances. Here's an article on MathWorld including many references.
2
The Coriolis effect is proportional to velocity: $\boldsymbol{ F}_C = -2 \, m \, \boldsymbol{\Omega \times v}$, where $\boldsymbol{\Omega}$ is the angular velocity (of the earth). As waves, tsunamis have very high velocities, you would think it would be enough to see the Coriolis effect, but in fact there is little effect. When in deep water, tsunamis move ...
2
Your adjective "large" in "large whirlpool" may be very misleading. If you mean hundreds of meters, no effect of the Coriolis force may be visible by a naked eye at this scale. The origin of the whirlpool had to be different. Quite generally, tsunami is all about waves, and if one has a wave, molecules of water are moving back and forth, in circular ...
2
It does. To convince yourself, remember that rising hot air does experience a Coriolis force, so I am quite sure that your bubble does too. Also, think of what the Coriolis acceleration is - it is an apparent acceleration due to the fact that you, the observer, are in a rotating reference frame, so your definition of "straight up" is actually a curve. When ...
2
You can think about it like this: It takes one day for the earth to perform a full rotation (about 86k seconds), on the other hand, it takes a few seconds for your sink to drain (lets say 10 seconds). So it takes 8600 times longer for the earth to do a full rotation than it takes the water to drain down the sink. It is not too hard to imagine that the ...
2
Sure coriolis force applies, but I think a much simpler intuitive explanation is conservation of angular momentum. Think of the spinning skater who pulls in his/her arms & legs, spinning faster. If you look down on the earth from the north star, you see a whole hemisphere covered with air rotating counterclockwise. At the pole, it's basically not ...
1
There is no significant difference between what happens in the east-west direction and what happens in the north-south one. Upon impact it will experience a pretty sudden acceleration due to the impact, and depending on the trajectory and the nature of where it hits the ground, it may not stick, but rebound, skid, or whatever...
1
The Coriolis "force" isn't a proper force. It's probably better called the Coriolis effect. It's named after a French mathematician and engineer. In a rotating frame of reference, like the earth's, when we differentiate a dynamical variable (like the position of an object -- here, Baumgartner) with respect to time we get two terms. The first is just the ...
1
You land slightly ahead of where you jumped. As mentioned in the comments, see here. The Coriolis effect only applies to things that are moving in the rotating reference frame. If the air is stationary in the rotating frame, it feels only the centrifugal force. There will be a pressure gradient, creating buoyancy, just like on Earth, but all the air will ...
1
A discussion by 'The Straight Dope' website http://www.straightdope.com/columns/read/149/do-bathtubs-drain-counterclockwise-in-the-northern-hemisphere references experimental work carried out but Ascher Schapiro in 1962, which concluded something like it all depends on the shape of container and how its stirred before being left to empty. Here is ...
1
The main effect is angular momentum (rotational inertia) in the water set up by various movements before you start observing, such as getting out of your bath. This results in the water level being lower near the centre of rotation than further away, setting up centripital forces which maintain the rotation. When the difference in levels is significant ...
Only top voted, non community-wiki answers of a minimum length are eligible | 1,618 | 7,182 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2013-20 | latest | en | 0.952162 |
https://socratic.org/questions/how-can-i-balance-this-equation-kclo3-kcl-o2 | 1,532,124,658,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591837.34/warc/CC-MAIN-20180720213434-20180720233434-00101.warc.gz | 757,619,298 | 90,309 | Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team
# How can I balance this equation? ____ KClO_3 -> ____ KCl + ____ O_2
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
205
Apr 29, 2016
You follow a systematic procedure to balance the equation.
#### Explanation:
${\text{KClO"_3 → "KCl" + "O}}_{2}$
A method that often works is to balance everything other than $\text{H}$ and $\text{O}$ first, then balance $\text{O}$, and finally balance $\text{H}$.
Another rule is to start with what looks like the most complicated formula.
The most complicated formula looks like ${\text{KClO}}_{3}$.
We put a $\textcolor{red}{1}$ in front of it and mark the $\textcolor{red}{1}$ to remind ourselves that the number is now fixed.
We can't change a coefficient unless we run into a roadblock (like having to use fractions).
My teacher never let me use fractions.
My solution when I hit a roadblock was to erase all the numbers and then start over again with a 2 as the starting coefficient.
$\textcolor{red}{1} {\text{KClO"_3 → "KCl" + "O}}_{2}$
Balance $\text{K}$:
We have 1 $\text{K}$ on the left, so we need 1 $\text{K}$ on the right. We put a $\textcolor{b l u e}{1}$ in front of the $\text{KCl}$.
$\textcolor{red}{1} {\text{KClO"_3 → color(blue)(1)"KCl" + "O}}_{2}$
Balance $\text{Cl}$:
$\text{Cl}$ is already balanced, with 1 $\text{Cl}$ on each side.
Balance $\text{O}$:
We have 3 $\text{O}$ atoms on the left and only $2$ on the right. We need 1½ ${\text{O}}_{2}$ molecules on the right. Uh, oh! Fractions!
We start over with a 2 as the coefficient.
$\textcolor{red}{2} {\text{KClO"_3 → color(blue)(2)"KCl" + "O}}_{2}$
Now we have 6 $\text{O}$ atoms on the left. To get 6 $\text{O}$ atoms on the right, we put a 3 in front of the ${\text{O}}_{2}$.
$\textcolor{red}{2} {\text{KClO"_3 → color(blue)(2)"KCl" + color(orange)(3)"O}}_{2}$
Every formula now has a fixed coefficient.
We should have a balanced equation.
Let’s check:
Left hand side: 2 $\text{K}$, 2 $\text{Cl}$, 6 $\text{O}$
Right hand side: 2 $\text{K}$, 2 $\text{Cl}$, 6$\text{O}$
All atoms balance.
The balanced equation is
${\text{2KClO"_3 → "2KCl" + "3O}}_{2}$
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
15
Oct 18, 2016
To balance a chemical equation is not different to solve a linear equations system. We assign a unknown variable to each of the coefficients of the reaction, and then solve the system.
#### Explanation:
Formally, the equation of a chemical reaction can be written as a mathematical equation that relates the number of atoms of each element involved in the reaction.
Because the atoms of the different elements are not transformed into each other (are talking about chemical reactions, not nuclear ones), the number of atoms of each element that is in the left side of the reaction (reactants) has to be equal to the number of atoms of the same element present on the right side of reaction (products).
Therefore, we can write a mathematical equation that equals both amounts. There will be as many equations as elements present in the reaction and have as many unknown variables as molecules of different substances involved.
Let's see how all this works with a real example. Consider the reaction that has been proposed in this question:
$K C l {O}_{3} \rightarrow K C l + {O}_{2}$
We calculate the coefficients of the reaction, ie numbers ${x}_{i}$ ranging front of each substance, so that the reaction becomes matched. I.e:
${x}_{1} K C l {O}_{3} \rightarrow {x}_{2} K C l + {x}_{3} {O}_{2}$
Then we must write one equation for each element present in the reaction. There are three different elements in the reaction, so we have three equations:
Potassium : ${x}_{1} = {x}_{2}$
Chlorine: ${x}_{1} = {x}_{2}$
Oxygen: $3 {x}_{1} = 2 {x}_{3}$
In this example, we obtain a linear equations system with three unknown variables and only two equations, because two of them are the same one:
Obviously, it is a system of linear equations compatible but undetermined. We need to get only one solution (there are infinite ones).
Then, if we assign (arbitrarily) ${x}_{3} = 1$, we obtain a solution:
${x}_{3} = 1 \rightarrow {x}_{1} = \frac{2}{3} \rightarrow {x}_{2} = \frac{2}{3}$
Mathematically, the solution thus obtained and would be acceptable, but from a chemical point of view, it is inconvenient to use fractional coefficients. So what we do is multiply all the coefficients obtained by a number such that we obtain integer values. In this case, we multiply by 3, and obtain:
$2 K C l {O}_{3} \rightarrow 2 K C l + 3 {O}_{2}$
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• 2 hours ago | 1,437 | 5,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 43, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2018-30 | longest | en | 0.83902 |
https://onlinejudge.org/board/viewtopic.php?f=55&t=70997 | 1,582,930,309,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147647.2/warc/CC-MAIN-20200228200903-20200228230903-00527.warc.gz | 480,195,796 | 7,296 | ## 12058 - Highway Monitor
Moderator: Board moderators
sith
Learning poster
Posts: 72
Joined: Sat May 19, 2012 7:46 pm
### 12058 - Highway Monitor
Hello!
I've got WA for this problem.
Here is my solution. What is wrong?
Code: Select all
``````import java.io.BufferedReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Set;
import java.util.StringTokenizer;
class Main {
public static void main(String[] args) {
String line;
try {
int caseCount = Integer.parseInt(line.trim());
for (int i = 1; i <= caseCount; i++) {
StringBuilder builder = new StringBuilder();
builder.append("Case #").append(i).append(": ");
int n = Integer.parseInt(tokenizer.nextToken());
int h = Integer.parseInt(tokenizer.nextToken());
int k = Integer.parseInt(tokenizer.nextToken());
Map<Integer, Set<Route>> cities = new HashMap<Integer, Set<Route>>();
for (int j = 0; j < h; j++) {
int x = Integer.parseInt(tokenizer.nextToken());
int y = Integer.parseInt(tokenizer.nextToken());
Route route = new Route(x, y);
}
PriorityQueue<Routes> routes = new PriorityQueue<Routes>();
for (Map.Entry<Integer, Set<Route>> integerSetEntry : cities.entrySet()) {
}
Set<Integer> result = new HashSet<Integer>();
Set<Route> coveredRoutes = new HashSet<Route>();
Routes poll;
while ((poll = routes.poll()) != null) {
if (coveredRoutes.containsAll(poll.routes)) {
continue;
}
else {
}
if (coveredRoutes.size() == h) {
break;
}
if (result.size() > k) {
break;
}
}
if (result.size() > k) {
System.out.println(builder.append("no").toString());
continue;
}
else {
builder.append("yes");
for (Integer city : result) {
builder.append(" ").append(city);
}
System.out.println(builder);
}
}
}
}
catch (IOException e) {
}
}
private static void addRoute(final Map<Integer, Set<Route>> cities, final int city, final Route route) {
Set<Route> routes = cities.get(city);
if (routes == null) {
routes = new HashSet<Route>();
cities.put(city, routes);
}
}
static class Routes implements Comparable<Routes> {
int city;
Set<Route> routes;
public int compareTo(final Routes o) {
return o.routes.size() - routes.size();
}
Routes(final int city, final Set<Route> routes) {
this.city = city;
this.routes = routes;
}
}
static class Route {
final int x;
final int y;
Route(final int x, final int y) {
this.x = Math.min(x, y);
this.y = Math.max(x, y);
}
@Override
public boolean equals(final Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
final Route route = (Route) o;
if (x != route.x) {
return false;
}
if (y != route.y) {
return false;
}
return true;
}
@Override
public int hashCode() {
int result = x;
result = 31 * result + y;
return result;
}
}
}
`````` | 700 | 2,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-10 | latest | en | 0.35245 |
http://ngss.nsta.org/DisplayStandard.aspx?view=pe&id=229 | 1,527,115,499,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865830.35/warc/CC-MAIN-20180523215608-20180523235608-00291.warc.gz | 218,988,951 | 22,137 | # Earth's Place in the Universe
### Students who demonstrate understanding can:
#### Performance Expectations
1. Develop and use a model of the Earth-sun-moon system to describe the cyclic patterns of lunar phases, eclipses of the sun and moon, and seasons.
Clarification Statement and Assessment Boundary
A Peformance Expectation (PE) is what a student should be able to do to show mastery of a concept. Some PEs include a Clarification Statement and/or an Assessment Boundary. These can be found by clicking the PE for "More Info." By hovering over a PE, its corresponding pieces from the Science and Engineering Practices, Disciplinary Core Ideas, and Crosscutting Concepts will be highlighted.
### Connections to Nature of Science
By clicking on a specific Science and Engineering Practice, Disciplinary Core Idea, or Crosscutting Concept, you can find out more information on it. By hovering over one you can find its corresponding elements in the PEs.
## Planning Curriculum
### Common Core State Standards Connections
#### ELA/Literacy
• SL.8.5 - Integrate multimedia and visual displays into presentations to clarify information, strengthen claims and evidence, and add interest. (MS-ESS1-1)
#### Mathematics
• 6.RP.A.1 - Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. (MS-ESS1-1)
• 7.RP.A.2 - Recognize and represent proportional relationships between quantities. (MS-ESS1-1)
• MP.4 - Model with mathematics. (MS-ESS1-1)
## Resources & Lesson Plans
• More resources added each week!
A team of teacher curators is working to find, review, and vet online resources that support the standards. Check back often, as NSTA continues to add more targeted resources.
• Exploring Lunar and Solar Eclipses via a 3-D Modeling Design Task is a detailed lesson plan published in the October 2016 issue of NSTA’s Science Scope journal. It is free to NSTA members but can be purchased for a nominal fee by non-memb ...
• Including Students presents a single, kinesthetic model that can be used to explain such varied concepts as rotation, revolution, phases of the Moon and seasons. The outlined activities are meant to be used throughout an entire Astronomy unit. ...
• The NASA Eclipse Web Site contains multiple pages of data which students can explore, analyze and use to explain phenomena. The site contains data on solar and lunar eclipses and on the transits of Venus and Mars. On the solar eclipse pag ...
• Seasons Interactive provides students an opportunity to investigate the existence of seasons. In this simulation, users cause the Earth to revolve around the Sun by toggling through a calendar year. Users can compare daylight duration (in hours ...
• Lunar Phases is an interactive simulation consisting of three scaffolding activities. In Activity One, students are tasked with determining which half of the Earth and Moon is lit as the Moon revolves around the Earth. The directions state that use ...
• In Eclipse Interactive, students investigate both lunar and solar eclipses by manipulating up to three independent variables: Moon's tilt from orbit, Earth-Moon distance and size of the Moon. By viewing the effects of changes to these variables, stud ...
• Seasons Interactive provides students with the opportunity to investigate how Earth's angle of inclination affects three factors: the angle of incoming sunlight, average daily temperatures and the Sun’s ecliptic path. Three preset values fo ...
• Do you have a great resource to share with the community? Click here.
• This simulation from the University of Nevada-Lincoln shows the geometry of the earth and sun over the course of a year, demonstrating how seasons occur. This simulator allows both orbital and celestial sphere representations of the seasonal motions.
• This is a set of multiple choice items developed by AAAS Project 2061 focused on the topic of the seasons. The items are different from many multiple choice science test items in that they assess students’ conceptual understanding (not just facts ...
• The activity is similar to the traditional activity where there is a bright light source in the classroom and students place a Styrofoam ball on a pencil. However, rather than having the teacher explain the model, students are instead asked to use a ...
• This simulation by the University of Nebraska at Lincoln demonstrates the correspondence between the moon's position in its orbit, its phase, and its position in an observer's sky at different times of day.
• This vocabulary matching engagement includes terms for tides and lunar phases and can be used as a formative assessment with middle school students.
• This activity description provides a kinesthetic model of tides: high and low, as well as spring and neap.
• This activity is designed to show students the relationship between ocean tides and the lunar movements by creating and examining graphs of the tides over months and comparing the lunar phases to the pattern.
• Complete Description of project and files for your use shared in an easy to navigate website.
• In this series of games, your students will learn about the Earth’s rotation and the resulting “migrations” of celestial bodies. The Sun, Moon, and Stars: Patterns of Apparent Motion learning objective — based on NGSS and state standards — delivers i...
• In this series of games, your students will learn about the what, how, and why of the Moon and its cycles. The Lunar Phases learning objective — based on NGSS and state standards — delivers improved student engagement and academic performance in your...
• In this series of games, your students will learn about the major phenomena caused by the sun, earth, and moon’s relative positions: solar eclipses, lunar eclipses, and the four seasons. The Eclipses and Seasons learning objective — based on NGSS and...
• To assess students' three dimensional learning it's necessary to have a task that will enable you to see what they can do.
• In this hands-on lesson, students measure the effect of distance and inclination on the amount of heat felt by an object and apply this experiment to building an understanding of seasonality. In Part 1, the students set up two thermometers at differe...
• In this activity, students investigate what causes the seasons by doing a series of kinesthetic modeling activities and readings.
• This activity engages learners to investigate the impact of Earth's tilt and the angle of solar insolation as the reason for seasons by doing a series of hands-on activities that include scale models. Students plot the path of the Sun's apparent move...
• This teaching activity is an introduction to how ice cores from the cryosphere are used as indicators and record-keepers of climate change as well as how climate change will affect the cryosphere.
• A computer animation on the reason for the seasons. Voice-over describes the motion of Earth around the sun to show how the sun's light impacts the tilted Earth at different times of the year, causing seasonal changes.
• This activity supports educators in the use of the activities that accompany the GLOBE Program's Earth System Poster 'Exploring Connections in Year 2007'. Students identify global patterns and connections in environmental data that include soil moist...
• An applet about the Milankovitch cycle that relates temperature over the last 400,000 years to changes in the eccentricity, precession, and orbital tilt of Earth's orbit.
• This animated visualization of precession, eccentricity, and obliquity is simple and straightforward and provides text explanations. It is a good starting place to show Milankovitch cycles.
• An interactive that illustrates the relationships between the axial tilt of the Earth, latitude, and temperature. Several data sets (including temperature, Sun-Earth distance, daylight hours) can be generated.
• This animation demonstrates the changing declination of the sun with a time-lapse animation. It shows how the shadow of a building changes over the course of a year as the declination of the sun changes.
• This Motions of the Sun Lab is an interactive applet from the University of Nebraska-Lincoln Astronomy Applet project.
• This three-part, hands-on investigation explores how sunlight's angle of incidence at Earth’s surface impacts the amount of solar radiation received in a given area. The activity is supported by PowerPoint slides and background information.
• This interactive activity, in applet form, guides students through the motion of the sun and how they relate to seasons.
Planning Curriculum gives connections to other areas of study for easier curriculum creation. | 1,743 | 8,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-22 | latest | en | 0.871744 |
http://www.fixya.com/support/t6843834-graph_y_35cos2_pi_53_37 | 1,510,981,647,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804610.37/warc/CC-MAIN-20171118040756-20171118060756-00120.warc.gz | 416,086,375 | 34,921 | Question about Texas Instruments TI-84 Plus Calculator
# How to graph y=-35cos2(pi)/53 +37 in ti 83 calculator ?how to set zoom and how to set windows for getting a visible curve?
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• Texas Instru... Master
Your expression does not contain an independent variable, so all you will get is a horizontal line passing through the point (0, 2.2456616) corresponding to the value 37-35cos(2pi/53).
As to the window setting, you can always use ZoomStd (Zoom standard).
But first you must put some independent variable in the definition of the function.
Posted on Nov 16, 2010
Hi,
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## Related Questions:
### Graphing cos 2x? i have a TI-86 but i think my problem is something that can be done on any TI calculator, i need to know how to graph cos 2x, all i get is a straight line
Your angle unit is set to degrees. If you do not change the Xmin and Xmax window dimensions to say -180 to -90 for Xmin, and 90 to 180 for Xmax you will not see much of the features of the graph; In the default interval Xmin, Xmax the curve is a staright line.
Sep 17, 2011 | Texas Instruments TI-83 Plus Calculator
### My TI-83 Plus calculator won't graph anything whatsoever. Did I need new software. Please help me.
First - Are you graphing within the proper window? Hit the zoom button and change it to zoom Standard within the graphing window.
Second - Are you inputting the equation correctly? Make sure you are inputting your equations in a y= format. If possible, please reply with a screenshot of your graph screen.
Feb 07, 2011 | Texas Instruments TI-83 Plus Calculator
### ERR: Window Range
Press ZOOM and select Zoom standard (ZoomStd).
Alternatively, press the WINDOW key to access the window range setup
Set Xmin=-10
Xmax=10
Ymin=-10
Ymax=10
These are the standard dimensions of the window.
You can change the dimensions as long as Xmin is less then Xmax, and Ymin is less than Ymax. Do not make Xres close to Xmax-Xmin.
Jan 10, 2010 | Texas Instruments TI-83 Plus Calculator
### How to set the window?
Hello,
Press [Window]
Set the values of Xmin, Xmax, Ymin, Ymax.
If the x values that the function takes are all positive, you may set Xmin =0. If the values of the function are all positive, you can set Ymin=0.
If you are drawing a sine or a cosine function, you know that the function values are between -1 and +1, so you cant restict Ymin an Ymax to say -1.2 and 1.2. Similarly if you draw trigonometric function no need to have a range of X values between -10 and 10: the function being periodic you can restrict values (in radians to -pi to +pi, ) or to -90 to +90 degrees.
You set the limits of the window by making use of your knowledge of the functions you are drawing.
Hope it helps.
Sep 19, 2009 | Texas Instruments TI-83 Plus Calculator
### When i try to graph it says error and wont let me graph
Hello,
Reset calculator to defaults
[2nd][MEM][7][2][2] Defaults set.
[Y=] on line Y1= enter your function then press [GRAPH]
You can draw several graphs on the same screen.
If nothing shows (except the axes) use [ZOOM] to Zoom in or zoom out or use [WINDOW] to change the values of Xmin, Xmax, Ymin, Ymax.
Hope it helps.
Hope it helps
Sep 09, 2009 | Texas Instruments TI-83 Plus Calculator
### Graph feature not working on my TI-83 Plus
Simple:
Hit this sequence:
"ZOOM" "0"
That is for the "Zoom Fit" option. The calculator does not like the window you have for the equation you've given it. With this option, it will automatically give you the best fit in the window for the equation!
Nov 14, 2008 | Texas Instruments TI-83 Plus Calculator
### TI 83 Plus
Try hitting "ZOOM" then "0". This is for the "Zoom Fit" option. It will automatically adjust your window range to best fit your equation!
Sep 27, 2008 | Texas Instruments TI-83 Plus Calculator
### TI-83 Plus...Problem:
try seting the zoom to standard then zooming out untill you find it im not sure if that will fix it but it may be of some help
Jan 19, 2008 | Texas Instruments TI-83 Plus Calculator
## Open Questions:
#### Related Topics:
74 people viewed this question
Level 3 Expert
Level 3 Expert | 1,216 | 4,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-47 | latest | en | 0.90408 |
https://math.stackexchange.com/questions/3274785/let-a-beginbmatrix-1-2-3-4-endbmatrix-then-deta3-6a25a3i-3 | 1,563,415,674,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.62/warc/CC-MAIN-20190718001934-20190718023934-00189.warc.gz | 433,217,139 | 36,267 | # Let $A=\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}$ then det$(A^3-6A^2+5A+3I)=3$
Let $$A=\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}$$ then det$$(A^3-6A^2+5A+3I)=3$$
det$$(A^3-6A^2+5A+3I)=$$det$$((A^2-5A-2I)(A-I)+2A+I)=$$det$$(2A+I)=3$$, Since a matrix satisfies its characteristic polynomial. Is this right?
• @J.W.Tanner Yes, is there a mistake? – rhaldryn Jun 26 at 10:30
Yes, this looks fine. Since $$A$$ satisfies its own characteristic polynomial, you have: $$\color{blue}{A^2-5A-2I=O}$$ and so, as you wrote: $$A^3-6A^2+5A+3I=\underbrace{\left(\color{blue}{A^2-5A-2I}\right)}_{\color{blue}{O}}\left(A-I\right)+2A+I=2A+I$$ which leaves you with the (easier) $$\det\left(2A+I\right)$$ and that is indeed $$3$$.
Here is another method. Assume that $$\lambda_i\in\mathbb{C}$$, $$i=1,2$$ are the two igenvalues of $$A$$. Let $$f(x)\in \mathbb{C}[x]$$ be the characteristic polynomial of $$A$$. Then $$f(x)=\det(A-xI)=x^2-5x-2.$$ The matrix $$A$$ can be diagonalized as $$A=P\begin{bmatrix} \lambda_1 & 0\\ 0& \lambda_2 \end{bmatrix}P^{-1},$$ where $$P$$ is an invertible matrix over $$\mathbb{C}$$. Let $$g(x)=x^3-6x^2+5x+3$$. We have $$g(x)=f(x)(x-1)+2x+1$$. It follows that $$\det(A^3-6A^2+5A+3I)=\det(g(A))=\det\left(g\left(\begin{bmatrix} \lambda_1 & 0\\ 0& \lambda_2 \end{bmatrix}\right)\right)=\det\left(\begin{bmatrix} g(\lambda_1) & 0\\ 0& g(\lambda_2) \end{bmatrix}\right)=g(\lambda_1)g(\lambda_2).$$ Note that $$f(\lambda_i)=0$$, $$i=1,2$$. Thus $$g(\lambda_i)=f(\lambda_i)(\lambda_i-1)+2\lambda_i+1=2\lambda_i+1$$. So $$\det(A^3-6A^2+5A+3I)=(2\lambda_1+1)(2\lambda_2+1)=4\lambda_1\lambda_2+2(\lambda_1+\lambda_2)+1.$$ According to Vieta theorem, we have $$\lambda_1\lambda_2=-2, \lambda_1+\lambda_2=5.$$ Consequently, $$\det(A^3-6A^2+5A+3I)=4\times (-2)+2\times 5 +1=3.$$ | 802 | 1,788 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 29, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-30 | latest | en | 0.563774 |
https://www.mekasmart.club/2020/03/water-cooled-chiller-design-data.html | 1,659,885,494,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00307.warc.gz | 773,116,426 | 33,428 | Water Cooled Chiller Design Data
Water Cooled Chiller Design Data
Water Cooled Chiller Design Data
I want to stress that this is simply design data. Every chiller is different and you should speak to your manufacturer for the relevant information. The results will vary from the real world and also with loading.
In the illustration above we have shown the main chiller components. The compressor , which is the driving force of the refrigerant around the system. The condenser which removes the unwanted heat from the system and send this to the cooling tower. The expansion valve which expands the refrigerant and controls the superheat into the compressor and the evaporator which collects the unwanted heat coming from the building and generates the chilled water.
We’re going to be looking at all the points on these two charts to see the pressure, temperature, enthalpy and entropy are around this system. The left chart is our Temperature v’s Entropy chart and the right chart is our Pressure v’s Enthalpy chart.
• Point 1 is just before the compressor, and the exit of the evaporator. That will be a low pressure, low temperature saturated / slightly superheated vapor.
• Point 2 is just after the compressor, before the condenser. That’s a high pressure, high temperature superheated vapor.
• Point 3 is just after the condenser, but before the expansion valve. That’s going to be a high pressure, medium temperature saturated liquid.
• Point 4 is just after the expansion valve but before the evaporator. This will be a low pressure, low temperature and it’s going to be a mix between a liquid and vapor.
Compressor
In this example the compressor is pushing a refrigerant with a flow rate of 16.5 kg/s (36.4 lbm/s). That motor is then consuming 425.9 kilowatts and the compressor is running 100% load. If the chiller runs at part load then the values will be different.
The refrigerant is being sucked in from the evaporator (Point 1) at around 356 kPa (3.56 Bar) and at a temperature of 5.5 ° C (41.9 ° F). The refrigerant enthalpy is 402 kJ/kg (173 BTUs/lbm). The entropy will be 1.73 kJ/kg.K (0.41 BTUs/lbm.F).
The compressor is compacting the refrigerant into a smaller space, and looking at our charts we know that the enthalpy is going to increase, the entropy is going to slightly increase and the pressure and temperature will massively increase.
When the refrigerant leaves (Point 2), it will be 915 kPa (9.15 bar). The temperature reaches 43.6 ° C (110.5 ° F). The enthalpy is now 426 kJ/kg.K (183 Btu/lbm) and the entropy is now 1.74 kJ/kg.K (0.042 Btu/lbm.F).
Remember the temperature of the refrigerant entering into the condenser has to be higher than the incoming condenser water temperature for heat transfer to occur. If they were the same temperature, then no heat transfer would occur and the chiller would do no cooling.
Condenser
The next part we’ll look at is the condenser. In this example the condenser water is flowing through the condenser at 116.6 L/s (247 cfm). The condenser water is coming into the condenser, from the cooling tower, at 29 ° C (84.2 ° F). The refrigerant will then transfer the buildings unwanted heat into the condenser water. This will increase the temperature of the condenser water, so when it leaves to go back to the cooling tower it will be around 35 ° C (95 ° F).
Now the reason the flow rate is higher in the condenser compares to the evaporator is because the condenser has to reject more heat. It also has to take the heat away from the compressor and other parts of the machine.
The refrigerant came from the compressor and entered the condenser at a pressure of 915 kPa (9.15 Bar), a temperature of 43.6 ° C (110.5 ° F) with an enthalpy of 426 kJ/kg.K (183 Btu/lbm) and an entropy of 1.74 kJ/kg.K (0.428 Btu/lbm.F).
Once the refrigerant has give away some of its energy to the circulating condenser water, it will now leave as a liquid at 36.1 ° C (97 ° F) but still at the same pressure as it entered. It’s entropy will have dropped to 1.17 kJ.kg.K (0.28 BTU/lbm.F) and the enthalpy increases to 250 kJ/kg.K (107.5 BTU/lbm). It then enters into the expansion valve.
Expansion Valve
The expansion valve controls the flow of refrigerant, it measures the superheat on the suction line of the chiller and then reacts to this by allowing or restricting refrigerant flow to maintain a certain value. The refrigerant is entering the expansion valve as a liquid and leaving as a vapour/liquid mixture.
It enters, in this example, at a temperature of 36.1 ° C (97 ° F), a pressure of 915 kPa (9.15 Bar), the entropy is 1.17 kJ.kg.K (0.28 BTU/lbm.F) and the enthalpy is 250 kJ/kg.K (107.5 BTU/lbm).
The refrigerant is expanded through a small orifice which sprays the refrigerant. It expands into a larger volume and decreases in pressure as a result, which allows it to drop in temperature as it’s now not packed so tightly. It will leave at a temperature of 5.5 ° C (41.9 ° F), a pressure of 356 kPa (3.56 Bar) and from the charts we know it will maintain the same enthalpy but the entropy will change slightly and it leaves at 1.20 kJ/kg.K (0.29 BTU/lbm.F).
Evaporator
The evaporator generates the cold “chilled water” which cycles around the building, providing air conditioning and collecting the buildings unwanted heat. This now warm chilled water returns to the evaporator and transfers this heat into the refrigerant, the chilled water then leaves cooler and cycles around the building, meanwhile the refrigerant boils and carry’s the thermal energy to the compressor.
In this example, the chilled water is flowing through the evaporator at around 99.5 Litres per second, which is around 210 cubic feet per minute. The chilled water enters the evaporator at around 12 ° C (53.6 ° F). After the chilled water has transferred it’s heat over to the refrigerant, it will leave the evaporator at around 6°C (42.8°F).
The refrigerant is picking up thermal energy but the temperature only changes slightly which confuses many people. The reason it doesn’t increase dramatically is because it is undergoing a phase change from a liquid to a vapour so the thermal energy is being used to break the bonds between the molecules. The enthalpy and entropy will increase and this is where the energy is going
When the refrigerant leaves it will be a slightly superheated vapour at 5.5 ° C (41.9 ° F), a pressure of 356 kPa (3.56 Bar) an entropy of 402 kJ/kg.K (173 Btu/lbm) and an enthalpy of 1.73 kJ/kg.K (0.41 btu/lbm.F).
The refrigerant then returns to the compressor to start the cycle all over again. | 1,703 | 6,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | longest | en | 0.853797 |
http://www.physicsforums.com/showthread.php?t=246100 | 1,369,241,619,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702019913/warc/CC-MAIN-20130516110019-00065-ip-10-60-113-184.ec2.internal.warc.gz | 660,116,471 | 8,390 | ## Induced voltages and step-up transformers
I have been trying, but I still have 3 problems I can't get.
1)A simple generator has a 600 loop square coil 18.0 cm. on a side. How fast (in rev/s.) must it turn in a 0.730 T. field to produce a 120-V. peak output?
I think I need to use the equation Emax=NBAw. N=600 B=.730 A=.0324 Emax=120
I found w to be 8.456 but I didn't know if it was in rev/min or rev/s. So I tried that answer and then tried dividing it by 60 for seconds, but neither answer worked. Any help?
2)A step-up transformer increases 50 V. to 110 V. What is the current in amperes in the secondary as compared to the primary? Assume 100 percent efficiency.
I know the primary voltage is 50V and the secondary voltage is 110V. And I think Vp/Vs=Ip/Is, but I'm not sure how to find the answer w/o one of the currents.
3)A transformer has 32 turns in the primary and 181 turns in the secondary. Assuming 100 percent efficiency, by what factor does it change the voltage?
I think this one is basically the same as number 2, and I just don't understand what I am supposed to do. I think I'm supposed to divide one by the other, but when i put those answers in, they're not right.
Recognitions:
Homework Help
Hi kitkat2950,
Quote by kitkat2950 I have been trying, but I still have 3 problems I can't get. 1)A simple generator has a 600 loop square coil 18.0 cm. on a side. How fast (in rev/s.) must it turn in a 0.730 T. field to produce a 120-V. peak output? I think I need to use the equation Emax=NBAw. N=600 B=.730 A=.0324 Emax=120 I found w to be 8.456 but I didn't know if it was in rev/min or rev/s. So I tried that answer and then tried dividing it by 60 for seconds, but neither answer worked. Any help?
The $\omega$ is angular frequency, which is not in rev/min or rev/s; what are the units of angular frequency?
Once you find $\omega$, you'll have to convert it to rev/s.
2)A step-up transformer increases 50 V. to 110 V. What is the current in amperes in the secondary as compared to the primary? Assume 100 percent efficiency. I know the primary voltage is 50V and the secondary voltage is 110V. And I think Vp/Vs=Ip/Is, but I'm not sure how to find the answer w/o one of the currents.
It's strange that the ask for the current in the secondary in amperes. When they say "as compared to the primary" that suggests that what they are looking for is the ratio of the currrents. Was there a diagram or anything else in the problem?
Your equation is not correct. If the transformer is 100% efficient, what is not lost by the transformer?
3)A transformer has 32 turns in the primary and 181 turns in the secondary. Assuming 100 percent efficiency, by what factor does it change the voltage? I think this one is basically the same as number 2, and I just don't understand what I am supposed to do. I think I'm supposed to divide one by the other, but when i put those answers in, they're not right.
What is the relationship between voltage and turns in a transformer? (Without knowing what numbers you used and got, it's difficult to determine what you might have done wrong.)
2. the ratio you want is: Vs/Vv=Ns/Np: Is/Ip=Np/Ns >>> Vs/Vp=Ip/Is >>> Is/Ip=Vp/Vs = 50/100= .5
Recognitions:
Homework Help
## Induced voltages and step-up transformers
Quote by annythewitch 2. the ratio you want is: Vs/Vv=Ns/Np: Is/Ip=Np/Ns >>> Vs/Vp=Ip/Is >>> Is/Ip=Vp/Vs = 50/100= .5
No, I don't believe that's the answer. You seem to have put in a wrong number.
Also, you have put in a few extra steps. There is 100% efficiency, so the power is the same on both sides, so you can immediately write down:
$$I_p V_p = I_s V_s$$
and get the ratio in the next step.
you are right. I mistyped the fraction. it should have been 50/110, and yeah, your way is easier. | 1,045 | 3,780 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2013-20 | longest | en | 0.969535 |
https://richardcoyne.com/2023/03/25/plain-guide-to-neural-networks/ | 1,686,088,703,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00713.warc.gz | 527,243,309 | 34,151 | # Just one neuron
I revisited our earlier (1990) article on neural networks “Spatial applications of neural networks in computer-aided design.” Neural networks were novel in architecture and CAD. What follows is an update of the part of that article in starting to explain how neural networks function.
## Neural network layers
Neural network (NN) models store information as numbers attached to nodes and arcs (links) in a network. Nodes are sometimes called “units” or “neurons” to reinforce the analogy with biological neural systems (i.e. brains). There are input nodes, output nodes and hidden nodes, all connected by directed arcs. Input, output and hidden nodes are generally organised in layers. The layers are stacked in the direction from inputs to outputs. Every node in a layer is connected to every node in the next layer. There can be several layers in a neural network.
A very helpful video by Grant Sanderson explains the process for recognising hand-written numerical integers 0-9.
He uses slick animated graphics to demonstrates NN processes. The input layer of the NN consists of the 784 individual pixel values from a 28×28 scan of a hand written integer. So for that exercise the input layer consists of 784 nodes. There are 2 hidden layers in his example, each with an arbitrary number of nodes (16 each) and the output layer, consists of 10 nodes, one for each integer (0-9).
My example that follows is much simpler than the network described by Sanderson, and I’ll focus here on what happens to a single node. To demonstrate operations on a single node, it is simplest to think of each node as receiving binary inputs that produce binary outputs. The output value of a node is the result of adding the products of the weights of arcs directed inwards to the node. Before the node delivers its output, the weighted sum is adjusted by a kind of bias function — sometimes called a threshold function.
In my simplified model here I start with a particular a threshold value. That will be an integer 0, 1, 2, 3, …. If the sum of the weighted inputs to a node is greater than the threshold then the node will “fire” (producing an output value of 1). Net values less than or equal to the threshold produce a 0 as output.
In a full NN, the output from one node provides the input to a node with which it is connected. The NN algorithm inspects each node in turn and passes output values through the network, effectively propagating activation values through the network to produce a configuration of output values in response to a given set of inputs.
Here I will consider just 3 input nodes and one output node that processes binary values. Each input node is connected to the output node with a directed arc of weight 1. The output node has a threshold value (θ) associated with it of 0.
To summarise so far, the node output is calculated by summing the product of each input node with each weight (to produce a net input) and comparing this with the threshold of the output node. If the weighted input is a greater value than the threshold then this mini network produces a 1 as output. If it is less than or equal to the threshold the network generates a 0 output value.
Following this procedures, as shown below the input pattern {1 0 1} produces 1 as output (the weighted sum is 2, which is greater than the threshold value 0).
The input pattern {0 1 0} also produces 1 (the weighted sum is 1, which is greater than the threshold value 0).
Suppose in this input case {0 1 0} we would rather the node generate a different response, e.g. produce a 0 as output instead of 1. By inspecting the node above it should be apparent that we wish to decrease the influence of weights from input units with value 1. The following algorithm accomplishes this:
1. Calculate the value of the output unit produced from the given input pattern.
2. Compare this calculated value with the preferred output presented to the node.
3. Inspect each of the input values in tum
• If the input value is 1, the calculated output is 1 and the preferred output is 0
• then subtract 1 from the weight of the arc emanating from that node and add 1 to the threshold of the output node.
Decreasing the weight and raising the threshold of the output unit adjusts the node to produce a 0 as output. This results in the revised network here.
If we now present this network with the input pattern {0 1 0} it produces the required output pattern of {0}. The training algorithm needs also to check that the previous input {1 0 1} still gives the desired output of { 1 }. As shown here the output is still {1}.
We also need a rule for adjusting weights and thresholds in the event that we require the output to be 1 but the NN calculates a 0.
• If the input value is 1, the predicted output is 0 and the required output is 1
• then add 1 to the weight of the arc emanating from that unit and subtract 1 from the threshold of the output node.
There is no change to weights and thresholds if the training and predicted outputs are the same. The resultant network above provides the required match for both input-output patterns.
## Scaling up
Full-fledged NNs operate with real numbers and not just integers or binaries, and a node will have many inputs. For a network as a whole there are two operations: that in which the system is trained on various output patterns, and the simulation or matching operation where we present the NN with the input only and it generates an output.
The training operation involves cycling through the set of given input-output patterns and adjusting weights and thresholds according to rules similar to those above. Training is a much longer process than simulation. The length of time it takes the training algorithm to calculate its weights and thresholds increases with the number of training examples presented. Running the network to produce an output consistent with a new input during simulation is a simpler operation and computation time is independent of the number of training examples.
In the training operation it is usual to cycle through the algorithm several times as the adjustment of weights and thresholds may cause the network to “forget” a pattern it has just stored via its weights and thresholds.
Where there is no overlap between the input patterns (that is, where the input patterns are linearly independent, as in my simple example) the algorithm is guaranteed to find a system of weights and thresholds to store all input and output pattern pairs. Where any input node has the same value (1 or 0) in more than one training pattern (the most usual case) there is no such certainty.
## Randomness improves accuracy
Stochastic methods improve the performance of the NN in the most general case. The difference between the net input and the threshold to an output unit actually carries some useful information. The differences serve not only to gauge whether the output should be 0 or 1, but it also gives some indication of the strength of conviction in the network in the output value. It is a measure of the probability that the output unit will “fire.”
In simulation we could therefore produce a graded output to indicate non-binary measures of certainty in the output. There is a convenient function for distributing the probability from 0 to 1, calculated from the differences between the net inputs and the output threshold. Here is the logistic function for mapping the difference between the input and the threshold to a unit (Diff) onto a probability value (p) depending on a constant T which determines the slope of
the curve. The function is
Where there are similarities in the input patterns, that is, where different input patterns have certain nodes in common, the probabilisitic values of the output nodes can accommodate the overlapping nature of the corresponding outputs.
## Getting stuck
One of the problems that a neural network may encounter is that during successive cycles the system may change certain weights in response to a particular input-output pair in one cycle, only to undo the effect in another cycle in response to another input-output pair. The system can become “locked” into a set of weights that is not optimal and which is influenced by the order in which input-output pairs are presented to the NN. One further sophistication in the training procedure seeks to address this problem.
A common technique is to introduce controlled randomness into the training process. As described above, when the NN trains, it adjusts weights and thresholds on the basis of the predicted value of an output node and the desired (i.e. training) output. Using the logistic function the training algorithm can calculate the likelihood of the predicted value being a 1 or 0 and during subsequent training iterations generate a 1 or 0 at random according to this probability. Surprisingly, this introduction of randomness over a large number of training cycles across a large network produces a more accurate result. Over many learning cycles it produces a smoother series of weights and thresholds. It allows the system to train more slowly. The analogy is often made between this process and thermodynamic annealing.
The constant factor T in the logistic function is analogous to temperature. The higher the temperature of the system the more random the predicted values at output units and the greater the means of escaping from a set of weights and thresholds that is sub-optimal. With this stochastic approach the difference between net values and thresholds serves as a good measure of the degree of certainty attached to the output.
The slope of the curve varies according to the constant T. Very small values of T produce a stepped function. To illustrate a more general input-output calculation, here is network with non-binary inputs, weights, threshold and output for T = 1.0.
## What hidden layers are for
Hidden layers of nodes in the network increases the number of parameters by which patterns can be reconstructed. With large numbers of hidden units there is more chance that the system will generalize on features rather than patterns of binary values with absolute locations, e.g. on a grid as in my previous post. Sanderson explains features in terms of the relative slopes and positions of various lines in hand drawn digits. In the case of the rudimentary floor plans in my previous post the features might include the relative positions of floor plan elements (courtyards, recesses, wings, etc).
With hidden units the algorithms for finding a set of weights and thresholds to account for input and output pairs must resolve a complex optimization problem. That’s for another time.
## References
• Coyne, Richard, and Arthur Postmus. “Spatial applications of neural networks in computer-aided design.” Artificial Intelligence in Engineering 5, no. 1 (1990): 9-22.
• Hinton, G. E., and T. J. Sejnowski. “Learning and relearning in Boltzmann machines.” In Parallel Distributed Processing: Explorations in the Microstructure of Cognition, Volume I, Foundations, edited by D. E. Rumelhart, and J. L. McClelland, 282-314. Cambridge, MA: MIT Press, 1986.
• Rumelhart, D. E., and J. L. (eds) McClelland. Parallel Distributed Processing: Explorations in the Microstructure of Cognition: Volume 1, Fundamentals. Cambridge, Mass.: MIT Press, 1987.
• McClelland, J. L., D. E. Rumelhart, and G. E. Hinton. “The appeal of parallel distributed processing.” In Parallel Distributed Processing: Explorations in the Microstructure of Cognition, Volume 1, Foundations, edited by D. E. Rumelhart, and J. L. McClelland, 3-44. Cambridge, MA: MIT Press, 1986.
• Sanderson, Grant. “Neural Networks: The basics of neural networks, and the math behind how they learn.” 3Blue1Brown, 2023. Accessed 12 February 2023. https://www.3blue1brown.com/topics/neural-networks
## Note
• Other methods are also used for calculating the probability of an output value than the logistic function, notably Tanh and ReLU (Rectified Linear Unit).
• Featured image is by MidJourney, prompted with the cover of the book by Rumelhart and McClelland and the words “neural network.” Here is the full image.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2,589 | 12,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-23 | longest | en | 0.92784 |
https://cstheory.stackexchange.com/questions/9031/how-many-tautologies-are-there/34383 | 1,631,941,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056297.61/warc/CC-MAIN-20210918032926-20210918062926-00239.warc.gz | 247,802,842 | 39,245 | # How many tautologies are there?
Given $m, n, k$, how many of $k$-DNFs with $n$ variables and $m$ clauses are tautology? (or how many $k$-CNFs are unsatisfiable?)
• A bit of motivation would help us believe that this is not just a random question. Nov 19 '11 at 9:34
• @AndrejBauer: I was reading about SAT solvers and their performance. Oct 22 '13 at 22:50
The answer depends on $k$, $m$, and $n$. Exact counts are generally not known, but there is a "threshold" phenomenon that for most settings of $k$, $m$, $n$, either nearly all $k$-SAT instances are satisfiable, or nearly all instances are unsatisfiable. For example, when $k=3$, it has been empirically observed that when $m < 4.27 n$, all but a $o(1)$ fraction of 3-SAT instances are satisfiable, and when $m > 4.27n$, all but a $o(1)$ fraction are unsatisfiable. (There are also rigorous proofs of bounds known.)
One starting point is "The Asymptotic Order of the k-SAT Threshold".
Amin Coja-Oghlan has also done a lot of work on these satisfiability threshold problems.
This is an extended comment to complement Ryan's answer, which deals with the thresholds where the number of clauses becomes large enough that the instance is almost surely unsatisfiable. One can also compute the much larger thresholds where the number of clauses forces unsatisfiability when it exceeds a function of $n$.
Note that some technical issues need to be addressed. If repeated clauses are counted in $m$, then $m$ can be made as large as desired without changing $n$. This would destroy most relationships between $m$ and $n$. So assume that $m$ is the number of distinct clauses. We need to decide on another detail, whether instances are encoded so that order of literals within a clause or order of clauses within an instance matter. Suppose this is not important, so two instances are regarded as equivalent if they contain the same clauses, and two clauses are equivalent if they contain the same literals. With these assumptions we can now bound the number of distinct clauses that can be expressed with $n$ variables. Each clause can have each variable occurring positively or negatively, or not at all, and then $m\le 3^n$.
First consider SAT without a restriction on $k$. What is the largest $m$ such that the instance is satisfiable? Without loss of generality we can suppose that the all-zero assignment is a solution. There are then $3^n-2^n$ different clauses consistent with this solution, each containing at least one negated literal. Hence $m\le 3^n-2^n$ for any satisfiable instance. The instance consisting of all clauses that each contain at least one negated literal has this many clauses, and is satisfied by the all-zero assignment. Further, by the pigeonhole principle any instance with at least $3^n-2^n+1$ clauses is unsatisfiable.
This yields $2^{3^n-2^n}$ different subsets of such clauses, each representing a distinct instance which is satisfied by some assignment. In comparison, the total number of different instances is $2^{3^n}$.
Now modifying the above for instances in which each clause has at most $k$ literals, there are $\sum_{i=0}^k \binom{n}{i}2^i$ distinct such clauses, and $\sum_{i=0}^k \binom{n}{i}$ clauses in which there are no negative literals, so $m\le \sum_{i=0}^k \binom{n}{i}(2^i - 1)$ for satisfiable instances, and any larger $m$ is unsatisfiable. There are then $2^{\sum_{i=0}^k \binom{n}{i}(2^i - 1)}$ instances satisfied by any particular assignment, out of the total of $2^{\sum_{i=0}^k \binom{n}{i}2^i}$ $k$-SAT instances.
• I also produced the same result back in 2008 ish. There are also complimentary functions for literals and variables such that if you assume no repetition of literals, variables or clauses then if more than x many or y many literals or variables occurs respectively then the given instance is not satisfiable. I would have to dig to find those two functions. +1 Apr 18 '16 at 21:34 | 988 | 3,920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-39 | longest | en | 0.948112 |
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# MIT OpenCourseWare http://ocw.mit.
edu
1.061 / 1.61 Transport Processes in the Environment
Fall 2008
The example problems ask the user to make qualitative statements about the dispersion expected in various flows and to predict the shape of a dispersing cloud. a process distinct from diffusion. and can be caused by boundaries (which create shear in the flow). Dispersion processes This section looks at the longitudinal spreading of clouds through dispersion. at which point they have a Gaussian concentration profile in the longitudinal direction. It is also shown that longitudinal dispersion is maximized when lateral diffusion is small. In this chapter. dead zones and absorption reactions (for example. to a porous medium). Dispersion is the spreading of a cloud by any process that causes different chemical particles to move at different velocities. dispersion is generally far more important than diffusion in the lengthening of chemical clouds. . In environmental systems. it is shown that dispersing clouds eventually reach a Fickian limit. The non-uniformity of particle velocity allows some particles to move quickly while others are held back.8.
accelerates the dilution of the cloud. In the top channel the side-wall boundary condition allows slip.y. and as a result this cloud spreads longitudinally more rapidly than the cloud in the uniform channel. However. must exist close to a boundary because the no-slip condition requires that the relative velocity of the fluid be zero at the boundary. u. C C C C C C C +u +v +w = Dx + Dy + Dz . spatial gradients of velocity.z). known as shear. Tracer is released at t = 0. Both channels have the same cross-sectional mean velocity. causing it to spread longitudinally more quickly than a cloud released into a uniform current (Figure 1). The spatial distribution of each cloud is shown above at time t1 and t2. Top-view of two channels. u � f(x. In many instances it is sufficient to integrate the domain over y and z. However. x = 0. with u = f(x. In the bottom channel there is a no-slip condition at the side-wall boundary. �u/�y. i. and the velocity profile is uniform across y. t x y z x y z x y z (1) if we include the spatial variation in velocity. Shear Dispersion. The distortion observed in the bottom channel of Figure 1 can be predicted from the full transport equation. the center of mass for both clouds advects at speed u. When shear is present. The transport models and concentration field solutions developed in previous sections assume that currents are spatially uniform. it is difficult to find simple analytical solutions to (1).y. Velocity shear in the bottom channel stretches the tracer cloud. when acted upon by lateral diffusion. and consider only transport in the stream-wise direction. After this integration the combined impact of differential advection and lateral diffusion is 1 . Figure 1.z). This 'differential advection' stretches the tracer cloud. In addition. differential advection creates lateral gradients in concentration (�C/�y in Figure 1) which. Note. creating lateral shear.e. different parts of a tracer cloud will advect at different speeds.1 8.
In general Dx << Kx. the longitudinal length-scale for the cloud is �x. C + u t C = Kx x 2 C 2 . It is also known as Taylor Dispersion. h u = (1 h)�0 udz . leaving us with the one-dimensional advection-dispersion equation. I. �u/�z. Integrating (1) over y and z yields the one-dimensional equation. (1) reduces to C C C C +u = Dx + Dz . x (3) Furthermore. to appreciate under what conditions the approximate equation (3) may hold. in honor of G. and uniform (�u/�x = 0). KX. in the limit of Fickian behavior. e. such that the former may be neglected. Specifically. we assume that the flow is strictly in the x-direction (w = v = 0). where �x = 2 Kx t . Kx = 1 2 x 2 t . In addition. It is important to understand the steps of this derivation. (4) From (4) we can estimate Kx in the field by observing the evolution of variance for a tracer cloud. which we term Shear Dispersion. is so small compared to the shear in the vertical. Thus. that the lateral shear is neglected at the outset. Deviations from the depth-average will be denoted by a prime. Under limiting conditions. 2 . Dx � f(x). Shear Dispersion behaves as a Fickian diffusion process. we describe the evolution of the crosssectionally averaged cloud using the same statistical description developed for diffusion in previous chapters. C + u t C = Dx x 2 C 2 + Kx x 2 C 2 . who first described this process in 1953. This allows us to assume a two-dimensional channel with �/�y = 0. x x z x z t (6) Next. With these assumptions. we decompose u and C into depth-averaged values denoted by an over-bar. observing the variance over time. �u/�y. x (2) in which brackets represent cross-sectional averages. we represent its impact in the cross-sectionally averaged (one-dimensional) transport equation with an additional coefficient of dispersion. Specifically. For simplicity we will consider a very wide channel for which the shear in the lateral.g. Taylor. utilizing scaling arguments to simplify the mathematics along the way. (5) We can find an analytical expression for KX by explicitly carrying out the integration that leads to (3). to be discussed shortly. and for simplicity we have assumed homogeneous diffusivity.2 expressed as an additional process creating longitudinal spreading.
noting that by definition c’ = 0 and u’ = 0. we have eliminated the vertical gradient terms by integrating over the vertical domain. (6). then the magnitude of both the third and fourth terms in (10) must be small compared to the second term. c = x c' . so that we might evaluate this new flux term. u' c' x . z u' z Dz (12) 3 . and we may drop the small terms. If this is true. (c + c' ) = c + c' ) + (u + u' ) ( x t c' c c c + u + u' = Dx x x t x x x Dx (c + c' ) + x z Dz (c' ) z . When these two processes are in balance. c'. As both (8) and (9) must both hold true. when �c/�x is large. (8) We take the depth-average of each term.u' = Dz x x z x z t (10) Now we place an important condition on the magnitude of the spatial fluctuation c'. differential advection dominates. creating c' faster than vertical diffusion can erase it. and concentration c’. Initially. are created by differential advection and destroyed by vertical diffusion. u’. (9) Relative to our original equation. relative to their depth-averaged values. �c'/�t goes to zero and (11) becomes. and note that by definition �c/�z = 0. With the next few steps we will find a solution for c'. c c' c' c' c' + u' + u' . In addition. But. z) We put the decomposed variables into (6).z) = c (x) + c’(x. eventually. that represents the flux associated with the correlation between the fluctuations in velocity. Specifically. a new term has appeared. reducing (10) to c' t + c = x differential advection u' c' z z vertical diffusion Dz (11) This equation describes how vertical fluctuations in concentration. we may subtract (9) from (8) to yield. we assume that c'<< c.3 u(z) = u + u’(z) (7) c(x. �c/�x becomes small enough that the rate at which c' is created by differential advection is balanced by the rate at which it is destroyed by vertical diffusion.
Differential advection distorts the cloud giving it a new shape by time t+dt. KX. and we can write the new term as u' c' x . noting that u' c' (z = 0) = u' c' (z = 0) = 0 . and so remains inside the integral. h 0 x 0 Dz 0 h 0 0 Dz 0 (14) This correlation is the advective flux associated with spatial fluctuations in the velocity field. Consider a tracer cloud whose leading edge is uniform at time t. creates the concentration perturbation seen at t + dt. because u' = 0 . DZ. Recall that by definition. u’. �c'/�z = 0 at z = 0 and h. c � f(z). from the first two terms in (11). Because we assume u � f(x). x 0 Dz 0 (13) Because the diffusivity. Now we return to (9) to evaluate the new term. it may be a function of vertical position. then u' � f(x). Then. Specifically. u' c' x . 1h 1 ch z 1 z cz 1 z u' c' = � u' � � u' dzdzdz = x � u' � � u' dzdzdz . we choose to model this flux using a transport coefficient. x (15) . This flux is proportional to the concentration gradient. Figure 2. i. just as we did for turbulent diffusion in the previous chapter. is the sum of turbulent and molecular diffusion. u' c' = � Kx where 4 c . and so behaves as a Fickian diffusion. The spatial deviation in velocity. The transport coefficient is called the longitudinal dispersion coefficient. Thus.e. c’ = -( C/ x ) u’ dt.4 We can solve (12) with a no-flux boundary condition at the bed and water surface. c' (z) = cz 1 z � � u' dzdz + c' (z = 0). The negative sign indicates that the flux is counter-gradient. We evaluate the correlation u' c' with (13).
(16) indicates that KX is inversely proportional to the cross-stream diffusivity. t x x x x (18) where DX in (18) and DZ in (16) represent the sum of molecular and turbulent diffusion. 2c c c +u = Kx 2 t x x (20) Solutions to this one-dimensional equation. As one might expect. This expression describes the contribution of vertical shear to longitudinal dispersion. This is an acceptable approximation for wide. straight channels. This is the differential advection component of shear dispersion. However. A similar term describes the contribution of lateral shear. which for practical purposes is just turbulent diffusion. These solutions can now be used for systems with shear-dispersion dominating longitudinal diffusion (as in 20). the diffusion term is often neglected in (18). If 5 . and that Dz ~ h. consistent with (19).e. Comparing (19) with the empirical estimate for longitudinal diffusion. (19) Note that (19) only accounts for velocity shear over depth (�u/�z) and not for lateral velocity shear (�u/�y). have already been described for instantaneous and continuous point releases. (16) shows that the coefficient of shear dispersion increases with the magnitude of the shear. (16) yields KX ~ h. considering that each integral is proportional to h. DX = 0. we see that KX is an order of magnitude larger.45u*h.93u*h. Finally. u' 2c c' = u' c' = -Kx 2 x x x (17) Using (17) in (9). The dependence KX ~ h is less straight-forward. Elder evaluated (16) for a logarithmic profile with DZ = f(z) and found KX = 5. remember that shear dispersion arises because velocities on adjacent streamlines are different so particles on different streamlines advect at different speeds and thus become separated (dispersed) longitudinally. To understand this somewhat unintuitive dependence. each particle is mixed over depth (or width). (16) The second expression in (16) is valid when DZ � f(z). experiencing the velocity of each streamline that is passes for some fraction of time. Now we can write the new term in (9) as. DZ (i. Because of this.5 . Kx is greatest when Dz is smallest!). we arrive at the one-dimensional equation. At the same time. 2 c c c c +u = Dx + Kx 2 . with Dx replaced by Kx. repeated here for convenience. expressed as the magnitude of the spatial deviations u'. and we have returned to equation (3).
(21) after the tracer is released. If they all effectively travel at the mean velocity. The tracer is introduced as a line distributed over the full depth at time t = 0. which acts at the rate. the mean concentration gradients are very large. even though shear is not zero. This means that the conditions (DZ) which yield the greatest KX will also require the longest time (longest distance from source) for shear-dispersion to take hold and for (20) to apply. this is observed to occur at time scale TDispersion = 0. at the rate �c’/�t = u’ C / x . such that the creation of c' is very rapid and pronounced. C / x . For vertical shear.g. creates perturbations in the concentration profile. we consider the evolution of a tracer cloud from its initial release. Early on we made the assumption that c'<< C . the longitudinal gradient of mean concentration. where Lz is the length-scale of the channel in the cross-stream direction (width or depth). Now it is important to recall the assumptions that allowed us to reach the advection-dispersion equation (20).6 the cross-channel diffusion is high. so that a nearly uniform concentration profile exists in the cross-stream direction. and shear dispersion will be zero. as shown in Figure 3. u'. Eventually. such that the two assumptions are closely linked. 6 . This is reflected in the geometric distortion of the cloud shape (t1. �c’/�t = DZ �2c’/�z2. 3). for (12) to be valid. such that c' << C . we specifically required that the rate at which c' is created exactly balance the rate at which c' is destroyed. decreases and the rate at which concentration perturbations are created slows. DZ. The process of shear-dispersion has reached its Fickian limit and the longitudinal distribution of the depth-averaged concentration is Gaussian. The concentration perturbations are dissipated by lateral diffusion.4 LZ2/DZ. As the cloud spreads. particles can accumulate more differential advection as they remain on individual streamlines for a longer period. In addition. Differential advection. the particles sample all the velocities in the profile so rapidly that they effectively travel at the mean channel velocity. c'. And recall from (16) that KX is also inversely proportional to DZ. As the cross-stream diffusion is reduced. Note that this time scale is inversely proportional to cross-stream diffusion. After this time (e. the rate at which the perturbations are created by differential advection comes into balance with the rate at which they are destroyed by diffusion such that (12) is valid. they do not disperse (separate) in the longitudinal direction. the water depth. When the cloud is first introduced. LZ = h. t2 in Figure 3) the concentration is close to uniform over depth. Fig. Note that a nearly uniform profile would automatically satisfy c'<< C . To understand when these assumptions are valid. From tracer studies.
This form of dispersion can reach a Fickian limit after every tracer particle has sampled a sufficient number of pore channels. At this time (t1) the longitudinal distribution of the depth-averaged concentration. is skewed. Eventually differential advection and diffusion balance and the cloud approachs uniform concentration over depth. For example. some flow paths are nearly linear. The time to traverse a given flow path depends on the geometry of the pores. 7 . and multiple flow paths exist between any two points. Together. except diffusion. and thus short. C . Other Mechanisms of Dispersion Dispersion is a broad term used to describe all processes. the existence of multiple discrete flow paths also creates differential advection that leads to dispersion. Initially (t1) differential advection dominates. After this time (t2) the longitudinal distribution of the depth-averaged concentration is Gaussian and the dispersion process has reached its Fickian limit. The flow paths in a porous medium are constrained to discrete. as described above. Flow through narrow pores is slower than flow through wider pores. the above processes contribute to a longitudinal spreading of tracer particles. that disperse a patch of tracer and diminish concentration. That is. with higher velocity at the center of the pore than at the grain surface. tracer particles released together will take different flow paths and get separated (dispersed) both laterally and longitudinally. visibly distorting the geometric shape of the cloud. each pore channel behaves like a small tube. intertwining pore channels. The pore-scale shear contributes to dispersion in the same manner as channel-scale shear. Also. and others are very tortuous (bending) and thus long.7 Figure 3. In flow through porous media. Typically dispersion coefficients reflect a combination of advection and diffusion processes that are difficult to model separately. Evolution of tracer cloud released at t = 0 as a line distributed over depth. In addition. This is called mechanical dispersion. the coupled effects of differential longitudinal advection and cross-channel diffusion create shear dispersion. | 3,804 | 16,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-47 | latest | en | 0.915659 |
https://crazyproject.wordpress.com/2011/06/14/as-an-f-vector-space-an-infinite-direct-sum-of-f-has-strictly-smaller-dimension-than-an-infinite-direct-power-of-f-over-the-same-index-set/ | 1,490,235,538,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186608.9/warc/CC-MAIN-20170322212946-00599-ip-10-233-31-227.ec2.internal.warc.gz | 784,050,482 | 19,846 | ## As an F-vector space, an infinite direct sum of F has strictly smaller dimension than an infinite direct power of F over the same index set
Let $F$ be a field. Prove that a vector space $V$ over $F$ having basis $B$ (regardless of the cardinality of $B$) is isomorphic as a vector space to $\bigoplus_B F$. Prove that $\prod_B F$ is also an $F$-vector space which has strictly larger dimension than that of $\bigoplus_B F$.
(So a free module on any set is isomorphic to a direct sum. We’ve never gotten around to proving this in the best possible generality, though, so we’ll just prove it for vector spaces here.)
Note that, by the universal property of free modules, the natural injection $B \rightarrow \bigoplus_B F$ which sends $b$ to the tuple which has 1 in the $b$th component and 0 elsewhere induces a vector space homomorphism $\varphi : V \rightarrow \bigoplus_B F$. This mapping is clearly surjective, and also clearly injective. So $V \cong_F \bigoplus_B F$.
Certainly $\prod_B F$ is an $F$-vector space which contains $\bigoplus_B F$, so that $\mathsf{dim}\ \bigoplus_B F \leq \mathsf{dim}\ \prod_B F$. Suppose these two dimensions are in fact equal. Identify $B$ with the usual basis of $\bigoplus_B F$. By this previous exercise, there is a basis $D$ of $\prod_B F$ which contains $B$, and as argued above, $\prod_B F \cong_F \bigoplus_D F$. By our hypothesis, in fact $B$ and $D$ have the same cardinality, and so there exists a bijection $\theta : B \rightarrow D$. Now $\theta$ induces a vector space isomorphism $\Theta : \bigoplus_B F \rightarrow \prod_B F$.
However, note that $|\prod_B F| = |F|^{|B|}$, while $|\bigoplus_B F| = |\bigcup_{T \subseteq \mathcal{P}(X), T\ \mathrm{finite}} \prod_T F|$ $= \sum_{|B|} |F|^{|T|}$ $= \sum_{|B|} |F|$ $= |B| \cdot |F|$. Since $|B| \cdot |F| < |F|^{|B|}$, we have a contradiction. Thus the dimension of $\bigoplus_B F$ is strictly smaller than that of $\prod_B F$.
• Josh Swanson On August 31, 2011 at 8:58 pm
How does |F| * |B| < |F|^|B| follow when |B| = |F|, can just use Cantor’s diagonalization argument, since |F| * |B| = max{|F|, |B|}.
That an infinite cartesian product increases the cardinality of a set is intuitive, at least.
• Josh Swanson On August 31, 2011 at 9:01 pm
My comment above was mangled. I guess the angle brackets were eaten as HTML. What I wrote was…
How does |F| * |B| less than |F|^|B| follow when |B| is smaller than |F|? When |B| is greater than or equal to |F|, we can just use Cantor’s diagonalization argument, since |F| * |B| = max{|F|, |B|}, so
|F| * |B| = |B| is less than 2^|B| is less than or equal to |F|^|B|
That an infinite cartesian product increases the cardinality of a set is intuitive, at least. | 843 | 2,722 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 37, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-13 | latest | en | 0.816987 |
https://encyclopedia2.thefreedictionary.com/Poisson+random+process | 1,537,905,174,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162385.83/warc/CC-MAIN-20180925182856-20180925203256-00046.warc.gz | 489,858,606 | 11,045 | # Poisson Process
(redirected from Poisson random process)
## Poisson process
[′pwä′sōn‚prä·səs]
(statistics)
A process given by a discrete random variable which has a Poisson distribution.
## Poisson Process
a stochastic process describing the moments at which certain random events occur. In a Poisson process, the number of events occurring within any fixed interval of time has a Poisson distribution, and the numbers of the events occurring in nonoverlapping intervals of time are independent.
Suppose μ(s, t) is the number of events whose moments of occurrence τi satisfy the inequalities 0 ≤ s < τit, and suppose λ (s, t) is the mathematical expectation of μ(s, t). In a Poisson process, for any 0 ≤ s1 < t1 ≤ s2 < t2 ≤ … ≤ sr < tr’ the random variables μ(s1, t1), μ(s2, t2), …, μ(sr, tr) are independent, and the equality μ(s, t) = η has probability
e-λ(s, t)[λ(s, t)]n/n!
In a homogeneous Poisson process, λ (s, t) = a(t – s), where a is the mean number of events in a unit of time and the distances τnn-1 between neighboring moments τn are independent and have an exponential distribution with density ae-at, t ≥ 0.
If there are many independent processes that describe the moments certain rare events occur, the total process yields in the limit a Poisson process under certain conditions.
The Poisson process is a convenient mathematical model that is often used in various applications of probability theory. In particular, it is used to describe a request flow in queuing theory—for example, calls arriving at a telephone exchange or ambulance trips in response to traffic accidents in a large city.
A generalization of the Poisson process is the Poisson random distribution of points in a plane or in space. The number of points here in any fixed region has a Poisson distribution (with mean proportional to the area or volume of the region), and the numbers of points in nonoverlapping regions are independent. This distribution is often used in calculations in such fields as astronomy, physics, ecology, and engineering.
### REFERENCE
Feller, W. Vvedenie v teoriiu veroiatnostei i ee prilozheniia, vols. 1–2. Moscow, 1967. (Translated from English.)
B. A. SEVAST’IANOV
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The case of the missing. Subglacial Lakes. Your Text Here. Your Text Here. Your Text Here. Your Text Here. Subglacial Lakes. Adrienne is exploring East Antarctica!!!!. She heard there was a lake in the area and she has been looking for it everywhere!. I hope you can help her find it!.
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2. Adrienne is exploring East Antarctica!!!! She heard there was a lake in the area and she has been looking for it everywhere! I hope you can help her find it!
3. She is fascinated by lakes! Maybe it is because she comes from Ohio where there are well over 1000 lakes! 6 1 2 7 5 3 4 9 8 OR maybe its because she likes to snorkel?
4. OR maybe it is because she likes to swim? Yes that’s Adrienne swimming in a hole cut into the Ross Ice Shelf in Antarctica. An ice shelf is a thick plate of floating ice that forms where a glacier flows off the land. credit: YoYo' Johnson for the Ice Stories Project, (c) Exploratorium, www.exploratorium.edu
5. Did you get an answer close to 152 cms? That would be right IF we were measuring JUST what we can see in the picture ….but with ice shelves there is more… From what you can see, how thick, from top to bottom, do you think the ice is behind Adrienne? Ladder rungs are generally one foot apart.* Can you use the ladder to estimate ice thickness (height) in that area? Report your answer in cms. *(1 inch is 2.54 cm) Let’s have fun with ice shelf math!
6. 10% ~ 90% of the Ice Shelf extends underwater! Water line 90% The Ross Ice Shelf is THICK! Scientists estimate it ranges from 15 to 50 meters above the water surface. Robin and Detlef are standing INSIDE the top 10% of the ice shelf ! Can you calculate in meters the depth of the ice sheet that lies below the water surface? 135 to 450 meters below the ice! Now estimate the TOTAL ice sheet thickness in meters including both the amounts above and below the water. 150 to 500 meters total
7. But ice shelves are not subglacial lakes. So what is a subglacial lake? What Exactly Are We Looking For? And where should we be looking?
8. ‘Sub’ + ‘Glacial’ = Subglacial A + B = C Simple formula right? Sub means ‘under’ Glacial refers to a glacier or ice sheet SO we are looking for a lake underneath a glacier…hmmm Like this submarine is ‘under’ water
9. But how far underneath? 15-50 m Although Adrienne did manage to swim in the Ross Sea through a cut out in the Ice Shelf, this was under 15-50 meters (49-164 ft.) of ice. 3 to 4 km The lakes we are looking for are under 3 to 4 kilometers (9843-13,123 ft.) of ice in the middle of a glacier! She won’t be swimming there!
10. How does water stay liquid under all that ice? Frozen lakes and ponds often have water beneath a cover of ice. Maybe you have even skated on one? But the Antarctic subglacial lakes Adrienne is looking for are are NOT like any of the lakes you might have seen! These lakes will be hard to find since they are covered by a HUGE ice sheet that spreads over most of the Antarctic continent. This ice sheet is millions of years old, and remember it is up to 4 kms thick!
11. Let's try some activities and see if we can find Adrienne’s missing lakes somewhere under all that ice! clicking on a photo below to start
More Related | 809 | 3,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-40 | latest | en | 0.946996 |
http://www.youtube.com/watch?v=0RPkODrcjkU | 1,369,143,030,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700014987/warc/CC-MAIN-20130516102654-00029-ip-10-60-113-184.ec2.internal.warc.gz | 816,455,617 | 27,476 | • 0:23 = epic face
• 250 different types of combination? Check your math.
This guy pressed 5 buttons for his password and there are 5 possible letters to choose. So, a permutation with repetition would mean that there were 5^5 possible combinations or 3,125 possible codes. Make your code 6 letters long and that jumps to over 15,000 possibilities.
• Sorry? I don't think those numbers mean what you think they do. They are just there so you could in theory have a password that's "918918918". In order to type that, you just type in EAEEAEEAE. You don't type E twice to get 9, or A to get 1, you just press it once.
• Try opening your eyes before trying to show off math skills on youtube. If you look at the card you can see that each letter "A-E" has two numbers under each so there are 9 numbers to combine.
• Bagsy my card doesn't need batteries to work.
• you fail to see the meaning in my comment --- if all your info is in ONE PLACE on ONE CARD -- and you LOSE THAT CARD...youre royally screwed.
• You mean just like if you lost your wallet? If you're the type to lose things just get the security one that was discussed in the video. You lose it, no problem!
• Who else noticed the big triangle symbol on the top of the card? hmm.. i bet £100m somebodys gonna makes a video about illuminati including this ._.
• soo...if someone loses that or misplaces it... they're screwed because all of their info is in one place
ingenious idea!
• It's acutally 10 buttons...look closer at it. An account digit can be 0-9, what you're doing is typing in the missing digits of the account (we'll call it a long PIN number), I would guess it's like texting on a non-QWERTY pad, if you wanted a "B" you hit "1" twice. Looks similar here, if you want a "1", hit the first button twice, a "4" hit the third button once, etc. | 472 | 1,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2013-20 | latest | en | 0.95245 |
http://www.topperlearning.com/forums/ask-experts-19/how-do-you-calculate-thomas-coefficient-and-seebeck-coeffici-physics-current-electricity-47290/reply | 1,487,990,951,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171646.15/warc/CC-MAIN-20170219104611-00176-ip-10-171-10-108.ec2.internal.warc.gz | 633,038,439 | 38,314 |
Question
Tue April 05, 2011
# How do you calculate Thomas coefficient and Seebeck coefficient?
Wed April 06, 2011
Dear student
If the temperature difference ÎT between the two ends of a material is small, then the thermopower of a material is conventionally (though only approximately, see below) defined as:
where ÎV is the thermoelectric voltage seen at the terminals. (See below for more on the signs of ÎV and ÎT.)
This can also be written in relation to the electric field E and the temperature gradient , by the equation:
Hope this helps.
Regards
Team
Topperlearning
Related Questions
Wed January 18, 2017
# A cell of emf E and internal resistance r is connected to two external resistance R1 and R2 and a perfect ammeter.The current in the circuit is measured in four different situation: (i)without any external resistance in the circuit (ii)with resistance R1 only (iii)with Ri and R2 in series combination (iv)with R1 and R2 in parallel combination The currents measured in the four cases are 0.42A,1.05A,1.4A and 4.2A,but not necessarily in that order.Identify the currents corresponding to the four cases mentioned above.
Thu November 24, 2016
| 291 | 1,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-09 | longest | en | 0.937081 |
https://www.hackmath.net/en/math-problem/8249 | 1,606,924,984,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00094.warc.gz | 692,153,975 | 12,023 | # Pupils
There are 27 pupils in the classroom. They can swim 21 and ski nine pupils. Neither swim nor ski three pupils. How many pupils can swim and ski?
Correct result:
x = 6
#### Solution:
$x=21+9-(27-3)=6$
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https://www.coursehero.com/file/20287415/HW1-Solutions/ | 1,547,731,518,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658981.19/warc/CC-MAIN-20190117123059-20190117145059-00503.warc.gz | 762,643,207 | 69,240 | ENGL
HW1_Solutions
# HW1_Solutions - IE:3610 Stochastic Modeling(Fall 2015...
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IE:3610 Stochastic Modeling (Fall 2015) Homework 1 Solutions 1. Problem 24-1 (a, b) from Chapter 24 (on ICON) A cube has its six sides colored red, white, blue, green, yellow, and violet. It is assumed that these six sides are equally likely to show when the cube is tossed. The cube is tossed once. (a) Describe the sample space. (b) Consider the random variable that assigns the number 0 to red and white, the number 1 to green and blue, and the number 2 to yellow and violet. What is the distribution of this random variable? Solution: (a) S = { Red, White, Blue, Green, Yellow, Violet } (b) X = 0 if Read or White 1 if Green or Blue 2 if Yellow or Violet Probability of either of the faces turning up is 1/6. That is, P(Red) = P(White) = P(Blue) = P(Green) = P(Yellow) = P(Violet) = 1 / 6. Then probability mass function for X is: P( X = 0) = P(Read S White) = P(Read) + P(White) = 1/6 + 1/6 = 1/3 P( X = 1) = P(Green S Blue) = P(Green) + P(Blue) = 1/6 + 1/6 = 1/3 P( X = 2) = P(Yellow S Violet) = P(Yellow) + P(Violet) = 1/6 + 1/6 = 1/3 Therefore, the probabilities are equally distributed. 2. Problem 24-2 (c) from Chapter 24 (on ICON) Suppose the sample space Ω consists of the four points { ω 1 , ω 2 , ω 3 , ω 4 } , and the associated probabilities over the events are given by P( ω 1 ) = 1 3 , P( ω 2 ) = 1 5 , P( ω 3 ) = 3 10 , P( ω 4 ) = 1 6 . Define the random variable X 1 by X 1 ( ω 1 ) = 1, X 1 ( ω 2 ) = 1, X 1 ( ω 3 ) = 4, X 1 ( ω 4 ) = 5, 1
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and the random variable X 2 by X 2 ( ω 1 ) = 1, X 2 ( ω 2 ) = 1, X 2 ( ω 3 ) = 1, X 2 ( ω 4 ) = 5.
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https://chemistry.stackexchange.com/questions/131785/why-is-the-molecular-structure-of-water-bent?noredirect=1 | 1,638,134,684,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00143.warc.gz | 249,130,932 | 42,098 | # Why is the molecular structure of water bent?
Water has the formula $$\ce{H2O}$$ and we can draw a Lewis structure with two lone pairs on the central oxygen. As a physics student and not a chemist, I think to myself, "Okay, there are two lone pairs, they will repel each other" and so we arrive at the conclusion of bent shape.
The issue in my mind is that I don't see why the lone electron pairs have to exist on the same side of the atom. Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)? If that were true, then there would be a resonance structure between the two states and we would get a linear geometry. Clearly I'm running around in circles here, please someone enlighten me!
• The lone pairs repel each other, but you should also throw the bonding electrons of the O-H bonds into the mix if you're approaching the problem from just electron repulsion.
– Zhe
Apr 14 '20 at 13:36
• You must account for repulsion between bonding electrons and non bonding electrons too (Edit: just as @zhe said). Therefore, each pair are repelling every other pair. So, every pair of electrons must have farthest possible distance from each other, which results in tetrahedral shape. And because lone pair repulsions are greater, it is distorted tetrahedral. Apr 14 '20 at 13:55
• You could have a look at the answers here chemistry.stackexchange.com/questions/14981/… which deal with Walsh diagrams. Apr 14 '20 at 14:50
• Does this answer your question? Are the lone pairs in water equivalent? Apr 14 '20 at 16:37
• a minor side point on terminology - in chemistry the term "resonance structures" is not used to describe structures in which atoms occupy different spatial positions (such as your example of water molecules with H atoms on opposite sides from each other). Resonance structures are electronic states. Crucially, molecules do not alternate between resonance structures. All resonance structures simultaneously contribute to the single electronic structure of the molecule. Apr 14 '20 at 17:52
I mean, there is a time and place for VSEPR, and this is probably as good a time as any, because all beginning chemistry students go through it. The actual model has already been explained multiple times, so I will only briefly say that according to this theory, there are four pairs of electrons around the central oxygen. In order to minimise electron-electron repulsions, these pairs adopt a tetrahedral arrangement around the oxygen. It does not matter which two are lone pairs and which two are connected to hydrogen atoms; the resulting shape is always bent.
What's worth bearing in mind (and hasn't been explained very carefully so far) is that VSEPR is a model that chemists use to predict the shape of a molecule. The truth is that there is no real way to predict the shape of a molecule, apart from solving the Schrodinger equation, which is not analytically possible for water. Everything else is an approximation to the truth. Some of these approximations are pretty accurate, such as the use of density functional theory. Some of them are extremely crude, and VSEPR falls into this category: it essentially treats electrons as classical point charges, and seeks to minimise the electrostatic repulsion between these point charges. As a physics student you should know better than to do this. Thus, while it predicts the correct result in this case, it is more in spite of the model rather than because of the model. And you should not be surprised to hear that in some slightly more complicated cases, VSEPR can predict entirely wrong outcomes.
As you learn more chemistry you will find that there are increasingly sophisticated ways of explaining molecular geometry. Most revolve around molecular orbital theory. For a qualitative method, you have Walsh diagrams which have been explained at Why does bond angle decrease in the order H2O, H2S, H2Se?. For a more rigorous method you would likely have to run some quantum chemical computations, e.g. Are the lone pairs in water equivalent?. Of course, the drawback of this is that it becomes more and more difficult to extract true chemical understanding from the numbers. Although it should also be said that you cannot extract any true chemical understanding from the VSEPR model.
What interests me more is the followup question:
Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)?
Because the Hamiltonian of the water molecule is invariant upon rotation, this means that indeed, any orientation of the water molecule is equally likely. However, this only refers to the orientation of the water molecule as a whole. It does not say anything about the internal degrees of freedom, such as the bond angle.
In the absence of any external force, the molecule is free to bend in whichever direction it likes, and most water molecules indeed do do this as they float through space or swim in a lake. But it will always be bent.
If that were true, then there would be a resonance structure between the two states and we would get a linear geometry.
If you were to think of a single particle in a double-well potential, say something with
$$V = \begin{cases} \infty & x < -b \\ 0 & -b \le x \le -a \\ \infty & -a < x < a \\ 0 & a \le x \le b \\ \infty & x > b \end{cases}$$
then because of the symmetry of your system, in every eigenstate of your system, the expectation value of $$x$$ would be $$\langle x \rangle = 0$$. This is quite similar to your argument. In the case of water, let's set the oxygen nucleus to be at the origin. Because it can point either up or down, the expectation value of the hydrogen nucleus position along the up-down axis would be exactly level with the oxygen atom, i.e. 0. In fact, don't stop there: it can point to the left or the right, and to the front or the back. So the hydrogen nucleus has a position expectation value of exactly $$(0, 0, 0)$$, i.e. right inside the oxygen nucleus.
Does that mean it's actually there, though? In our contrived double-well system, it's patently impossible for the particle to be at $$x = 0$$, because $$V = \infty$$ there. If you were to measure its position, you would never find it at $$x = 0$$; you would only find it in the left-hand side $$[-b, -a]$$, or the right-hand side $$[a,b]$$. Just because the particle has an expectation value of $$\langle x \rangle = 0$$ does not mean that it is physically there, or that $$x = 0$$ is somehow its equilibrium state. You're confusing an expectation value with a genuine eigenstate (which is what a resonance structure is).
In exactly the same way, if you ever were to measure the properties of water (and bear in mind that practically every interaction with a water molecule is, in effect, a measurement), we would find that it is indeed always bent.
• "Solving the Schrödinger equation" is of course also just an approximation to the truth. In a very basic sense one could even argue that its one that is not in principle any better than Lewis + VSEPR. It is in fact a misunderstanding to imply a truth behind a model. This is in my point of view the cause of almost all missunderstandings in Theiretical Chemistry. A model shall predict the phenomena. It does not make sense to say a chemical model (that is a" theory" really) is wrong. A theory is always correct. The question is only its range of validity. Apr 15 '20 at 17:38
• @Rudi_Birnbaum "A theory is always correct." -- I have a theory. My theory supposes that my theory is incorrect. Apr 17 '20 at 18:35
• I like your placing of ideas into a spectrum of accuracy. However, your discussion of the "good" approximations are slightly off. DFT, as a theory, is exact. It is the approximation of the unknown universal functional that makes its implementation inexact. Also, "solving the Schrödinger equation" and DFT are typically associated with electronic structure theory, unless you are treating atoms quantum mechanically as well, in which case you can only predict the expectation value of position. Thus, a pin-point description of atomic locations necessitates an approximation (Born-Oppenheimer). Apr 17 '20 at 19:53
• @LordStryker: That's not a theory in the sense of epistemology. It could be a hypotheses. Apr 17 '20 at 22:04
• @jezzo thanks for your comment! I’m aware of your points, but am slightly loathe to include them because I feel like I already digressed too much. But the comment can and should stay as a useful reminder to other readers about technical accuracy :) Apr 18 '20 at 5:25
Lewis structures are dots around atoms on 2-dimensional paper. Although H-O-H is planar, when you throw in the lone pairs, you have to think three-dimensionally. How would oxygen be hybridized, in readiness to accept two incoming hydrogens, each with an electron?
Oxygen could hybridize as sp2p, with 120 degree angles between bonds to hydrogen and a 90 degree angle between the p2 lone pair and the sp2 lone pair. Or as sp3, with 109 degrees between all the bonds. In water, we have 104.5 degrees https://en.wikibooks.org/wiki/Structural_Biochemistry/Water. 104.5 is between 90 and 109, so there must be some accommodation between the repulsion of lone pairs, which would push toward sp3 hybridization, and perhaps some repulsion from the electrons in the O-H bonds, which would tend to spread the H-O-H angle toward 120 degrees (sp2) and push one of the lone pairs more completely into the p2 orbital.
I just can't decide which is the more powerful force; perhaps the water molecule can't either, so it just compromises.
• This reasoning is backwards. Hybridisation is a model of description, hence it always follows the molecular structure. For more see: Are the lone pairs in water equivalent? As an additional note, while the phrasing 'oxygen is hybridised' is unfortunately very common, it is incorrect. Atoms cannot be hybridised, only orbitals can be constructed as hybrid orbitals. The repulsion of these lone pairs is also a lot more involved than what is presented here. Apr 14 '20 at 15:21
Yet another way to think about the fact that $$\ce{H2O}$$ does not have the full symmetry theoretically possible, is that the number of electrons does not properly match the highest possible symmetry point group (which has the odd name $$D_{\infty\mathrm{h}}$$. In this sense it can be seen as simple example of symmetry breaking (its however not a simple first order Jahn-Teller distortion).
• Why is $D_\mathrm{\infty h}$ an odd name? Please use the chemical construct $\ce{H2O}$ instead of the concoction H$_2$O which may have plenty of unwanted side effects. If you want to know more about MathJax, please have a look here and here. Apr 16 '20 at 22:11
• I am writing to a Physicist which possibly never came in contact to the Schönflies notation. They usually prefer other ones like Hermann-Mauguin or more mathematically oriented ones like Coxeter. Apr 17 '20 at 5:15
I don't see why the lone electron pairs have to exist on the same side of the atom.
Nobody says they exist "on the same side of the atom". In the picture showing Lewis structures of waters, the lone pairs are shown on opposite sides in the left panel , and on the same side on the right.
These two structures are identical. A Lewis structure does not make a statement about the geometry of a molecule.
If you look at a 3D model of water where the lone pairs are shown according to sp3-hybridization in the valence bond view of things, it depends on the orientation of the molecule whether it looks like "on opposite sides" or "on the same side" (the elongated shape - bunny ears - of the lone pairs is exaggerated; they should add up to a roughly spherical electron density). (Source: https://www.biotopics.co.uk/jsmol/watersingle.html)
In the molecular orbital view (where the molecular orbitals share the symmetry of the molecule) the two lone pairs have distinct shapes (each panel shows one lone pair, orientation of the molecule distinct from figure above).
This was calculated using molcalc.org. As these two molecular orbitals are of similar energy, you could make linear combinations of them to arrive at orbitals similar to the valence bond picture.
So what is the real picture? Looking at hydrogen bond geometries, either one describes the directions from which hydrogen acceptors on other molecules would be located, so both models are consistent with experimental data.
There are eight valence electrons on the oxygen that have similar energies then there is a gap and two electrons very close to oxygen core follow. These eight electrons form four spin-pairs and those point in directions that more or less minimise the repulsion. These directions are called "tetrahedral" in Chemistry language, since they point from the center (= oxygen atom core) to the vertices of an tetrahedron. Now the two protons are connected with two of those "electron pairs" and two are "alone". Hence you obtain the bent shape of H$$_2$$O.
(This is kind of trivialized version of the VSEPR model.)
• This kind of trivialised version of VSEPR is unfortunately as wrong as the VSEPR explanation for water itself. Water is indeed one of the most popular examples for when VSEPR breaks down. Apr 14 '20 at 15:23
• @Martin-マーチン to add: I would make a distinction between when the VSEPR algorithm breaks down, and when the premises of the VSEPR model fail. The algorithm itself works surprisingly well on H2O: if one assumes that the model is correct (i.e. two equivalent LP + two equivalent BP + LP repel more strongly than BP), it predicts a bond angle slightly smaller than 109.5°, in line with experiment. The algorithm fails for H2S, even if one makes the necessary assumptions. The premises fail for H2O. But then again it is arguable as to whether the premises are ever true... Apr 14 '20 at 17:36
• Not sure about which premises you are talking. Obviously VSEPR works fine, you can even predict a compression of the tetrahedral angle if you include the refined rule the the sp3 LP requires more space than the sp3 bond. The VSEPR model in that sense is one of chemistries most important models at all I would say, since it has a huge predictive power. In the sense of the amount of data you need and the amount of molecular structures you get proprerly described by that. Its paper and pencil and the next better one is the MO and that you have to buy with a huge complexity. Apr 15 '20 at 17:19 | 3,363 | 14,747 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-49 | latest | en | 0.93679 |
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