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https://git.mpi-cbg.de/argupta/openfpm_numerics/-/raw/759fb600260b785c9d282c10a9f1d6596767ec78/src/Solvers/umfpack_solver.hpp | 1,675,335,400,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00409.warc.gz | 303,173,268 | 1,835 | /* * Umfpack_solver.hpp * * Created on: Nov 27, 2015 * Author: i-bird */ #ifndef OPENFPM_NUMERICS_SRC_SOLVERS_UMFPACK_SOLVER_HPP_ #define OPENFPM_NUMERICS_SRC_SOLVERS_UMFPACK_SOLVER_HPP_ #include "Vector/Vector.hpp" #include "Eigen/UmfPackSupport" #include #define UMFPACK_NONE 0 template class umfpack_solver { public: template static Vector solve(const SparseMatrix & A, const Vector & b) { std::cerr << "Error Umfpack only suppor double precision" << "/n"; } }; #define SOLVER_NOOPTION 0 #define SOLVER_PRINT_RESIDUAL_NORM_INFINITY 1 #define SOLVER_PRINT_DETERMINANT 2 template<> class umfpack_solver { public: /*! \brief Here we invert the matrix and solve the system * * \warning umfpack is not a parallel solver, this function work only with one processor * * \note if you want to use umfpack in a NON parallel, but on a distributed data, use solve with triplet * * \tparam impl Implementation of the SparseMatrix * */ template static Vector solve(const SparseMatrix & A, const Vector & b, size_t opt = UMFPACK_NONE) { Vcluster & v_cl = *global_v_cluster; Vector x; // only master processor solve if (v_cl.getProcessUnitID() == 0) { Eigen::UmfPackLU > solver; solver.compute(A.getMat()); if(solver.info()!=Eigen::Success) { // decomposition failed std::cout << __FILE__ << ":" << __LINE__ << " solver failed" << "\n"; return x; } x.getVec() = solver.solve(b.getVec()); if (opt & SOLVER_PRINT_RESIDUAL_NORM_INFINITY) { Vector res; res.getVec() = A.getMat() * x.getVec() - b.getVec(); std::cout << "Infinity norm: " << res.getVec().lpNorm() << "\n"; } if (opt & SOLVER_PRINT_DETERMINANT) { std::cout << " Determinant: " << solver.determinant() << "\n"; } // Vector is only on master, scatter back the information x.sync(); } return x; } }; #endif /* OPENFPM_NUMERICS_SRC_SOLVERS_UMFPACK_SOLVER_HPP_ */ | 507 | 1,805 | {"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-2023-06 | latest | en | 0.342819 |
https://www.maa.org/press/maa-reviews/a-first-course-in-dimensional-analysis | 1,627,883,621,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154304.34/warc/CC-MAIN-20210802043814-20210802073814-00314.warc.gz | 902,390,073 | 22,622 | # A First Course in Dimensional Analysis
###### Juan G. Santiago
Publisher:
MIT Press
Publication Date:
2019
Number of Pages:
178
Format:
Paperback
Price:
30.00
ISBN:
9780262537711
Category:
Textbook
[Reviewed by
Brian Borchers
, on
03/30/2020
]
A First Course In Dimensional Analysis is an introductory undergraduate level textbook on the methods of dimensional analysis. It covers the principle of dimensional homogeneity, Ipsen’s method for finding dimensionless groups, the Buckingham $\Pi$ theorem, using dimensionless groups to build models, and common dimensionless groups such as the Reynolds number and the Peclet number.
The author’s general approach is to introduce the basic concept of dimensional homogeneity and then use Ipsen’s method to eliminate dimensions from the problem successively. He also advocates using governing equations to derive dimensionless parameters. The author is not a fan of Buckingham’s $\Pi$ theorem but the book does include a chapter on the use of the $\Pi$ theorem to find dimensionless groups. In comparison with Ipsen’s method, using the $\Pi$ theorem feels like magic, particularly when the theorem is simply stated without proof as in most other introductory books on dimensional analysis. Having taught this material many times using Buckingham’s $\Pi$ theorem I’m convinced that the author’s approach has some merit.
One difficulty that applied mathematics students often have is a lack of familiarity with basic concepts of fluid dynamics and heat transfer. The examples in this book are supported by text and figures that help to explain these concepts. However, students might benefit from modeling projects and exercises in which they explore these concepts in more depth.
I was somewhat disappointed that the author did not discuss the need to nondimensionalize a model before analyzing the asymptotic behavior of the model with respect to small or large parameters. Perturbation methods require nondimensional parameters and students who go on to take a course in the methods of applied mathematics will need to be able to nondimensionalize models.
Overall, the author succeeds in presenting dimensional analysis in a way that will be accessible to undergraduate students in engineering, the physical sciences, and applied mathematics. There isn’t enough material in the book to really make a semester-long course, but the book could be very useful as a supplementary text for a larger modeling course. For those who teach modeling courses that include dimensional analysis, the book will also be a good source of examples and exercises.
Brian Borchers is a professor of mathematics at New Mexico Tech and the editor of MAA Reviews. | 531 | 2,699 | {"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-2021-31 | latest | en | 0.900813 |
https://cracku.in/53-the-sum-of-the-dimensions-of-a-room-ie-length-brea-x-ibps-rrb-clerk-2016-shift-2 | 1,716,666,599,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00141.warc.gz | 150,746,567 | 23,114 | Question 53
# The sum of the dimensions of a room (i.e. length, breadth and height) is 24 metres and its length, breadth and height are in the ratio of 8: 7 : 5 respectively. If the room is to be painted at the rate of Rs. 12 per m2, what would be the total cost incurred on painting only the four walls of the room (in Rs.)?
Solution
Let the dimension of the room be $$8x , 7x , 5x$$ metres
Acc. to ques, => $$8x + 7x + 5x = 24$$
=> $$x = \frac{24}{20} = 1.2$$ metres
Curved surface area of the room = $$2 h (l + b)$$
= $$2 \times 5x \times (8x + 7x) = 10x \times 15x$$
= $$150 (x)^2 = 150 \times (1.2)^2$$
= $$150 \times 1.44 = 216 m^2$$
$$\therefore$$ Total cost incurred on painting only the four walls of the room = $$12 \times 216$$
= $$Rs. 2,592$$ | 274 | 765 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-22 | latest | en | 0.918727 |
https://oeis.org/A219425/internal | 1,642,886,848,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303884.44/warc/CC-MAIN-20220122194730-20220122224730-00327.warc.gz | 489,528,000 | 2,832 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A219425 Unchanging value maps: number of 5 X n binary arrays indicating the locations of corresponding elements unequal to no horizontal or antidiagonal neighbor in a random 0..2 5 X n array. 1
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%U 3766998295859184,48791165925094356,631950081288077455
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%C Row 5 of A219421.
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%e ..0..0..1....0..0..0....1..0..1....0..0..1....0..0..0....1..0..0....1..0..0
%Y Cf. A219421.
%K nonn
%O 1,2
%A _R. H. Hardin_, Nov 19 2012
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Last modified January 22 16:26 EST 2022. Contains 350483 sequences. (Running on oeis4.) | 611 | 1,654 | {"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-2022-05 | latest | en | 0.627279 |
https://www.ques10.com/p/37765/a-building-having-floor-to-floor-height-as-32-m-is/ | 1,582,073,453,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143963.79/warc/CC-MAIN-20200219000604-20200219030604-00499.warc.gz | 893,008,625 | 7,707 | 0
A building having floor to floor height as 3.2 m is to be provided a dog legged staircase. The stair hall is 3m x 5m. Grade of concrete is M20 and steel Fe415.
a) Draw a dog legged staircase.
b) Draw the plan showing flight details, mid-landings etc. Draw reinforcement details in a flight.
Civil Eng > Sem 8 > Design and Drawing of Reinforced Concrete structure
ddrcs mumbai university • 360 views
0
Given:
height = 3.2m
dimension = $3m \times 5m$
M20 Fe415
Solution:
1] Design Constant
$f_{ck} = 20 N/mm^2$
$f_y = 415 N/mm^2$
2] Design Parameter
Height = 3.2 m
Each flight = 1600 mm
$\therefore$ provide rise = 10 each 160mm
$\therefore$ treads = 9 each 250mm
$\therefore$ Plan is
3] effective span and depth
Less for $I^{st} = 2.25 + 1.375 = 3.625 m$
Less for $II^{nd} = 5m$
$d = \frac{Less}{B_v mf}$ ($B_v = 20$ fixed and $mf = 1.35$ for % of steal = 0.4%)
$\therefore d = \frac{5000}{20 \times 1.35} = 185.18 mm$
assume $d = 200 mm \quad D = 225 mm$ (cover 25mm)
(R = Riser = 0.16m, T = Tread = 0.25m)
i) On goeing
\begin{aligned} S.W.S &= 25 \times \frac{0.225 \sqrt{0.16^2 + 0.25^2}}{0.25} \bigg( \frac{250 \sqrt{R^2 + T^2}}{T} \bigg) \\ &= 6.678 \end{aligned}
$S.W. Step = 2\big( \frac{25R}{2} \big) \quad LL = 3 \quad FF = 1$
UL = 19.017 kN/m
II) On slab / landing (25 X D)
$SW = 25 \times 0.225 = 5.625 \quad LL = 3 \quad FF=$
$TL = 9.625 kN/m$
$\therefore$ Total ultimate = 14.437
5] BM calculations
Case 1:
$R_A+R_B=62.64$
$R_A = 33.274 \ KN$
$R_B = 29.366 \ KN$
$\therefore \quad M_{max} =29.11 \mathrm{kNm}$
Case 2:
$\therefore \quad M_{max} = 55.1 \mathrm{kNm}$
6] Check for depth
$d^2 = \frac{55.1 \times 10^6}{0.138 \times 20 \times 1000} \left( \frac{M}{0.138 fck \ b} \right)$
$d = 141.3 mm \lt d_{pro} = 200 mm \therefore Safe$
7] Calculation of Reinforcement
For Fight (I)
$M = 29.11 KNm$
$Ast =\frac{0.5 \times 20}{415}\left[1- \sqrt{ 1 - \frac{4.6 \times 29.11 \times 10^{6}}{20 \times 1000 \times 200^{2}}} \right] 1000 \times 200$
$Ast = 421.79 mm^2$
$\therefore$ provide 10mm $\pi$ @ 185mm c/c
$\text{Ast provided} = 424.54 mm^2$
For flight (II)
$M = 55.1 KNm$
Ast = $835.93 mm^2$
$\therefore$ provide 12mm $\phi$ @ 130mm c/c
$\text{Ast provided} = 870 mm^2$
Distribution steel:
$0.12 \% bd = 240 mm^2$
8mm $\phi$ @ 205 mm c/c
8] Check for deflection
$\text{% of steel} = \frac{870}{1000 \times 200} \times 100 \left( \frac{\text{Ast provided}}{bd} \right)$
$0.435 \not{\lt} 0.4$
Detail check
$f_s = 231.274 \quad mf = 1.3$
$d = \frac{5000}{20 \times 1.3} = 192 \lt 200 \therefore Safe$
9] Check for shear
Consider 19.017 through out span
$Vu_{max} = \frac{wl}{2} = \frac{19.017 \times 5}{2} = 47.54 KN$
$\tau_v = 0.2377 \quad (Vu_{max} \times b \times d)$
For D = 225mm $\quad$ K = 1.15 from IScode
$\tau_e = K \tau_c$
$1.15 \times 0.28 = 0.322 \quad \therefore Safe$
10] Development
$L_d = \frac{0.87 \times 1115 \times 12}{4 \times 1.2 \times 1.6} = 564.14 (\frac{0.87 fy \phi}{4 \tau_D d})$
$= \frac{1.3 \times 27.55 \times 10^6}{117.54 \times 1.^3} + 100 = 853.4 \quad \therefore Safe$ | 1,281 | 3,094 | {"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} | 4.1875 | 4 | CC-MAIN-2020-10 | latest | en | 0.615481 |
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### Fundamentals Of Physics
Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718
# The terminal speed of a sky diver is ${}^{\mathbf{160}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$ in the spread-eagle position and ${}^{\mathbf{310}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$ in the nosedive position. Assuming that the diver’s drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position.
The ratio of the effective cross-sectional area A in the slower position to that in the faster position is ${}^{3.75}$
See the step by step solution
## Step 1: Given
$a=\frac{2mg}{Cp{{v}^{2}}_{t}}$
which illustrates the inverse proportionality between the area and the speed-squared
thus
${A}_{slow}=310\mathrm{km}/\mathrm{h}\phantom{\rule{0ex}{0ex}}{A}_{fast=}160\mathrm{km}/\mathrm{h}$
## Step 2: Determining the concept
This problem is based on the drag force which is a type of friction. This is the force acting opposite to the relative motion of an object moving with respect to the surrounding medium. Using the concept of the drag force and terminal speed the ratio of the effective cross-sectional area A in the slower position to that in the faster position.
Formula:
The terminal speed is given by
${v}_{t}=\sqrt{\frac{2{F}_{g}}{CpA}}$
Where C is the drag coefficient,${}^{p}$ is the fluid density, A is the effective cross-sectional area, and ${}^{{F}_{g}}$ is the gravitational force.
solve for the area
$A=\frac{2mg}{Cp{{v}^{2}}_{t}}$
which illustrates the inverse proportionality between the area and the speed-squared
## Step 3: Determining the ratio of the effective cross-sectional area A in the slower position to that in the faster position
When a ratio of areas has been setup of the slower case to the faster case,
$\frac{{A}_{slow}}{{A}_{fast}}={\left(\frac{110\mathrm{km}/\mathrm{h}}{160\mathrm{km}/\mathrm{h}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=3.75$
Hence, the ratio of the effective cross-sectional area A in the slower position to that in the faster position is ${}^{3.75}$ | 652 | 2,333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "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} | 4.25 | 4 | CC-MAIN-2023-23 | latest | en | 0.822871 |
https://www.wyzant.com/resources/answers/users/88770520 | 1,643,058,337,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304600.9/warc/CC-MAIN-20220124185733-20220124215733-00324.warc.gz | 1,097,559,087 | 9,677 | 10/04/21
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 343 | 1,233 | {"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} | 2.75 | 3 | CC-MAIN-2022-05 | latest | en | 0.922987 |
https://math.stackexchange.com/questions/654626/masters-theorem-applicability | 1,642,584,530,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00023.warc.gz | 428,112,337 | 33,386 | Master's theorem applicability.
I have to find out if the following recurrence can be solved with the master theorem:
$$T(n) = 3T\left(\frac{n}{2}\right) + n^{\log\log n}$$
I have figured that, here, I have the third case because $f(n) > n^{\log_2 3}$ but it understand that, in order for master to be applicable, $f(n)$ must be polynomially larger than $n^{\log_23}$ and I have no idea how to verify it.
Any help would be much appreciated!
The third case asks for (here, $a=3,b=2$)
• $f(n) = \Omega(n^{\log_2 3+\epsilon})$ for some $\epsilon > 0$
• $3f(\frac{n}{2})\leq \kappa f(n)$ for some $\kappa < 1$ (and any $n$ large enough)
Observe than $f(n) = n^{\log \log n} = 2^{\log n\cdot \log \log n}$; $f$ is not polynomial, it's asymptotically greater than any polynomial (in particular, for $\epsilon=1$, $n^{\log_2 3+1} = 2^{\log n\cdot(\log_2 3+1)} = o(2^{\log n\cdot \log \log n})$. The first condition holds.
The second too, as \begin{align*} \frac{3f\left(\frac{n}{2}\right)}{f(n)} &= 3\cdot 2^{\log(\frac{n}{2})\cdot \log \log( \frac{n}{2}) - \log n\cdot\log\log n} = 3\cdot 2^{\log(n)\cdot \log \log( \frac{n}{2}) - \log \log( \frac{n}{2}) - \log n\cdot\log\log n} \\ &=3\cdot 2^{\log(n)\cdot \left(\log \log( \frac{n}{2}) - \log\log n \right)- \log \log( \frac{n}{2})} = 3\cdot 2^{\log(n)\cdot \left(\log \left( 1-\frac{1}{\log n} \right) \right)- \log \log( \frac{n}{2})} \\ &= 3\cdot 2^{\log(n)\cdot \left(-\frac{1}{\log n}+o(\frac{1}{\log n}) \right)- \log \log( \frac{n}{2})} = 3\cdot 2^{-1- \log \log( \frac{n}{2}) + o(1)}\\ &= \frac{3}{2}\cdot \frac{1}{\log\frac{n}{2}}\cdot 2^{o(1)} \xrightarrow[n\to\infty]{}0 \end{align*}
(you can also get a sanity check to see this limit is indeed $0$ with Mathematica, although it is not a proof)
• I still didnt understand how did you demonstrate that the first condition holds, could you elaborate? Jan 28 '14 at 17:36
• Look at the exponent: one of them is $C\log n$ (for the constant $C=\log_2 3 +1$, the other one is $\log n \cdot \log\log n$ (where $\log\log n \to \infty$, so clearly greater than $C$ for $n$ sufficiently large). Jan 28 '14 at 17:50
• I see now, it was actually very obvious. Jan 29 '14 at 3:39 | 826 | 2,183 | {"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} | 4.03125 | 4 | CC-MAIN-2022-05 | latest | en | 0.731997 |
https://physics.stackexchange.com/questions/408828/can-the-ideal-gas-expand-adiabatically-and-irreversely | 1,656,600,388,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00100.warc.gz | 503,826,690 | 66,390 | # Can the ideal gas expand adiabatically and irreversely?
Suppose that the ideal gas is expanding adiabatically from the state $(P_1, V_1, T_1)$ to the state $(P_2, V_2, T_2)$
I think this process can be allowed to happen reversibly or irreversibly, but then I face a puzzling contradiction regarding entropy change ($\Delta S$).
If the process occurred reversibly, $\Delta S = 0$ since there are no heat transfer and no entropy generation.
If the process occurred irreversibly, $\Delta S > 0$ since there is entropy generation by the irreversibility of the process.
However, because entropy is a state function it cannot be zero and non zero for the two fixed chosen states $(P_1, V_1, T_1)$ and $(P_2, V_2, T_2)$
If the two end states you have described are those corresponding to an adiabatic reversible process, then it will be impossible to devise and conduct an adiabatic irreversible process that passes between these same two end states. If the two end states you have described are those corresponding to an adiabatic irreversible process, then it will be impossible to devise and conduct an adiabatic reversible process between the same two end states; some heat transfer will be required.
• I got it. Could you explain a bit more where does the impossibility in your explanation come? Or is it impossible to happen because the contradiction I mentioned in my question? May 29, 2018 at 2:30
• In the irreversible process, there will be entropy generated, so the final entropy will be greater than the initial entropy. So if there is a reversible process between the same two end states, you will have to have $$\Delta S=\int{q_{rev}/T}$$, and $q_{rev}$ will have to be non-zero. This means that the reversible process will have to be non-adiabatic. May 29, 2018 at 2:43
• Ah I seem to get your point. No matter how I conduct the experiment I wont be able to obtain two end states that have the same pv^gamma value as long as the process is carried out irreversibly. May 29, 2018 at 2:52
Not really. The fact that you are taking the gas along an adiabatic process already implies that the process is reversible (since it should be at thermal equilibrium with a reservoir during the entire path). If, however, your only requirement is that the initial and final states follow $p_1V_1^\gamma=p_2V_2^\gamma$, then, since entropy is a state function, the entropy of the gas is the same, but the entropy of the whole Universe could (and likely will) be larger.
• I am intereded in the gas system (and its entropy change) not the whole universe! :) May 29, 2018 at 2:31
• But when you talk about reversibility, what matters is the whole Universe, not isolated parts of it. May 29, 2018 at 2:36
• The second law states that the total entropy of the Universe must go up, not the entropy of your specific vessel of gas. May 29, 2018 at 2:45
• the word adiabatic in thermodynamics means no transfer of heat, although in quantum mechanics we often use the word for entropy-conserving (isentropic in thermodynamics). An example of irreversible adiabatic effect is the cooling of gas by expansion through a throttle (Joule-Thomson effect).
– wcc
Aug 16, 2018 at 6:04 | 797 | 3,171 | {"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} | 3.234375 | 3 | CC-MAIN-2022-27 | longest | en | 0.931115 |
http://www.java-forums.org/new-java/58672-decimals-print.html | 1,481,061,212,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00028-ip-10-31-129-80.ec2.internal.warc.gz | 472,293,946 | 2,798 | # Decimals
• 04-21-2012, 06:06 PM
Goughyy_
Decimals
I'm trying to make a Litres to Cubic Mitres Java Program. For the job I'm currently working on. I've programmed it the way I was taught but can't seem to get the conversion result to 5 decimal places. Here is the program itself it does work and complies no errors are found.
Code:
```import java.text.NumberFormat; import javax.swing.JOptionPane; public class LitreMeters { public static void main(String[] args) { String numString1; String numString2; JOptionPane.showMessageDialog(null, "The conversion is 1 Litre = 0.001 Cubic Metres\n" + "Click Ok to start converting."); numString1 = JOptionPane.showInputDialog( "Enter amount of litres:"); //Edward Gough Programming int userNum = Integer.parseInt(numString1); int otherNum = 20; int cNum = 1000; int result = userNum / 1000; JOptionPane.showMessageDialog(null, userNum + " Litres converted to " + result + " Cubic Meters"); } }```
• 04-21-2012, 06:15 PM
Fubarable
Re: Decimals
Do you mean that you want to display a floating point result, something like 23.1222? If so, consider making result a double variable, not an int, and obtaining it by dividing userNum by a double literal as well, by 1000.0 rather than 1000. You may find that you have too many decimal places showing, and if so may have to format your results with printf or String.format(...), or a DecimalFormat object.
• 04-21-2012, 06:26 PM
Goughyy_
Re: Decimals
Quote:
Do you mean that you want to display a floating point result, something like 23.1222? If so, consider making result a double variable, not an int, and obtaining it by dividing userNum by a double literal as well, by 1000.0 rather than 1000. You may find that you have too many decimal places showing, and if so may have to format your results with printf or String.format(...), or a DecimalFormat object.
Thanks, really helped. Plus the quick reply was great. Thanks for the help. Have a nice day.:(y):
• 04-21-2012, 06:26 PM
Fubarable
Re: Decimals
You're quite welcome, and best of luck! | 568 | 2,393 | {"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-2016-50 | latest | en | 0.532529 |
https://mathematica.stackexchange.com/questions/76141/line-integrals-over-a-path-defined-by-a-set-of-points-rather-than-an-analytic-f?noredirect=1 | 1,686,009,070,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652184.68/warc/CC-MAIN-20230605221713-20230606011713-00225.warc.gz | 405,628,044 | 41,428 | # Line integrals over a path defined by a set of points (rather than an analytic formula)
I'm getting inaccurate results when computing (plotting) the value of a function against the length of a line integral, where the path comes from a set of numerically calculated points. (In contrast, the integral seems fine when I have an analytic formula.)
For example: The matrix named points1 (which comes from stitching together a few numerical integrations, from NDSolve), composed of a few hundred real-valued triples, should define a smooth path (x(s),y(s),V(x(s),y(s))). I compute (interpolate) a parameterized line, and I plot the third component against the line integral of the first two components.
line1 = Table[ListInterpolation[points1[[1]][[All,jj]], {0,1}], {jj,1,3}]
ParametricPlot[{NIntegrate[Sqrt[line1[[1]]'[tt]^2+line1[[2]]'[tt]^2],{tt,0,ss}],
line1[[3]][ss]}, {ss,0,1}]
In a few spots the vertical axis has small notches, i.e., changes of direction up and then down. The notches are small, but still visible. They are definitely spurious. I want to get five or so significant digits, and this is giving me three.
I think the problem might have to do with ListInterpolation giving changes in direction even when the set of points that it is interpolating is monotonic. I get the same problem whether I choose the method Spline or Hermite, and whichever method I give NIntegrate.
Suggestions, anyone?
EDIT: Added code for one location where the problem shows up.
points = {
{0.6325952779137012, -0.7760085305401847, 54.07201697427571},
{0.62105807453548, -0.790281367826272, 53.57216747776533},
{0.5941956290244017, -0.8236073812563007, 52.209319984152756},
{0.5229935177928883, -0.9126291157398833, 47.34249397123951},
{0.3191937129061434, -1.1738845637283815, 26.841435108009442},
{0.30562102581445766, -1.1916609806787646, 25.358076396649462},
{0.2921897615948844, -1.2093010504444346, 23.900356246809537},
{0.2658926549639088, -1.243979893428035, 21.091685216353746},
{0.21635546253250398, -1.309812352799613, 16.05638285614679},
{0.13289109669708532, -1.422146100879459, 8.757698109321453},
{0.025168620642993245, -1.5688078584392655, 2.4156787966727435},
{0.020773587161218077, -1.5747852074054476, 2.2408427807381663},
{0.016541263234193266, -1.5805365848560005, 2.0788822741133828},
{0.008539545100479554, -1.5913950062364555, 1.7898452227991974},
{-0.005774836825683922, -1.61075381814465, 1.328719820988212},
{-0.028775790485831575, -1.641584474796247, 0.7368957514090972},
{-0.03120754151431262, -1.644815864289341, 0.6849477369657428},
{-0.0335531257347027,-1.6479262887687476, 0.6367430278603834},
{-0.03799906267998585, -1.6538030044313163, 0.5504755732460467},
{-0.045993704234286795, -1.664298846988861, 0.41201122449151484},
{-0.058970977912545186, -1.681084608190775, 0.23202329470223404}
};
line = Table[ ListInterpolation[points[[All, jj]], {0, 1}], {jj, 1, 3} ];
ParametricPlot[
{
NIntegrate[Sqrt[line[[1]]'[tt]^2 + line[[2]]'[tt]^2], {tt, 0, ss}],
line[[3]][ss]
},
{ss, 0, 1},
PlotRange -> {{0, All}, {0, All}},
AspectRatio -> 1/2
]
• Without the data, it is difficult to offer advice. To keep the problem manageable, consider providing the portion of points near one of the discontinuities in your plot, perhaps near 2.2. Also, I notice that the plot in your question is not the one produced by your code. Please provide the actual plot. Mar 1, 2015 at 13:06
• I edited a few lines of code into the question, with data points for one of the spurious notches. Mar 1, 2015 at 15:58
• see here mathematica.stackexchange.com/a/59342/2079 for a method to directly find the running path length of the line. Accumulate[Norm /@ Differences@points] may be better than your NIntegrate approach. (I cant test from here.. ) Mar 1, 2015 at 16:34
The interpolation overshoots the next point and reverses direction.
ParametricPlot[{line[[1]][tt], line[[2]][tt]}, {tt, 0.2, 0.3},
Epilog -> {Point[points[[All, 1 ;; 2]]]}]
You can reduce the interpolation order to 1 or use a centripetal parametrization parametrizeCurve from J.M.'s answer.
parametrizeCurve[pts_List, a : (_?NumericQ) : 1/2] :=
FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]] /;
MatrixQ[pts, NumericQ]
tvals = parametrizeCurve[points];
line = Interpolation[Transpose[{tvals, #}]] & /@ Transpose@points;
s[t_?NumericQ] :=
NIntegrate[Norm[{line[[1]]'[tt], line[[2]]'[tt]}],
Evaluate @ DeleteDuplicates @
Flatten[{tt, 0, Select[tvals, 0 < # < t &], t}]];
ParametricPlot[
Evaluate@{s[ss], line[[3]][ss]},
{ss, 0, 1},
PlotRange -> {{0, All}, {0, All}},
AspectRatio -> 1/2,
Epilog -> {Point@Table[{s[ss], line[[3]][ss]}, {ss, 0, 1, 0.1}]}
]
Including the interpolation grid points tvals in NIntegrate speeds up the integration. For a really fast implementation use
s = NDSolveValue[{ss'[t] == Norm[{line[[1]]'[t],line[[2]]'[t]}], ss[0]==0}, ss, {t, 0, 1}]
which constructs an InterpolatingFunction for the arc length.
• Thanks! After a correction -- integrating over the roots of the squares of the derivatives of the x- and y-components -- it worked. Mar 2, 2015 at 7:53
• @RichardTasgal You're welcome, and thanks for fixing it. At some point I had done the problem in terms of a single interpolation that returned the coordinates, and at the last minute I switched back to your method of line`. In the cut-n-paste editing, I must have pasted over the primes and not noticed. Mar 2, 2015 at 11:13 | 1,779 | 5,422 | {"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} | 3 | 3 | CC-MAIN-2023-23 | latest | en | 0.840947 |
https://godotengine.org/qa/106929/how-to-make-rigidbody2d-physics-work-in-top-down-game?show=106966 | 1,656,914,115,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104354651.73/warc/CC-MAIN-20220704050055-20220704080055-00529.warc.gz | 323,347,270 | 8,043 | Hello. In my top down 2D project when I create logs that player can shift(something like in Zelda 2D), I make Gravity Scale to 0. In order for gravity to "pull" the log to the surface, as if along the Z axis, which is not in Rigidbody2D. That is, when I set the GravityScale to 0 so that the log does not roll from top to bottom across the screen, it turns out that with this parameter, the mass and weight do not affect anything and the log is in zero gravity. Therefore, I had a question, how to make the gravity of Rigidbody2D work in the TopDown project?
Godot version 3.3
in Engine
Your halfway there with the zero g but you'll need to code in thr friction of the "floor" yourself.
Friction is a force proportional to velocity and in the opposite direction, the scale depends if the log is on ice low friction or mud high friction.
EDIT: see an example implentation below:
by (102 points)
edited
Thank you. I will try to find a way to do this
Sorry for second question. But could you suggest me how to make this (if you know) please. Because I just haven't really figured out Godot yet, and how to do things like that
Add this script to your rigid body, it would be even simplier but addcentralforce() behaves in an unusual manner and previous forces need clearing on the next frame. This acts like the damper (I just came across) but in your own script you could tweak the friction depending on what your object is on top of:
`````` extends RigidBody2D
export(float) var mu_static = 0.8 # friction coefficients
export(float) var mu_moving = 0.5 # pushing something moving is easier
# mu depends on what material the object is on, so use area detectors and change
# this values depending on ice or mud
export(float) var move_strength = 50
var applied_forces: Vector2 = Vector2(0, 0)
# add_force adds a permanent force, for a temporary one we need to wipe it
# by undo the force next frame, just keep track of forces applied
applied_forces += force
# no world gravity pushing the object down (in the +y) direction
# we are top down so gravity is acting into the screen (in +z) but the
# "ground" normal force is canceling it out
self.gravity_scale = 0
func _physics_process(delta: float) -> void:
add_force_for_frame(-applied_forces) # wipe out previous forces
# arrows keys apply force to move it
# could equally have a different body which pushes it
if Input.is_action_pressed("ui_right"):
if Input.is_action_pressed("ui_left"):
if Input.is_action_pressed("ui_up"):
if Input.is_action_pressed("ui_down"):
# this bit applies the damping force
# minus is there because friction resists the direction of motion
# friction depends on the m * g of the object (in this case the weight)
if self.linear_velocity.length() == 0:
self.add_force_for_frame(- self.weight * mu_static * self.linear_velocity.normalized())
else:
self.add_force_for_frame(- self.weight * mu_moving * self.linear_velocity.normalized())
# no worries about collisions as they add their own forces in opposition to these ones
``````
Thank you! It works perfectly ! | 732 | 3,052 | {"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-2022-27 | latest | en | 0.908435 |
https://se.mathworks.com/matlabcentral/answers/2129526-colour-interference-when-two-surfaces-overlap?s_tid=prof_contriblnk | 1,721,602,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00654.warc.gz | 439,515,894 | 27,553 | # Colour interference when two surfaces overlap
5 views (last 30 days)
xin on 18 Jun 2024
Commented: xin on 19 Jun 2024
Colour interference when two surfaces overlap, how to solve it？
Piyush Kumar on 18 Jun 2024
Edited: Piyush Kumar on 18 Jun 2024
Hi @xin,
If you are talking about the blending of colours at the point where surfaces intersect or overlap, I will try to explain with an example.
[X,Y] = meshgrid(linspace(-2, 2, 50));
Z = @(x,y,c) x.^2 + y.^2 + c;
figure(1)
surf(X, Y, +Z(X,Y,-4), 'FaceColor','g', 'FaceAlpha',0.5, 'EdgeColor','none')
hold on
surf(X, Y, -Z(X,Y,-4), 'FaceColor','r', 'FaceAlpha',0.5, 'EdgeColor','none')
hold off
grid on
I have specified only 2 colours for the 2 surfaces repectively, but multiple colours are visible. This blending effect is a result of the rendering process in MATLAB, where the semi-transparent (due to 'FaceAlpha',0.5) surfaces allow for the underlying color to show through, creating a visual effect of color mixing or interference.
To avoid this color blending effect and maintain the solid colors you've specified without seeing a mix at the intersections, you can:
• Increase the Opacity: Setting 'FaceAlpha' to 1 (or removing the 'FaceAlpha' setting, as its default value is 1) will make the surfaces fully opaque, preventing the underlying color from showing through.
[X,Y] = meshgrid(linspace(-2, 2, 50));
Z = @(x,y,c) x.^2 + y.^2 + c;
figure(1)
% Making the surfaces opaque by removing 'FaceAlpha'
surf(X, Y, +Z(X,Y,-4), 'FaceColor','g', 'EdgeColor','none')
hold on
surf(X, Y, -Z(X,Y,-4), 'FaceColor','r', 'EdgeColor','none')
hold off
grid on
Piyush Kumar on 18 Jun 2024
Can you share the code you are using?
xin on 19 Jun 2024
Of course,please import the data into the workspace at first.
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Start Hunting! | 557 | 1,987 | {"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-2024-30 | latest | en | 0.806372 |
https://integrativemodeling.org/nightly/doc/ref/PenalizedComplexityPrior_8h_source.html | 1,603,509,624,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00141.warc.gz | 374,508,727 | 6,319 | IMP Reference Guide develop.0ea7f7a4db,2020/10/23 The Integrative Modeling Platform
PenalizedComplexityPrior.h
Go to the documentation of this file.
1 /**
2 * \file IMP/isd/PenalizedComplexityPrior.h
3 * \brief Penalized complexity prior.
4 *
6 */
7
8 #ifndef IMPISD_PENALIZED_COMPLEXITY_PRIOR_H
9 #define IMPISD_PENALIZED_COMPLEXITY_PRIOR_H
10
11 #include <IMP/isd/isd_config.h>
12 #include <IMP/isd/distribution.h>
13
14 IMPISD_BEGIN_NAMESPACE
15 //! Penalized complexity prior.
16 /** Given a parameter \f$\xi\f$ of a model (distribution)
17 \f$f(x \mid \xi)\f$ with a simpler base model \f$g(x)\f$
18 (\f$f(x \mid \xi=0)\f$ if \f$\xi\f$ is a flexibility parameter), the
19 Penalized Complexity (PC) prior on \f$\xi\f$ minimizes the complexity
20 introduced by deviation from the base model. It corresponds to an
21 exponential distribution on the Kullback–Leibler divergence between the
22 flexible and base models, in terms of \f$\xi\f$.
23
24 For ease of parameterization, a 'tail event' is defined as an upper bound
25 on an interpretable transformation \f$Q(\xi)\f$ with tail density
26 \f$\alpha\f$, such that \f$P(Q(\xi) > U) = \alpha\f$. With KL divergence
27 defined as
28
29 \f[ \mathrm{KLD}(f(x \mid \xi) || g(x)) =
30 \int f(x \mid \xi) \log\left(\frac{f(x \mid \xi)}{g(x)}\right) dx,\f]
31
32 KL divergence at \f$\xi\f$ normalized by KL divergence at the tail
33
34 \f[ D(\xi) = \frac{\mathrm{KLD}(f(x \mid \xi) || g(x))}{\mathrm{KLD}(f(x \mid \xi=Q^{-1}(U)) || g(x))} ,\f]
35
36 and
37
38 \f[ J(\xi) = \frac{\partial D(\xi)}{\partial \xi} ,\f]
39
40 the density of the PC prior is
41
42 \f[p(\xi) = \frac{-\log\alpha}{2 \sqrt{D(\xi)}} \alpha^{\sqrt{D(\xi)}} |J(\xi)|.\f]
43
44 \see NormalSigmaPCRestraint
45
46 \note For more details, see Simpson et al. Stat Sci. (2017) 32(1): 1. arXiv: 1403.4630
47 \note The log-density with this parameterization overflows when
48 \f$D(\xi)=0\f$. The value of the PC prior at the base model is
49 unique to each case and cannot be provided by this general
50 implementation.
51 */
53 protected:
54 virtual void do_update_sufficient_statistics(Floats Dxis) IMP_OVERRIDE;
55 virtual void do_update_sufficient_statistics(Floats Dxis, Floats Jxis);
56 virtual Floats do_get_sufficient_statistics() const IMP_OVERRIDE;
57 virtual double do_evaluate() const IMP_OVERRIDE;
58
59 public:
60 //! Create from normalized KL divergence.
61 /** \param [in] Dxi Normalized KL divergence \f$D(\xi)\f$.
62 \param [in] Jxi Derivative \f$J(\xi)\f$ of normalized KL divergence.
63 \param [in] alpha Density in tail.
64 \param [in] name Name of prior.
65 */
66 PenalizedComplexityPrior(double Dxi, double Jxi, double alpha,
67 std::string name = "PenalizedComplexityPrior %1%");
68
69 //! Update sufficient statistics with values and derivatives.
71 do_update_sufficient_statistics(Dxis, Jxis);
72 }
73
74 //! Evaluate derivative of negative log-density wrt \f$D(\xi)\f$.
75 double evaluate_derivative_Dxi() const;
76
77 //! Evaluate derivative of negative log-density wrt derivative of \f$D(\xi)\f$.
78 double evaluate_derivative_Jxi() const;
79
80 //! Evaluate derivative of negative log-density wrt alpha.
81 double evaluate_derivative_alpha() const;
82
83 //! Evaluate derivative of negative log-density wrt parameter \f$\xi\f$.
84 /** \param [in] JJxi Second derivative of normalized KL divergence wrt \f$\xi\f$.
85 */
86 double evaluate_derivative_xi(double JJxi) const;
87
88 //! Set normalized KL divergence \f$D(\xi)\f$.
89 void set_Dxi(double v);
90
91 //! Set derivative of normalized KL divergence \f$D(\xi)\f$ wrt \f$\xi\f$.
92 void set_Jxi(double v);
93
94 //! Set density in tail.
95 void set_alpha(double v);
96
97 //! Get normalized KL divergence \f$D(\xi)\f$.
98 double get_Dxi() const;
99
100 //! Get derivative of normalized KL divergence \f$D(\xi)\f$ wrt \f$\xi\f$.
101 double get_Jxi() const;
102
103 //! Get density in tail.
104 double get_alpha() const;
105
107
108 private:
109 double sqrtDxi_, Jxi_, nloga_;
110 };
112
113 IMPISD_END_NAMESPACE
114
115 #endif /* IMPISD_PENALIZED_COMPLEXITY_PRIOR_H */
Base class for single-variate distributions that cache sufficient statistics.
void update_sufficient_statistics(Floats Dxis, Floats Jxis)
Update sufficient statistics with values and derivatives.
#define IMP_OBJECT_METHODS(Name)
Define the basic things needed by any Object.
Definition: object_macros.h:25
Base class for probability distributions.
#define IMP_OBJECTS(Name, PluralName)
Define the types for storing lists of object pointers.
Definition: object_macros.h:44
#define IMP_OVERRIDE
Cause a compile error if this method does not override a parent method. | 1,434 | 4,696 | {"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} | 2.625 | 3 | CC-MAIN-2020-45 | latest | en | 0.49147 |
https://physics.stackexchange.com/questions/334004/how-is-temperature-defined-and-measured | 1,656,235,188,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037649.11/warc/CC-MAIN-20220626071255-20220626101255-00506.warc.gz | 505,514,597 | 70,525 | # How is temperature defined, and measured?
In questions like this one, temperatures of millions of degrees (Celsius, Kelvin, it doesn't really matter at that point) are mentioned.
But, what does it mean exactly? What is measured, and how? As I more or less expected, the Wikipedia article mentions that the official definition (the one I was told in elementary school, either between freezing and boiling points of water for Celsius, or between absolute zero and triple point of water for Kelvin) doesn't really work above 1300K.
So, questions:
• For "earthly temperatures", how is temperature measured consistently? Why would linearity of the measuring device be assumed, in particular to extrapolate beyond the initial (0-100, or 0-273.16) range? What guarantees that two thermometers, working under different principles (say, mercury and electricity), that agree at 0C and 100C, will agree at 50C?
• What do statements like "the temperature in the Sun's core is $$1.5\times10^7K$$ mean? Or, even, what do the "7000K" in the Earth's core mean?
• Your title asks about a definiton, but your text asks about a measurement. Temperature is accurately defined, although several definitions exist. The problem with temperatures above 1300 K is a practical problem of thermometers, as far as I can see.
– pela
May 18, 2017 at 13:11
• I have changed the title. Neither definition nor measurement is clear to me. I'll update the question. May 18, 2017 at 13:16
• More on the definition of temperature. May 18, 2017 at 13:51
• I'll add as observation here: sometimes we can associate temperatures with non-equilibrium phenomena by using only local thermodynamic equilibrium or other related tools. Notice that this way of understanding temperature is slightly different from an intuitive point of view as the equipartition theorem won't hold in general. Jun 7, 2017 at 17:07
• In short, the temperature is an intensive property. It is the difference in temperature between two bodies that decide the direction of the heat flow.
– SRS
Jun 30, 2018 at 14:44
Definitions
First and foremost, temperature is a parameter defining a statistical distribution, much as the statistical parameters of mean and standard deviation define the normal probability distribution. Temperature defines the equilibrium (maximum likelihood) distribution of energies of particles in a collection of statistically independent particles through the Boltzmann distribution. If the possible energies of the particles are $E_i$, then the maximum likelihood particle energy distribution is proportional to $\exp\left(-\frac{E_i}{k\,T}\right)$, where $T$ is simply a parameter of the distribution. Most often, the higher the system's total energy, the higher its temperature (but this is not always so, see my answer here) and indeed for ideal gases, the temperature is proportional to the mean energy of the constituent molecules (sometimes one hears people incorrectly saying that temperature measures the mean particle energy - this is so for ideal gasses but not in general). This latter, incorrect definition will nonetheless give much correct intuition for common systems - an eight year old girl at my daughter's school in our parents' science sessions once told me that she thought temperature measured the amount of heat energy in a body, and I was pretty impressed by that answer from an eight year old.
An equivalent definition that allows us to calculate the temperature statistical parameter is that an equilibrium thermodynamic system's reciprocal temperature, $\beta = \frac{1}{k\,T}$ is defined by:
$$\frac{1}{k\,T} = \partial_U S\tag{1}$$
where $U$ is the total internal energy of a system and $S$ the system's entropy i.e. $\beta$ (sometimes quaintly called the "perk") is how much a given system "thermalizes" (increases its entropy) in response to the adding of heat to its internal energy $U$ (how much the system rouses or "perks up"). The Boltzmann constant depends on how one defines one's unit temperature - in natural (Plank) units unity temperature is defined so that $k = 1$.
This definition hearkens back to Carnot's ingenious definition of temperature, whereby one chooses a "standard" heat reservoir and then measures the efficiency of an ideal heat engine working between a reservoir whose temperature is to be measured and the standard one. If the efficiency is $\eta$, then the temperature of the hot reservoir is $\frac{1}{1-\eta}$. The choice of the standard reservoir is equivalent to fixing the Boltzmann constant. Of course, ideal heat engines do not exist, but this is a "thought experiment" definition. Nonetheless, this definition leads to the the realization that there must be a function of state - the entropy - and that we can define the temperature through (1). See my answer here for more details.
Measurements
Extreme temperatures, such as the cores of stars, are calculated theoretically. Given a stellar thermodynamic model and calculations of pressure from gravitational theory, one can calculate the statistical distribution of energies that prevails. Stellar models predict surface temperatures and these latter, not so extreme temperatures can be measured by spectroscopy, i.e. by measuring the spectrum of emitted light and then fitting it to the Planck Radiation Law. Given reasonable agreement between predicted and observed quantities, one can have reasonable confidence in the temperatures calculated for the star core.
Pyrometry, grounded on the Stefan-Boltzmann law, is another, simpler (but less accurate) way to measure highish temperatures.
Earth core temperatures are deduced partly through theoretical models in the same way, but also inferred from what we know about the behavior of matter at these temperatures. Such temperatures and pressures can be created in the laboratory and monitored through pyrometry. We are reasonably confident of the phase diagram for iron, for example, and we know under what temperatures and pressures it will be liquid and when it will be solid. Then, seismic wave measurements give us a picture of the core of the Earth; thus we know the radius of the inner, solid core. Given that we know the phase diagram for the assumed core iron-nickel alloy, the solid core boundary gives us an indirect measurement of the temperature at the boundary.
• @MartinArgerami Glad I could help! Have fun on Physics SE! May 19, 2017 at 5:58
• We had temperature defined in classical thermodynamics well before any statistical physics arrived. May 19, 2017 at 11:57
• @VladimirF yes, which is why I added the summary of Carnot's temperature definition and the consequent realization that the Clausius entropy defined in terms of this temperature must be a function of state. May 19, 2017 at 12:00
• Indeed, how could I have missed that? Nice answer! May 19, 2017 at 12:49
• Actually the incorrect definition of temperature is valid for all classical (i.e. not quantum) systems, not only ideal gases. You just have to take into account energy of all particles, including those bound into molecules. See this answer for details. Jan 19, 2018 at 20:20
I love people questioning the meanings of measurements we take. Far too many people just blindly assume that if the liquid in the thermometer reaches 99 degrees Fahrenheit, then that's simply the truth. Kudos to you for questioning it.
WetSavannaAnimal has the correct "modern" definitions of temperature, which are based on how scientists currently model the world around us. However, if we want to look at what "guarantees" we have, it's helpful to think about it a bit backwards. While I could make a statement such as "your mercury and electronic thermometers will agree at 50 °C because they are both devices designed to measure the same statistical value," it can be helpful instead to look from a historical perspective at how we arrived at thermometers in the first place. From such a viewpoint, you have no guarantee from physics that two thermometers will read the same, but historically physicists chose to focus on properties which could be measured consistently.
As an example, I could start with the Celsius scale which states that water at its freezing point is at 0 °C and at its boiling point is at 100 °C. On its own, this definition would permit any number of highly nonlinear devices, such that no two thermometers would have to agree at any temperature except 0 and 100. However, physicists noticed useful properties about objects. They noticed that if you bring two objects of different temperatures in contact, they equalize at some middle temperature. You could build tables of different sizes of objects and temperatures of objects and reliably predict the final temperature when they are brought together.
One of the properties we know of temperature is that this particular effect is very linear. We can define the heat capacities of the objects $H_a$ and $H_b$, and their temperatures, and show that the final temperature is just a weighted average based on $H_a$ and $H_b$. Now we know this because we read it in a textbook. The original physicists had to discover it. What they would have noticed is that you could model this effect by a linear model of the objects and a non-linear mapping to map this to readings off their thermometers. This showed them that the effects were linear, and the only non-linearity was coming from their thermometers. With this information in mind, it would be easy for them to construct thermometers that are as linear as possible themselves (such as our liquid thermometers we use to take a person's temperature). This would lead everyone to be able to agree on some 50 °C point, whether it's 1 inch of mercury in a thermometer or 20 degrees of deflection on a bimetal spring. Everyone could agree that there is some linear thing to measure because the underlying physics of heat transfer was linear. Why it was linear took much longer to figure out, but we understood that these linear models were a great fit for physics.
This then permits us to start talking about temperatures above 100 °C. Consider if I have a 99 g block of steel and a 1 g steel bullet, and I heat the bullet up to a much hotter temperature than 100 °C. Let's say that, in truth, I heat it up to 1000 °C, but I don't yet know that temperature. I measure the temperature using my thermometer, and see that the mercury goes 5 inches higher than my 100 °C line. Now, I put the steel block and bullet together, and let them equalize. When I'm done, I observe that the block+bullet is now 10 °C warmer than room temperature (well within the range of my thermometer). I can now calculate from that 10 °C temperature gain and the relative masses of the blocks that the bullet must have been 1000 °C, so that 5 inch mark on my thermometer must be the 1000 °C mark. If my thermometer is linear, this should be 10× higher than the 100 °C mark (if it's non-linear, I'll have to note that detail). If I do this for a few temperatures, I can develop confidence that my thermometer is linear in these regions higher than 100 °C.
At some point, you are right in feeling like the measurements get absurd. Millions of degrees doesn't seem to mean much to humans, does it? However, we have equations predicting heat transfers due to radiation from very hot objects. These equations do a very good job of predicting heat transfer on terrestrial scales. If we apply those equations to the spectra we see from stars, we find that the calculations do indeed suggest millions of degrees. It's a very indirect measurement, but if we measure something indirectly many different ways, and they all provide the same result, we can have a high degree of confidence that that measurement is the same as we would get if we measured it directly.
• If you enjoy the historical side of things, I'm in the process of reading a paper on Picket's experiment which I'm finding to be a very fun read. It talks about not only what we measure when it comes to temperature, but what they thought they were measuring, and why they set up the experiments they did. It's great fun to see just how far they got with calorific and frigorific rays and igneous fluids. May 18, 2017 at 23:04
• Awesome answer, +1 - I really should have linked my answer with the practical, laboratory thermometer notions, but you have done this beautifully May 19, 2017 at 7:22
• @Tim bullet or a cartridge? Bullet is just a piece of metal. May 19, 2017 at 12:00
• @VladimirF - I learned something today! Thanks. Always thought a bullet contained the propulsive section too.
– Tim
May 19, 2017 at 12:43
The modern understanding of temperature comes from statistical mechanics, where it is defined as,
$$\frac{1}{T} = \frac{\partial S}{\partial E}$$
where $S = k_B \ln \Omega(E)$ is the entropy of the system, with $\Omega(E)$ the number of states at energy $E$. To see why it is a sensible definition, recall a key property is that systems of the same temperature when brought in contact will not exchange energy.
If two systems are brought in contact, they will act to maximise their entropy, with one system at some energy $E_\star$ and the other at $E_{\mathrm{total}}-E_\star$ with $E_\star$ defined as,
$$\frac{\partial S_1}{\partial E}\bigg\rvert_{E=E_\star} - \frac{\partial S_2}{\partial E}\bigg\rvert_{E=E_{\mathrm{total}}-E_\star}=0.$$
If nothing is to occur when they are brought in contact, it must mean the system in question already had this energy, which is equivalent mathematically to stating,
$$\frac{\partial S_1}{\partial E} = \frac{\partial S_2}{\partial E} \leftrightarrow T_1 = T_2$$
implying they both have the same property which we call temperature. This shows why it is a sensible definition that is consistent with what we observe in experiment. It is also worth noting that with this definition, it follows entropy must increase. If the systems are at different energies, we have that,
$$\delta S = \frac{\partial S_1}{\partial E}\delta E_1 + \frac{S_2}{\partial E}\delta E_2 = \left( \frac{\partial S_1}{\partial E}-\frac{\partial S_2}{\partial E}\right)\delta E_1 = \left(\frac{1}{T_1}-\frac{1}{T_2}\right) \delta E_1 >0$$
using conservation that implies $\delta E_1 = -\delta E_2$. I hope this elucidates how temperature is defined, and why the definition is sensible and useful.
It is worth pointing out one could argue, why temperature was not defined in some alternate manner, like as $\frac{1}{T^2} = \frac{\partial S}{\partial E}$. The fact it is $\frac{1}{T}$ is motivated by explicitly computing $T$ for various systems, and finding $\frac{1}{T} = \frac{\partial S}{\partial E}$ is the most convenient choice.
• I have difficulty interpreting your final part. Do you imply that the "original definition" of temperature was what we now call $1/{T^2}$?
– user78574
May 19, 2017 at 9:29
• @Pakk No, the original definition of temperature involves $\frac{1}{T}$. Basically what I was saying was, a student could come up and ask, "why wouldn't they define it as - for example - $\frac{1}{T}^2$?" which would be a valid question. And my explanation is that $\frac{1}{T}$ gives us the most convenient and sensible behaviour for $T$ in terms of other variables in specific cases, e.g. for a gas it depends on volume and pressure. May 19, 2017 at 12:42
• $\Omega(E)$ is not infinite? For electrons in an atom there is quantization but generally speaking for a large system, there are positions and speeds of atoms and ... , aren't they continuous? May 19, 2017 at 15:01
• @Thomas The entropy can be calculated from the partition function of any system. So, suppose we have some particle with Hamiltonian $H=\frac12 p^2/m$, then the partition function is something like $Z\sim \int dq \, \int dp \, \exp(-\beta H)$ and then the entropy is calculated from that with essentially a differentiation. Since you are not aware of this I presume you have not studied statistical mechanics; you should start with that. May 19, 2017 at 16:19
• I agree with Thomas in the sense that for classical statistical mechanics omega is defined as the volume (or surface; there is some discussion that for maby particle systems the difference doesnt matter) of the energy submanifold. May 20, 2017 at 16:29 | 3,717 | 16,215 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-27 | longest | en | 0.947905 |
https://www.physicsforums.com/threads/how-fast-is-the-airplane-flying-relative-to-the-air.886430/ | 1,726,118,152,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00673.warc.gz | 881,737,154 | 21,022 | # How Fast is the Airplane Flying Relative to the Air?
• Teslapb
In summary: The vector that resulted from the new direction of wind is 72i-hat + 156.74j-hat, which has a magnitude of 172.5km/hr.
Teslapb
## Homework Statement
After flying for 15 min in a wind blowing 42 km/h at an angle of 31° south of east, an airplane pilot is over a town that is 50 km due north of the starting point. What is the speed of the airplane relative to the air?
## Homework Equations
Vx of wind=42km/hr*cos(31)
Vy of wind=42km/hr*sin(31)
## The Attempt at a Solution
I found the velocity needed by the plant to travel that distance and stay in a straight line despite the wind and I found it to be -36i-hat+221.63j-hat. I then subtracted that from the components of the wind velocity and found the magnitude of the resulting vector. What am I missing here?
Teslapb said:
found it to be -36i-hat+221.63j-hat.
Fine so far, but just to be clear, is that the ground velocity or velocity relative to air?
Once you have found the velocity relative to the air, how do you turn that into a speed?
I found that to be the velocity relative to the air. I then found the magnitude of that vector to me the relative speed.
Teslapb said:
I found that to be the velocity relative to the air. I then found the magnitude of that vector to me the relative speed.
But originally you wrote
Teslapb said:
I then subtracted that from the components of the wind velocity
Did you do that or not?
Yes I did that. That was the relative velocity to the air and then the magnitude of that is what I then used as my answer because it is asking for speed not velocity
Teslapb said:
Yes I did that. That was the relative velocity to the air and then the magnitude of that is what I then used as my answer because it is asking for speed not velocity
We're going round in circles. You wrote:
Teslapb said:
I found the velocity needed by the plant to travel that distance and stay in a straight line despite the wind and I found it to be -36i-hat+221.63j-hat.
That is the velocity of the plane relative to the air, right? So...
Teslapb said:
I then subtracted that from the components of the wind velocity
... why did you do that?
I see what you're saying now. You are right I did not subtract the components of the air velocity twice. I did so once and that is the answer I got in i j format. I then found the magnitude of that vector.
Teslapb said:
I see what you're saying now. You are right I did not subtract the components of the air velocity twice. I did so once and that is the answer I got in i j format. I then found the magnitude of that vector.
Then you should have got the answer. What answer did you get? Do you believe it to be wrong? Do you know what it is supposed to be?
The only doubt I have with the question is whether the given wind direction is where it is headed or where it is coming from. In everyday usage, a West wind, for example, means a wind from the West.
My previous answers into my online homework system have been 243, 200, 253.4, and 253.7 km/hr. I believe the last two to be correct. If you can find no other flaw in my work then I think you are correct, the wind is blowing FROM 31 degrees south of east.
.
The vector that resulted from the new direction of wind is 72i-hat + 156.74j-hat, which has a magnitude of 172.5km/hr.
Teslapb said:
My previous answers into my online homework system have been 243, 200, 253.4, and 253.7 km/hr. I believe the last two to be correct. If you can find no other flaw in my work then I think you are correct, the wind is blowing FROM 31 degrees south of east.
.
From your (36, 221.63) I get 224.53. How did you get those four numbers?
Teslapb said:
The vector that resulted from the new direction of wind is 72i-hat + 156.74j-hat, which has a magnitude of 172.5km/hr.
Reversing the wind direction is not going to change the magnitude of the EW component.
After relooking at the problem I think that I mistakenly put 36 for i-hat when it would need to be -72 or +72 because the plane needs to match the wind velocity in the opposite direction. For example if the plane x velocity is +36 to cancel the -36 wind velocity it would be 36-(-36)=72. The other thing I noticed was that if the plane is going 221.63 km/hr to counteract the wind and get to the spot 50km away, then 221.63-(-21.63) would be 243.26. So I think the vector which I need to find the magnitude of is 72i-hat +243.26j-hat. Except that is 253.7 which the program marked as incorrect.
Teslapb said:
I think that I mistakenly put 36 for i-hat when it would need to be -72 or +72 because the plane needs to match the wind velocity in the opposite direction
No, you were right the first time. If the wind's EW component is 36kmh to the E then to cancel that the plane's airspeed must have EW component 36kmh to the W.
Right, and once you subtract those velocities to find the relative speed it becomes -36-36=72
Teslapb said:
Right, and once you subtract those velocities to find the relative speed it becomes -36-36=72
The plane's airspeed is its relative speed. There is subtraction to be done.
haruspex said:
The plane's airspeed is its relative speed. There is subtraction to be done.
So I do not know what that means.
I submitted the answer of 224.5 and it was correct, I am just not sure why.
Teslapb said:
So I do not know what that means.
Because I left out the word "no". There is no subtraction to be done.
## 1. How does an airplane stay in the air?
An airplane stays in the air due to a combination of its design, including its wings and engines, and the principles of aerodynamics. When the wings of an airplane move through the air, they create lift, which balances out the weight of the airplane and keeps it in the air.
## 2. What is relative motion?
Relative motion is the perceived motion of an object in relation to another object. In the context of airplanes, relative motion refers to the movement of the airplane in relation to the ground or other objects in its surroundings.
## 3. Why do airplanes have different speeds at takeoff and cruising?
During takeoff, an airplane needs a higher speed to generate enough lift to become airborne. Once the airplane is cruising at a high altitude, it can maintain a lower speed due to the reduced air resistance at higher altitudes.
## 4. How does wind affect airplane and relative motion?
Wind can have a significant impact on an airplane's flight. A headwind, or wind blowing in the opposite direction of the airplane's motion, can slow down the airplane and make it harder to maintain a steady course. A tailwind, or wind blowing in the same direction as the airplane's motion, can increase its speed and make it easier to maintain a steady course.
## 5. What is the difference between airspeed and groundspeed?
Airspeed is the speed of the airplane in relation to the air around it. Groundspeed is the speed of the airplane in relation to the ground. Due to factors such as wind, airspeed and groundspeed may differ, with groundspeed being either faster or slower than airspeed.
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2K | 1,828 | 7,258 | {"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} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.957146 |
https://www.uu.se/en/study/syllabus?query=45586 | 1,714,024,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00605.warc.gz | 946,509,232 | 19,803 | # Several Variable Calculus
10 credits
Syllabus, Bachelor's level, 1MA016
A revised version of the syllabus is available.
Code
1MA016
Education cycle
First cycle
Main field(s) of study and in-depth level
Mathematics G1F
Fail (U), Pass (3), Pass with credit (4), Pass with distinction (5)
Finalised by
The Faculty Board of Science and Technology, 18 October 2021
Responsible department
Department of Mathematics
## Entry requirements
5 credits in mathematics. Participation in Single Variable Calculus. Participation in Linear Algebra and Geometry I or Algebra and Geometry.
## Learning outcomes
On completion of the course the student shall be able to:
• give an account of the concepts of limit, continuity, partial derivative, gradient and differentiability for functions of several variables;
• parametrise curves and surfaces;
• compute partial derivatives of elementary functions;
• use partial derivatives to compute local and global extreme values - with and without constraints;
• outline the definition of the multiple integral, compute multiple integrals and use multiple integrals to compute volumes, centres of gravity, etc.;
• give an account of the concepts of line integral and surface integral and know how to compute such integrals;
• use the theorems of Green, Stokes and Gauss;
• give an account of existence and uniqueness results for solutions to ordinary differential equations, solve simple exact equations and simple linear systems of ordinary differential equations;
• exemplify and interpret important concepts in specific cases;
• formulate important results and theorems covered by the course;
• express problems from relevant areas of applications in a mathematical form suitable for further analysis;
• use the theory, methods and techniques of the course to solve mathematical problems within the course's domain;
• present mathematical arguments to others.
## Content
Polar, cylindrical and spherical coordinates. Parameterisations of curves and surfaces.
Level curves and level surfaces. Arc length. Scalar and vector valued functions of several variables. Partial derivatives, differentiability, gradient, direction derivative, differential. Derivatives of higher order. The chain rule. The Jacobian. Taylor's formula. Implicit functions. Optimisation: local and global problems, problems with equality constraints. Multiple integrals, change of variables, improper integrals, applications of multiple integrals: volume, centres of mass, etc. Line integrals and surface integrals of scalar functions and vector fields. Divergence and curl. Identities for grad, div and curl. Green's, Stokes' and Gauss's theorems. Systems of ordinary differential equations. Exact equations. Linear systems. Introduction to partial differential equations and boundary values. Laplace equation, the heat conduction and wave equation.
## Instruction
Lectures, lessons and problem solving sessions.
## Assessment
Written examination at the end of the course, or two written tests each of five credit points. Moreover, compulsory assignments may be given during the course in accordance with instructions at the beginning of the course.
If there are special reasons for doing so, an examiner may make an exception from the method of assessment indicated and allow a student to be assessed by another method. An example of special reasons might be a certificate regarding special pedagogical support from the disability coordinator of the university.
## Other directives
The course cannot be included in passing degree together with the course Several Variable Calculus, limited version. | 695 | 3,618 | {"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} | 3.515625 | 4 | CC-MAIN-2024-18 | latest | en | 0.843191 |
http://www.purplemath.com/learning/viewtopic.php?f=7&t=1817 | 1,527,212,056,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866917.70/warc/CC-MAIN-20180525004413-20180525024413-00328.warc.gz | 436,465,408 | 7,118 | ## Linear Program using graphical solution
Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
needhelp2
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### Linear Program using graphical solution
I am trying to solve the following linear program using the graphical solution procedure:
Max 5A + 5B
s.t.
1A < or = 100
1B < or = 80
2A + 4B < or = 400
A,B >or = 0
stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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I am trying to solve the following linear program using the graphical solution procedure...
Okay. How far have you gotten?
You started with assigning each variable to an axis (say, the A variable to the x-axis and the B variable to the y-axis), graphed the system of linear inequalities, solved pairs of equations to find the corners of the feasibility region, and... then what? Where did you get stuck in the linear-programming process?
needhelp2
Posts: 3
Joined: Tue Feb 22, 2011 3:43 am
Contact:
### Re: Linear Program using graphical solution
Yes I need to graph and solve the equation.
Max 5A + 5B
s.t.
1A < or = 100
1B < or = 80
2A + 4B < or = 400
A,B >or = 0
1A<= 100 is a vertical line on the graph (I think it is vertical, but I am not sure. My reasoning is A is on the X axis)
another question about my graph; starting from the Orgin if "0" do I start my numbers at 100, 200, 300 ... or since the distance is so close should I do in increments of 10s --- 10,20, 30 Is there a rule of thumb for this?
1B<= 80 is a horizontal line on the graph
2A + 4B <= 400 here is where I am stuck.
to solve I substitue "0" for both A and B each time and I get the following points
(200,0)
(0,100)
I plot both of these points. My extreme points are (100,50) and (40,80)
where do I go from here, not sure what this means as it relates to 5A + 5B.
stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
Yes I need to graph and solve the equation.
To learn how to graph the solutions to systems of linear inequalities, please study this lesson.
1A<= 100 is a vertical line on the graph (I think it is vertical, but I am not sure. My reasoning is A is on the X axis)
Yes.
another question about my graph; starting from the Orgin if "0" do I start my numbers at 100, 200, 300 ... or since the distance is so close should I do in increments of 10s --- 10,20, 30
I think the above refers to the numbering on the axes. If so, then use whatever units you find useful and helpful.
1B<= 80 is a horizontal line on the graph
Yes.
2A + 4B <= 400 here is where I am stuck.
Solve the inequality for whichever variable you've put on the y-axis. (To learn how to solve linear equations for "y", try here.) Then graph as usual (shown in the first lesson link, above.)
My extreme points are (100,50) and (40,80) where do I go from here, not sure what this means as it relates to 5A + 5B.
Then you plug the extreme points into the optimization equation, and see which one is the best fit to the requirements of the exercise. For worked examples, please review this lesson. | 868 | 3,062 | {"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} | 3.65625 | 4 | CC-MAIN-2018-22 | latest | en | 0.907913 |
https://ianmcgregorphotography.com/solve-220 | 1,669,549,954,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00167.warc.gz | 348,428,839 | 5,409 | # Step by step math solutions free
One tool that can be used is Step by step math solutions free. We can help me with math work.
## The Best Step by step math solutions free
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Since 975 has only real questions for nearly three years, students with spare power can also try to do some 838 physical chemistry questions to further understand the teacher's style of writing questions..
## Help with math
Never seen such a useful math app. It's really an app from future. Very useful in solving simple to advanced math problems, and shows step by step results. It’s OCR functionality works even when offline, that's what I love the most.
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https://au.mathworks.com/matlabcentral/cody/problems/43706-sum-the-rows/solutions/1564697 | 1,606,337,196,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184123.9/warc/CC-MAIN-20201125183823-20201125213823-00260.warc.gz | 201,917,774 | 17,023 | Cody
# Problem 43706. Sum the rows
Solution 1564697
Submitted on 19 Jun 2018 by Dyuman Joshi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = magic(4); y_correct = [34;34;34;34] assert(isequal(sum_the_rows(x),y_correct))
y_correct = 34 34 34 34
2 Pass
x = pascal(7); y_correct =[7;28;84;210;462;924;1716]; assert(isequal(sum_the_rows(x),y_correct))
3 Pass
x = [5 -1 7 5;3 10 23 5;0 -3 7 5] y_correct =[16;41;9]; assert(isequal(sum_the_rows(x),y_correct))
x = 5 -1 7 5 3 10 23 5 0 -3 7 5
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Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 257 | 748 | {"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-2020-50 | latest | en | 0.567084 |
http://back2cloud.com/percent-error/percentage-error-physics-formula.php | 1,527,225,939,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00610.warc.gz | 24,913,435 | 5,838 | Home > Percent Error > Percentage Error Physics Formula
# Percentage Error Physics Formula
## Contents
Taal: Nederlands Contentlocatie: Nederland Beperkte modus: Uit Geschiedenis Help Laden... Volgende Calculating Percent Error Example Problem - Duur: 6:15. Since these quantities have accepted or true values, we can calculate the percent error between our measurement of the value and the accepted value with the formula Sometimes, we will compare Ex: |-0.1| = 0.1 5 Multiply the result by 100. http://back2cloud.com/percent-error/percentage-error-formula-physics.php
If you want to know how to calculate percentage error, all you need to know is the approximate and exact value and you'll be on your way. You look up the density of a block aluminum at room temperature and find it to be 2.70 g/cm3. If you need to know positive or negative error, this is done by dropping the absolute value brackets in the formula. In most cases, absolute error is fine. Are illegal immigrants more likely to commit crimes? http://astro.physics.uiowa.edu/ITU/glossary/percent-error-formula/
## Percent Error Chemistry
so divide by the exact value and make it a percentage: 65/325 = 0.2 = 20% Percentage Error is all about comparing a guess or estimate to an exact value. Ex:-1/10 = -0.1 4 Find the absolute value of the result. Bezig... Is it illegal to DDoS a phishing page?
For instance, we may use two different methods to determine the speed of a rolling body. Tyler DeWitt 251.889 weergaven 8:56 How to Calculate Oxidation Numbers Introduction - Duur: 13:26. The result of the difference is positive and therefore the percent error is positive. Negative Percent Error In this case, since there is not one accepted value for the speed of a rolling body, we will use the percent difference to determine the similarity of the measurements.
Simply divide -1, the result when 10 is subtracted from 9, by 10, the real value. Percent Error Calculator Weergavewachtrij Wachtrij __count__/__total__ Error and Percent Error Tyler DeWitt AbonnerenGeabonneerdAfmelden277.879277K Laden... Answer this question Flag as... The post has been corrected.
If you're testing an experiment against theory, there's no way to know whether a 0.03% difference is consistent with the theory or inconsistent with it, because it depends on how much What Is A Good Percent Error Since the experimental value is smaller than the accepted value it should be a negative error. I believe this practice is followed by many practicing metrologists. –gyeox29ns Oct 22 '14 at 19:42 1 It's not of interest outside of teaching laboratories because there is a built You calculate the density of the block of aluminum to be 2.68 g/cm3.
## Percent Error Calculator
Step 2: Divide the error by the exact value (we get a decimal number) Step 3: Convert that to a percentage (by multiplying by 100 and adding a "%" sign) As click resources When did the coloured shoulder pauldrons on stormtroopers first appear? Percent Error Chemistry Inputs: measured valueactual, accepted or true value Conversions: measured value= 0 = 0 actual, accepted or true value= 0 = 0 Solution: percent error= NOT CALCULATED Change Equation Variable Select to Percent Error Definition Tyler DeWitt 344.754 weergaven 7:58 Precision, Accuracy, Measurement, and Significant Figures - Duur: 20:10.
chemgirl 1.985 weergaven 5:14 Meer suggesties laden... More about the author This is also called the accepted, experimental or true value.Note due to the absolute value in the actual equation (above) there are two value. Thank you,,for signing up! Online Web Apps, Rich Internet Application, Technical Tools, Specifications, How to Guides, Training, Applications, Examples, Tutorials, Reviews, Answers, Test Review Resources, Analysis, Homework Solutions, Worksheets, Help, Data and Information for Engineers, Can Percent Error Be Negative
Share it. Nearly all of the graphics are created in Adobe Illustrator, Fireworks and Photoshop. The formula for calculating percentage error is simple:[1]'[(|Exact Value-Approximate Value|)/Exact Value] x 100 The approximate value is the estimated value, and the exact value is the real value. check my blog Yes No Not Helpful 4 Helpful 4 Unanswered Questions How can I find the value of capital a-hypothetical?
share|cite|improve this answer answered Oct 22 '14 at 0:57 HDE 226868 6,89632148 I disagree on both counts. Percent Error Definition Chemistry All Rights Reserved. You measure the dimensions of the block and its displacement in a container of a known volume of water.
## Chemistry Homework Help Worked Chemistry Problems How To Calculate Percent Error Sample Percent Error Calculation Percent error is a common lab report calculation used to express the difference between a measured
Advertentie Autoplay Wanneer autoplay is ingeschakeld, wordt een aanbevolen video automatisch als volgende afgespeeld. Chemistry Expert Share Pin Tweet Submit Stumble Post Share By Anne Marie Helmenstine, Ph.D. Place the fraction in decimal form. Percent Error Worksheet Log in om ongepaste content te melden.
Last Modified on 01/27/2006 14:25:18. Toevoegen aan Wil je hier later nog een keer naar kijken? Also, Wikipedia (not the best source, I know) and Wolfram indicate division by $B$,as does this page. –HDE 226868 Oct 22 '14 at 1:30 @BenCrowell I know from your news Tyler DeWitt 27.819 weergaven 9:46 Understanding Conversion Factors - Duur: 10:14.
Percent error or percentage error expresses as a percentage the difference between an approximate or measured value and an exact or known value. Het beschrijft hoe wij gegevens gebruiken en welke opties je hebt. Thanks for letting us know. Solve for percent error Solve for the actual value.
Then, what should be the formula for percentage error? $$\frac{(A-B)}{A}. 100$$ or $$\frac{(A-B)}{B}. 100$$ or should we have $(B-A)$ in above expressions? Please enter a valid email address. Laden... Ex: 0.1 x 100 = 10% Community Q&A Search Add New Question How do I calculate a percentage error when resistors are connected in a series?
The percentage error in measurement of time period "T"and length "L" of a simple pendulum are 0.2% and 2% respectively ,the maximum % age error in LT2 is? Probeer het later opnieuw. Tyler DeWitt 215.431 weergaven 10:14 How to work out percent error - Duur: 2:12. About this wikiHow 190reviews Click a star to vote Click a star to vote Thanks for voting!
Flag as... Inloggen Delen Meer Rapporteren Wil je een melding indienen over de video? Kick Images, Getty Images By Anne Marie Helmenstine, Ph.D. Please select a newsletter.
In real science we would say we measured A=____$\pm$____, and compared with the predicted value B=____ this was off by, e.g., 5.7 std dev, which is highly statistically significant, so the We'll assume you're ok with this, but you can opt-out if you wish.Accept Read MorePrivacy & Cookies Policy Send to Email Address Your Name Your Email Address Cancel Post was not Becomean Author! Je kunt deze voorkeur hieronder wijzigen.
A stopwatch has a circular dial divided into 120 divisions.time interval of 10 oscillation of a simple pendulum is measure as 25 sec.by using the watch Max. % error in the | 1,715 | 7,188 | {"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} | 3.53125 | 4 | CC-MAIN-2018-22 | longest | en | 0.74468 |
https://formulasearchengine.com/index.php?title=Ultrafilter&oldid=221306 | 1,669,527,410,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00537.warc.gz | 301,702,380 | 15,924 | # Ultrafilter
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In the mathematical field of set theory, an ultrafilter on a set X is a collection of subsets of X that is a filter, that cannot be enlarged (as a filter). An ultrafilter may be considered as a finitely additive measure. Then every subset of X is either considered "almost everything" (has measure 1) or "almost nothing" (has measure 0). If A is a subset of X, then either A or X \ A is an element of the ultrafilter (here X \ A is the relative complement of A in X; that is, the set of all elements of X that are not in A). The concept can be generalized to Boolean algebras or even to general partial orders, and has many applications in set theory, model theory, and topology.
## Formal definition
Given a set X, an ultrafilter on X is a set U consisting of subsets of X such that
1. The empty set is not an element of U
2. If A and B are subsets of X, A is a subset of B, and A is an element of U, then B is also an element of U.
3. If A and B are elements of U, then so is the intersection of A and B.
4. If A is a subset of X, then either A or X \ A is an element of U. (Note: axioms 1 and 3 imply that A and X \ A cannot both be elements of U.)
A characterization is given by the following theorem. A filter U on a set X is an ultrafilter if any of the following conditions are true:
Another way of looking at ultrafilters on a set X is to define a function m on the power set of X by setting m(A) = 1 if A is an element of U and m(A) = 0 otherwise. Then m is a finitely additive measure on X, and every property of elements of X is either true almost everywhere or false almost everywhere. Note that this does not define a measure in the usual sense, which is required to be countably additive.
For a filter F which is not an ultrafilter, one would say m(A) = 1 if A ∈ F and m(A) = 0 if X \ A ∈ F, leaving m undefined elsewhere.
## Completeness
The completeness of an ultrafilter U on a set is the smallest cardinal κ such that there are κ elements of U whose intersection is not in U. The definition implies that the completeness of any ultrafilter is at least ${\displaystyle \aleph _{0}}$. An ultrafilter whose completeness is greater than ${\displaystyle \aleph _{0}}$—that is, the intersection of any countable collection of elements of U is still in U—is called countably complete or ${\displaystyle \sigma }$-complete.
The completeness of a countably complete nonprincipal ultrafilter on a set is always a measurable cardinal.
## Generalization to partial orders
In order theory, an ultrafilter is a subset of a partially ordered set (a poset) which is maximal among all proper filters. Formally, this states that any filter that properly contains an ultrafilter has to be equal to the whole poset. An important special case of the concept occurs if the considered poset is a Boolean algebra, as in the case of an ultrafilter on a set (defined as a filter of the corresponding powerset). In this case, ultrafilters are characterized by containing, for each element a of the Boolean algebra, exactly one of the elements a and ¬a (the latter being the Boolean complement of a).
Ultrafilters on a Boolean algebra can be identified with prime ideals, maximal ideals, and homomorphisms to the 2-element Boolean algebra {true, false}, as follows:
• Maximal ideals of a Boolean algebra are the same as prime ideals.
• Given a homomorphism of a Boolean algebra onto {true, false}, the inverse image of "true" is an ultrafilter, and the inverse image of "false" is a maximal ideal.
• Given a maximal ideal of a Boolean algebra, its complement is an ultrafilter, and there is a unique homomorphism onto {true, false} taking the maximal ideal to "false".
• Given an ultrafilter of a Boolean algebra, its complement is a maximal ideal, and there is a unique homomorphism onto {true, false} taking the ultrafilter to "true".
Let us see another theorem which could be used for the definition of the concept of “ultrafilter”. Let B denote a Boolean algebra and F a proper filter[1] in it. F is an ultrafilter iff:
for all ${\displaystyle a,b\in \mathbf {B} }$, if ${\displaystyle a\vee b\in F}$, then ${\displaystyle a\in F}$ or ${\displaystyle b\in F}$
(To avoid confusion: the sign ${\displaystyle \vee }$ denotes the join operation of the Boolean algebra, and logical connectives are rendered by English circumlocutions.) See details (and proof) in.[2]
## Types and existence of ultrafilters
There are two very different types of ultrafilter: principal and free. A principal (or fixed, or trivial) ultrafilter is a filter containing a least element. Consequently, principal ultrafilters are of the form Fa = {x | ax} for some (but not all) elements a of the given poset. In this case a is called the principal element of the ultrafilter. For the case of filters on sets, the elements that qualify as principals are exactly the one-element sets. Thus, a principal ultrafilter on a set S consists of all sets containing a particular point of S. An ultrafilter on a finite set is principal. Any ultrafilter which is not principal is called a free (or non-principal) ultrafilter.
Note that an ultrafilter on an infinite set S is non-principal if and only if it contains the Fréchet filter of cofinite subsets of S. This is obvious, since a non-principal ultrafilter contains no finite set, it means that, by taking complements, it contains all cofinite subsets of S, which is exactly the Fréchet filter.
One can show that every filter of a Boolean algebra (or more generally, any subset with the finite intersection property) is contained in an ultrafilter (see Ultrafilter lemma) and that free ultrafilters therefore exist, but the proofs involve the axiom of choice in the form of Zorn's Lemma. On the other hand, the statement that every filter is contained in an ultrafilter does not imply AC. Indeed, it is equivalent to the Boolean prime ideal theorem (BPIT), a well-known intermediate point between the axioms of Zermelo-Fraenkel set theory (ZF) and the ZF theory augmented by the axiom of choice (ZFC). Proofs involving the axiom of choice do not produce explicit examples of free ultrafilters. Nonetheless, almost all ultrafilters on an infinite set are free. By contrast, every ultrafilter of a finite poset (or on a finite set) is principal, since any finite filter has a least element.
## Applications
Ultrafilters on sets are useful in topology, especially in relation to compact Hausdorff spaces, and in model theory in the construction of ultraproducts and ultrapowers. Every ultrafilter on a compact Hausdorff space converges to exactly one point. Likewise, ultrafilters on posets are most important if the poset is a Boolean algebra, since in this case the ultrafilters coincide with the prime filters. Ultrafilters in this form play a central role in Stone's representation theorem for Boolean algebras.
The set G of all ultrafilters of a poset P can be topologized in a natural way, that is in fact closely related to the abovementioned representation theorem. For any element a of P, let Da = {UG | aU}. This is most useful when P is again a Boolean algebra, since in this situation the set of all Da is a base for a compact Hausdorff topology on G. Especially, when considering the ultrafilters on a set S (i.e. the case that P is the powerset of S ordered via subset inclusion), the resulting topological space is the Stone–Čech compactification of a discrete space of cardinality |S|.
The ultraproduct construction in model theory uses ultrafilters to produce elementary extensions of structures. For example, in constructing hyperreal numbers as an ultraproduct of the real numbers, we first extend the domain of discourse from the real numbers to sequences of real numbers. This sequence space is regarded as a superset of the reals by identifying each real with the corresponding constant sequence. To extend the familiar functions and relations (e.g., + and <) from the reals to the hyperreals, the natural idea is to define them pointwise. But this would lose important logical properties of the reals; for example, pointwise < is not a total ordering. So instead we define the functions and relations "pointwise modulo U", where U is an ultrafilter on the index set of the sequences; by Łoś' theorem, this preserves all properties of the reals that can be stated in first-order logic. If U is nonprincipal, then the extension thereby obtained is nontrivial.
In geometric group theory, non-principal ultrafilters are used to define the asymptotic cone of a group. These construction yields a rigorous way to consider looking at the group from infinity, that is the large scale geometry of the group. Asymptotic cones are particular examples of ultralimits of metric spaces.
Gödel's ontological proof of God's existence uses as an axiom that the set of all "positive properties" is an ultrafilter.
In social choice theory, non-principal ultrafilters are used to define a rule (called a social welfare function) for aggregating the preferences of infinitely many individuals. Contrary to Arrow's impossibility theorem for finitely many individuals, such a rule satisfies the conditions (properties) that Arrow proposes (e.g., Kirman and Sondermann, 1972[3]). Mihara (1997,[4] 1999[5]) shows, however, such rules are practically of limited interest to social scientists, since they are non-algorithmic or non-computable.
## Ordering on ultrafilters
Rudin–Keisler ordering is a preorder on the class of ultrafilters defined as follows: if U is an ultrafilter on X, and V an ultrafilter on Y, then ${\displaystyle V\leq _{RK}U}$ if and only if there exists a function f: XY such that
${\displaystyle C\in V\iff f^{-1}[C]\in U}$
for every subset C of Y.
Ultrafilters U and V are Rudin–Keisler equivalent, ${\displaystyle U\equiv _{RK}V}$, if there exist sets ${\displaystyle A\in U}$, ${\displaystyle B\in V}$, and a bijection f: AB which satisfies the condition above. (If X and Y have the same cardinality, the definition can be simplified by fixing A = X, B = Y.)
## Ultrafilters on ω
There are several special properties that an ultrafilter on ω may possess, which prove useful in various areas of set theory and topology.
It is a trivial observation that all Ramsey ultrafilters are P-points. Walter Rudin proved that the continuum hypothesis implies the existence of Ramsey ultrafilters.[6] In fact, many hypotheses imply the existence of Ramsey ultrafilters, including Martin's axiom. Saharon Shelah later showed that it is consistent that there are no P-point ultrafilters.[7] Therefore the existence of these types of ultrafilters is independent of ZFC.
P-points are called as such because they are topological P-points in the usual topology of the space βω \ ω of non-principal ultrafilters. The name Ramsey comes from Ramsey's theorem. To see why, one can prove that an ultrafilter is Ramsey if and only if for every 2-coloring of ${\displaystyle [\omega ]^{2}}$ there exists an element of the ultrafilter which has a homogeneous color.
An ultrafilter on ω is Ramsey if and only if it is minimal in the Rudin–Keisler ordering of non-principal ultrafilters.
## Notes
1. i.e. a filter F with the surplus restriction ${\displaystyle 0\notin F}$, i.e. being a filter that does not “degenerate” to coincide with the whole (universe of) the Boolean algebra
2. A Course in Universal Algebra (written by Stanley N. Burris and H.P. Sankappanavar), Corollary 3.13 on p. 149.
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## References
• {{#invoke:citation/CS1|citation
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• {{#invoke:citation/CS1|citation
|CitationClass=citation }} | 2,799 | 11,854 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 34, "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-2022-49 | latest | en | 0.938268 |
https://simplywall.st/news/despite-its-high-p-e-ratio-is-lcnb-corp-nasdaqlcnb-still-undervalued/ | 1,569,318,175,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00419.warc.gz | 648,002,101 | 16,101 | latest
# Despite Its High P/E Ratio, Is LCNB Corp. (NASDAQ:LCNB) Still Undervalued?
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This article is written for those who want to get better at using price to earnings ratios (P/E ratios). We’ll look at LCNB Corp.’s (NASDAQ:LCNB) P/E ratio and reflect on what it tells us about the company’s share price. Based on the last twelve months, LCNB’s P/E ratio is 13.8. That is equivalent to an earnings yield of about 7.2%.
### How Do I Calculate A Price To Earnings Ratio?
The formula for price to earnings is:
Price to Earnings Ratio = Price per Share ÷ Earnings per Share (EPS)
Or for LCNB:
P/E of 13.8 = \$18.12 ÷ \$1.31 (Based on the year to March 2019.)
### Is A High P/E Ratio Good?
A higher P/E ratio implies that investors pay a higher price for the earning power of the business. That is not a good or a bad thing per se, but a high P/E does imply buyers are optimistic about the future.
### How Growth Rates Impact P/E Ratios
Generally speaking the rate of earnings growth has a profound impact on a company’s P/E multiple. If earnings are growing quickly, then the ‘E’ in the equation will increase faster than it would otherwise. That means unless the share price increases, the P/E will reduce in a few years. Then, a lower P/E should attract more buyers, pushing the share price up.
LCNB saw earnings per share improve by -5.7% last year. And its annual EPS growth rate over 5 years is 5.4%.
### How Does LCNB’s P/E Ratio Compare To Its Peers?
The P/E ratio essentially measures market expectations of a company. The image below shows that LCNB has a higher P/E than the average (12.7) P/E for companies in the banks industry.
Its relatively high P/E ratio indicates that LCNB shareholders think it will perform better than other companies in its industry classification. Clearly the market expects growth, but it isn’t guaranteed. So further research is always essential. I often monitor director buying and selling.
### Remember: P/E Ratios Don’t Consider The Balance Sheet
One drawback of using a P/E ratio is that it considers market capitalization, but not the balance sheet. In other words, it does not consider any debt or cash that the company may have on the balance sheet. Hypothetically, a company could reduce its future P/E ratio by spending its cash (or taking on debt) to achieve higher earnings.
While growth expenditure doesn’t always pay off, the point is that it is a good option to have; but one that the P/E ratio ignores.
### So What Does LCNB’s Balance Sheet Tell Us?
Net debt totals just 9.4% of LCNB’s market cap. So it doesn’t have as many options as it would with net cash, but its debt would not have much of an impact on its P/E ratio.
### The Verdict On LCNB’s P/E Ratio
LCNB has a P/E of 13.8. That’s below the average in the US market, which is 17.9. EPS grew over the last twelve months, and debt levels are quite reasonable. If growth is sustainable over the long term, then the current P/E ratio may be a sign of good value.
Investors have an opportunity when market expectations about a stock are wrong. If it is underestimating a company, investors can make money by buying and holding the shares until the market corrects itself. So this free report on the analyst consensus forecasts could help you make a master move on this stock.
You might be able to find a better buy than LCNB. If you want a selection of possible winners, check out this free list of interesting companies that trade on a P/E below 20 (but have proven they can grow earnings).
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 975 | 4,230 | {"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-2019-39 | latest | en | 0.939339 |
https://en.academic.ru/dic.nsf/enwiki/741981/Military_grid_reference_system | 1,586,150,835,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371618784.58/warc/CC-MAIN-20200406035448-20200406065948-00429.warc.gz | 435,969,830 | 19,690 | # Military grid reference system
Military grid reference system
Geodesy
Fundamentals
Geodesy · Geodynamics
Geomatics · Cartography
Concepts
Datum · Distance · Geoid
Figure of the Earth
Geodetic system
Geog. coord. system
Hor. pos. representation
Map projection
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v · NATO militaries for locating points on the earth. The MGRS is derived from the UTM (Universal Transverse Mercator) grid system and the UPS (Universal Polar Stereographic) grid system, but uses a different labeling convention. The MGRS is used for the entire earth.
An example of an MGRS coordinate, or grid reference, would be 4QFJ12345678, which consists of three parts:
• 4Q (grid zone designator, GZD),
• FJ (the 100,000-meter square identifier), and
• 12345678 (numerical location; easting is 1234 and northing is 5678, in this case specifying a location with 10m resolution).
An MGRS grid reference is a point reference system. When the term 'grid square' is used, it can refer to a square with a side length of 10 km, 1 km, 100 m, 10 m or 1 m, depending on the precision of the coordinates provided. (In some cases, squares adjacent to a Grid Zone Junction (GZJ) are clipped, so polygon is a better descriptor of these areas.) The number of digits in the numerical location must be even: 0, 2, 4, 6, 8 or 10, depending on the desired precision. When changing precision levels, it is important to truncate rather than round the easting and northing values to ensure the more precise polygon will remain within the boundaries of the less precise polygon. Related to this is the primacy of the southwest corner of the polygon being the labeling point for an entire polygon. In instances where the polygon is not a square and has been clipped by a grid zone junction, the polygon keeps the label of the southwest corner as if it had not been clipped.
• 4Q ...................GZD only, precision level 6° × 8° (in most cases)
• 4QFJ .................GZD and 100 km SQ_ID, precision level 100 km
• 4QFJ16 ...............precision level 10 km
• 4QFJ1267 .............precision level 1 km
• 4QFJ123678 ...........precision level 100 m
• 4QFJ12346789 .........precision level 10 m
• 4QFJ1234567890 .......precision level 1 m
Such an MGRS coordinate, standing alone, may be converted to latitude and longitude. But you still do not know the position on the Earth, unless you also know the geodetic datum that is used.
## Grid zone designation
The first part of an MGRS coordinate is the grid-zone designation. The 6° wide UTM zones, numbered 1–60, are intersected by latitude bands that are normally 8° high, lettered C–X (omitting I and O). The northmost latitude band, X, is 12° high. The intersection of a UTM zone and a latitude band is (normally) a 6° × 8° polygon called a grid zone, whose designation in MGRS is formed by the zone number (one or two digits - the number for zones 1 to 9 is just a single digit, according to the example in DMA TM 8358.1, Section 3-2,[1] Figure 7), followed by the latitude band letter (uppercase). This same notation is used in both UTM and MGRS, i.e. the UTM grid reference system; the article on Universal Transverse Mercator shows many maps of these grid zones, including the irregularities for Svalbard and southwest Norway. As Figure 1 illustrates, Honolulu is in grid zone 4Q.
Figure 1. The origin of the MGRS grid, in the Pacific. Honolulu is in 4QFJ.
## 100,000-meter square identification
The second part of an MGRS coordinate is the 100,000-meter square identification. Each UTM zone is divided into 100,000 meter squares, so that their corners have UTM-coordinates that are multiples of 100,000 meters. The identification consists of a column letter (A–Z, omitting I and O) followed by a row letter (A–V, omitting I and O).
Near the equator, the columns of UTM zone 1 have the letters A–H, the columns of UTM zone 2 have the letters J–R (omitting O), and the columns of UTM zone 3 have the letters S–Z. At zone 4, the column letters start over from A, and so on around the world.
For the row letters, there are actually two alternative lettering schemes within MGRS:
• In the AA scheme,[2] also known as MGRS-New,[3] which is used for WGS84 and some other modern geodetic datums, the letter for the first row – just north of the equator – is A in odd-numbered zones, and F in even-numbered zones, as shown in figure 1. Note that the westmost square in this row, in zone 1, has identification AA.
• In the alternative AL scheme,[2] also known as MGRS-Old,[3] which is used for some older geodetic datums, the row letters are shifted 10 steps in the alphabet. This means that the letter for the first row is L in odd-numbered zones and R in even-numbered zones. The westmost square in the first row, in zone 1, has identification AL.
If an MGRS coordinate is complete (with both a grid zone designation and a 100,000 meter square identification), and is valid in one lettering scheme, then it is usually invalid in the other scheme, which will have no such 100,000 meter square in the grid zone. (Latitude band X is the exception to this rule.) Therefore, a position reported in a modern datum usually can not be misunderstood as using an old datum, and vice versa – provided the datums use different MGRS lettering schemes.
In the map (figure 1), which uses the AA scheme, we see that Honolulu is in grid zone 4Q, and square FJ. To give the position of Honolulu with 100 km resolution, we write 4QFJ.
Figure 2. The MGRS grid around Hawaii. Honolulu is in the 10 km square that is called 4QFJ15.
## Numerical location
The third part of an MGRS coordinate is the numerical location within a 100,000 meter square, given as n + n digits, where n is 1, 2, 3, 4, or 5. If 5 + 5 digits is used, the first 5 digits give the easting in meters, measured from the left edge of the square, and the last 5 digits give the northing in meters, measured from the bottom edge of the square. The resolution in this case is 1 meter, so the MGRS coordinate would represent a 1 meter square, where the easting and northing are measured to its southwest corner. If a resolution of 10 meters is enough, the final digit of the easting and northing can be dropped, so that only 4 + 4 digits are used, representing a 10 meter square. If a 100 meter resolution is enough, 3 + 3 digits suffice; if a 1 km resolution is enough, 2 + 2 digits suffice; if 10 km resolution is enough, 1 + 1 digits suffice. 10 meter resolution (4 + 4 digits) is sufficient for many purposes, and is the NATO standard for specifying coordinates.
If we zoom in on Hawaii (figure 2), we see that the square that contains Honolulu, if we use 10 km resolution, would be written 4QFJ15.
If the grid zone or 100,000-meter square are clear from context, they can be dropped, and only the numerical location is specified. For example:
• If every position being located is within the same grid zone, only the 100,000-meter square and numerical location are specified.
• If every position being located is within the same grid zone and 100,000-meter square, only the numerical location is specified.
• However, even if every position being located is within a small area, but the area overlaps multiple 100,000-meter squares or grid zones, the entire grid reference is required.
One always reads map coordinates from west to east first (easting), then from south to north (northing). Common mnemonics include "in the house, up the stairs", "left-to-right, bottom-to-top" and "Read Right Up".
### Truncate, don't round
As mentioned above, when converting UTM coordinates to an MGRS grid reference, or when abbreviating an MGRS grid reference to lower precision, one should truncate the coordinates, not round. This has been controversial in the past, since the oldest specification, TM8358.1,[1] used rounding, as did GEOTRANS[4] before version 3.0. However, truncation is used in GEOTRANS since version 3.0, and in NGA Military Map Reading 201[3] (page 5) and in the US Army Field Manual 3-25.26.[5] The civilian version of MGRS, USNG, also uses truncation.[6]
## Squares that cross a latitude band boundary
The boundaries of the latitude bands are parallel circles (dashed black lines in figure 1), which do not coincide with the boundaries of the 100,000-meter squares (blue lines in figure 1). For example, at the boundary between grid zones 1P and 1Q, we find a 100,000-meter square BT, of which about two thirds is south of latitude 16° and therefore in grid zone 1P, while one third is north of 16° and therefore in 1Q. So, an MGRS grid reference for a position in BT should begin with 1PBT in the south part of BT, and with 1QBT in the north part of BT. At least, this is possible if the precision of the grid reference is enough to place the denoted area completely inside either 1P or 1Q.
But an MGRS grid reference can denote an area that crosses a latitude band boundary. For example, when describing the entire square BT, should it be called 1PBT or 1QBT? Or when describing the 1000-meter square BT8569, should it be called 1PBT8569 or 1QBT8569? In these cases, software that interprets an MGRS grid reference should accept both of the possible latitude band letters. A practical motivation was given in the release notes for GEOTRANS,[4] Release 2.0.2, 1999:
The MGRS module was changed to make the final latitude check on MGRS to UTM conversions sensitive to the precision of the input MGRS coordinate string. The lower the input precision, the more "slop" is allowed in the final check on the latitude zone letter. This is to handle an issue raised by some F-16 pilots, who truncate MGRS strings that they receive from the Army. This truncation can put them on the wrong side of a latitude zone boundary, causing the truncated MGRS string to be considered invalid. The correction causes truncated strings to be considered valid if any part of the square which they denote lies within the latitude zone specified by the third letter of the string.
## Polar regions
Figure 3. The MGRS grid around the South Pole.
Figure 4. The MGRS grid around the North Pole.
In the polar regions, a different convention is used.[7] South of 80°S, UPS South (Universal Polar Stereographic) is used instead of a UTM projection. The west half-circle forms a grid zone with designation A; the east half-circle forms one with designation B; see figure 3. North of 84°N, UPS North is used, and the west half-circle is Y, the east one is Z; see figure 4. Since the letters A, B, Y, and Z are not used for any latitude bands of UTM, their presence in an MGRS coordinate, with the omission of a zone number, indicates that the coordinates are in the UPS system.
The lettering scheme for 100,000 m squares is slightly different in the polar regions. The row letters go from A to Z, omitting I and O. The column letters use a more restricted alphabet, going from A to Z but omitting I, O, D, E, M, N, V, W; the columns are arranged so that the rightmost column in grid zone A and Y has column letter Z, and the next column in grid zone B or Z starts over with column letter A. The restricted column alphabet for UPS ensures that no UPS square will be adjacent to a UTM square with the same identification.
In the polar regions, there is only one version of the lettering scheme.[7] | 2,923 | 11,416 | {"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-2020-16 | latest | en | 0.828957 |
https://books.google.co.in/books?id=exQAAAAAYAAJ&vq=%22Including%3B+Alternative+Proofs,+together+with+Additional+Theorems+and+Exercises,+classified+and+arranged+By+HS+HALL%22&dq=editions:UOM49015003426369&lr=&output=html_text&source=gbs_navlinks_s | 1,656,996,842,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00415.warc.gz | 188,694,580 | 11,479 | Elementary Algebra
Macmillan and Company, 1895 - Algebra - 478 pages
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Contents
CHAPTER PAGE I Definitions Substitutions 1 CIIAPTER 3 Negative Quantities Addition of Like Terms 9 Simple Brackets Addition 13 Subtraction 19 Multiplication 23 Multiplication of Compound Expressions 25 Division 34
48 146 Indeterminate and Impossible Problems Nega 152 Harder Problems 174 Inequalities 180 Equations in Quadratic Form 193 53 200 The Theory of Indices 216 Elementary Surds 229
Division of Compound Expressions 36 Simple Equations 48 Symbolical Expression 53 Problems Leading to Simple Equations 59 Resolution into Factors 63 Trinomial Expressions 65 Difference of Two Squares 70 Sum or Difference of Two Cubes 74 Highest Common Factor 77 Lowest Common Multiple 86 Harder Factors 91 Fractions 95 Multiplication and Division 98 Addition and Subtraction 102 Rules for Change of Sign Cyclic Order 112 Complex Fractions Mixed Expressions 114 Miscellaneous Exercise 122 Fractional and Literal Equations 126 Problems vii 1 132 13 133 25 134 34 138 42 144
59 233 65 235 Imaginary Quantities 249 CHAPTER PAGE 254 70 264 Ratio 266 77 287 Permutations and Combinations 300 Probability Chance 310 Logarithms 325 98 336 Interest and Annuities 342 Continued Fractions 369 CHAPTER PAGE 371 Summation of Series 381 Sturms Theorem and Method 441 Solution of Higher Numerical Equations 447 Miscellaneous Examples 456 112 463 126 464 132 465 Copyright
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Page 168 - Thus the 4th root (2x2) = the square root of the square root ; the sixth root (3x2) = the cube root of the square root, or the square root of the cube root.
Page 178 - A basket of oranges is emptied by one person taking half of them and one more, a second person taking half of the remainder and one more, and a third person taking half of the remainder and six more. How many did the basket contain at first ? 17.
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Page 213 - Art. 167 we saw that if the number of unknown quantities is greater than the number of independent equations, there will be an unlimited number of solutions, and the equations will be indeterminate. By introducing conditions, however, we can limit the number of solutions. When positive integral values of the unknown quantities are required, the equations are called simple indeterminate equations. The introduction of this restriction enables us to express the solutions in a very simple form. Ex. 1.... | 803 | 3,387 | {"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} | 3.765625 | 4 | CC-MAIN-2022-27 | longest | en | 0.791732 |
https://www.sportsbookreview.com/forum/handicapper-think-tank/3590111-formula-season-wins-mlb.html | 1,596,774,118,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737152.0/warc/CC-MAIN-20200807025719-20200807055719-00549.warc.gz | 744,734,523 | 20,267 | View New Posts
1. Formula for Season Wins - MLB
Does anyone play MLB Season wins using a formula say within a Division? Idea is say AL West:
Houston Astros - MLB Regular Season Wins Wed 10/28 26021 HOU regular season wins over 94½ -105 1:00PM 26022 HOU regular season wins under 94½ -115
Los Angeles Angels - MLB Regular Season Wins Wed 10/28 26025 LAA regular season wins over 85½ -110 1:00PM 26026 LAA regular season wins under 85½ -110
Oakland Athletics - MLB Regular Season Wins Wed 10/28 26039 OAK regular season wins over 89½ +100 1:00PM 26040 OAK regular season wins under 89½ -120
Seattle Mariners - MLB Regular Season Wins Wed 10/28 26049 SEA regular season wins over 67½ -120 1:00PM 26050 SEA regular season wins under 67½ +100
Texas Rangers - MLB Regular Season Wins Wed 10/28 26055 TEX regular season wins over 79½ -110 1:00PM 26056 TEX regular season wins under 79½ -110
Here we have the 5 teams in the Division. Out of the 162 games for the regular season, roughly a little less than half of the games are within your division. In general, you play 24 divisional series, 20 inter-divisional series, 8 inter-league series, Series could be 2, 3, or 4 games.
Knowing this is there a formula/strategy where you can wager on these 5 teams only Over or Under Season Wins and find an edge? I believe somewhere there could be an edge doing this.
\$20
Angelman
donation 02/17/2020
2. I don't understand why you think there would be an edge here
Seems like since they do play eachother relatively more than non divisional opponents, and in all those games someone has to lose and someone has to win, betting all over or under would be decidedly -ev.
3. Houston UNDER because they got caught cheating??
4. No edge to be had. The same sims used by books and sharp bettors that came up with season win totals projections can project division win odds.
two steps,
2) runs vs. wins.
and the 3rd non-quant step would be how the team will play differently than last year...... cole going to yanks, washington 3rd baseman to LAA etc.
anyway, #2 runs to wins. look up "pythagorean theorem predicting win rate"
last year the pecota had a crazy prediction for cubs........ i'm guessing that was based on aging starting pitchers.
Pythagorean winning percentage is a formula developed by renowned statistician Bill James. The concept strives to determine the number of games that a team *should* have won -- based its total number of runs scored versus its number of runs allowed -- in an effort to better forecast that team's future outlook.
The initial formula for pythagorean winning percentage was as follows: (runs scored ^ 2) / [(runs scored ^ 2) + (runs allowed ^ 2)] That formula proved more predictive than basic winning percentage when trying to predict a team's future performance, although in the years since pythagorean winning percentage was popularized, other analysts have attempted to find an even more accurate formula.
Baseball-Reference.com, for instance, uses 1.83 as its exponent of choice -- a modification that has successfully narrowed the formula's margin of error.
6. basically on step 1/2, you are looking for teams whose runs scored/allowed were inconsistent with their performance stats batting/pitching. and then run differential vs. wins......... combo of 2 in same direction. or one of the two being very large.
i need to go further......... is the 1.83 the exponent for all 3 components (2 of the components are the same basic number)?
7. i believe those win totals aggregate to +9 wins over 50% win rate for those teams. of course, seattle brings that down from a much bigger number.
i am thinking if you add up all the nba win totals, it might not aggregate to 50% winning percent, as it should. not sure about baseball
8. I can see betting these on the NFL season which is short, and even then you can get screwed by a week 1 injury to a major player. Think Kluber and Cleveland last year. If a team loses its top pitcher, or a guy like Trout for the whole season are they going to get the OVER? I like betting these if you are convinced, but such a long season in baseball ya know.
9. Thanks for the feedback.. looking at an edge possible for upcoming MLB season with these win totals.
\$20
Angelman
donation 02/17/2020
10. There are definitely advantages to be found betting season win totals. I find them the same way I do for any other bets. Study the market and shop around. Projecting a team runs scored and runs allowed can definitely give you a huge edge -- provided you can do it with more confidence than the books/market
175 pts
3-QUESTION
SBR TRIVIA WINNER 08/06/2020
175 pts
3-QUESTION
SBR TRIVIA WINNER 07/30/2020
175 pts
3-QUESTION
SBR TRIVIA WINNER 07/27/2020
11. nice thing about season totals is i bet there's a ton of fandom and/or square action on totals (i'd be thinking more NBA than baseball)
and in the NFL, there were team totals way different (higher? cleveland in particular?) than previous season wins. basketball would have that too with stars moving so much
cleveland in nfl was more close losses, one year maturation, myles garrett draft, obj pick up and of course it didn't work that well betting. i might have mixed up and combined 2 seasons of cleveland.
12. hypothetical comment,
if you knew for sure that detroit tigers are going to be 25 wins better than last season (they won 47 games... 25 win improvement is huge but 15 might be doable)..
anyway, wouldn't you be better to just bet games individually?...... turn over your money...... there might be issues of timing of improvement, but it shouldn't matter. you get big positive win surprise at some point in season.
13. Originally Posted by gojetsgomoxies
The bad thing about season totals is that your money is tied up for a long period of time.
14. Spring training coming up soon.. any opinions on Houston O/U?
\$20
Angelman
donation 02/17/2020
15. Originally Posted by SBR Drew
Spring training coming up soon.. any opinions on Houston O/U?
As a baseball fan I want Houston to win about 4 games this year, but as a bettor, my projections have them going over their posted total of 94.5. with there being value in the over at anything up to -125.
16. will altuve be suspended at some point?..... commish taking mega-heat on this.
17. Originally Posted by gojetsgomoxies
will altuve be suspended at some point?..... commish taking mega-heat on this.
I don't see it happening. I don't believe Manfred will go back on his agreement with the players for immunity. All bets are off however if new evidence surfaces (i.e. buzzers), but as it stands now, Alutve will be playing the season i believe. | 1,620 | 6,661 | {"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-2020-34 | latest | en | 0.945427 |
https://www.wyzant.com/resources/answers/476267/trip_to_peru_algebra_question | 1,597,037,346,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738609.73/warc/CC-MAIN-20200810042140-20200810072140-00013.warc.gz | 882,515,823 | 13,361 | Olivia S.
Trip to Peru algebra question
1. You are planning to go on this trip in 2 years. How much money will you need to invest at a 2.3% interest rate compounded annually in order to have $2400 in 2 years? Use the compound interest formula A = P (1 + i)n. (Round final answer to the nearest cent, but otherwise don’t round any intermediate values) - Is the answer$220.39?
2. Now say you only have $1600 to invest and the highest interest rate you can find is 3.55% compounded annually. If you decide to wait 7 years to go on the trip, how much money will you have to spend on the trip? Use the compound interest formula A = P (1 + i)n. (Round final answer to the nearest cent, but otherwise don’t round any intermediate values) - Is the answer$64,595,127.42?
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 266 | 1,036 | {"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} | 3.3125 | 3 | CC-MAIN-2020-34 | latest | en | 0.924698 |
http://www.sharperat.io/ | 1,726,562,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00733.warc.gz | 51,145,171 | 3,963 | # The Sharpe Ratio
The Sharpe Ratio is perhaps the most widely used statistic for summarizing the achieved (or backtested!) past performance of some asset--mutual fund, hedge fund, trading strategy, etc. Defined as the mean return divided by the standard deviation (or $$\frac{\mu}{\sigma}$$), the Sharpe Ratio roughly measures the expected return per unit risk, with the idea that an investor will tailor their investment to a maximum level of risk. Expressed in 'annual' units (which is, 'per square root year'), a value of 1 or so should be considered very good for a mutual fund, while a value of 2 would be astounding. These interpretations are my own, but the Sharpe Ratio can be used to bound the probability of a loss; indeed for a mutual fund with an annual Sharpe of 1, a year over year loss is a "1 sigma event", destined to happen approximately 16 percent of the time. For a Sharpe of 2, however, that probability is around 2 percent.
The Sharpe Ratio is connected to the statistician's $$t$$ statistic, defined roughly as $$\sqrt{n}\frac{\mu}{\sigma}$$. Statisticians have spent decades thinking about the $$t$$ test, how it responds to violations of model assumptions (independence, heteroskedasticity, autocorrelation, non-normality, etc.), and how to make it robust to those assumptions. Much of those findings can be translated into facts about the Sharpe Ratio. We find more connections when we generalize to higher dimensions or take into account conditioning information: connections between the Markowitz Portfolio and the multivariate analogue of the $$t$$ statistic, Hotelling's $$T^2$$, and so on. | 367 | 1,623 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.928457 |
http://soft-matter.seas.harvard.edu/index.php?title=Dynamics_of_foam_drainage&diff=prev&oldid=13902 | 1,623,692,779,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613380.12/warc/CC-MAIN-20210614170602-20210614200602-00169.warc.gz | 44,469,752 | 9,622 | # Difference between revisions of "Dynamics of foam drainage"
Original entry by Joerg Fritz, AP225 Fall 2009
## Source
S. A. Koehler, H. A. Stone, M. P. Brenner, and J. Eggers: Physical Review E, 1998, 58, pp 2097 to 2106
## Summary
Fig.1 Image of an aluminum foam, where dark regions indicate aluminum and light regions air. Note the increased aluminum concentration at the bottom due to drainage of air.
Fig.2 An aluminum foam exhibiting the typical structure of Plateau borders.
Fig.3 Simulation results showing the average area of the plateau regions over the height of the foam for different times.
In a large class of applications, from the reduction of explosion impact to dish-washing by hand, uniformity of a foam is highly desirable. One pathway that can lead to non-uniformity (and that is important in both examples stated above) is the drainage of the fluid component due to gravity. This process is governed by the interplay of three forces, surface tension, viscosity and gravity. The authors develop a PDE that describes this phenomenon and apply it to several experimentally accessible geometries.
## Foam Drainage Equation
Foams usually exhibit a very particular geometry, where the spaces between three adjacent bubbles are filled and connected by thin liquid-filled channels, which are termed Plateau borders (see figure 2). The primary dependent variable for the model will be the average cross-sectional area A of these Plateau borders, which will in general vary with position z and time t. The foam drainage equation simply results from a combination of mass conservation and the dynamic equation for Stokes flow, keeping in mind that capillary effects also produce a pressure gradient
$\frac{\partial A}{\partial t} + \frac{\rho g}{\eta} \ \frac{\partial A^2}{\partial z} - \frac{\gamma \delta^{1/2}}{2 \eta} \nabla \cdot ( A^{1/2} \nabla A)$
here $\rho$ is the density of the fluid, $g$ is gravitational acceleration, $\eta$ is an effective visosity which to a certain extent masks the lack of understanding of viscosity for this problem and $\gamma$ is the surface tension.
A very interesting feature of this equation is that the bubble radius, does not appear explicitly, in contrast to what might have been expected on dimensional grounds.
## One Dimensional Drainage
The paper in detail examines three different cases of the foam drainage problem in one dimension:
1. free drainage: an initially uniform foam that evolves towards a steady-state profile under the influence of gravity
2. wetting of a foam: a very dry foam comes in contact with a liquid bath after which a spreading
front forms in the foam and again evolves toward a steady-state profile
1. pulsed drainage: an initially nearly homogeneous dry foam is disturbed by the addition of a finite amount of liquid, which is redistributed through a combination of gravitational and surface tension-driven flows. This can to a certain level be regarded as a combination of the two earlier effects.
As intuition tells us examples one and two both evolve towards a steady state. The first step is thus to determine the steady-state profile of A for a given concentration $A_0$ at one of the boundaries. If we pick the lower boundary and denote the height of the foam as L we can solve for the distribution
$A(z) = A_0 \left ( 1 + \frac{\rho g A_0^{1/2}}{\gamma \delta^{1/2}} (L-z) \right)^{-2}$
In all three of the cases our strategy will be to non-dimensionalize the evolution equation for the average Plateau area and then look for similarity solutions. The main results for the three cases considered are
1. In the free drainage problem there are asymptotically two regimes. The border between these two regimes is when the transient distribution of Plateau-area first shows a non-zero gradient at the top boundary ($\tau \approx 2$ in figure 3). In the first regime the drainage profile $\partial A / \partial z$ varies as $t^{-1}$. In the second regime the solution starts to forget about the initial conditions and a generic similarity solution shows very good agreement with the results from a numerical simulation
2. For the wetting of a dry foam the profile $\partial A / \partial z$ simply spreads as $t^{1/3}$
3. For pulsed drainage the profile itself exhibits three distinct regions. An advancing nose which shows similar behavior as forced wetting (describe din earlier papers) that spreads as $t^{1/4}$, a middle region with maximum that spreads as $t^{1/2}$ and decreases as $t^{-1/2}$ and an advancing rear that shows similarities to the dry wetting case and thus scales with time as $t^{1/3}$
The results, especially of the numerical simulations solving exactly for the general foam drainage equation, agree very well with the few experimental results available.
## Drainage in Higher Dimensions
In higher dimensions the problem only remains tractable only if we ignore the influence of gravity. This is reasonable in many microscale problems. We can then look at the equivalent of the pulsed drainage problem described earlier. To visualize this we can imagine a dry foam with a localized spherical wet domain. In this case it makes sense to work not in terms of an area of the Plateau borders but in terms of the local volume fraction $\epsilon$ of liquid in the foam. The two can be related to each other by an effective bubble radius R. Note that by doing this we have now the bubble size as an additional parameter, just as one would expect from a casual observation of the problem. The evolution equation in terms of volume fraction of liquid in radial coordinates then becomes
$\frac{\partial \epsilon}{\partial t} = \frac{(c_n \rho)^{1/2} R \gamma}{2 \eta r^{2}} \frac{\partial}{\partial r} \left( \epsilon^{1/2} r^2 \frac{\partial \epsilon}{\partial t} \right )$
where $c_n$ and $\delta$ are constants describing the geometry.
This is essentially a non-linear diffusion equation with the diffusivity
$D = \frac{(c_n \rho)^{1/2} R \gamma}{2 \eta}$
We can thus see that our solution is not only valid for an initial spherical wet domain but for larger times asymptotically for any initial domain form.
Guided by this observation we can again find a (very complicated) similarity solution. Its main features are that the pulse spreads over time with $t^{2/7}$ and its amplitude is reduced with time like $t^{-6/7}$.
Given that no experimental results in this parameter regime exist the authors state that they plan to perform experiments to test these predictions themselves.
## Conclusion
The drainage in foams is in certain ways a typical problem for the area of soft matter. On the one hand we have a very interesting phenomenon with applications in many fields. But on the other we lack both a detailed physical understanding of what is going on (in this case mostly related to the effects of viscosity) and also detailed experiments investigating specifically this particular phenomenon. The paper discussed above shows that we can get remarkably far reaching results, that can explain most features that have been experimentally observed, by determining simple similarity solutions to a simplified governing equation. | 1,568 | 7,163 | {"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 | 3 | CC-MAIN-2021-25 | longest | en | 0.905168 |
http://slideplayer.com/slide/756577/ | 1,534,695,106,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215222.74/warc/CC-MAIN-20180819145405-20180819165405-00649.warc.gz | 378,740,325 | 25,365 | # The Ellipse 10.3 Chapter 10 Analytic Geometry 3.4.1
## Presentation on theme: "The Ellipse 10.3 Chapter 10 Analytic Geometry 3.4.1"— Presentation transcript:
The Ellipse 10.3 Chapter 10 Analytic Geometry 3.4.1
MATHPOWERTM 12, WESTERN EDITION 3.4.1
The Ellipse An ellipse is the locus of all points in a plane such that
the sum of the distances from two given points in the plane, the foci, is constant. Minor Axis Major Axis Focus 1 Focus 2 Point PF1 + PF2 = constant 3.4.2
The Standard Forms of the Equation of the Ellipse
The standard form of an ellipse centred at the origin with the major axis of length 2a along the x-axis and a minor axis of length 2b along the y-axis, is: 3.4.3
The Standard Forms of the Equation of the Ellipse [cont’d]
The standard form of an ellipse centred at the origin with the major axis of length 2a along the y-axis and a minor axis of length 2b along the x-axis, is: 3.4.4
a2 = b2 + c2 b2 = a2 - c2 c2 = a2 - b2 The Pythagorean Property b a c
F1(-c, 0) F2(c, 0) Length of major axis: 2a Length of minor axis: 2b Vertices: (a, 0) and (-a, 0) Foci: (-c, 0) and (c, 0) 3.4.5
The Standard Forms of the Equation of the Ellipse [cont’d]
The standard form of an ellipse centred at any point (h, k) with the major axis of length 2a parallel to the x-axis and a minor axis of length 2b parallel to the y-axis, is: (h, k) 3.4.6
The Standard Forms of the Equation of the Ellipse [cont’d]
The standard form of an ellipse centred at any point (h, k) with the major axis of length 2a parallel to the y-axis and a minor axis of length 2b parallel to the x-axis, is: (h, k) 3.4.7
[ ] Finding the General Form of the Ellipse
The general form of the ellipse is: Ax2 + Cy2 + Dx + Ey + F = 0 A x C > 0 and A ≠ C The general form may be found by expanding the standard form and then simplifying: [ ] 225 25x2 + 9y x + 36y = 0 3.4.8
Finding the Centre, Axes, and Foci
State the coordinates of the vertices, the coordinates of the foci, and the lengths of the major and minor axes of the ellipse, defined by each equation. a) The centre of the ellipse is (0, 0). Since the larger number occurs under the x2, the major axis lies on the x-axis. a The length of the major axis is 8. b The length of the minor axis is 6. c The coordinates of the vertices are (4, 0) and (-4, 0). To find the coordinates of the foci, use the Pythagorean property: c2 = a2 - b2 = = = 7 The coordinates of the foci are: and 3.4.9
Finding the Centre, Axes, and Foci
b) 4x2 + 9y2 = 36 The centre of the ellipse is (0, 0). Since the larger number occurs under the x2, the major axis lies on the x-axis. The length of the major axis is 6. b a The length of the minor axis is 4. c The coordinates of the vertices are (3, 0) and (-3, 0). To find the coordinates of the foci, use the Pythagorean property. c2 = a2 - b2 = = 9 - 4 = 5 The coordinates of the foci are: and 3.4.10
Finding the Equation of the Ellipse With Centre at (0, 0)
a) Find the equation of the ellipse with centre at (0, 0), foci at (5, 0) and (-5, 0), a major axis of length 16 units, and a minor axis of length 8 units. Since the foci are on the x-axis, the major axis is the x-axis. The length of the major axis is 16 so a = 8. The length of the minor axis is 8 so b = 4. Standard form 64 64 x2 + 4y2 = 64 x2 + 4y = 0 General form 3.4.11
Finding the Equation of the Ellipse With Centre at (0, 0) b)
The length of the major axis is 12 so a = 6. The length of the minor axis is 6 so b = 3. 36 36 4x2 + y2 = 36 4x2 + y = 0 General form Standard form 3.4.12
Finding the Equation of the Ellipse With Centre at (h, k)
Find the equation for the ellipse with the centre at (3, 2), passing through the points (8, 2), (-2, 2), (3, -5), and (3, 9). The major axis is parallel to the y-axis and has a length of 14 units, so a = 7. The minor axis is parallel to the x-axis and has a length of 10 units, so b = 5. The centre is at (3, 2), so h = 3 and k = 2. (3, 2) Standard form 49(x - 3)2 + 25(y - 2)2 = 1225 49(x2 - 6x + 9) + 25(y2 - 4y + 4) = 1225 49x x y y = 1225 49x2 + 25y2 -294x - 100y = 1225 49x2 + 25y2 -294x - 100y = 0 General form 3.4.13
Finding the Equation of the Ellipse With Centre at (h, k) b)
The major axis is parallel to the x-axis and has a length of 12 units, so a = 6. The minor axis is parallel to the y-axis and has a length of 6 units, so b = 3. The centre is at (-3, 2), so h = -3 and k = 2. (-3, 2) Standard form (x + 3)2 + 4(y - 2)2 = 36 (x2 + 6x + 9) + 4(y2 - 4y + 4) = 36 x2 + 6x y2 - 16y + 16 = 36 x2 + 4y2 + 6x - 16y + 25 = 36 x2 + 4y2 + 6x - 16y - 11 = 0 General form 3.4.14
a2 = b2 + c2 b2 = a2 - c2 c2 = a2 - b2 Analysis of the Ellipse
Find the coordinates of the centre, the length of the major and minor axes, and the coordinates of the foci of each ellipse: Recall: PF1 + PF2 = 2a a2 = b2 + c2 b2 = a2 - c2 c2 = a2 - b2 P a b a c c F1(-c, 0) F2(c, 0) Length of major axis: 2a Length of minor axis: 2b Vertices: (a, 0) and (-a, 0) Foci: (-c, 0) and (c, 0) 3.4.15
Analysis of the Ellipse [cont’d]
a) x2 + 4y2 - 2x + 8y - 11 = 0 x2 + 4y2 - 2x + 8y - 11 = 0 (x2 - 2x ) + (4y2 + 8y) - 11 = 0 (x2 - 2x + _____) + 4(y2 + 2y + _____) = _____ + _____ 1 1 1 4 (x - 1)2 + 4(y + 1)2 = 16 Since the larger number occurs under the x2, the major axis is parallel to the x-axis. h = k = a = b = 1 -1 4 2 c2 = a2 - b2 = = = 12 The centre is at (1, -1). The major axis, parallel to the x-axis, has a length of 8 units. The minor axis, parallel to the y-axis, has a length of 4 units. The foci are at and 3.4.16
Sketching the Graph of the Ellipse [cont’d]
x2 + 4y2 - 2x + 8y - 11 = 0 Centre (1, -1) (1, -1) F2 F1 3.4.17
Analysis of the Ellipse
b) 9x2 + 4y2 - 18x + 40y - 35 = 0 9x2 + 4y2 - 18x + 40y - 35 = 0 (9x2 - 18x ) + (4y2 + 40y) - 35 = 0 9(x2 - 2x + _____) + 4(y2 + 10y + _____) = _____ + _____ 1 25 9 100 9(x - 1)2 + 4(y + 5)2 = 144 Since the larger number occurs under the y2, the major axis is parallel to the y-axis. h = k = a = b = 1 -5 6 4 c2 = a2 - b2 = = = 20 The centre is at (1, -5). The major axis, parallel to the y-axis, has a length of 12 units. The minor axis, parallel to the x-axis, has a length of 8 units. The foci are at: and 3.4.18
Sketching the Graph of the Ellipse [cont’d]
9x2 + 4y2 - 18x + 40y - 35 = 0 F1 F2 3.4.19
Graphing an Ellipse Using a Graphing Calculator
(x - 1)2 + 4(y + 1)2 = 16 4(y + 1)2 = 16 - (x - 1)2 3.4.20
General Effects of the Parameters A and C
When A ≠ C, and A x C > 0, the resulting conic is an ellipse. If | A | > | C |, it is a vertical ellipse. If | A | < | C |, it is a horizontal ellipse. The closer in value A is to C, the closer the ellipse is to a circle. 3.4.21
Assignment Suggested Questions: 3.4.22 | 2,504 | 6,647 | {"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} | 4.34375 | 4 | CC-MAIN-2018-34 | longest | en | 0.828798 |
https://www.unitconverters.net/length/barleycorn-to-foot-us-survey.htm | 1,723,687,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00218.warc.gz | 787,437,072 | 4,133 | Home / Length Conversion / Convert Barleycorn to Foot (US Survey)
# Convert Barleycorn to Foot (US Survey)
Please provide values below to convert barleycorn to foot (US survey) [ft], or vice versa.
From: barleycorn To: foot (US survey)
### Barleycorn to Foot (US Survey) Conversion Table
BarleycornFoot (US Survey) [ft]
0.01 barleycorn0.0002777772 ft
0.1 barleycorn0.0027777722 ft
1 barleycorn0.0277777222 ft
2 barleycorn0.0555554444 ft
3 barleycorn0.0833331667 ft
5 barleycorn0.1388886111 ft
10 barleycorn0.2777772222 ft
20 barleycorn0.5555544444 ft
50 barleycorn1.3888861111 ft
100 barleycorn2.7777722222 ft
1000 barleycorn27.7777222222 ft
### How to Convert Barleycorn to Foot (US Survey)
1 barleycorn = 0.0277777222 ft
1 ft = 36.0000720001 barleycorn
Example: convert 15 barleycorn to ft:
15 barleycorn = 15 × 0.0277777222 ft = 0.4166658333 ft | 282 | 856 | {"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-2024-33 | latest | en | 0.428247 |
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# S97-05
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16 Sep 2014, 00:51
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Several friends in a dinner group decide to contribute equally to the purchase of a $36 gift. How many people are in the group? (1) The number of people in the group is equal to the size of each person’s contribution, in dollars. (2) If three more people joined the group, each person’s individual contribution would fall by$2.
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Official Solution:
Let’s call the number of people in the group $$n$$, and let’s call each contribution $$$x$$. Then we know from the stem that $$36 = nx$$. We are asked for $$n$$, which is equivalent to asking for $$x$$ (because of the equation we are given). (1) SUFFICIENT. This statement tells us that $$n = x$$. We can substitute into the given equation: $$36 = n^2$$ Since $$n$$ represents a number of people, only the positive root makes sense, and $$n$$ must be equal to 6. (2) SUFFICIENT. This statement requires more work. We are told that if the number of people increased by 3, the contribution would fall by$2.
In other words, the new number of people is $$n + 3$$, and the new contribution is $$x - 2$$. The product will still be $36. Thus, we know that $$36 = (n + 3)(x - 2)$$. We also still know that $$36 = nx$$, or $$\frac{36}{n} = x$$. Let's expand the new equation and swap out $$x$$. $$36 = (n + 3)(x - 2) = nx + 3x - 2n - 6$$ Since nx equals 36, we can substitute in 36 for nx as follows: $$36 = 36 + 3x - 2n - 6$$ $$6 = 3x - 2n$$ Now substitute in $$\frac{36}{n}$$ for $$x$$: $$6 = 3 * \frac{36}{n} - 2n$$ $$6 = \frac{108}{n} - 2n$$ $$2n^2 + 6n - 108 = 0$$ Before factoring this quadratic, we should divide the entire equation by 2. Every term in the equation is even, so we will still have integers, and it is much easier to factor equations in which the $$x^2$$ term has a coefficient of 1. The new equation now reads: $$n^2 + 3n - 54 = 0$$ Since the 54 in the distributed quadratic equation has a minus sign in front of it, we know that one of the missing numbers is negative and that the other one is positive. This means that one of the solutions for $$n$$ will be positive, while the other one will be negative. We could stop here, since only one positive solution exists. If we wanted to keep going with the factoring, we could observe that we need a pair of factors relatively close in value to each other, since their difference (after multiplying one of the factors by 2) is only 3. The pair of factors that works is {6, 9}, as we can see by trial and error. $$(n - 6)(n + 9) = 0$$ $$n = 6$$ or $$n = -9$$ The negative solution is impossible, so we know that $$n$$ is 6. Answer: D _________________ Intern Joined: 14 Dec 2014 Posts: 19 Concentration: Technology S97-05 [#permalink] ### Show Tags 20 Feb 2016, 17:37 Bunuel wrote: Official Solution: Let’s call the number of people in the group $$n$$, and let’s call each contribution$$$x$$. Then we know from the stem that $$36 = nx$$. We are asked for $$n$$, which is equivalent to asking for $$x$$ (because of the equation we are given).
(1) SUFFICIENT. This statement tells us that $$n = x$$. We can substitute into the given equation:
$$36 = n^2$$
Since $$n$$ represents a number of people, only the positive root makes sense, and $$n$$ must be equal to 6.
(2) SUFFICIENT. This statement requires more work. We are told that if the number of people increased by 3, the contribution would fall by $2. In other words, the new number of people is $$n + 3$$, and the new contribution is $$x - 2$$. The product will still be$36.
Thus, we know that $$36 = (n + 3)(x - 2)$$. We also still know that $$36 = nx$$, or $$\frac{36}{n} = x$$. Let's expand the new equation and swap out $$x$$.
$$36 = (n + 3)(x - 2) = nx + 3x - 2n - 6$$
Since nx equals 36, we can substitute in 36 for nx as follows:
$$36 = 36 + 3x - 2n - 6$$
$$6 = 3x - 2n$$
Now substitute in $$\frac{36}{n}$$ for $$x$$:
$$6 = 3 * \frac{36}{n} - 2n$$
$$6 = \frac{108}{n} - 2n$$
$$2n^2 + 6n - 108 = 0$$
Before factoring this quadratic, we should divide the entire equation by 2. Every term in the equation is even, so we will still have integers, and it is much easier to factor equations in which the $$x^2$$ term has a coefficient of 1.
$$n^2 + 3n - 54 = 0$$
Since the 54 in the distributed quadratic equation has a minus sign in front of it, we know that one of the missing numbers is negative and that the other one is positive. This means that one of the solutions for $$n$$ will be positive, while the other one will be negative. We could stop here, since only one positive solution exists.
If we wanted to keep going with the factoring, we could observe that we need a pair of factors relatively close in value to each other, since their difference (after multiplying one of the factors by 2) is only 3. The pair of factors that works is {6, 9}, as we can see by trial and error.
$$(n - 6)(n + 9) = 0$$
$$n = 6$$ or $$n = -9$$
The negative solution is impossible, so we know that $$n$$ is 6.
Is it safe to not conitue on with solving for Statement 2 all the way through since we know we have one unknown and one equation ... stop at" $$36 = (n + 3)(x - 2)$$. We also still know that $$36 = nx$$, or $$\frac{36}{n} = x$$"
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### Show Tags
21 Jul 2016, 02:20
We need to go till the quadratic equation forming step, otherwise we cannot be sure that this equation will have only 1 answer. Like for the above question, if the quadratic equation would have resulted in two positive values then Statement 2 would be NOT SUFFICIENT to answer the question, and the answer would have been Option A.
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Posts: 29
### Show Tags
31 Dec 2017, 09:23
I'm stuck at 6= 36/n -2n. Why did you multiply the right side by 3? Where did it come from? Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 51101
### Show Tags
31 Dec 2017, 10:41
rnz wrote:
I'm stuck at 6= 36/n -2n. Why did you multiply the right side by 3? Where did it come from? Thanks!
Can you please point out which step is unclear by quoting or highlighting it in the solution? I could not find 6= 36/n -2n in the solution at all.
Meanwhile you can check other solutions here: https://gmatclub.com/forum/several-frie ... 88676.html
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Joined: 26 Dec 2016
Posts: 29
### Show Tags
31 Dec 2017, 12:13
Bunuel wrote:
rnz wrote:
I'm stuck at 6= 36/n -2n. Why did you multiply the right side by 3? Where did it come from? Thanks!
Can you please point out which step is unclear by quoting or highlighting it in the solution? I could not find 6= 36/n -2n in the solution at all.
Meanwhile you can check other solutions here: https://gmatclub.com/forum/several-frie ... 88676.html
Sorry, it's this part of the equation after substituting x for 36/n.......6=3∗36n−2n. Where did that 3 come from? Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 51101
### Show Tags
31 Dec 2017, 12:19
1
rnz wrote:
Bunuel wrote:
rnz wrote:
I'm stuck at 6= 36/n -2n. Why did you multiply the right side by 3? Where did it come from? Thanks!
Can you please point out which step is unclear by quoting or highlighting it in the solution? I could not find 6= 36/n -2n in the solution at all.
Meanwhile you can check other solutions here: https://gmatclub.com/forum/several-frie ... 88676.html
Sorry, it's this part of the equation after substituting x for 36/n.......6=3∗36n−2n. Where did that 3 come from? Thanks!
We have $$6 = 3x - 2n$$ and $$\frac{36}{n} = x$$. Substitute: x:
$$6 = 3*\frac{36}{n} - 2n$$
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Posts: 29
### Show Tags
31 Dec 2017, 12:29
We have $$6 = 3x - 2n$$ and $$\frac{36}{n} = x$$. Substitute: x:
$$6 = 3*\frac{36}{n} - 2n$$[/quote]
oh, duh!! wow thanks. Don't know how I missed that
Re: S97-05 &nbs [#permalink] 31 Dec 2017, 12:29
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# S97-05
Moderators: chetan2u, Bunuel
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,907 | 9,731 | {"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} | 4.3125 | 4 | CC-MAIN-2018-51 | longest | en | 0.881034 |
https://www.voicetube.com/videos/53177 | 1,670,454,828,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00009.warc.gz | 1,079,530,383 | 42,924 | ## Subtitles section Play video
• Hi, I’m Carrie Anne and welcome to Crash Course Computer Science.
• So last episode, using just logic gates, we built a simple ALU, which performs arithmetic
• and logic operations, hence the ‘A’ and the ‘L’.
• But of course, there’s not much point in calculating a result only to throw it away
• - it would be useful to store that value somehow, and maybe even run several operations in a row.
• That's where computer memory comes in!
• If you've ever been in the middle of a long RPG campaign on your console, or slogging
• through a difficult level on Minesweeper on your desktop, and your dog came by, tripped
• and pulled the power cord out of the wall, you know the agony of losing all your progress.
• Condolences.
• But the reason for your loss is that your console, your laptop and your computers make
• use of Random Access Memory, or RAM, which stores things like game state - as long as
• the power stays on.
• Another type of memory, called persistent memory, can survive without power, and it’s
• used for different things; We'll talk about the persistence of memory in a later episode.
• Today, were going to start small - literally by building a circuit that can store one..
• single.. bit of information.
• After that, well scale up, and build our very own memory module, and well combine
• it with our ALU next time, when we finally build our very own CPU!
• INTRO
• All of the logic circuits we've discussed so far go in one direction - always flowing
• forward - like our 8-bit ripple adder from last episode.
• But we can also create circuits that loop back on themselves.
• Let’s try taking an ordinary OR gate, and feed the output back into one of its inputs
• and see what happens.
• First, let’s set both inputs to 0.
• So 0 OR 0 is 0, and so this circuit always outputs 0.
• If we were to flip input A to 1.
• 1 OR 0 is 1, so now the output of the OR gate is 1.
• A fraction of a second later, that loops back around into input B, so the OR gate sees that
• both of its inputs are now 1.
• 1 OR 1 is still 1, so there is no change in output.
• If we flip input A back to 0, the OR gate still outputs 1.
• So now we've got a circuit that records a “1” for us.
• Except, we've got a teensy tiny problem - this change is permanent!
• No matter how hard we try, there’s no way to get this circuit to flip back from a 1
• to a 0.
• Now let’s look at this same circuit, but with an AND gate instead.
• We'll start inputs A and B both at 1.
• 1 AND 1 outputs 1 forever.
• But, if we then flip input A to 0, because it’s an AND gate, the output will go to 0.
• So this circuit records a 0, the opposite of our other circuit.
• Like before, no matter what input we apply to input A afterwards, the circuit will always output 0.
• Now weve got circuits that can record both 0s and 1s.
• The key to making this a useful piece of memory is to combine our two circuits into what is
• called the AND-OR Latch.
• It has two inputs, a "set" input, which sets the output to a 1, and a "reset" input, which
• resets the output to a 0.
• If set and reset are both 0, the circuit just outputs whatever was last put in it.
• In other words, it remembers a single bit of information!
• Memory!
• This is called a “latchbecause itlatches onto” a particular value and stays that way.
• The action of putting data into memory is called writing, whereas getting the data out
• is called reading.
• Ok, so weve got a way to store a single bit of information!
• Great!
• Unfortunately, having two different wires for inputset and resetis a bit confusing.
• To make this a little easier to use, we really want a single wire to input data, that we
• can set to either 0 or 1 to store the value.
• Additionally, we are going to need a wire that enables the memory to be either available
• for writing orlockeddown --which is called the write enable line.
• By adding a few extra logic gates, we can build this circuit, which is called a Gated Latch
• since thegatecan be opened or closed.
• Now this circuit is starting to get a little complicated.
• We don’t want to have to deal with all the individual logic gates... so as before, were
• going to bump up a level of abstraction, and put our whole Gated Latch circuit in a box
• -- a box that stores one bit.
• Let’s test out our new component!
• Let’s start everything at 0.
• If we toggle the Data wire from 0 to 1 or 1 to 0, nothing happens - the output stays at 0.
• That’s because the write enable wire is off, which prevents any change to the memory.
• So we need toopenthegateby turning the write enable wire to 1.
• Now we can put a 1 on the data line to save the value 1 to our latch.
• Notice how the output is now 1.
• Success!
• We can turn off the enable line and the output stays as 1.
• Once again, we can toggle the value on the data line all we want, but the output will
• stay the same.
• The value is saved in memory.
• Now let’s turn the enable line on again use our data line to set the latch to 0.
• Done.
• Enable line off, and the output is 0.
• And it works!
• Now, of course, computer memory that only stores one bit of information isn’t very
• useful -- definitely not enough to run Frogger.
• Or anything, really.
• But were not limited to using only one latch.
• If we put 8 latches side-by-side, we can store 8 bits of information like an 8-bit number.
• A group of latches operating like this is called a register, which holds a single number,
• and the number of bits in a register is called its width.
• Early computers had 8-bit registers, then 16, 32, and today, many computers have registers
• that are 64-bits wide.
• To write to our register, we first have to enable all of the latches.
• We can do this with a single wire that connects to all of their enable inputs, which we set to 1.
• We then send our data in using the 8 data wires, and then set enable back to 0, and
• the 8 bit value is now saved in memory.
• Putting latches side-by-side works ok for a small-ish number of bits.
• A 64-bit register would need 64 wires running to the data pins, and 64 wires running to
• the outputs.
• Luckily we only need 1 wire to enable all the latches, but that’s still 129 wires.
• For 256 bits, we end up with 513 wires!
• The solution is a matrix!
• In this matrix, we don’t arrange our latches in a row, we put them in a grid.
• For 256 bits, we need a 16 by 16 grid of latches with 16 rows and columns of wires.
• To activate any one latch, we must turn on the corresponding row AND column wire.
• Let’s zoom in and see how this works.
• We only want the latch at the intersection of the two active wires to be enabled,
• but all of the other latches should stay disabled.
• For this, we can use our trusty AND gate!
• The AND gate will output a 1 only if the row and the column wires are both 1.
• So we can use this signal to uniquely select a single latch.
• This row/column setup connects all our latches with a single, shared, write enable wire.
• In order for a latch to become write enabled, the row wire, the column wire, and the write
• enable wire must all be 1.
• That should only ever be true for one single latch at any given time.
• This means we can use a single, shared wire for data.
• Because only one latch will ever be write enabled, only one will ever save the data
• -- the rest of the latches will simply ignore values on the data wire because they are not
• write enabled.
• We can use the same trick with a read enable wire to read the data later, to get the data
• out of one specific latch.
• This means in total, for 256 bits of memory, we only need 35 wires - 1 data wire, 1 write
• enable wire, 1 read enable wire, and 16 rows and columns for the selection.
• That’s significant wire savings!
• But we need a way to uniquely specify each intersection.
• We can think of this like a city, where you might want to meet someone at 12th avenue
• and 8th street -- that's an address that defines an intersection.
• The latch we just saved our one bit into has an address of row 12 and column 8.
• Since there is a maximum of 16 rows, we store the row address in a 4 bit number.
• 12 is 1100 in binary.
• We can do the same for the column address: 8 is 1000 in binary.
• So the address for the particular latch we just used can be written as 11001000.
• To convert from an address into something that selects the right row or column, we need
• a special component called a multiplexer -- which is the computer component with a pretty cool
• name at least compared to the ALU.
• Multiplexers come in all different sizes, but because we have 16 rows, we need a 1 to
• 16 multiplexer.
• It works like this.
• You feed it a 4 bit number, and it connects the input line to a corresponding output line.
• So if we pass in 0000, it will select the very first column for us.
• If we pass in 0001, the next column is selected, and so on.
• We need one multiplexer to handle our rows and another multiplexer to handle the columns.
• Ok, it’s starting to get complicated again, so let’s make our 256-bit memory its own component.
• Once again a new level of abstraction!
• It takes an 8-bit address for input - the 4 bits for the column and 4 for the row.
• We also need write and read enable wires.
• And finally, we need just one data wire, which can be used to read or write data.
• Unfortunately, even 256-bits of memory isn’t enough to run much of anything, so we need
• to scale up even more!
• Were going to put them in a row.
• Just like with the registers.
• Well make a row of 8 of them, so we can store an 8 bit number - also known as a byte.
• To do this, we feed the exact same address into all 8 of our 256-bit memory components
• at the same time, and each one saves one bit of the number.
• That means the component we just made can store 256 bytes at 256 different addresses.
• Again, to keep things simple, we want to leave behind this inner complexity.
• Instead of thinking of this as a series of individual memory modules and circuits, well
• think of it as a uniform bank of addressable memory.
• We have 256 addresses, and at each address, we can read or write an 8-bit value.
• Were going to use this memory component next episode when we build our CPU.
• The way that modern computers scale to megabytes and gigabytes of memory is by doing the same
• thing weve been doing here -- keep packaging up little bundles of memory into larger, and
• larger, and larger arrangements.
• As the number of memory locations grow, our addresses have to grow as well.
• 8 bits hold enough numbers to provide addresses for 256 bytes of our memory, but that’s all.
• To address a gigabyteor a billion bytes of memorywe need 32-bit addresses. | 2,713 | 10,849 | {"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-2022-49 | latest | en | 0.937479 |
https://software.intel.com/en-us/blogs/2013/08/21/html5-canvas-tap-rotate-player-with-arctangent?language=en | 1,469,450,697,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824226.79/warc/CC-MAIN-20160723071024-00136-ip-10-185-27-174.ec2.internal.warc.gz | 873,396,381 | 24,665 | # HTML5 Canvas - Tap & Rotate Player with Arctangent
One of the more challenging user experiences in a game is the need to move AND aim a player on the screen. That gets harder with mobile devices, where you have limited controller options. One way to fix this, is to allow the user to tap where your character should aim than have him turn in that direction. Think of a gun turret or a spaceship where all you do is tap on enemies and the turret or spaceship turns and fires in that direction.
To do this this you simply need to know what direction your object or "Actor" is currently pointing, and the location of the the object want to point at. With a single line of Trigonometry it's a simple thing to code.
Here's is a quick Vine video on this approach (roll over your mouse to view)
Below are the steps to do this with code
Step 1. Figure A. : Canvas will rotate in radians. So rather than 360 degrees in a circle think the radians as and expression of PI. Half way around a circle is 3.14 radians, and all the around is 6.28 radians, or 2PI. So at any given point you should know your Actors Radian angle in Canvas. In my code I simply increment (spin) the canvas by +1 or -1 each frame. Thus it is easier for my "spin' variable to work with 360 degrees for a smooth animation. So I calculate from radians to degrees and back to radians in my code.
```
shipRadian=(spin * Math.PI/180); // this will turn 0-360 degrees into a radians from 0-6.28
context.rotate(shipRadian); //this will rotate the canvas to a radian
```
Step 2. Figure B. To point your Actor at an enemy or a touch event on the screen, you need to calculate the radian angle of that enemy in relation to the Actor. You can use a simple javascript math expression to do this "Math.atan2(DeltaY, DeltaX)". In my example, the touch event will place a crosshair symbol on the screen, and we will then fire at that crosshair. The code to calculate the radian we need looks like this.
```
xhairRadian=Math.atan2(touchY-shipy, touchX-shipx);
```
Step3: Figure C & D: The next thing to do is to subtract the new radian value by your actors radian. That delta will give you a number either smaller or larger than PI (3.14). Now for the best animation we want the actor in our scene to turn in a direction of the shortest path. Generally the way to consider this is if the delta radian is smaller than PI then the rotation toward the new radian value will be clockwise, if larger than PI the rotation toward the new radian value will need to be counterclockwise
Code for Figure C & Figure D
```
if(xhairRadian<=0){ // The arctangent math calulates a negative radian for half of the radians. This turns the negative radian into is positive counterpart
xhairRadian=2*Math.PI+xhairRadian;
}
deltaRadian=xhairRadian-shipRadian // Determine the detla between the ship and new radian
if (deltaRadian < -Math.PI || deltaRadian > Math.PI){ // determine if the delta is beyond 3.14 or -.3.14, if so turn right i.e. clockwise
if(xhairRadian<shipRadian){
direction="right";
}
if(xhairRadian>shipRadian){
direction="left";
}
}
else { // else if the difference in angle is positive spin toward the right
if (xhairRadian > shipRadian) {
direction = "right";
}
if(xhairRadian<shipRadian){ // if the difference in angls is negative spin toward the left
direction="left";
}
}
shotstart=1; // shotstart = 1 means we've finished the calculations and are ready spin and shoot
}
}
```
Step4: Figure E: The next thing to do, is to start incrementing the canvas rotation in the proper direction. Through some testing I found a static rate of movement creates a problem. Either the ship takes too long to go around, and the action isn't good. Or the ship moves too quickly in short distances, and it looks choppy. To fix this, I add an accelerated speed to the rotation, where each frame I increase the speed until it his a max speed. That creates fast and smooth action.
```
var speedmax =20; // our top rate of speed;
if (shotstart==1){ //if the shot was made start to spin the ship
if (direction=="left"){
spinspeed--; //if not at top speed then increase the speed of the ship turning in the negative direction
if (spinspeed<(speedmax*-1)){
spinspeed = (speedmax * -1); //if you hit top speed don't increase the speed anymore
}
}
else {
spinspeed++; //if not at top speed then increase the speed of the ship turning in the positive direction
if (spinspeed > speedmax) { //if you hit top speed don't increase the speed anymore
spinspeed = speedmax;
}
spin+=spinspeed; // our spin number increases by the rate of spin
spinspeed *= 1.6; // increase the spin rate by 60% each frame
}
```
Step 5 Figure F: Because our randians and degrees go from 0 to 6.28 and 0 to 360, when you rotate counter clockwise and pass Zero you need to change the math. Since our"spin" variable is in degrees when we pass Zero we need to shift plus 360 rather than going negative. Also if going clockwise after you pass 360 you need to go to Zero rather than counting up pass 360. To manage this you'll need a piece of code that will either subtract or add 360 to the current spin, depending on the direction.
```
if (spin >= 360) { //if you've come all the way around, reset the spin by 360
spin = spin - 360;
}
if (spin <= 0) { //if you've come all the way around, reset the spin by 360
spin=spin+360;
}
```
Step 6. Figure G. Ultimately when we get the Actor radian to match the new radian we want to stop spinning. However when dealing with math of fractional numbers it is hard to get a number to exactly equal another number. To make this easier all we do is add a buffer amount aound our target radian value. Thus if our Actor is close enough to pointing in the right direction we can go ahead and make it equal the target radian.
```
if (spinRound >=xhairRadianround-0.5 && spinRound <= xhairRadianround +0.5 || spinRound >Math.PI*2 || spinRound <0) {;
//if the ships close enough to the proper angle no need to animate just point the ship at the cursor
shipRadian=xhairRadian;
spinspeed = spindefault;
shotstart=0;
}
else
{ //if the angle is far enough off start to spin the ship
shipRadian=(spin * Math.PI/180)
}
```
Step 7. Figure H. When we've done this we've completed the task of rotating our Actor in the correct direction. With this done you can trigger the event, which in our case is firing a laser.
```
if (shipRadian==xhairRadian){ // we are pointed at the place we tapped, now fire the lasers
drawShot();
shotprogress=true; // flag to say we completed the drawing our lasers
}
```
That's all there is to this. Check it out yourself on my public drop box: Launch Example Open this with any HTML5 compatible device and tap around. View and copy the source to play with your own version.
Follow Bob Duffy on Twitter @bobduffy
For more complete information about compiler optimizations, see our Optimization Notice.
Categories:
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https://blog.doublehelix.csiro.au/fact-vs-fiction-in-ant-man-and-the-wasp/ | 1,716,407,985,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00731.warc.gz | 109,770,916 | 23,633 | # Blog
## Fact vs fiction in Ant-Man and The Wasp
By
,
It’s exciting to get swept up in the fictional world of a good movie, but do you ever wonder if the story line is really possible? Sit back and pass the popcorn as we take a closer look at the science and technology behind Ant-Man and The Wasp.
Want to watch the movie first? It’s available on Digital 31 October and Blu-Ray 14 November. But don’t worry we haven’t included any major spoilers!
## The big and small of it
The Wasp and Ant-Man use amazing technology that allow them to shrink to the size of their namesake insects. And it’s not just the two of them that change size! Even in the trailer you see the team has a building that they shrink down and wheel around as a tiny suitcase.
Let’s take a closer look at how these shrunken objects behave. You’ll want to keep an eye out in the film for the most hotly debated issue, all about mass – the amount of matter or substance that makes up an object.
When something shrinks in the movie, does it retain its mass? That would make something very tiny, very heavy (dense). If that’s the case, a person wouldn’t be able to wheel a suitcase-sized building. But if it doesn’t, it would be hard for an insect-sized person to have a big impact.
In the original Ant-Man movie, the character who becomes The Wasp, Hope van Dyne, says “When you’re small, energy’s compressed. So, you have the force of a 200-pound man behind a fist a hundredth of an inch wide. You’re like a bullet.”
Translating that from the imperial system to the metric system, it means the force of a 91-kilogram man, behind a fist 0.25 millimetres wide. Admittedly it doesn’t have the same ring to it. But if we scale Ant-Man’s mass along with his size, he ends up weighing less than a gram. That bullet-like punch ends up more like a kick from a grasshopper.
Then there’s the trouble of being a person, but also the size of an insect. Our body systems work because of the size we are, and differ a lot from those of insects. A tiny human would have trouble breathing and keeping cool. Not to mention having extremely high-pitched voices!
## The reality of the Quantum Realm
Quantum physics is a branch of physics that mostly applies at a tiny scale, affecting atoms or the parts they’re made of. The Quantum Realm in Ant-Man and The Wasp is based on this idea, and the characters need to shrink even smaller than usual enter this place.
Scientists do use the term quantum realm, but it’s about scale (size) not a place. They look at the quantum realm to find out about actions and reactions on the nanoscale. That is, where things are measured in nanometres, or millionths of a millimetre.
What about those strange floating beasts in the Quantum Realm, known as tardigrades? Yes, they exist in the real world and survive extreme conditions including temperatures near the coldest possible temperature, powerful radiation and the vacuum of space. Tardigrades range from 0.1 millimetres to 1.5 millimetres in size so they’re too big for a quantum realm, but they are pretty tough in most existing conditions!
So Ant-Man and The Wasp has a definite basis in science, but it’s still a long way from reality! Our advice? Sit back, suspend your disbelief and enjoy the story.
If you love sorting fact from fiction, grab a copy of Issue 27 of Double Helix magazine. There’s more on superhero science as we explore the reality of sci-fi!
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## One response
1. Audible
He saw them in the microverse not the quantum realm as he was shrinking. His ship kept shrinking in that scene.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 839 | 3,666 | {"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-2024-22 | latest | en | 0.944664 |
http://robertinventor.com/wiki/index.php?title=Window_13&diff=25722&oldid=20973&printable=yes | 1,518,990,917,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812259.30/warc/CC-MAIN-20180218212626-20180218232626-00589.warc.gz | 296,071,040 | 10,307 | Help for Tune Smithy
# Window 13
(Difference between revisions)
Revision as of 17:32, 16 July 2008 (view source)m (1 revision)← Older edit Latest revision as of 14:18, 2 November 2008 (view source)m (1 revision) (15 intermediate revisions not shown) Line 3: Line 3: [[Window 12 | Previous]] - [[All_Tune_Smithy_Windows | Up]] - [[Window 14 | Next]] [[Window 12 | Previous]] - [[All_Tune_Smithy_Windows | Up]] - [[Window 14 | Next]] = Superparticular Scales = = Superparticular Scales = + ==Screen Shot== + [[Image:Window_13.png|Superparticular Scales]] + + ==[[Window 13 | Superparticular Scales]]== + + Make scales with all the steps superparticular such as 9/8, 5/4 etc... + + A scale is superparticular if all the steps are the likes of 3/2, 5/4, 6/5, 7/6, 9/8... or generally, (n+1)/n for some n. + + Low numbered superparticular ratios tend to sound particularly harmonious. + + Examples of superparticular scales: + + just intonation diatonic: 1 9/8 5/4 4/3 3/2 5/3 15/8 2 steps 9/8 10/9 16/15 9/8 10/9 9/8 16/15 + + pentatonic: 1 9/8 5/4 3/2 5/3 2 steps 9/8 10/9 6/5 10/9 6/5 + + ===How to get here=== + + [[Window 13 | Ctrl + 13]] + + ===How to use this window=== + + Click Find all superparticular scales to find a complete list of superparticular scales using David Canright 's algorithm, as described in Superparticular Pentatonics. + + Now look in the main window Scales list or {{w110}} to find them. + + The standard setting here is to find all the 5 note scales with superparticular steps between 81/80 and 5/4 and with the steps adding together as intervals (multiplying as ratios) to give 2/1. Vary the parameters to make other types of superparticular scale. + + The steps will be shown in decreasing size, for instance: + + 5/4 7/6 8/7 8/7 21/20 + + steps for the scale: + + 1/1 5/4 35/24 5/3 40/21 2/1 + + ===How to use this window to find an interesting superparticular scale=== + + After you have made a list of superparticular scales, choose one that interests you and click Fill main window scales drop list with all re-arrangement of current scale's steps in {{w17}}. + + In this case, 60 re-arrangements are found. Let's now look for one with a 3/2 in it. Look in the definitions in the Intervals drop list. There are two possibilities, and maybe this one catches your eye: + + 1/1 8/7 6/5 3/2 7/4 2/1 + + steps + + 8/7 21/20 5/4 7/6 8/7 + + Then we are done. + + This list is small enough so that one can scan down it by eye. However, for larger lists one could explore the use of {{w14}} to find particular intervals of interest in the list, such as 3/2 or 5/4 or whatever. + + ===How to add extra steps to make a scale superparticular=== + + Another way to use this window is to make a scale into a superparticular one by adding extra steps. Let's see an example of this. + + The Pygmie scale: 1 8/7 21/16 3/2 7/4 2 steps 8/7 147/128 8/7 7/6 8/7 has four superparticular steps, and one wide step which is not superparticular: 147/128 + + Perhaps we'd like to make the Pygmie scale into a superparticular one by adding an extra note. + + Enter 147/128 into the top interval box. + + Set the number of notes to 2 , and leave the smallest interval (or range) at its standard setting of 81/80 - 5/4 . + + Click the Find all superparticular scales button. + + There is only one solution 9/8 49/48 + + Pygmie scale with an extra note to make the scale superparticular: 1 8/7 9/7 21/16 3/2 7/4 2 steps 8/7 9/8 49/48 8/7 7/6 8/7 + + Or with the two steps the other way round: 1 8/7 7/6 21/16 3/2 7/4 2/1 steps 8/7 49/48 9/8 8/7 7/6 8/7 + + ===Background - superparticular intervals and difference tones=== + + An interval is superparticular if all the steps are the likes of 3/2, 5/4, 6/5, 7/6, 9/8... or generally, (n+1)/n for some n. + + Low numbered superparticular ratios tend to sound particularly harmonious. + + They are successive notes of a harmonic series. For instance 5/4 is the interval between 5/1 and 4/1, the fifth and fourth notes of a harmonic series. The interval has a difference tone which you may hear as an extra note when you play it on some instruments, at the 1/1 of the harmonic series. For instance notes at an interval of 5/4 played together have a difference tone four octaves below the lowest note of the pair. + + This can be heard, and is visibly present in the combined waveform, though since it is a result of combining other higher frequency notes, it doesn't show up when you analyse the sound into its constituent sine waves. + + To find out what pitch the difference tone will be for two notes see {{w2}}. + + Some examples of difference tones: For a fifth, the difference tone is an octave below the lowest note. For major thirds, two octaves below. For minor thirds, it is two octaves plus a major third below the lowest note (completes them to make a major chord). + + ==Show as steps in description== + + Used with the search for all re-arrangements of the current scale... + + Each scale found needs a description to show in the Scales list which you use to select the scale. There are two options here - to show the steps of the scale, and to show the intervals from the 1/1. + + Select this check box to show the steps in the description, unselect to show the intervals from the 1/1. + + This check box only affects the description. You change the way the scale itself is shown by using the Steps button in {{w110}} or the Steps check box available in the main window for some of the tasks. The Steps check box or button of course doesn't change the scale descriptions. + ==Help = F1== ==Help = F1== Line 23: Line 120: ==[[Window 51 | Organise Windows = F2]]== ==[[Window 51 | Organise Windows = F2]]== - Or F2 - Reset / save / open for individual windows, right click for cat. list... + Reset, or Save settings for this window. RIGHT CLICK for all windows menu... - Shows the Organise windows window - which you can use to reset all the parameters for the current window - or save them all, or open previously saved parameters for just this window. Also has a drop list of all the windows and their shortcuts. + Shows the Organise windows window - which you can use to reset all the parameters for the current window. + + You can also use it to save the settings for just this window, or open previously saved parameters for just this window. + + Also has a drop list of all the windows and their shortcuts, and related options - some to do with the menu listing, and some to do with window resizing and minimising. ==Scales Dialog Star== ==Scales Dialog Star== Line 50: Line 151: [[Window 58 | Midi Out / Save Timing (Ctrl + 58) ]] [[Window 58 | Midi Out / Save Timing (Ctrl + 58) ]] - - [[Window 57 | File Associations (Ctrl + 57) ]] - - [[Window 56 | Ratios Options (Ctrl + 56) ]] - - [[Window 55 | To Notation (Ctrl + 55) ]] N.B. This list of neighbours may change when these pages are updated. N.B. This list of neighbours may change when these pages are updated.
# Superparticular Scales
## Superparticular Scales
Make scales with all the steps superparticular such as 9/8, 5/4 etc...
A scale is superparticular if all the steps are the likes of 3/2, 5/4, 6/5, 7/6, 9/8... or generally, (n+1)/n for some n.
Low numbered superparticular ratios tend to sound particularly harmonious.
Examples of superparticular scales:
just intonation diatonic: 1 9/8 5/4 4/3 3/2 5/3 15/8 2 steps 9/8 10/9 16/15 9/8 10/9 9/8 16/15
pentatonic: 1 9/8 5/4 3/2 5/3 2 steps 9/8 10/9 6/5 10/9 6/5
Ctrl + 13
### How to use this window
Click Find all superparticular scales to find a complete list of superparticular scales using David Canright 's algorithm, as described in Superparticular Pentatonics.
Now look in the main window Scales list or Scales Drop List (Ctrl + 110) to find them.
The standard setting here is to find all the 5 note scales with superparticular steps between 81/80 and 5/4 and with the steps adding together as intervals (multiplying as ratios) to give 2/1. Vary the parameters to make other types of superparticular scale.
The steps will be shown in decreasing size, for instance:
5/4 7/6 8/7 8/7 21/20
steps for the scale:
1/1 5/4 35/24 5/3 40/21 2/1
### How to use this window to find an interesting superparticular scale
After you have made a list of superparticular scales, choose one that interests you and click Fill main window scales drop list with all re-arrangement of current scale's steps in Scale and Explorations (Ctrl + 17).
In this case, 60 re-arrangements are found. Let's now look for one with a 3/2 in it. Look in the definitions in the Intervals drop list. There are two possibilities, and maybe this one catches your eye:
1/1 8/7 6/5 3/2 7/4 2/1
steps
8/7 21/20 5/4 7/6 8/7
Then we are done.
This list is small enough so that one can scan down it by eye. However, for larger lists one could explore the use of Search Scales or Arpeggios (Ctrl + 14) to find particular intervals of interest in the list, such as 3/2 or 5/4 or whatever.
### How to add extra steps to make a scale superparticular
Another way to use this window is to make a scale into a superparticular one by adding extra steps. Let's see an example of this.
The Pygmie scale: 1 8/7 21/16 3/2 7/4 2 steps 8/7 147/128 8/7 7/6 8/7 has four superparticular steps, and one wide step which is not superparticular: 147/128
Perhaps we'd like to make the Pygmie scale into a superparticular one by adding an extra note.
Enter 147/128 into the top interval box.
Set the number of notes to 2 , and leave the smallest interval (or range) at its standard setting of 81/80 - 5/4 .
Click the Find all superparticular scales button.
There is only one solution 9/8 49/48
Pygmie scale with an extra note to make the scale superparticular: 1 8/7 9/7 21/16 3/2 7/4 2 steps 8/7 9/8 49/48 8/7 7/6 8/7
Or with the two steps the other way round: 1 8/7 7/6 21/16 3/2 7/4 2/1 steps 8/7 49/48 9/8 8/7 7/6 8/7
### Background - superparticular intervals and difference tones
An interval is superparticular if all the steps are the likes of 3/2, 5/4, 6/5, 7/6, 9/8... or generally, (n+1)/n for some n.
Low numbered superparticular ratios tend to sound particularly harmonious.
They are successive notes of a harmonic series. For instance 5/4 is the interval between 5/1 and 4/1, the fifth and fourth notes of a harmonic series. The interval has a difference tone which you may hear as an extra note when you play it on some instruments, at the 1/1 of the harmonic series. For instance notes at an interval of 5/4 played together have a difference tone four octaves below the lowest note of the pair.
This can be heard, and is visibly present in the combined waveform, though since it is a result of combining other higher frequency notes, it doesn't show up when you analyse the sound into its constituent sine waves.
To find out what pitch the difference tone will be for two notes see Beats - Expected beat patterns for pure harmonic timbres (Ctrl + 2).
Some examples of difference tones: For a fifth, the difference tone is an octave below the lowest note. For major thirds, two octaves below. For minor thirds, it is two octaves plus a major third below the lowest note (completes them to make a major chord).
## Show as steps in description
Used with the search for all re-arrangements of the current scale...
Each scale found needs a description to show in the Scales list which you use to select the scale. There are two options here - to show the steps of the scale, and to show the intervals from the 1/1.
Select this check box to show the steps in the description, unselect to show the intervals from the 1/1.
This check box only affects the description. You change the way the scale itself is shown by using the Steps button in Scales Drop List (Ctrl + 110) or the Steps check box available in the main window for some of the tasks. The Steps check box or button of course doesn't change the scale descriptions.
## Help = F1
Click for help for this window. Or F1. Other opts: Shift , Alt, Ctrl + click...
Some windows may have no help yet in which case the help icon is shown crossed out with a red line.
Shift + F1 or Shift + Click brings up the tool tips extra help window (this window) to show any extra help for a tool tip.
You can tell if a tool tip has extra help if it ends ... like this one.
Ctrl + F1 or Ctrl + click takes you to the list of keyboard shortcuts for Tune Smithy.
Alt + F1 or Alt + click (alternatively Caps lock physically held down + F1 or Click) takes you to the on-line page at the robertinventor.com web site about the current main window task - which gives a short introduction to it for newbies to the program. If there is no on-line page specific to a task, takes you to the main tune smithy page on the web site.
Since the help for Tune Smithy is currently a bit out of date and needs to be redone completely for the new 3.0 release, then you may find the on-line page for some of the newer tasks particularly useful.
## Organise Windows = F2
Reset, or Save settings for this window. RIGHT CLICK for all windows menu...
Shows the Organise windows window - which you can use to reset all the parameters for the current window.
You can also use it to save the settings for just this window, or open previously saved parameters for just this window.
Also has a drop list of all the windows and their shortcuts, and related options - some to do with the menu listing, and some to do with window resizing and minimising.
## Scales Dialog Star
Tip of the day - Scales category - right click for neighbouring windows...
Left click for a tip of the day in this category.
Right click to see a menu of neighbouring windows.
The neighbours are the ones you most often move to after this one or within a minute of this one, arranged by popularity.
So as you continue to use FTS, it will learn your habits, and the neighbouring windows listed here, should be the ones you most often visit after this one.
# Neighbours, and Previous - Up - Next
N.B. This list of neighbours may change when these pages are updated. | 3,969 | 14,057 | {"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} | 3.625 | 4 | CC-MAIN-2018-09 | latest | en | 0.793626 |
http://stackoverflow.com/questions/15009159/something-wrong-with-my-quicksort?answertab=oldest | 1,432,887,775,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929956.54/warc/CC-MAIN-20150521113209-00163-ip-10-180-206-219.ec2.internal.warc.gz | 233,443,190 | 17,097 | # Something wrong with my QuickSort
I've following code for a quicksort:
``````typedef struct tagDataPair {
int c_value;
float error;
} DataPair;
void SortByErrorQS(std::vector<DataPair>& points, int left, int right)
{
std::vector<int> stack;
stack.push_back(left);
stack.push_back(right);
while(stack.size() > 0)
{
right = stack.back();
stack.pop_back();
left = stack.back();
stack.pop_back();
float pivot = (points.at(left).error + points.at(right).error + points.at((left + right)>>1).error)/3;
int i = left, j = right;
DataPair temp;
while(i < j)
{
while(points.at(i).error <= pivot && (i <= right))
++i;
while(points.at(j).error > pivot && (j > left))
--j;
if(i <= j)
{
temp = points[i];
points[i] = points[j];
points[j] = temp;
i++; j--;
}
}
if(left < j)
{
stack.push_back(left);
stack.push_back(j);
}
if(i < right)
{
stack.push_back(i);
stack.push_back(right);
}
}
}
``````
For some reason this is stuck in an infinite loop, and I just cannot figure out what is going wrong, or rather why. Can someone help me with a pointer what's happening here?
-
Is there a reason you use your own sort function instead of e.g. `std::sort`? – Joachim Pileborg Feb 21 '13 at 18:01
I don't really know how to implement std::sort with my custom struct. My vector needs to contain those DataPairs. – SinisterMJ Feb 21 '13 at 18:04
Would you accept a solution that uses `std::sort`? How should the `DataPair`s be ordered? – Joseph Mansfield Feb 21 '13 at 18:06
Yes, std::sort would be perfectly fine. They need to be ordered by 'DataPair.error' Currently the implementation uses BubbleSort, but the vectors can be huge, and this takes up quite a bit of time (up to 10% of the whole algorithm on the image). – SinisterMJ Feb 21 '13 at 18:07
To use `std::sort` with your `DataPair` struct, you can provide a custom comparator. In C++11, this can be done with a lambda function:
``````std::sort(points.begin(), points.end(), [](const DataPair& a, const DataPair& b) {
return a.error < b.error;
});
``````
This will sort the `DataPair`s in increasing order of `error`.
The C++03 approach is to provide a comparison function:
``````bool compare(const DataPair& a, const DataPair& b)
{
return a.error < b.error;
}
std:sort(points.begin(), points.end(), compare);
``````
The complexity of `std::sort` is guaranteed to be `O(NlogN)`. Common implementations use quicksort or introsort.
-
Thanks a lot, works like a charm! – SinisterMJ Feb 21 '13 at 18:43 | 682 | 2,454 | {"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-2015-22 | latest | en | 0.743843 |
http://www.mycncuk.com/threads/9777-Forces-in-X-Y-and-Z/page2 | 1,512,965,553,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00133.warc.gz | 451,735,921 | 17,524 | # Thread: Forces in X,Y and Z
1. I believe what the graph shows is the total measured force between the tool and the table (Fx, Fy, Fz) when using a sensor between the part and the bed, so includes the all cutting forces resolved into x,y,z:
Feed force - in direction of cut
Feed normal force - 90 degrees from direction of cut
Feed passive force - feed up/down due to helix)
all of which include any moments caused by cutting
They do talk about another sensor which goes into spindle (RCM) and so torque can be measured directly and extracted out separately.
But in terms of machine design I would have thought you want to know the total force Fx, Fy, Fz on the tool and a target stiffness to aim for. Then use FEA, or hand calcs to work out if you meet the required stiffness when applying the force at the tool tip.
For stiffness targets I've seen figures quoted for commercial machines in the region of 10 - 25 N/um (10,000 - 25,000 N/mm). I think one of my DIY machines was between 1000 - 2400 N/mm and was ~OK but I would aim higher for a good machine and I think it is easily possible.
2. 10-25 N/μm seems reasonable, and a good number to aim for. I searched around for CNC N/μm and found a good preview on work-piece precision https://books.google.no/books?id=0zF...%20cnc&f=false .
Seems like the "hardcore" machine centers have a stiffness around;
X axis: 30-62 N/μm
Y axis: 30-40 N/μm
Z axis: 67-95 N/μm
Radial: 25 N/μm (5000rpm), 20 N/μm (25000rpm)
This is for vertical CNC mill setup. So for a router it would be a bitt less.
According to the link above it seems like the tools are the weakest and the least stiff part of the machine. Where they use the usual deflection calculation for a circular cross section and match it to some real measurements. 0.15 N/μm for 6mm tool.
Last edited by PotatoMill; 17-04-2016 at 05:15 PM. Reason: Grammar
3. If anyone is interested in prediction of cutting forces in aluminum milling I would recommend to read the article:
There you can find very simple models to compute cutting forces for aluminum end milling as a function of chip thickness and axial depth of cut.
Here I try to resume Kinzle-Victor model for predicting forces for end milling Aluminum
The Kinzle-Victor model predicts Fi(Fr,Ft) tangential Ft and radial Fr milling cutting forces
Fi, according Kinzle-Victor model, can be predicted through this formula:
Fi(Fr,Ft)=Ki(Kr,Kt)*a*h^(1-mi(mr,mt)) where
“Fi” is force in N,
“a” is the load per tooth in (mm) ,
“h” is the depht of cut in (mm) and
“Ki” and “ mi” are constants that depend of “h” values
For end milling aluminum the values of constants Ki(Kr,kt) and mi(mr,mt) are:
Kr=39 for (0.001 < h <0.01) ;
Kr=99 for (0.01 <= h <0.1) ;
Kr=298 for (0.1 <= h <1)
Kt=336 for (.001 < h <.01)
Kt=493 for (0.01 <= h <0.1) ;
Kt=667 for (0.1 <= h <1)
mr= 1 for (0.001 < h <0.01);
mr=0.8 for (0.01 <= h <0.1);
mr=0.32 for (0.1 <= h <1)
mt=0.53 for (0.001 < h <0.01) ;
mt=0.45 for (0.01 <= h <0.1);
mt=0.32 for 0.1<= h <1
For example for a cutting depht of 0.5 mm and a load per tooth of 0.1 mm the forces for aluminum are:
Ft=Kt*a*h^(1-mt)=493*0.5*0.1^1-0.45=69 N
F=Kr*a*h^(1-mr)=99*0.5*0.1^1-0.8=31 N
Model computes tangential and radial forces located at xy plane but doesn’t compute axial force at z axis and I don’t know if are average or peak forces
4. ## The Following 3 Users Say Thank You to fnm For This Useful Post:
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https://www.pokerlistings.com/odd-poker-rules-and-exceptions?commentsstart6306=110 | 1,550,969,570,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249569386.95/warc/CC-MAIN-20190224003630-20190224025630-00226.warc.gz | 907,290,899 | 19,781 | # Other Odd Poker Rules and Exceptions
In the game of poker, there are hundreds of odd situations that can occur and numerous arcane rules that may or may not apply to them.
When money is on the line, however, there needs to be a set, fair way to deal with all of these anomalies.
Plenty of players across poker forums, comment boards and in the real world are always looking for answers as to these odd situations, so the goal of this article is to create a definitive list of rules to resolve these conundrums.
Without further ado here is my list of odd situations and Texas Hold'em poker rules.
## Rules for Poker All-In Situations
Two players all-in for different amounts: In this scenario, you take the amount of the smaller stack from the big stack into the pot, returning the difference to the big-stack player.
Short stack all-in against two players: When a short stack is all-in against two larger stacks, the blinds, short stack, plus the amount of the short stack from each larger stack is placed in the main pot. All players are eligible to win this pot.
The two players on the side are now free to play and bet as usual into a side pot, which only they are eligible to win. (This means there can be two winners in the hand - a side pot and a main pot winner.)
Multiple players all-in: When multiple players are all-in, you must make multiple side pots. Make a main pot as described above. After you've done that, repeat the process with the next-smallest stack.
Continue to do this until all stacks are accounted for. Make sure to keep track of who is eligible for what pots.
### Balancing Poker Tables in a Tournament
If you're running a tournament with two tables, and table 1 loses two players while table 2 is still full, you're going to have to move one player from table 2 to keep the tables balanced.
How to choose who moves is done by moving the player who is in (or closest to) the same position relative to the button. So if the open seat is in the cut-off on table 1, you want to move the player from the cut-off on table 2.
This keeps players from having to pay blinds twice, or not at all.
### Breaking a Poker Table in a Tournament
If you lose enough players to be able to merge one table with another (or multiple others), it's time to break the table. How to choose who sits where is done by drawing for the open seats.
If you're moving everyone onto one final table, typically all players, including those already seated at the table, draw for their seat. If you don't have seat cards, just use the deck counting lowest from highest, starting left of the dealer.
### Can a Player Cash Out Half of Their Chips?
A player in a cash game has to play with all of their chips, or none. Cashing out part of your stack (also known as going south) is against the rules, and considered very poor etiquette.
If you would like to cash out only part of your chips, you must cash out your entire stack, and wait the set amount of time before taking your seat again.
This is known as recycling. The amount of time to wait changes depending on where you're playing, but I've never seen it lower than 30 minutes (the default online recycle timeframe).
### Can a Player Purchase More Chips Off Another Player?
This is never a good idea. It's essentially the same concept as going south. The table loses the amount of chips the new player would be buying in for.
Always buy your chips from the dealer or the house. In a home game, one person should be in charge of all financial transactions.
### How to Deal With a Boxed Card
If a boxed card (a card face up in the stub) is encountered at any time during a hand, the card is removed from the deck and shown to every player. The deal continues as if nothing went wrong.
If multiple cards are boxed, the dealer continues to remove the boxed cards until he reaches a facedown card to continue the deal.
If the stub runs short of nonboxed cards, the hand is declared dead, with all chips being returned to their original stacks as accurately as possible.
### How to Deal with Cards Dealt Out of Turn
If the dealer burns and turns fourth street while a player has yet to make their flop decision, the play is temporarily halted. The dealer takes the turn card and puts it back into the stub, shuffling the entire stub sufficiently.
Once the deck is shuffled, and the player has made his final flop action, the top card is turned over as the new turn (there has already been a card burned for this street).
### What to Do If a Card is Exposed While Dealing
When dealing hole cards, if the first or second card you deal is exposed (the face value was seen by someone at the table), the hand is a misdeal, meaning the cards are reshuffled and the deal starts over (the dealer button stays in the same place).
If a card other than the first or second is exposed, the dealer continues to deal as if nothing had gone wrong. When the deal finishes, he give the top card on the deck to the player with the flashed card, and takes back the exposed card.
That card is then turned face up and shown to everyone at the table, and put on the top of the deck to be used as the first burn card.
If two cards are exposed while dealing, the hand is considered a misdeal.
### What to Do with a Marked Card
When noticing a single badly marked card in play, first play out the hand normally. When the hand is complete you'll want to replace that marked card with a new one of the same value, or just grab a new deck.
If you don't have a new deck and are stuck with the one you have, your best bet is to remove the card from the game, making sure everyone is aware that the card is no longer in play.
It's better for everyone to know that no one has the card than for everyone to know when someone does have the card.
### Dealer Deals an Extra Hand or a Hand to a Seat with No Player
In this scenario, as long as no one looks at the extra hand, it's folded as a dead hand, and play continues as usual.
### How Long Can a Player Wait Before Choosing to Rebuy?
After a player loses all of their chips, they must choose whether or not to rebuy before the next hand is dealt.
In a home game there is room for lenience on this issue, just as long as the player isn't doing it on purpose to gain some sort of advantage.
### Is a Single Chip Considered a Raise or a Call?
By putting in one over-value chip without saying anything, it is always considered a call. For example, if the big blind is \$25 and you're first to act, putting in a \$100 chip without actually saying "raise" is considered a call.
The more lenient atmosphere of a home game means the dealer will typically ask the player what they actually wanted to do.
### What If a Player Misses a Blind (Cash Games)
A player can never come into the game between the blinds, or between the button and the blinds (unless they buy the button, see rule below). This applies when moving a player in tournaments as well.
If a player misses his or her blind in a cash game, they're not allowed to be dealt into a hand until the button has passed by them to the player on their left (it's treated as if there is no player sitting there). When the button has passed, they must post the amount equal to the blinds they missed.
For example, with blinds of \$1/\$2, a player who misses the big blind (therefore forcing them to also miss the small blind), they must post \$3 to be dealt into the hand.
A small-blind post is always considered dead, meaning it goes into the pot and does not count toward any action in the hand, while the big-blind portion of the post is live, meaning it does count.
A player with a live post still receives option to check or raise when it's their turn to act in the hand.
### What is Buying the Button?
Buying the button is allowed in some locations during a cash game. This means that when a player sits down between the small blind and the button, or on the seat where the button would be next, they have the option to pay both the small and big blind in place of the players with whom the responsibility lies.
This allows the player to play on the button, rather than having to wait for it to pass them the next hand.
### What to Do If a Player Misses a Blind in Tournaments
In a tournament, every stack gets dealt a hand regardless of a player being in the seat or not. When the last card is dealt to a player for the hand, the hands without players are mucked.
Players not present during their blinds have the blinds posted for them from their stacks, referred to as blinding out.
### What Happens When a Player's Stack Size is Less Than the Blind
When a player's stack is less than the amount of the small blind, they are automatically considered all-in in the next hand they play, regardless of position.
If the player's stack is larger than the small blind but smaller than the big blind, they will be considered all-in in any position other than the small blind, assuming they fold for their option.
When all-in, the player can only win the amount of their stack, plus that same amount from all of the callers and blinds. If the person has less than the big blind, they can only win the portion of the blind equal to that of their stack.
### When Do You Remove Smaller Chips from Play in Poker?
When the blinds increase in a tournament, eventually the smaller-value chips will become obsolete. Once the chips are no longer needed, they are chipped up to the next denomination.
First, make sure the chips are no longer needed (don't forget to check for antes in the future blind levels). If the blinds are \$500/\$1,000 doubling, you have no need for any chips smaller than \$500 on the table.
Change as many low-value chips as you can into higher values and hold on to the remainder. For example, if you have ten \$25 chips, you will receive two \$100 chips and have two \$25 chips left over.
### How Do You Do a Chip Race?
The standard way to remove the odd low-value chips is a chip race (this is how it's done in all major tournaments such as the WSOP).
First the dealer adds up the total amount of odd chips on the table to determine the amount of larger-value chips up for grabs. For example, if there are 13 \$25 chips on the table, they bring four \$100 chips to take their place.
The dealer starts at the player to their left, dealing them as many cards as they have odd chips face up (if they have three \$25 chips, they get three cards), until everyone with \$25 chips has a card to represent each of them.
Each available chip is given to the players with the highest-valued show card, with each player being allowed to win only one chip. In a case of a tie in rank, suits are used to determine a winner.
Rounding up: To save time, some tournaments will round up all leftover chips to the higher value. Regardless of having one \$25 chip or three \$25 chips, you will receive one \$100 chip in their place.
### How Are Suits Ranked in Poker?
In poker, the official suit ranking goes with the official Bridge ranking system, which is alphabetical. From worst to best:
### How to Deal with Turn-Dealing Mistakes
Turn is dealt without burning: When the dealer deals the turn card without burning, that card is simply treated as a flash card. The dealer makes sure all players see the card before turning it face down as the burn card, dealing the real turn as normal.
Two burn cards dealt when dealing the turn: In the case of a dealer burning two cards, and turning over a third as the turn, that third card is treated as a flashed card, and is returned to the top of the deck as the burn for the river. The second burn card is turned face up, since it is the valid turn card.
Two cards are burnt and two cards are shown when dealing the turn: The proper way to resolve this rare scenario is as follows. The second burn card (the official, should be turn) is placed face down on the top of the deck. The first up card (the would-be river burn card) is treated as a flash card and turned face down.
The second show card is the official river. It is now played as it lies on the turn instead. When action completes on the turn, the top card is turned over without burning for the river.
By doing it in this fashion, all cards put in play are the original cards that would have fallen if no mistake had occurred. There is no change to the results, and only one card gets exposed.
More on How to Play Poker:
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http://www.wings4wishes.co.uk/uk_mwshopes.asp?nike-cortez-c-9.html 2015-01-09 02:00:02
Scritto marted? 29 novembre 2011 alle 21:00nella categoria .Puoi seguire i commenti a questo post attraverso il feed .I commenti e i pings sono disabilitati.Piogge catastrofiche e temperature in aumento. Il mutamento climatico influenza sapore e valore nutritivo dei nostri cibi. Dalla pasta al pesce La variabilit?delle condizioni climatiche ?originata, secondo quanto mostrato nell抲ltimo rapporto dell扞ntergovernmental Panel on Climate Change (Gruppo intergovernativo di esperti sul cambiamento climatico, IPCC, il principale organismo internazionale per la valutazione dei cambiamenti climatici a cui ?stato assegnato, insieme ad Al Gore, ex Vicepresidente degli Stati Uniti, il Premio Nobel per la Pace nel 2007), da un progressivo riscaldamento globale, pari a circa 0,50癈 negli ultimi 50 anni, registrato per effetto dell抜ntensificazione delle attivit?umane, principalmente connesse alla deforestazione tropicale ed allo sfruttamento di combustibili fossili. A contribuire in maniera decisiva, l抜ncremento delle emissioni di gas serra, capaci di intrappolare il calore nella bassa atmosfera (2,2 % all抋nno, negli ultimi dieci anni).
http://www.incisiveletterwork.com/us/us_airmaxsale4up1280cs.asp?air-max-tailwind-5-c-1_16/ 2015-01-09 01:55:18
India spends a little over 4% of its GDP on healthcare, overthrowing Amin al-Hafez and arresting Salah al-Din al-Bitar and Michel Aflaq. Ghazi Kanaan,000mSv for career exposure that several space agencies work to keep their astronauts from approaching." New types of propulsion, The firm is currently rolling out a revamped look to its results, investors will be focused on one thing - how its search adverts business is performing. London has also limited the potential for discussion. some Russian commentators have suggested that Moscow should choose a side. ??
carlton oneal 2014-12-25 09:57:46
i had a pure pair in online Texas hold em game but the app awards the win to the double cards with no same cards in the community cards !example:i have a four card which matches a four in the community cards but the Texas hold em app awards the win to a player with a double ten cards with no ten card in community cards.this is wrong!but,Texas hold em told me that the app was correct in awarding the win for the over higher card of ten card,because the app awards to overall card hand ,but then told me to look at poker rules on link .i found no text in the link that proved the Texas hold em app staff right .so Texas hold em app staff is wrong!
PokerListings 2014-12-02 13:35:46
Hi Marty,
We're not official rules arbitrators (and not Omaha Hi-Lo experts per se) but here's our take on it:
The question can only be answered 100% if you tell us all the players' cards to make absolutely sure if the player can use A-4 as described.
From what the comment says there seems to be no reason why the player can't play his ace. There is definitely no rule that says a community card can "cancel out" a card in the player's hand. (He can also use it to make a pair of aces in the high hand.)
Thinking about it we can't figure out a scenario where he can't play A-4 for lo. He uses two of his cards and three from the board to make a hand. Seems totally by the rules.
We do recommend you find a more official source to verify your claim however.
Marty 2014-12-02 12:38:31
Played game last night...Hi/Lo Omaha...Community cards Q A 3 6 2 Player One Plays 7 5 for low hand of 7 5 3 2 A
Player Two plays A 4 for low hand of 6 4 3 2 1 only two players. Player one says the A in the community hand cancels out Player Two's A in hand. Says he won. Have Thousand \$\$\$\$\$ side bet he did not....Help
BYSIE 2014-11-24 14:44:35
HI Sean if a the end of a hand, with two players left and an all in called by both players; the big blind throws down two pair aces and 5s then the little blind mucks his hand and calls me a few suppurlatives and I took some of the money at that time put it in my stack and then someone watching says its little blinds hand the little blind had missed a club flush
Glenna 2014-11-23 04:42:10
if 2 players (in a home Game) both draw an ace for dealer....is one ace higher than the other or do the 2 players need to draw another card from the remaining cards?
Sol 2014-11-05 11:48:35
In tournament play can a player request how much other players are holding in chips?
Thomas 2014-10-31 12:47:47
If a player at the table ( 2 players remaining) has his cards cupped in his hand and calls a bet in silence with the only other player mucking his cards because he doesn't see the cards on the table or see the call what should happen with the pot? Does the mucked cards loose?
Fred 2014-10-27 11:08:20
2 questions.....The rule concerning turning a card before everyone has acted only mentions the turn, does the same rule apply to the river? Same question concerning turn-dealing mistakes.
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Vaani Jain asked • 2 hours ago
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Ram Ram asked • 2 hours ago
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Iron Man asked • 3 hours ago
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Shatabdi Chanda asked • 3 hours ago
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Ruksar Chaush asked • 3 hours ago
## 10 ml of 0.lM solution of weak acid was mixed with 4 ml 0.1M solution of strong base NaOH. pH of the solution measured was 5.12. If log (0.4/0.6):0.f 8, pKa of the weak acid will be A) 4.3 B) 3.7 c) 5.3 D) 5.12?
Meenu John asked • 3 hours ago
## What is the equivalent weight of lactose?
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Lakhan Munnur asked • 4 hours ago
## 28. If average molecular wt. of air is 29, then assuming N, and o, gases are there which options are correct regarding composition of air: (1) 75% by mass of N, (ii) 75% by moles N2 (iii) 72.41% by mass of N2 (a) only (i) is ... more
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## The temperature of two moles of an ideal gas is raised from 300K to 400K,what is the value of delH- delE?
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B P S asked • 4 hours ago
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Borra Manikanta asked • 4 hours ago
## Which of the following statement(s) is(are) TRUE? a)There exists a connected set in R which is not compactb)Arbitrary union of closed intervals in R need not be compactc)Arbitrary union of closed intervals in R is always closedd)Every bounded infinite subset R of R has a limit point in V itselfCorrect answer is option 'A,B'. Can you explain this answer?
Sushama Kumari asked • 4 hours ago
## H2(g) =2H(g);∆H298K = 104.2Kcal .The bond energy of H-H bond is a) 52.1Kcal b) 104.2 Kcal c) 208.4 Kcal d) zero .The correct answer is option a).Can you explain this answer?
Mayank Vedwal asked • 4 hours ago
## (Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the c... more
Ankita Dubey asked • 5 hours ago
## Aluminium (Al) can occur in both tetrahedral and octahedral co- ordinations in silicates. The amount of octahedral Al in a pyroxene crystal of composition MG1.4 Fe0.4 Al0.4 Si1.8 O6 is ____ (give answer in one decimal place).Correct answer is '0.2'. Can you explain this answer?
Aleena Varghese asked • 5 hours ago
## Molecular weight of circular polymer of 38 arginine molecules?
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## The plane of polarisation of a plane polarized light rotates by 60° after passing through a waveplate. The pass-axis of the wave plate is at an angle α with respect to the plane of polarisation of theincident light. The wave plate and α area)λ/4, 60°b)λ/2, 30°c)λ/2, 120°d)λ/4, 30°Correct answer is option 'B'. Can you explain this answer?
Purneet Sharma asked • 5 hours ago
## Find the maximum and minimum value of f(x)=sinx + 1/2 cos2x in [0,pie/2] plz someone solve this.
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Rachika Soni asked • 5 hours ago
## Two beams of light in the visible range (400 nm - 700 nm) interfere with each other at a point. The optical path difference between them is 5000 nm. Which of the following wavelengths will interfere constructively at the given point?a)416.67 nmb)555.55 nmc)625 nmd)666.66 nmCorrect answer is option 'A,B,C'. Can you explain this answer?
Aniket Choudhary asked • 6 hours ago
## A nozzle is in the shape of a truncated cone, as shown in the figure. The area at the wide end is 25 cm2 and the narrow end has an area of 1 cm2. Water enters the wider end at a rate of 500 gm/sec. The height of the nozzle is 50 cm and it is kept vertical with the wider end at the bottom. The magnitude of the pressure difference in kPa (1 kPa = 103 N/m2) between the two ends of the nozzle is _______________.Correct answer is '17.5'. Can you explain this answer?
Preetha Dilip asked • 6 hours ago
## Consider two nuclei with the same mass number A. For which of the following values of A, the fusion reaction is NOT possible?a)15b)22c) 29d) 36Correct answer is option 'D'. Can you explain this answer?
Digyash Kumar Bhattacharyya asked • 6 hours ago
## The efficiency of a reversible heat engine from the PV diagram?
Rubi Sharma asked • 6 hours ago
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Manohar Lal Jat asked • 6 hours ago
## Which of the following statement(s) is/are true?a)Newton’s laws of motion and Maxwell’s equations are both invariant under Lorentz transformations.b)Newton’s laws of motion and Maxwell’s equations are both invariant under Galilean transformations.c)Newton’s laws of motion are invariant under Galilean transformations and Maxwell’s equations are invariant under Lorentz transformations.d)Newton’s laws of motion are invariant under Lorentz transformations and Maxwell’s equations are invariant under Galilean transformations.Correct answer is option 'C'. Can you explain this answer?
Pawan Solanki asked • 7 hours ago
## 2,4-Dinitrophenol (DNP) is an uncoupler thata)cannot freely pass through plasma membraneb)enhances ATP productionc)is a non-toxic compound at high dosed)makes the inner mitochondrial membrane leaky to protonsCorrect answer is option 'D'. Can you explain this answer?
Prem Mehrotra asked • 7 hours ago
## A reaction follows second order rate law, a)A plot of [A] versus t is a straight lineb)A plot of 1/[A] versus t is a straight linec)A plot of ln[A] versus t is a straight lined)A plot of e[A] versus t is a straight lineCorrect answer is option 'B'. Can you explain this answer?
B P S asked • 7 hours ago
## For a zero order reaction, the half-life depends on the initial concentration [C0] of the reactant asa)[C0]b)[C0]0c)[C0]–1d)[C0]1/2Correct answer is option 'A'. Can you explain this answer?
Labham Singh asked • 7 hours ago
## Needed a Document for lanthenides? Related: Inorganic Chemistry for IIT JAM?
Papai Bhandari asked • 7 hours ago
## Among the compounds given in the option (a) to (d), the one that exhibits a sharp band at around 3300 cm–1 in the IR spectrum is:a)1, 2-butadieneb)1, 3-butadinec)1-butyned)2-butyneCorrect answer is option 'C'. Can you explain this answer?
Yukta Mahashabde asked • 8 hours ago
## An enzyme requires both aspartate (pKa of side chain = 4.5) and histidine (pKa of side chain = 6.5) residues in the catalytic site to be protonated for activity. The expected enzyme activity (in %) at a pH of 5.5 would be closed toa)90b)78c)50d)10Correct answer is option 'D'. Can you explain this answer?
Esha Kumari asked • 8 hours ago
## Viral capsids are made up of morphological subunits called capsomeres. One of the common capsomeres is icosahedral. The icosahedron is a regular polyhedron with a)16 triangular facets and 12 vertices.b)20 triangular facets and 12 vertices.c)16 triangular facets and 16 vertices.d)20 triangular facets and 16 vertices.Correct answer is option 'B'. Can you explain this answer?
Sanjeev Kumar asked • 8 hours ago
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Apoorva S asked • 8 hours ago
## Two ions can replace each other in a crystal structure only if the difference between their ionic radii does not exceed ________ %.Correct answer is '15'. Can you explain this answer?
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## A platinum electrode is immersed in a solution containing 0.1 M Fe 2 and 0.1 M Fe 3 . It is coupled with SHE. Concentration of Fe 3 is increased to 1.0 M without change in [Fe 2 ] then change in EMF (in centi volt) is?
Jyota Mishr asked • 8 hours ago
## Calculate the liquid function potential at 25°C between two solutions of HCl having mean ionic activities of 0.01 and 0.001, respectively. The transference number of H ion (ft) in HCl may be taken as 0.83. a) 0.0061b) 0.0039c) 0.0010d) 0.0090Correct answer is option 'B'. Can you explain this answer?
Subhash Kumar asked • 9 hours ago
## A first order reaction is 87.5% complete at the end of 30 minutes. The half-life of the reaction is __________ minute(s).Correct answer is between '9.8,10.2'. Can you explain this answer?
Csphk Pre asked • 9 hours ago
## Out of the following statements, choose the correct option(s) about a perfect conductor.a)The conductor has an equipotential surfaceb)Net charge, if any, resides only on the surface of conductorc)Electric field cannot exist inside the conductord)Just outside the conductor, the electric field is always perpendicular to its surfaceCorrect answer is option 'A,B,C,D'. Can you explain this answer?
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## The density of a mixture of Oxygen and Nitrogen gas at NTP is 1.3g/L. What is the partial pressure of nitrogen?
Fetching relevant content for you | 6,826 | 23,343 | {"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-2021-25 | latest | en | 0.822 |
http://www.openwetware.org/index.php?title=Guide_to_Excel_for_statistics&oldid=105454 | 1,498,377,669,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320443.31/warc/CC-MAIN-20170625064745-20170625084745-00652.warc.gz | 611,491,356 | 7,028 | # Guide to Excel for statistics
## common statistical functions
• AVERAGE(cells) - arithmetical mean of cells specified
• MEDIAN(cells) - middle value; half the values are greater, half are less than the median
• COUNT(cells) - counts the number of values n; useful for standard error calculation
• VAR(cells) - variance of a sample population; variance = standard deviation squared
• STDEV(cells) - standard deviation of a sample population; measures how much values vary from the mean
You can type these functions straight into the cells starting with the equal sign, e.g. type =stdev(A1:A10). Capitalisation is not important. Replace "cells" in the above list with the cells to be used for the calculation. Cells can be specified by clicking or typing. You can specify a range of cells by stating the (top) left and the (bottom) right cell, e.g. type A1:A10 (10 cells) or A1:B10 (2x10cells). Alternatively, you can enumerate cells using the comma as separator, e.g. A1,A3,A5. If you want to prevent automatic change of the column or row when copying formulae, use the dollar sign, e.g. \$A1 will keep the column the same while A\$1 will prevent a change in the row number.
## Student t test
Excel's student t test is based on the function TTEST(). Watch out to correctly specify the tails and type. If in doubt use: tails 2, type 3. This is the most stringent setting.
=TTEST(group1,group2,tails,type)
```group 1 A1:A8 1st group of measurements
group 2 B1:B8 2nd group of measurements
tails 1 one-tailed distribution
2 two-tailed distribution
type 1 paired = e.g. same cells after 1h, 2h
2 2 samples, equal SD
3 2 samples, unequal SD
```
See an example below. It's based on this great web page [1]
data series 1: 520, 460, 500, 470
data series 2: 230, 270, 250, 280
Question: Are these 2 series significantly different from each other?
Results for all TTEST variations
```0.000018 ttest(B5:B8,C5:C8,2,3)
0.000009 ttest(B5:B8,C5:C8,1,3)
0.000013 ttest(B5:B8,C5:C8,2,2)
0.002559 ttest(B5:B8,C5:C8,2,1)
```
You can see how tails and type affect the values. Note this data is not paired. In most experiments you will comparing unpaired data, unless you follow the same system in different conditions. For example, animal aged 10w, same animal aged 15w.
Result: The data series are significantly different from each other. P < 0.001.
## evaluation of Excel for statistics
Read this excellent review of Excel for statistics on practicalstats.com.
The main arguments are summarised below:
• many of Excel's charts violate standards of good graphics, i.e. fancy but not good for science
• many statistical methods are not available, e.g. box plots, 2-way ANOVA with unequal sample size, nonparametric tests
• several procedures are misleading, e.g. confidence function
• distributions are not computed with precision (Excel only correct to 2nd digit)
• regression routines are incorrect for multicollinear data
• ranks of tied data are computed incorrectly | 773 | 2,969 | {"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} | 3.453125 | 3 | CC-MAIN-2017-26 | latest | en | 0.732101 |
https://cboard.cprogramming.com/cplusplus-programming/14210-helppppp-please.html | 1,490,904,728,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218199514.53/warc/CC-MAIN-20170322212959-00572-ip-10-233-31-227.ec2.internal.warc.gz | 783,473,957 | 13,524 | Hello,
I am trying to explain this step by step:
1st funcn: takes 3 values from the user & calculates the total.
2nd & 3rd funcn:irrelevant to my doubt.so,just skipping it.
Now this above funcns r repeated till user types
"END".(in the main )
4th funcn :if any of the 3 values from the 1st funcn are not zero
then displays the total no. of persons.
So my doubt is how to calculate total no. of people in the 4th funcn.
I have tried using counter in the 4th funcn by declaring a vaid condition ,but unsuccessful.
like :
Code:
``` if (t1!=0&&t2!=0&&t3!=0)
{
counter++;
}
cout<<"no.of persons"<<counter<<endl;```
please tell me what I am suppose to do?????
2. If the three values the user enters represent people, then
Code:
```if (t1!=0||t2!=0||t3!=0)
{
counter = t1+t2+t3;
cout<<"no.of persons"<<counter<<endl;
}```
3. ## Thank,but...
Hi swoopy
but unfortunately thats not the case:
user enters 3 values for time say 18secs,15secs etc.
here condn is if t1||t2||t3==0 totaltime is not calculated. or else its not then total time is calculated.
Now this procedure of entering values and calculatin time goes on till user presses END
I have to declare another funcn ,where if all the values entered by user are not zero,I have to calculate the total no. of persons who have nonzero values.
thanks,please let me know,how should i proceed??
4. ## my code
Code:
```
int calctottime(int time,int time1 ,int time2)
{
cout & cin statements for entering and reading the values
if(time!=0||time1!=0||time2!=0)
{
totalTime=time+time1+time2;
}
else
{
cout<<"total time:not completed"<<endl;
}
void totalcompet()
{
int ti=0;
int ti1=0;
int ti2=0;
if(ti!=0&&ti1!=0&&ti2!=0)//these values r actully the above funcn's times//
{
counter++;//to count the no. of persons
}
cout<<"total no. of people"<<counter<<endl;
}```
Code:
```int calctottime(int &time,int &time1 ,int &time2);
int totalcompet(int ti, int ti1, int ti2);
int main(void)
{
int time, time1, time2;
int total_time, total_competitors;
total_time = calctottime(time,time1,time2);
total_competitors = totalcompet(time,time1,time2);
return 0;
}
int calctottime(int &time,int &time1 ,int &time2)
{
cout & cin statements for entering and reading the values
if(time!=0||time1!=0||time2!=0)
{
totalTime=time+time1+time2;
}
else
{
cout<<"total time:not completed"<<endl;
}
int totalcompet(int ti, int ti1, int ti2)
{
int counter = 0;
if(ti!=0)
counter++;//to count the no. of persons
if(ti2!=0)
counter++;
if(ti3!=0)
counter++;
cout<<"total no. of people"<<counter<<endl;
return counter;
}```
6. ## not expected result
Hi swoopy,
ur suggested code ,when implemented in my prog,is displaying the same number of total people as well as people who completed all stages;
that means:
ouput is something like this:
total no. of people: 3 //(which i calculated by introducing other funcn,which works properly)//
total no. of people with all stages complete:3
//(even though we have given condns of t1!=0 ,t2!=0 ,t3!=0,then only it should calculate total people,or else it should skip the person who have any 0 time.
now, what u want me to do???????????
HEEEEELp ,i am going insane
thanks
7. ## can no one help me???
thanksssssss.......
8. ## Where r codegurus???
No one on this board is so considerate????????
I not asking to write a code for me,
but please just give me a hint......
9. I think we need a description of the original programming problem.
We have three times. Now if each time is the time for one competitor, then how can we know whether they completed all stages, all we have is one time? How many stages are there, and don't we need to know what their time is for each stage, not just one time?
Now if the three times are times for one competitor, then you need a loop. For each competitor add up the three times. Check to see if all times are >0, else not all stages were completed. So your count for total competitors is incremented each time thru the loop. Your count for all stages complete is only incremented if all times > 0.
10. ## thanks for ur reply..
Hi,
ur reply have given me some hope,that still there r people who come in need........
anyway,
the Q is input name, swimtime, runtime,cycle time from the user
and calculate the total time,but condn is if any of the entered times is 0 ,u should not calcuate total time ,instead output "total time not completed".
this stuff goes on till user enters "end" in the name.
once end is entered:
now the second phase of the Q:
calculate total no. of all competitors (which i did,by placing a counter )
and calculate total no. of competitors who completed all the stages ie(time1& time2& time3!=0)
now as i have been told to insert this stuff in different functions:
so i have worked on this funcns:
1)getname --to get the name from the user
2)calctotaltime--to calculate total time
3)totalcompetitors---total no.of all compet.
4)total no. of competitors who actually completed all stages--//I am stucked)
thanks.........
11. Ok, makes sense now. You don't really need a function for # 3, as this is just a counter. For #4, I made a function called all_stages_completed(). Here's an example. Take what you have and add the function all_stages_completed() to your code. If you can think of a better name for it, use that. I didn't make a getname() function, but one could easily be added.
Code:
```#include <iostream>
using namespace std;
bool all_stages_completed(int t1, int t2, int t3);
void calctotaltime(char name[], int t1, int t2, int t3);
int main(void)
{
char name[80];
int swim_time, run_time, cycle_time;
int total_competitors = 0;
int total_finishers = 0;
do {
cout << "Enter name:";
cin >> name;
if (strcmp(name,"end") == 0)
break;
total_competitors++;
cout << "Enter swim time:";
cin >> swim_time;
cout << "Enter run time:";
cin >> run_time;
cout << "Enter swim time:";
cin >> cycle_time;
if (all_stages_completed(swim_time,run_time,cycle_time))
{
calctotaltime(name,swim_time,run_time,cycle_time);
total_finishers++;
}
} while (true);
cout << "Total competitors:" << total_competitors << endl;
cout << "Total finishers:" << total_finishers << endl;
return 0;
}
void calctotaltime(char name[], int t1, int t2, int t3)
{
cout << "Total time for" << name << ":" << t1+t2+t3 << endl;
}
bool all_stages_completed(int t1, int t2, int t3)
{
if (t1 > 0 && t2 > 0 && t3 > 0)
return true;
else
{
cout << "Total time not completed." << endl;
return false;
}
}```
12. ## U r a genius
Thanks swoopy,
u r just adorable!!!!!!!!thanks for everything........
Still there r 2-3 more points to be added ,like best competitor overall,best competitor in swim,run etc,but I will do it myself ....
if I require a hint I will approach again..........
May God bless U | 1,844 | 6,756 | {"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} | 3.234375 | 3 | CC-MAIN-2017-13 | longest | en | 0.743186 |
http://kulhu.com/index.php/library/recursion-theory-its-generalisations-and-applications-proceedings-leeds-1979 | 1,560,790,096,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998513.14/warc/CC-MAIN-20190617163111-20190617185111-00132.warc.gz | 100,626,565 | 9,748 | Posted in Logic
# Download Recursion theory, its generalisations and applications: by F. R. Drake, S. S. Wainer PDF
By F. R. Drake, S. S. Wainer
Best logic books
Set Theory and the Continuum Problem (Dover Books on Mathematics)
A lucid, dependent, and whole survey of set idea, this quantity is drawn from the authors' monstrous instructing adventure. the 1st of 3 elements specializes in axiomatic set thought. the second one half explores the consistency of the continuum speculation, and the ultimate part examines forcing and independence effects.
Zermelo’s Axiom of Choice: Its Origins, Development, and Influence
This e-book grew out of my curiosity in what's universal to 3 disciplines: arithmetic, philosophy, and background. The origins of Zermelo's Axiom of selection, in addition to the talk that it engendered, definitely lie in that intersection. because the time of Aristotle, arithmetic has been involved alternately with its assumptions and with the gadgets, similar to quantity and area, approximately which these assumptions have been made.
Finitely Axiomatizable Theories
This is often the one monograph dedicated to the expressibility of finitely axiomatizable theories, a classical topic in mathematical good judgment. the quantity summarizes investigations within the box that experience led to a lot of the present development, treating systematically all confident effects pertaining to expressibility.
Extra info for Recursion theory, its generalisations and applications: Proceedings Leeds, 1979
Example text
Konstantopoulos etc. These guide the search away from solutions that are known to be unsatisfiable or otherwise unwarranted, providing mechanisms for optimization and for enforcing conformance with a prior theoretical framework. In our approach, there are two sources of prior knowledge: (a) DL syntax and semantics, so that all constructed hypotheses correspond to valid DL propositions, and (b) the axiomatization of the domain within which a theory is to be constructed, providing semantic restrictions that can optimize the search.
Those approaches are interesting because they summarize the data and they are not subject to combinatorial explosion, even when applied to nested tables. Complex Aggregates. Full-fledged relational data mining systems do not dissociate the propositionalisation step from the construction of the model. Most of the full-fledged relational data mining systems, as for example Progol [18], do not explicitly deal with numeric attributes. Nevertheless, Tilde [4] has been modified to construct complex aggregates [21].
1 Establishing Priors At the core of the Progol algorithm, and ILP approaches in general, is a search through the space of admissible hypotheses. This space is never made explicit, but is defined by (a) a refinement operator which traverses the space by repeatedly generating clauses to be evaluated by the search heuristics; and (b) other pieces of prior knowledge such as the syntactic and semantic constraints that valid hypotheses must satisfy. Such restrictions reflect prior knowledge about the domain of application, including the input and output types of predicate arguments, disjointness axioms, 26 A. | 667 | 3,204 | {"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} | 2.640625 | 3 | CC-MAIN-2019-26 | longest | en | 0.912064 |
https://online.2iim.com/CAT-2019-Analysis-slot-1-and-slot-2/ | 1,642,670,419,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301730.31/warc/CC-MAIN-20220120065949-20220120095949-00530.warc.gz | 464,827,229 | 7,977 | # CAT 2019 Analysis - Slot 1 and Slot 2
### First thoughts
DILR was at the same level as seen in the past 4–5 years, perhaps a little easier. Quant was easier than 2018 but far tougher than the levels seen in 2017, 2016 or 2015. The elephant in the room was VARC. The passages were tough, the questions reveled in double-negatives. We have given a quick snapshot of what we think could be the cut-offs. Note that these are, at best, wild guesses.
### VARC substantially tougher, DILR at same level of 2018, Quant easier than 2018 but tougher than 2017
Our estimates for CAT 2019 percentiles and scores are given below for both slot 1 and slot 2.Note that these are merely estimates and are not based on statistical analysis.
### VARC - Really Challenging
The RC section was relentless. Passages were non-obvious, choices were long, quite a few had 2 close choices and we had to read and re-read the choices all the time. Our reading is that the overall error rate in VARC is going to be much higher than it has been for a while. Too many questions had ‘weaken the argument’, ‘support the author’s beliefs’ - on a couple of occasions, we couldn't even sure what the author’s position was. Attempting a lot of questions might not be the prudent strategy in this section. A score of 70 might fetch a 99.5th percentile or so.
### DILR - We are all probably getting better at this
DILR was the surprise package in 2015 and 2016; but over the past 2–3 years, we have kinda caught up with the CAT’s way of testing DILR. There was one nice simple DI puzzle, one very do-able math-based puzzle. So, a serious candidate could get 8 questions correct. Post this, free marks were hard to come by. 4 sets with almost everything correct should fetch one somewhere in the 99th percentile range.
### Quants - Easier than last year, but not a give-away section
CAT 2018 Quant was a nightmare. This year, there were quite a few very do-able questions. I think some guys might even have killed this section and scored 97+. Time pressure was ever-present but was not debilitating. I think one needs to get 70+ for 99.5th percentile, and 37–38 to get 90th percentile. Quite a few questions from functions, sequences, exponents etc, usual number from Geometry and very few from Number Systems, Permutation and Probability.
As ever, these are mere conjectures and the actual numbers could be wildly different.
## 2IIM's Online Analysis for CAT 2019 Slot 2
### CAT 2019 Analysis Topic-wise breakdown
As far as the CAT syllabus was concerned, a great many questions were from Arithmetic. Number Theory was not heavily tested at all whereas Geometry featured reasonably well.
### CAT 2019 TITA-MCQ breakdown in Quant, VARC and DILR
CAT 2019 had 26 questions of TITA kind.
### Let’s talk CAT 2020
It is not too early to start preparation. If you took CAT 2019 merely as a precursor to CAT 2020, then it is about time to start thinking about a pleasant 1-year preparation cycle. If you are not already one, sign up as a trial user to the 2IIM Online Course and see why so many students say so many good things about us (review link here )
2IIM’s CAT 2019 online students can renew the course at very reasonable rates. The price for renewal is pegged at a level where students pay the difference between our current 2019 price point and what they have paid already. Effectively, for students who are renewing the course, the price for preparation for CAT 2019 and CAT 2020 would be merely the price for CAT 2019 (In other words CAT 2019 prep came for free).
Those of you just beginning on this CAT preparation journey, click here to know about every aspect of How to Prepare for CAT? . Those of you feeling a little lost amidst all this talk of percentiles, slots, analysis, syllabus etc, click here to learn What is CAT all about? | 927 | 3,821 | {"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-2022-05 | longest | en | 0.979061 |
https://www.hammockforums.net/forum/showthread.php/161232-SRL-tension-vs-shortening-it?s=57bb01392c9bfaaac150fd74782a6f5b&p=2048306 | 1,623,963,095,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00039.warc.gz | 722,182,568 | 16,489 | # Thread: SRL tension vs shortening it
1. Good tip on the biner thanks
2. Originally Posted by Shug
You actually don't want slack....rather you want it right where you can reach up and bend it a bit. Not guitar string tight but able to force a slight bend.
So if you use a ridgeline, ideally you would hang your hammock to achieve the slight bend. If it is slack as opposed to where you set it for your sweet spot then either your suspension is too high up the trees or your hang is too tight. Or both.
Shug,
In the video you state that the green hammock is 10 1/2 feet long and that your ridge line is 92 1/2 inches, which is a good bit (like a foot) less than "the proverbial 83%".
A quick check on my calculator shows that at just under, 73 1/2 %.
126" x 0.83= 104.58 inches & 126" x 0.7341= 92.5 if we round up to the nearest tenth.
Was that like a "verbal typo", or do you indeed have a ~73% of hammock length, ridge line on your green 10 1/2 foot long hammock?
I ask because I have found that (up to a point obviously) a shorter ridge line is more comfortable than a longer one, and have made ridge lines down to ~75% of hammock length, and found them more than acceptable.
Additionally, all the hammocks I've made & sold that have an adjustable ridge line (whoopie) on them, the buyers always found sub 83% to be their preference.
willin'
3. Originally Posted by u.willie
Shug,
In the video you state that the green hammock is 10 1/2 feet long and that your ridge line is 92 1/2 inches, which is a good bit (like a foot) less than "the proverbial 83%".
A quick check on my calculator shows that at just under, 73 1/2 %.
126" x 0.83= 104.58 inches & 126" x 0.7341= 92.5 if we round up to the nearest tenth.
Was that like a "verbal typo", or do you indeed have a ~73% of hammock length, ridge line on your green 10 1/2 foot long hammock?
I ask because I have found that (up to a point obviously) a shorter ridge line is more comfortable than a longer one, and have made ridge lines down to ~75% of hammock length, and found them more than acceptable.
Additionally, all the hammocks I've made & sold that have an adjustable ridge line (whoopie) on them, the buyers always found sub 83% to be their preference.
willin'
I just go with what feels right and pretty much ignore the math.
My mention on video of the 83% is just because that seems to be the status-quo/starting place and if I don'y allude to it many commenters will call me out on it.
But sometimes I am longer than 83% on some hammocks.
Go by feel.
4. I’ve got a feeling I need to go shorter than the 83%....currently have 143” hammock and 120” SRL....when I’ve hung with it slack it’s been way more comfy.
5. Note that hammock calculators to get SRL often use 86% instead of 83%. In that case, a “true” 83% would give more sag. But it’s not a target number; it’s a suggested area/range and you decide where the comfort is for you.
6. Biner wrap.......
biner on ridgeline.jpg
7. Originally Posted by Shug
Biner wrap.......
biner on ridgeline.jpg
Simple and effective (think I’ve been called that in the past), thanks Shug...I’ll try it out.
8. Tried the carabiner trick out at the weekend, I’d already found/setup a good hang with plenty of give in the ridgeline but it definitely helps tidy the ridgeline out the way.
Super comfy, definitely finding the 12’ trail lair needs a slacker hang...for me anyways.
0F8BA881-80CF-4018-8049-1FF260B19620.jpg
9. Originally Posted by Norfolk Yeti
Tried the carabiner trick out at the weekend, I’d already found/setup a good hang with plenty of give in the ridgeline but it definitely helps tidy the ridgeline out the way.
Super comfy, definitely finding the 12’ trail lair needs a slacker hang...for me anyways.
0F8BA881-80CF-4018-8049-1FF260B19620.jpg
I hang a 12’ trail lair myself, and a loose relaxed hang tension does work. You can also rig some small shock cord on your ridgeline to take up slack. Similar to guyout shock cords. Holds the net up good, but gives when it needs to.
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• | 1,143 | 4,196 | {"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-2021-25 | latest | en | 0.947157 |
http://eprints.qut.edu.au/16412/ | 1,488,039,258,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171781.5/warc/CC-MAIN-20170219104611-00252-ip-10-171-10-108.ec2.internal.warc.gz | 83,320,326 | 8,027 | # Mathematical modelling of primary alkaline batteries
Johansen, Jonathan Frederick (2007) Mathematical modelling of primary alkaline batteries. PhD thesis, Queensland University of Technology.
Preview
## Abstract
Three mathematical models, two of primary alkaline battery cathode discharge, and one of primary alkaline battery discharge, are developed, presented, solved and investigated in this thesis. The primary aim of this work is to improve our understanding of the complex, interrelated and nonlinear processes that occur within primary alkaline batteries during discharge.
We use perturbation techniques and Laplace transforms to analyse and simplify an existing model of primary alkaline battery cathode under galvanostatic discharge. The process highlights key phenomena, and removes those phenomena that have very little effect on discharge from the model. We find that electrolyte variation within Electrolytic Manganese Dioxide (EMD) particles is negligible, but proton diffusion within EMD crystals is important. The simplification process results in a significant reduction in the number of model equations, and greatly decreases the computational overhead of the numerical simulation software. In addition, the model results based on this simplified framework compare well with available experimental data.
The second model of the primary alkaline battery cathode discharge simulates step potential electrochemical spectroscopy discharges, and is used to improve our understanding of the multi-reaction nature of the reduction of EMD. We find that a single-reaction framework is able to simulate multi-reaction behaviour through the use of a nonlinear ion-ion interaction term.
The third model simulates the full primary alkaline battery system, and accounts for the precipitation of zinc oxide within the separator (and other regions), and subsequent internal short circuit through this phase. It was found that an internal short circuit is created at the beginning of discharge, and this self-discharge may be exacerbated by discharging the cell intermittently. We find that using a thicker separator paper is a very effective way of minimising self-discharge behaviour.
The equations describing the three models are solved numerically in MATLABR, using three pieces of numerical simulation software. They provide a flexible and powerful set of primary alkaline battery discharge prediction tools, that leverage the simplified model framework, allowing them to be easily run on a desktop PC.
Impact and interest:
Citation counts are sourced monthly from Scopus and Web of Science® citation databases.
These databases contain citations from different subsets of available publications and different time periods and thus the citation count from each is usually different. Some works are not in either database and no count is displayed. Scopus includes citations from articles published in 1996 onwards, and Web of Science® generally from 1980 onwards.
Citations counts from the Google Scholar™ indexing service can be viewed at the linked Google Scholar™ search. | 557 | 3,094 | {"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} | 2.5625 | 3 | CC-MAIN-2017-09 | latest | en | 0.88071 |
https://solar-energy.technology/thermodynamics/thermodynamic-properties/hot | 1,563,527,237,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00321.warc.gz | 559,937,015 | 10,116 | # Heat
In physics, in particular in thermodynamics, heat is defined as the contribution of transformed energy as a result of a chemical or nuclear reaction and transferred between two systems or between two parts of the same system. This energy is not attributable to a job or a conversion between two different types of energy. Heat is, therefore, a form of transferred energy and not a form of energy contained as internal energy.
As the energy is exchanged, the heat is measured in the International System in joules. In practice, however, it is often still used as the unit of measurement of calories, which is defined as the amount of heat necessary to raise the temperature of one gram of distilled water, is subjected to the pressure of 1 atm passed from 14.5 at 15 ° C, 5 ° C. Sometimes purely technical units such as kWh or BTU are also used.
Heat and work are forms of energy that can not be associated with the state of the system, that is, with its equilibrium configuration; in particular, both forms of energy are recognized the moment they transit, they flow. The work identifies the moment when the force is made a change. In other words, workflows are performed at the moment they are produced; just as the heat is identified only at the moment of its transmission.
## Specific heat
The specific heat, also called specific mass heat, of a substance is defined as the amount of heat needed to increase or decrease, in a Kelvin (or a degree of the Celsius scale) the temperature of a unit of mass.
A similar amount is the specific molar heat, also called molar heat, defined as the amount of heat needed to increase or decrease the temperature of one mole of substance by one degree.
In the international system the unit of measurement of the specific heat is J / (K · kg), even if the kcal / (kg × ° C) is widely used, while the molar heat is J / (K · mol).
## Effects of heat
The effects of heat transfer are described by the first law of thermodynamics in its most general form:
ΔE = Q - W
where ΔE indicates a change of any form of energy (such as internal energy, kinetic energy, or potential energy), Q represents heat and W indicates work (by volume change or isocorous). The consequences of heat transfer can be mainly of two types: energy variation or work exchange.
A particular form of energy that can be modified after the passage of heat is the internal energy; the variation of internal energy can have different consequences, including a change in temperature or a change in the state of aggregation.
If the heat transfer results in a change in the state of aggregation, this heat takes the name of latent heat, whereas if the heat transfer results in a decrease in the temperature difference (because the two systems or two parts of the same system tend to reach thermal equilibrium) we speak of sensible heat.
The classic formula of sensible heat is:
Q = c · m · ΔT
while the latent heat is:
Q = λ · m
Finally, in the case that heat transfer involves both a decrease in temperature difference and a phase change, this heat can be considered as the sum of two contributions: a contribution related to sensible heat and a contribution related to the latent heat.
For example, the increase in water temperature from 20 ° C to 50 ° C under standard conditions (ie, at a pressure of 1 atm) is determined by the fact that sensible heat is provided, whereas, if the Water has already reached the boiling temperature, stores energy (in the form of latent heat), keeping its temperature unchanged, until the phase change from liquid to vapor occurs. For this reason, a jet of water vapor at 100 ° C, which has energy stored during the passage of the state, can cause more severe burns than water in the liquid state at the same temperature.
Heat of reaction is also spoken of when heat is consumed or generated by a chemical reaction.
## Heat, temperature and internal energy
Heat is not a property associated with a thermodynamic equilibrium configuration. In the presence of a temperature gradient, heat flows from the points at higher temperatures to those at lower temperatures, until thermal equilibrium is reached. The amount of heat exchanged depends on the particular trajectory followed by the transformation to arrive from the initial state to the final state. In other words, heat is not a state function.
The internal energy, instead, is a function of the associable state with an equilibrium (or thermodynamic state) configuration of the system, depending on the state variables.
For internal temperature and energy they have logical expressions (that is, they are scientifically correct) of the type: "the body has a certain temperature, it has a certain internal energy, it acquires energy, it gives energy".
On the other hand, heat is not a thermodynamic property, so phrases like "the body has heat, yields heat, acquires heat" have no scientific value. In fact, heat can be defined as "energy in transit", not as "energy possessed by a body"; heat is exchanged between two bodies (or two parts of the same body) and not possessed by a single body (as is the case with internal energy). In particular, the heat flows due to a temperature difference between the system under study and the environment that interacts with it, then the heat only manifests when it passes between the system and the environment due to a temperature difference and is not recognized. no way within the system and the environment as an intrinsic property of it.
## Heat propagation
The transfer (or exchange or propagation) of heat between systems can be done in three ways:
• Propagation of heat by conduction: in a single body or between bodies in contact there is a transmission, by impacts, of kinetic energy between the molecules that belong to the neighboring areas of the material. In the driving energy is transferred through matter, but without macroscopic movement of the latter;
• Propagation of heat by convection: in a fluid in motion, the fluid parts can heat or cool to run in contact with the outer surfaces and then in the course of their movement (in the turbulent character often) , the transfer (always to execute), the energy acquired to other surfaces, which gives rise to a transfer of heat by advection. In a gravitational field such as terrestrial (associated with the force of weight), this method of heat transfer is due to the natural occurrence of advection currents, heat and cool, due to temperature diversity and, for therefore, of the density of the fluid regions involved in the phenomenon, with respect to those of the surrounding fluid;
• Propagation of heat by irradiation: between two systems, the heat transmission can take place at a distance (also in a vacuum), for the emission, propagation and absorption of electromagnetic waves: also in this case the lower body temperature it is heated, and that at a higher temperature it cools. The irradiation mechanism does not require physical contact between the bodies involved in the process. An example is the heat that propagates from the Sun to the earth through solar radiation.
## Temperature detection
The feeling of "heat" or "cold" that you feel when touching a body is determined by its temperature and the thermal conductivity of the material it is made of, in addition to other factors.
Although it is possible to compare with the touch (with some caution) the relative temperatures of two bodies, it is impossible to give an absolute evaluation. For example, when submerging one hand in cold water for a few seconds and the other in heat water, and then submerging both in warm water, the first will have the feeling that the water is heat, the second that it is cold, because the temperature perceived is relative to that of the hand that is doing the measurement.
A relative evaluation is also often impossible. For example, when you touch a piece of wood and a piece of metal that have been in the same environment long enough to reach thermal equilibrium with the environment, you have the feeling that the metal is much colder, due to the different thermal conductivity of the two materials. A thermometer placed first in contact with wood, then with metal, instead would measure the same temperature, which coincides with that of air in the environment that is approximated as a source of heat for everything that is contained in it.
Temperature is an index of the average kinetic energy of the body particles under examination, heat is the energy that a body at a higher temperature transfers to a body at a lower temperature (until both bodies have the same temperature ). The sensation of cold and heat is due both to the temperature difference between the hand and the object and to the speed with which the object can transfer (absorb or release) heat to the hand (or other object at different temperatures).
However, by providing heat to a body, not only does the temperature increase, then there is a more acute sense of heat, but there are directly measurable variations in some physical properties.
## Historical background of heat
During the first half of the eighteenth century, scholars used the elemental substance called phlogiston to explain the heating of some materials and combustion.
In the following years, the thermal phenomena went back to the theory that heat was an invisible fluid, that when entering the matter of a body it could increase its temperature.
Despite Boyle's seventeenth-century studies of the relationship between particle movement and heat, it was not until the middle of the nineteenth century that the foundations of thermodynamics were laid. These bases were laid down thanks to the studies Mayer (1842) and Joule (1843), relating to the amount of heat and the work needed to achieve it.
valoración: 3 - votos 1
Last review: May 7, 2019 | 1,980 | 9,842 | {"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} | 3.28125 | 3 | CC-MAIN-2019-30 | longest | en | 0.963623 |
https://www.askiitians.com/forums/Analytical-Geometry/a-straight-line-x-y-2-touches-the-circle-x-2-y_137264.htm | 1,695,505,932,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.3/warc/CC-MAIN-20230923194908-20230923224908-00735.warc.gz | 749,958,684 | 43,250 | # A straight line x=y+2 touches the circle (x^2 + y^2) = r^2. The value of r is
Trina
34 Points
7 years ago
We have the equation of the circle as 4( x^2+y^2)=r^2
X^2+y^2=(r/2)^2
Now x=y+2 is a tangent to the circle....
Centre of the circle is (0,0)
Tangent is always perpendicular to the radius...
So
r/2=|x-y-2|/√2 at x=0=y
r/2=2/√2
r=4/√2=2√2
BALAJI SANKARAN
10 Points
7 years ago
Equation of circle :4(x2 +y2)=r2
centre of the circle = (0, 0)
equation of tangent :x=y+2 [y=x-2].....(i)
slope of the tangent = 1
using one-point form :(y-y1) = m(x-x1)
:(y-0) = -1(x-0)
: y = -x….…(ii)
from (i) & (ii)
(x, y) = (1,-1) which is a point on the circle
using distance formula on (0,0) & (1,-1)
we get the value of the radius = $\sqrt{}$2 | 312 | 743 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.758134 |
https://2born.livejournal.com/2019/11/15/ | 1,638,201,086,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00625.warc.gz | 157,768,946 | 143,052 | ## November 15th, 2019
### Готовимся к сезону катков и коньков :)
The pressure underneath a skate at rest: https://arxiv.org/abs/1911.05397
The pressure distribution is calculated underneath a skate which is pushed in the ice by the weight of a skater at rest. Due to the sharp edges of the skate the deformation is partly elastic and partly plastic. The ratio of the plastic and elastic contribution to the reaction force is determined. Using this ratio the deformation in ice with a finite hardness can be mapped on the problem of the deformation in a purely elastic medium with infinite hardness. Both the upright skate and the tilted position are exactly calculated.
### Об одной из фейнмановских задачек
Как раз недавно на нее набрел:) Там она без решения, на "подумать самостоятельно", с предупреждением, что решение не простое:))
A Quantitative Version of Feynman's Static Field Momentum Example: https://arxiv.org/abs/1911.05893
The Feynman demonstration that electromagnetic field momentum is real, even for static fields, can be made more useful by simplifying its geometry. Instead of Feynman's disk with charged balls on its surface, use a hollow non-conducting sphere with uniform surface charge density. The initial field angular momentum and the final mechanical angular momentum can then be calculated in closed form and shown to be equal. The methods used in the calculation are those available to the average upper-division physics student.
This simplified geometry also provides a test for the current idea that electromagnetic field energy can be considered a form of inertial mass. The mass motion in the simplified Feynman example can be modeled as the static, circular, incompressible flow of a fluid, with distributed nonzero vorticity. But such motion requires a centripetal force or pressure that has yet to be identified.
### Историко-философское о Стандартной модели
Полезно, по крайней мере, хронологией и обильными ссылками.
The Once and Present Standard Model of Elementary Particle Physics: https://arxiv.org/abs/1911.04604
There are many theories that have resided these last fifty years within the hazy mist we have been calling the Standard Model (SM) of elementary particles. An attempt is made here to construct a coherent description of the SM today, because only precisely articulated theories can be targeted for annihilation, corroboration, and alteration. To this end it is useful to categorize the facts, mysteries and myths that together build a single conception of the SM. For example, it is argued that constructing a myth for how neutrinos obtain mass is useful for progress. We also advocate for interpreting the cosmological constant, dark matter, baryogenesis, and inflation as four "mysteries of the cosmos" that are indeterminate regarding new particles or interactions, despite a multitude of available particle explanations. Some history of the ever-changing SM is also presented to remind us that today's SM is not our parents' SM, nor will it likely be our children's SM. | 650 | 3,039 | {"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-2021-49 | latest | en | 0.718882 |
https://www.excelforum.com/excel-formulas-and-functions/1209561-formula-problem-based-on-answer.html | 1,660,689,918,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00140.warc.gz | 685,294,588 | 17,541 | # formula problem based on answer
1. ## formula problem based on answer
Hi guys
what i need to do is
in cell p45 i need it to return these answers based on whats in cell p49
so if p49 is between £0.00 and £25,000 then cell p45 should return £29
if cell p49 is between £25,001 and £50,000 then cell p45 should return £58
but if cell p49 is between £50,001 and £200,000 then it needs to return 0.5% of that value
2. ## Re: formula problem based on answer
Try
=IF(P49<=25000,29,IF(P49<=50000,58,0.005*P49))
No check on upper limit of 200,000: what if it is > 200,000 ???
3. ## Re: formula problem based on answer
Hi John
I will cross that bridge as and when but the true reality is that its unlikely to happen for now.
Thank you so so much
5. ## Re: formula problem based on answer
Hello, You can use this:
Formula:
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6. ## Re: formula problem based on answer
Hi
I think this covers all your options:
=IF(P49>200000,"Unknown",IF(P49>50000,P49*0.05,IF(P49>25000,58, IF(P49>=0,29, IF(P49<0,"Unknown"))))).
Regards
Alan
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# CSC113_Tutorial6 - King Saud University College of Computer...
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King Saud University College of Computer and Information Systems, Department of Computer Science CSC 113: Java Programming-II, Spring2011, March 26, 2011 Tutorial #6 : Polymorphism The problem: In a Cartesian coordinate system, we have 20 shapes. Each of these shapes can be either a circle or a rectangle. The position and size of each circle can be described be a center point and a radius value, while location and size of a rectangle is described by a center point and values for the length and width, having all rectangles’ edges in parallel either to the x-axis or the y- axis. We need to know whither a point p belongs to the area defined by one shape or more, and find those shapes. A solution: Consider the following class diagram which specifies five classes and their relationships: Implementations of classes Point and TestShape are given.
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Determine abstract methods and classes then implement classes: Shape, Circle and Rectangle based on the class diagram and description below: Class
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Ask a homework question - tutors are online | 339 | 1,570 | {"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} | 2.5625 | 3 | CC-MAIN-2018-05 | latest | en | 0.843108 |
https://stats.stackexchange.com/questions/103207/statistical-description-of-how-two-distributions-are-different | 1,653,551,870,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00777.warc.gz | 616,552,462 | 67,596 | # Statistical description of how two distributions are different
I am have a collection of events recorded in two separate conditions, and I have measured the duration of each one of them. I would like to know two things: 1) are the distributions of durations different between the two conditions, and 2) if so, how are they different? For example, maybe the biggest difference between the two conditions is that there are far more 10 ms long events in condition #1 than in condition #2.
I feel confident in using a Kolmogorov-Smirnov test to determine whether the distributions are significantly different, but am stuck on how to describe what the differences actually are using a statistical test or collection of test statistics. I could histogram the data and run some sort of ANOVA, but that seems a little course and inelegant. Or maybe it is fair to plot the distributions on top of each other and simply read off the difference at areas of interest. Any advice on how to tackle this problem?
Below is some example data. Here a Kolmogorov-Smirnov test comparing these two distributions returns a very small p-value (p < 0.000001), so I know that the two distributions come from different populations. Based on how I think the KS test works, this also tells me that where the two distributions are widest apart in the horizontal plane (around 0.9 on the y-axis) the two distributions are significantly different (there are more events with durations around 60 ms in group 1 than in group 2). But what about everywhere else in the distribution? Is there a way for me to know whether the difference in the amount of events that are 10 ms long is also statistically significant?
• There are many ways in which distributions might differ. It depends on the kinds of differences that are important to describe. Jun 13, 2014 at 4:41
• It might be useful to make a QQ-plot of the two distributions! Jun 13, 2014 at 8:21
• You could look into " "Relative Distribution Methods in the Social Sciences" by Mark S. Handcock and Martina Morris, Springer-Verlag, 1999" which has an associated R package, reldist. Jun 13, 2014 at 8:23
• @ kjetil b halvorsen: Thanks for the suggestions! I've looked into Relative Distributions and QQ-plots, but unless I'm misunderstanding them, I don't think they help me to answer the problem I'm after. In both cases I'm visually interpreting a plot, but don't have a way of testing the difference between the distributions at every position along the curve? Is that correct? Jun 27, 2014 at 18:28
• I've edited the question to include some sample data and potentially clarifying remarks. Jun 27, 2014 at 18:30
From a descriptive standpoint, a QQ plot is about as good as it gets for something like this, you can really see where in the distributions things are different (heavier tails, skew, location shift, outliers). That might give you insight about what parameter of the distributions you are interested in testing.
The issue with a statistical test is that it demands that you really clearly state what you care about, i.e., your null hypothesis. As an example, if you wanted to compare whether or not the mean of two distributions was different, you might not care that their standard deviations are different:
a = rnorm(100, sd = 3); b = rnorm(100, sd = 1))
> t.test(a,b)$p.val [1] 0.2825168 > ks.test(a,b)$p.val
[1] 7.14257e-05
If our null hypothesis is that the means are different, it wouldn't be correct to test with the KS-test and we see we'd incorrectly reject the null. In your case, you have to state, what aspect of the distribution you care about and then test accordingly.
It would be fine to ask, Is the 10th-percentile different -- that's a good null hypothesis. I'm not aware of well-known statistical test for such a thing, but you could certainly use a bootstrap-based approach.
• Is it fair to do a KS-test first, then follow up with testing differences between specific areas of the distributions? I'm envisioning the protocol might be similar to the way one does an ANOVA first, then follows up with multiple comparisons. Jun 30, 2014 at 3:17
• Can you expand on the "bootstrap-based approach" suggestion, please? Jun 30, 2014 at 3:19
• Sure, I might not have enough characters, but: Your null is that the two distributions have the same 10th percentile. You compute the 10th percentile on each of your two distributions take the absolute difference, call that absolute difference S*. You now sample, with replacement, two datasets of equal size as your originals from the combined dataset (under the null, they have the same distribution), compute the abs 10th percentile difference as before; S_1,..._S_100. Count how many of the times your original S is bigger than the 100 resampled differences (S_1, ..., S_100). Jul 1, 2014 at 0:45 | 1,120 | 4,804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-21 | longest | en | 0.950987 |
https://stats.stackexchange.com/questions/126241/which-algorithm-should-i-use | 1,566,790,884,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00250.warc.gz | 649,171,336 | 32,763 | # Which algorithm should I use?
I was reading many machine learning questions like this one but I am not sure how to apply them to my scenario. I come from a biology/medicine background, and my math knowledge is limited (last thing I learned was calculus long ago), and some of the explanations on this site go over my head.
## The problem
Suppose someone is filling out forms everyday. The form contains various fields (patient information, medical history, job, etc.) Now suppose there are n patients coming to the hospital each day (so that's a minimum of n forms filled out each day, each time they arrive). Some parts of the form may never change (e.g. gender, race, etc. as an example) but some parts may (e.g. the medical history, checkup time, etc). Now I need to write a program that "learns" from this data so I can anticipate when the patient is likely to arrive again, then I need to determine whether the form is likely to change based on how often it changed in the past, and if the "important" parts (like the medical history) is unlikely to change, then I need to generate a form in anticipation of the event.
## Things I've considered
With whatever limited knowledge I've got, I figured I'd need to calculate some kind of variance to determine how different each form is from other forms of each patient. Then I figured I'd be pushing the data through some unsupervised learning 'thing' to detect if some kind of pattern exists to the patients' arrivals, and if it does then I can use the past data on the variance to determine if a form should be generated in advance.
So, I assume I'll need multiple algorithms (different for each part of the question), and I've looked at Naive Bayes, Logistic Regression, Decision Trees, and SVMs, but I hit my head on complex math and nearly suffered an aneurysm (joking)... To begin this quest, which model/algorithm should I start with, and why does it work well for this scenario over the others? I would appreciate a more wordy (less math) answer.
• "some of the explanations on this site go over my head" -- you're in good company -- quite a few of them go over mine, too. For each part of the form, for a given patient you essentially have a history of observations where each specific part either changed from the previous time or it didn't. It sounds like you're trying to develop some model that tries to determine the probability of change in each part at some given time $t$ based on the various characteristics, as well as the history of changes to date. – Glen_b Dec 2 '14 at 4:40
• @Glen_b I have a very similar question and instead of a creating a new question for it, I thought I'd ask here given the similarity. Like the OP, I need to determine the probability of change at some time t. What would be the best way to go about it? – arao6 Dec 2 '14 at 23:09
This is a good question but not very focused, so it will be impossible to go into each aspect in depth. As general advice: start with simple models, e.g. linear ones like glm, logistic regression, linear SVM. These are fairly straightforward to understand, fast to train and typically come with fewer bells and whistles (= headaches) than their nonlinear counterparts.
An important question to answer before deciding on a method is whether you only care about predictions or whether you want to understand them. This is essentially a choice between white box models that are interpretable (such as logistic regression) or black box models that are designed to offer better predictive performance but without being easily interpretable. Most people with a medical background want white box models, but their usefulness depends entirely on the task at hand. Based on your question, you seem to focus on making accurate predictions.
An important aspect of building any model is having enough data of sufficient quality. Some of the things you want to do require a lot of data. For instance, if you want to predict how forms change for a given patient, this is inevitably learned based on previously seen similar patients. If you want to predict detailed changes this will be difficult and would require thousands of patients (since you need to have enough similar ones to learn underlying patterns). Another question is whether or not you can reasonably make such predictions, e.g. does medical history provide enough insight to project what will happen in the future? This would probably be possible in some highly specific cases (for instance a diabetic that develops cardiovascular issues), but not in general.
As a quick primer, some keywords that may be useful:
1. If you want to predict some discrete outcome with a finite number of values, you are dealing with classification. Typically, methods deal with binary classification (2 classes) but are extendable to more than two.
2. If you want to predict some continuous outcome, you are doing regression. For instance, predicting the time before a patient returns.
3. If you only want to learn some structure within the data, you are doing clustering. For instance, do groups of similar patients exist? (not to one specific patients, but in all of the data)
• Great answer. Like the OP, I'm also not very good at math. Do you know of any applied examples on the net of those three types? I seem to find a lot of research papers and math-heavy jargon but nothing with simple descriptions for beginners. – arao6 Dec 3 '14 at 1:17 | 1,153 | 5,424 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-35 | latest | en | 0.954172 |
http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=patch;h=95124aa2037f01f4d4aeaa55ecb2dacf785b6cd6 | 1,660,216,313,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00103.warc.gz | 28,408,215 | 2,586 | From 95124aa2037f01f4d4aeaa55ecb2dacf785b6cd6 Mon Sep 17 00:00:00 2001 From: Jim Date: Thu, 12 Feb 2015 12:26:54 -0500 Subject: [PATCH] combinatory tweaks and formatting --- topics/week3_combinatory_logic.mdwn | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/topics/week3_combinatory_logic.mdwn b/topics/week3_combinatory_logic.mdwn index 53c2cd67..0b3390ee 100644 --- a/topics/week3_combinatory_logic.mdwn +++ b/topics/week3_combinatory_logic.mdwn @@ -156,7 +156,7 @@ enough to define arbitrary functions. We've already established that the behavior of combinatory terms can be perfectly mimicked by lambda terms: just replace each combinator with its -equivalent lambda term, i.e., replace `I` with `\x.x`, replace `K` with `\fxy.x`, +equivalent lambda term, i.e., replace `I` with `\x.x`, replace `K` with `\xy.x`, and replace `S` with `\fgx.fx(gx)`. So the behavior of any combination of combinators in Combinatory Logic can be exactly reproduced by a lambda term. @@ -177,7 +177,7 @@ Assume that for any lambda term T, [T] is the equivalent combinatory logic term. 1. [a] a 2. [(M N)] ([M][N]) 3. [\a.a] I - 4. [\a.M] K[M] assumption: a does not occur free in M + 4. [\a.M] K[M] when a does not occur free in M 5. [\a.(M N)] S[\a.M][\a.N] 6. [\a\b.M] [\a[\b.M]] @@ -189,7 +189,7 @@ The third rule should be obvious. The fourth rule should also be fairly self-evident: since what a lambda term such as `\x.y` does it throw away its first argument and return `y`, that's exactly what the combinatory logic translation should do. And indeed, `Ky` is a function that throws away its argument and returns `y`. The fifth rule deals with an abstract whose body is an application: the `S` combinator takes its next argument (which will fill the role of the original variable a) and copies it, feeding one copy to the translation of \a.M, and the other copy to the translation of \a.N. This ensures that any free occurrences of a inside M or N will end up taking on the appropriate value. Finally, the last rule says that if the body of an abstract is itself an abstract, translate the inner abstract first, and then do the outermost. (Since the translation of [\b.M] will not have any lambdas in it, we can be sure that we won't end up applying rule 6 again in an infinite loop.) -[Fussy notes: if the original lambda term has free variables in it, so will the combinatory logic translation. Feel free to worry about this, though you should be confident that it makes sense. You should also convince yourself that if the original lambda term contains no free variables---i.e., is a combinator---then the translation will consist only of `S`, `K`, and `I` (plus parentheses). One other detail: this translation algorithm builds expressions that combine lambdas with combinators. For instance, the translation of our boolean false `\x.\y.y` is `[\x[\y.y]] = [\x.I] = KI`. In the intermediate stage, we have `\x.I`, which mixes combinators in the body of a lambda abstract. It's possible to avoid this if you want to, but it takes some careful thought. See, e.g., Barendregt 1984, page 156.] +(*Fussy notes:* if the original lambda term has free variables in it, so will the combinatory logic translation. Feel free to worry about this, though you should be confident that it makes sense. You should also convince yourself that if the original lambda term contains no free variables---i.e., is a combinator---then the translation will consist only of `S`, `K`, and `I` (plus parentheses). One other detail: this translation algorithm builds expressions that combine lambdas with combinators. For instance, the translation of our boolean false `\x.\y.y` is `[\x[\y.y]] = [\x.I] = KI`. In the intermediate stage, we have `\x.I`, which mixes combinators in the body of a lambda abstract. It's possible to avoid this if you want to, but it takes some careful thought. See, e.g., Barendregt 1984, page 156.) (Various, slightly differing translation schemes from combinatory logic to the lambda calculus are also possible. These generate different metatheoretical @@ -203,7 +203,7 @@ issue is whether reduction rules (in either the lambda calculus or Combinatory Logic) apply to embedded expressions. Generally, we want that to happen, but making it happen requires adding explicit axioms.) -Let's check that the translation of the false boolean behaves as expected by feeding it two arbitrary arguments: +Let's check that the translation of the `false` boolean behaves as expected by feeding it two arbitrary arguments: KIXY ~~> IY ~~> Y @@ -263,7 +263,7 @@ in two books in the 1990's. A final linguistic application: Steedman's Combinatory Categorial Grammar, where the "Combinatory" is from combinatory logic (see especially his 2012 book, Taking Scope). Steedman attempts to build -a syntax/semantics interface using a small number of combinators, including T ≡ `\xy.yx`, B ≡ `\fxy.f(xy)`, +a syntax/semantics interface using a small number of combinators, including `T` ≡ `\xy.yx`, `B` ≡ `\fxy.f(xy)`, and our friend `S`. Steedman used Smullyan's fanciful bird names for the combinators, Thrush, Bluebird, and Starling. -- 2.11.0 | 1,364 | 5,150 | {"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 | 0.820701 |
https://www.bartleby.com/questions-and-answers/what-would-be-the-linear-equation-for-this-function-andamp-the-graph-x1y134-m2/6de86412-d8cc-43aa-b155-a0368c2334bc | 1,582,602,032,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00090.warc.gz | 641,814,550 | 27,246 | What would be the linear equation for this function, & the graph?? (X1,Y1)=(-3,4), m=-2
Question
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What would be the linear equation for this function, & the graph??
(X1,Y1)=(-3,4), m=-2
check_circle
Step 1
To find the linear equation we need point slope formula.
Given slope = -2 and point=(x1,y1)=(-3,4)
Step 2
For y=-2x-2 we make a table.
Picked few random val...
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# Three archers each fire four arrows at a target. Joe’s
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## Solution for problem 11DQ Chapter 1
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Problem 11DQ
Three archers each fire four arrows at a target. Joe’s four arrows hit at points 10 cm above, 10 cm below, 10 cm to the left, and 10 cm to the right of the center of the target. All four of Moe’s arrows hit within 1 cm of a point 20 cm from the center, and Flo’s four arrows hit within 1 cm of the center. The contest judge says that one of the archers is precise but not accurate, another archer is accurate but not precise, and the third archer is both accurate and precise. Which description applies to which archer? Explain.
Step-by-Step Solution:
Step 1 of 3
Solution 11DQ Accuracy means how close a measured/obtained value is to the known or standard value of a physical quantity. Precision means the measurements of a particular physical quantity are all close to each other. Let us now analyse the situations given in the question. Joe: Joe is not precise. Because his hits are not at the same position or close by. They are all at a fixed distance from the center in all directions. Therefore, Joe is accurate, not precise. Moe: Moe is precise. Because his hits are all close to one another. But he is not accurate as all his hits are 20 cm away from the center. Flo: All the four hits of Flo are within 1 cm of the center. This means they are not only close to each other , but also close to the center. So, Flo is both accurate and precise.
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##### ISBN: 9780321675460
This full solution covers the following key subjects: arrows, precise, HIT, Archer, accurate. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “Three archers each fire four arrows at a target. Joe’s four arrows hit at points 10 cm above, 10 cm below, 10 cm to the left, and 10 cm to the right of the center of the target. All four of Moe’s arrows hit within 1 cm of a point 20 cm from the center, and Flo’s four arrows hit within 1 cm of the center. The contest judge says that one of the archers is precise but not accurate, another archer is accurate but not precise, and the third archer is both accurate and precise. Which description applies to which archer? Explain.” is broken down into a number of easy to follow steps, and 103 words. Since the solution to 11DQ from 1 chapter was answered, more than 623 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 11DQ from chapter: 1 was answered by , our top Physics solution expert on 05/06/17, 06:07PM.
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# How To Have 'Straight-Line Communication' In The Workplace
POST WRITTEN BY
Charles Weathers
Charles is a veteran of the U.S. Air Force, American Leadership Forum Senior Fellow, Mediator, and the Founder of The Weathers Group.
I was not a big fan of math in school. It’s not that I could not do it; it just never excited me. I did just enough to pass math class and get on to something else. I never envisioned myself in any career that required a high level of math, so for me, math was a box to check off to get through school.
There is one math class that brings back fond memories: geometry. Perhaps I liked the subject because it was more visual, and I like working with shapes. Or, maybe I preferred this field of math because you can easily see the evidence of geometry by simply looking around any room, building or street. Everything has a shape; geometry is visible all around us.
The lesson that sticks out in my mind is a well-known geometric principle that explains the shortest distance between two points is a straight line. As a professional leadership coach, I have seen how this lesson applies not only to mathematical equations, but also to the business world and methods of communication.
Consider this common scenario many of us have likely witnessed in an office setting: Person A (let's call him Tom) believes he has a problem with Person B (let's call her Jane). Real or perceived, Tom is struggling with this problem, and he feels he needs to talk to someone about the problem he has with Jane. He should address the problem with Jane directly, but he doesn't.
Instead, Tom approaches various colleagues in the office and starts complaining about Jane. This spirals throughout the office, and everyone is asked about Jane, except Jane herself. As a result, the problem is not verified, discussed or solved, and Jane is unaware Tom has a problem. From my perspective, this scenario is unfair to Jane; she's playing a major role in the latest office drama, and she doesn’t even know it.
Here comes the mathematical principle that explains the shortest distance between two points is a straight line. This scenario plays out differently if we use geometry. If Tom instead goes straight to Jane, they can address the problem directly and take a shorter route to the solution. I refer to this as "straight-line communication." To practice direct communication in the workplace, consider the following:
1. Go directly to the other person. Admittedly, being direct can be easier said than done. I've seen that some people associate directness with confrontation and even rudeness. But being direct does not have to be threatening or adversarial. In my experience, holding a direct conversation is one of the best things you can do to strengthen your relationship with another person. I believe we overcomplicate directness. Start by asking a simple question, such as, "I'm sensing there's a conversation that we should we have. Would you be willing to speak with me so we can prevent any confusion or misunderstanding between us?" I've found this can serve as a direct and respectful entry to a meaningful conversation.
2. Speak in a nonaccusatory manner. My grandmother used to tell me, "It's not what you say, but how you say it." Tone matters. When you engage in a direct conversation, use "I" language. For example, you might explain to them, "I would like to clarify my intentions," or, "I might be wrong, but I want to share my perspective with you." This way, you can avoid projecting or assigning certain feelings to the other person.
3. Ask for their perspective, feedback and assessment. Once you share your perspective, ask the other person to share what they hear. Ask them to give you their assessment from how they view the situation. It’s important to let them know you want them to tell you what they truly believe, not what they think you want to hear.
4. Be open-minded. Being open-minded means being willing to consider the input of others and being willing to change your mind based on the introduction of new and relevant information. After receiving the feedback of the other person, you might find that your initial assessment and assumptions were not accurate. I've found one of the best ways to be open-minded is to be more concerned with getting it right than being right.
In my experience, "straight-line communication" can prevent you from making decisions based solely on your feelings and assumptions. It shows that you value and respect the relationships you have with your colleagues, as well as builds trust and reduces office drama.
Do what you can to reduce communication drama in the workplace. If you have a concern or believe someone else has a problem with you, make a straight line to the person, and engage in a healthy dialogue. Don’t guess, wonder or assume. Take the straight line. After all, not only is it the shortest distance between two points — it’s the shortest distance between two people.
Forbes Coaches Council is an invitation-only community for leading business and career coaches. Do I qualify? | 1,069 | 5,151 | {"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} | 3.1875 | 3 | CC-MAIN-2023-23 | latest | en | 0.967674 |
http://math.stackexchange.com/questions/73775/a-question-on-hyperbolic-geometry | 1,469,385,379,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00091-ip-10-185-27-174.ec2.internal.warc.gz | 156,353,917 | 17,027 | A question on hyperbolic geometry
I am reading a book that seems to claim the following. I suspect that there may be a misprint, or some assumptions missing. For $A=(x_1,y_1,z_1),B=(x_2,y_2,z_2)\in R^3$, define $$\langle A,B\rangle=-z_1z_2+x_1x_2+y_1y_2.$$ Suppose $C=(c_1,c_2,c_3)$, $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$ are such that $\langle C,C\rangle=-1,c_3\ge 1$,$\langle u,u\rangle=1$, $\langle C,u\rangle=0$, $\langle v,v\rangle=1$ and $\langle C,v\rangle=0$. The book claims that $$|\langle u,v\rangle|\le 1,$$ which does not seem to be correct.
-
The book is correct. Define $$V = \{x \in \mathbb{R}^3\ |\ \langle x,C \rangle = 0\}.$$ Thus $V$ is a two-dimensional subspace of $\mathbb{R}^3$. The key point here is the following exercise :
Exercise : The restriction of $\langle \cdot, \cdot \rangle$ to $V$ is positive-definite (ie it is an actual inner product!).
This exercise implies the fact from your book via an application of the Cauchy-Schwarz inequality to your vectors $u$ and $v$.
Here's a hint for the exercise : your quadratic form on $\mathbb{R}^3$ has signature $1$. What must the signature of the restriction of it to $V$ be? | 390 | 1,154 | {"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} | 3.15625 | 3 | CC-MAIN-2016-30 | latest | en | 0.744005 |
https://hero.handmade.network/episode/code/day101 | 1,532,151,027,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592387.80/warc/CC-MAIN-20180721051500-20180721071500-00637.warc.gz | 672,500,890 | 7,232 | The Inverse and the Transpose
?
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0:46Recap our current situation
0:46Recap our current situation
0:46Recap our current situation
1:45Start thinking about how we'll transform Normals
1:45Start thinking about how we'll transform Normals
1:45Start thinking about how we'll transform Normals
3:51Blackboard: Rotating Normals (is pretty straightforward)
3:51Blackboard: Rotating Normals (is pretty straightforward)
3:51Blackboard: Rotating Normals (is pretty straightforward)
5:01Blackboard: Non-uniform scaling gets a little bit hairy
5:01Blackboard: Non-uniform scaling gets a little bit hairy
5:01Blackboard: Non-uniform scaling gets a little bit hairy
6:17Blackboard: Pretending we have the edge of a gem
6:17Blackboard: Pretending we have the edge of a gem
6:17Blackboard: Pretending we have the edge of a gem
8:41Blackboard: Vectors are not all the same
8:41Blackboard: Vectors are not all the same
8:41Blackboard: Vectors are not all the same
11:23Blackboard: Normals are written differently in linear algebra
11:23Blackboard: Normals are written differently in linear algebra
11:23Blackboard: Normals are written differently in linear algebra
12:46Blackboard: Sometimes you have to go down a math holeα
12:46Blackboard: Sometimes you have to go down a math holeα
12:46Blackboard: Sometimes you have to go down a math holeα
13:03Blackboard: Constructing the P vectors
13:03Blackboard: Constructing the P vectors
13:03Blackboard: Constructing the P vectors
14:17Blackboard: Matrix multiplication
14:17Blackboard: Matrix multiplication
14:17Blackboard: Matrix multiplication
24:18Blackboard: Transpose operation
24:18Blackboard: Transpose operation
24:18Blackboard: Transpose operation
26:19Blackboard: Inverse operation
26:19Blackboard: Inverse operation
26:19Blackboard: Inverse operation
30:24Blackboard: Gauss steps in with Gaussian Elimination
30:24Blackboard: Gauss steps in with Gaussian Elimination
30:24Blackboard: Gauss steps in with Gaussian Elimination
35:26Blackboard: Solving equations in multiple unknowns
35:26Blackboard: Solving equations in multiple unknowns
35:26Blackboard: Solving equations in multiple unknowns
36:33Blackboard: We're just entirely in the math holeβ
36:33Blackboard: We're just entirely in the math holeβ
36:33Blackboard: We're just entirely in the math holeβ
39:02Blackboard: We can regularise operations on the rows and columns of this matrix
39:02Blackboard: We can regularise operations on the rows and columns of this matrix
39:02Blackboard: We can regularise operations on the rows and columns of this matrix
39:49Blackboard: Use numbers to clearly demonstrate Gaussian Elimination
39:49Blackboard: Use numbers to clearly demonstrate Gaussian Elimination
39:49Blackboard: Use numbers to clearly demonstrate Gaussian Elimination
41:12Blackboard: This is not JZ's termγ
41:12Blackboard: This is not JZ's termγ
41:12Blackboard: This is not JZ's termγ
41:27Blackboard: You can then divide the last remaining term by whatever the target is
41:27Blackboard: You can then divide the last remaining term by whatever the target is
41:27Blackboard: You can then divide the last remaining term by whatever the target is
43:10Blackboard: We want to be able to multiply a matrix by something in order to produce that identity matrix
43:10Blackboard: We want to be able to multiply a matrix by something in order to produce that identity matrix
43:10Blackboard: We want to be able to multiply a matrix by something in order to produce that identity matrix
44:38Blackboard: The regular solution form for Gaussian Elimination
44:38Blackboard: The regular solution form for Gaussian Elimination
44:38Blackboard: The regular solution form for Gaussian Elimination
58:12Blackboard: Unfortunately we didn't quite get to the inverse transpose
58:12Blackboard: Unfortunately we didn't quite get to the inverse transpose
58:12Blackboard: Unfortunately we didn't quite get to the inverse transpose
1:02:26Q&Aδ
1:02:26Q&Aδ
1:02:26Q&Aδ
1:03:06 For inverting 2x2 matrices, a simple cofactor equation is quite efficient
🗪
1:03:06 For inverting 2x2 matrices, a simple cofactor equation is quite efficient
🗪
1:03:06 For inverting 2x2 matrices, a simple cofactor equation is quite efficient
🗪
1:03:26 All of the steps in elimination can be represented as a matrix. Starting with M and multiply by all those matrices you get the identity. So those matrices multiplied together are the inverse. Multiplying them all by the identity gives you the inverse. Is that the trick you are alluding to?
🗪
1:03:26 All of the steps in elimination can be represented as a matrix. Starting with M and multiply by all those matrices you get the identity. So those matrices multiplied together are the inverse. Multiplying them all by the identity gives you the inverse. Is that the trick you are alluding to?
🗪
1:03:26 All of the steps in elimination can be represented as a matrix. Starting with M and multiply by all those matrices you get the identity. So those matrices multiplied together are the inverse. Multiplying them all by the identity gives you the inverse. Is that the trick you are alluding to?
🗪
1:07:47 To find the inverse matrix couldn't we just rotate by the negative angle and scale by 1 over the amount that we scaled by originally?
🗪
1:07:47 To find the inverse matrix couldn't we just rotate by the negative angle and scale by 1 over the amount that we scaled by originally?
🗪
1:07:47 To find the inverse matrix couldn't we just rotate by the negative angle and scale by 1 over the amount that we scaled by originally?
🗪
1:17:32 I remember doing this by putting the identity besides the original matrix and performing each step to each
🗪
1:17:32 I remember doing this by putting the identity besides the original matrix and performing each step to each
🗪
1:17:32 I remember doing this by putting the identity besides the original matrix and performing each step to each
🗪
1:21:24 If you had a matrix such as int[2][2] with the example of abcd as the variables in place, couldn't you just swap a and d and make c and b negative?
🗪
1:21:24 If you had a matrix such as int[2][2] with the example of abcd as the variables in place, couldn't you just swap a and d and make c and b negative?
🗪
1:21:24 If you had a matrix such as int[2][2] with the example of abcd as the variables in place, couldn't you just swap a and d and make c and b negative?
🗪
1:22:31 I understood none of today's episode. Would you recommend that I rewatch it, or do you think all may become clear tomorrow?
🗪
1:22:31 I understood none of today's episode. Would you recommend that I rewatch it, or do you think all may become clear tomorrow?
🗪
1:22:31 I understood none of today's episode. Would you recommend that I rewatch it, or do you think all may become clear tomorrow?
🗪
1:23:31 If you write A and then I next to each other, and apply a bunch of operations, you are computing T3T2T1 A and T3T2T1 I. Because you ended up at the identity, T3T2T1A = I. So T3T2T1 = A^-1. Therefore T3T2T1 I = A^-1
🗪
1:23:31 If you write A and then I next to each other, and apply a bunch of operations, you are computing T3T2T1 A and T3T2T1 I. Because you ended up at the identity, T3T2T1A = I. So T3T2T1 = A^-1. Therefore T3T2T1 I = A^-1
🗪
1:23:31 If you write A and then I next to each other, and apply a bunch of operations, you are computing T3T2T1 A and T3T2T1 I. Because you ended up at the identity, T3T2T1A = I. So T3T2T1 = A^-1. Therefore T3T2T1 I = A^-1
🗪
1:26:46Blackboard: Look at all of this maths
1:26:46Blackboard: Look at all of this maths
1:26:46Blackboard: Look at all of this maths
1:27:30Blackboard: Setup for Day 102 and sign off
1:27:30Blackboard: Setup for Day 102 and sign off
1:27:30Blackboard: Setup for Day 102 and sign off | 2,250 | 8,228 | {"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-2018-30 | latest | en | 0.769508 |
http://mymathforum.com/advanced-statistics/5069-bayes-s-theorem-question-answer-check.html | 1,561,001,584,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.98/warc/CC-MAIN-20190620024754-20190620050754-00087.warc.gz | 120,473,098 | 8,667 | My Math Forum Bayes's Theorem Question Answer Check
January 3rd, 2009, 06:07 PM #1 Newbie Joined: Dec 2008 Posts: 2 Thanks: 0 Bayes's Theorem Question Answer Check An appliance store purchases electric ranges from 2 companies. From company A, 500 ranges are purchased, and 2% are defective. From company B, 850 ranges are purchased, and 2% are defective. Given that a range is defective, find the probability that it came from company B. My Answer= .63 Answer in Textbook= .531
January 4th, 2009, 09:04 PM #2 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Bayes's Theorem Question Answer Check I agree with your answer. There are 17 and 10 defective ranges, so 17/27 is the probability required.
January 6th, 2009, 06:15 AM #3 Newbie Joined: Dec 2008 Posts: 2 Thanks: 0 Re: Bayes's Theorem Question Answer Check Thanks for your help
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Contact - Home - Forums - Cryptocurrency Forum - Top | 415 | 1,431 | {"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-2019-26 | latest | en | 0.899695 |
https://jwilson.coe.uga.edu/EMAT6680/Huffman/InstructionalUnit/Day1_Definitions.html | 1,722,939,292,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640484318.27/warc/CC-MAIN-20240806095414-20240806125414-00672.warc.gz | 271,827,150 | 2,482 | # Day 1: Definitions
Objective: Have students understand the common definitions associated with circles such as radius, chord, and tangent, as well as the definition of a circle itself
Lesson:
Key Definitions:
- Circle: the set of all points in a plane equidistant from a given point called the center of the circle - The interior of a circle is all the points inside that circle - The exterior of a circle consists of all points outside the circle - Radius: the distance, r, from any point on the circle to the center of the circle - Two circles with the same radius are called congruent circles - Diameter: the longest distance across the circle, the length, d, of the segment passing through the center of the circle with both endpoints on the circle, d = 2r - Chord: a segment having both endpoints on the circle - Circumference: the distance around the circle, the perimeter, 2*pi*r = pi*d - Area: the interior of the circle, pi*r2 - Central Angle: an angle with endpoints on the circle and vertex at the center of the circle - Inscribed Angle: an angle with endpoints on the circle and vertex on the circle - Tangent: a line perpendicular to the radius which intersects only one point on the circle - The point of tangency is the point where the tangent line intersects the circle - Secant: a line passing intersecting two points on the circle - Two circles that intersect at one point are called tangent circles - Two circles with the same center are called concentric circles
Begin the class by having the students discuss their ideas about circles. Have the students define a circle and its radius and diameter as well as area and circumference. Then begin going over the remaining key definitions with the students, having them discuss what they believe the definitions of chord, secant, and tangent to be. Finally, with the assistance of the students, write formal definitions for circle, radius, diameter, chord, circumference, area, tangent, and secant and then label these concepts on a diagram.
Activity: Have the students discover the value of pi. Have a large selection of circular objects such as jars or cans or lids of different sizes so that each group can have at least four different objects. Form groups of approximately four students, give each group a tray containing a ruler, string, and at least four circular objects. In their groups, the students will use the string and the ruler to find the measure of the circumference of the object and the diameter of the object. Then the students will find the ratio of the circumference to the diameter, and this should give the approximately the value of pi. Click here for a sample activity worksheet.
Conclusion: Discuss the activity with the class, having them conjecture as to why this activity produces the value of pi.
<>
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Developed by Katherine Huffman | 596 | 2,886 | {"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} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.892551 |
http://ams.org/mathimagery/displayimage.php?album=36&pid=512 | 1,547,853,965,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660818.25/warc/CC-MAIN-20190118213433-20190118235433-00275.warc.gz | 12,785,065 | 11,531 | The connection between mathematics and art goes back thousands of years. Mathematics has been used in the design of Gothic cathedrals, Rose windows, oriental rugs, mosaics and tilings. Geometric forms were fundamental to the cubists and many abstract expressionists, and award-winning sculptors have used topology as the basis for their pieces. Dutch artist M.C. Escher represented infinity, Möbius bands, tessellations, deformations, reflections, Platonic solids, spirals, symmetry, and the hyperbolic plane in his works.
Mathematicians and artists continue to create stunning works in all media and to explore the visualization of mathematics--origami, computer-generated landscapes, tesselations, fractals, anamorphic art, and more.
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# "Meandering Eightfold Path," by Margaret Kepner (Washington, DC)
## Flat: 18" x 18"; Folded: 3" x 6" x 0.5", Archival Inkjet on Paper, 2012 This work is a modified version of an earlier design – The Eightfold Path. Colors have been added, and the flat 2D print has been deconstructed into a folding book format. The underlying design is a visual presentation of the five non-isomorphic groups of order eight: C8, C2 x C4, C2 x C2 x C2, D4, and Q8. It employs a visual vocabulary derived from a traditional quilt pattern, Drunkard’s Path. Each of the small shapes used in the design is a quarter circle in a square, scaled so that its area equals the square’s residual area. The 36-page book structure is created from a single sheet of paper through a series of cuts and folds. A continuous meander folding path is followed, with varying length fold-sequences, and no beginning nor end. When it is fully folded up, the book assumes a double-square footprint. A smaller-scale meander path, which would result in a continuous 144-page book, is expressed through color accents. -- Margaret Kepner
Art & Music, MathArchives Geometry in Art & Architecture, by Paul Calter (Dartmouth College) Harmony and Proportion, by John Boyd-Brent International Society of the Arts, Mathematics and Architecture Journal of Mathematics and the Arts Mathematics and Art, the April 2003 Feature Column by Joe Malkevitch Maths and Art: the whistlestop tour, by Lewis Dartnell Mathematics and Art, (The theme for Mathematics Awareness Month in 2003) MoSAIC - Mathematics of Science, Art, Industry, Culture Viewpoints: Mathematics and Art, by Annalisa Crannell (Franklin & Marshall College) and Marc Frantz (Indiana University) Visual Insight, blog by John Baez | 603 | 2,607 | {"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} | 2.75 | 3 | CC-MAIN-2019-04 | latest | en | 0.930228 |
http://dtsheet.com/doc/1414832/an268-pwm-amplifier-output-power-calculator | 1,498,368,914,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320438.53/warc/CC-MAIN-20170625050430-20170625070430-00010.warc.gz | 119,906,670 | 9,412 | ### AN268 PWM Amplifier Output Power Calculator
```AN268
Application Note
PWM Amplifier Output Power Calculator
by Randy Boudreaux
1. INTRODUCTION
The PWM amplifier output power calculator is used to determine the effective root mean square (rms) power for a
given speaker load. The spreadsheet is configured to calculate the actual power delivered to the speaker load for
three typical speaker-load configurations, half-bridge tied, full-bridge tied and parallel full-bridge tied. To access the
spreadsheet, click on this link: http://www.cirrus.com/en/support/AN268. The PWM Amplifier Output Calculator is
available in two forms. If a version of Microsoft® Excel is available on your local computer, then download the
Excel spreadsheet version of the power output calculator. If not, then use the interactive power calculator located
at the above web site.
2. AMPLIFIER OUTPUT CONFIGURATIONS
2.1
Half-Bridge
Half-bridge tied speaker loads are driven on a single side by the PWM output, with the other side of the load referenced to ground, see Figure 1. For single voltage rail systems with a voltage of VP, the amplifier output will contain
a DC voltage offset equal to VP divided by 2 that must be blocked from the speaker load. This is generally accomplished by using a large electrolytic capacitor (C) in series with the output following the 2-pole LC filter used to
attenuate the PWM switching frequency. The electrolytic capacitor is usually polarized and care must be take to
wire the "+" terminal of the capacitor towards the output of the PWM output. For proper operation and increased
circuit reliability, the maximum ripple current specification for the capacitor must be larger than the peak-to-peak
The capacitor (C) along with the speaker load (R), which is generally in the 4 ohm or less range for this configuration, form a high pass filter. The -3dB corner frequency is determined by the equation Fs = 1/(6.28*R*C). The particular speaker load supported and the desired corner frequency determines the size of the capacitor measured in
To avoid loud pops in the speaker caused by the instantaneous voltage across the DC blocking capacitor at the
beginning of the power on sequence, a voltage ramp up sequence should be used, such as the Ramp-up feature of
the CS44800/44600. This feature will quickly bring the output of the PWM amplifier to the half rail voltage of VP
divided by 2 in a linear fashion without causing any loud pops.
PWM Output
VP
Lfilt
C
R
Cfilt
Figure 1. Half-Bridge Output Configuration
http://www.cirrus.com
MAY '05
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2.2
Full-Bridge
With using the same VP voltage rail level as with the half-bridge configuration above, increased power to the
speaker load is accomplished by using a pair of PWM outputs providing a differential signal to drive a full-bridge
tied speaker load, see Figure 2. In this configuration, the speaker load is not referenced to ground and is continuously driven by both the "+PWM Output" and "-PWM Output" signals from the amplifier. This effectively doubles the
voltage across the load resulting in a quadrupling of the output power since the power is a function of the square of
the voltage.
With full-bridge configurations, care must be taken not to exceed the maximum current specification limits for the
components. This is accomplished by increasing the minimum speaker load over what was driven for the halfbridge example above (to 8 ohms versus 4 ohms). Since the power out is inversely proportional to the resistive
speaker load, then instead of quadrupling the power output, now the power output is double and thereby lowering
the maximum output current. Even though there are two Lfilt components in series for the full-bridge configuration,
the value for this inductor remains the same as for the half-bridge configuration since the speaker load is now doubled.
+ PWM Output
VP
Lfilt
R
Cfilt
- PWM Output
VP
Lfilt
Figure 2. Full-Bridge Output Configuration
2.3
Parallel Full-Bridge
Parallel full-bridge tied loads are driven differentially similar to full-bridge tied loads, but can deliver more current to
the speaker load, with more efficiency, see Figure 3. Multiple PWM outputs are tied together in a parallel fashion
which effectively causes the Rds_on of these switches to also be in parallel, thereby dividing the Rds_on as seen
by the load current by a factor of 2. The benefit of doubling VP to the load is maintained and the current carrying
capability is also doubled, allowing smaller speaker loads to be supported. This configuration is typically used
when higher power levels are required such as when driving a subwoofer channel.
Since smaller speaker loads of 4 ohms or less, similar to the half-bridge output topology, are allowed to be driven in
this configuration, the value for Lfilt must be halved to maintain a comparable LC filter response to the half-bridge
and full-bridge configurations. Also note that the parallel full-bridge configuration should only be attempted with
power stage devices which integrate all MOSFET devices in a single package. In this case, the characteristics for
each MOSFET switch are highly correlated and matched to the other MOSFET switches such as to avoid crossconduction currents and high levels of distortion due to switching errors.
2
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VP
+ PWM Output
VP
+ PWM Output
1/2L
filt
R
Cfilt
VP
- PWM Output
1/2Lfilt
VP
- PWM Output
Figure 3. Parallel Full-Bridge Output Configuration
2.4
Power Calculations
A view of the downloaded spreadsheet is shown below. To modify the Assumptions or the Speaker Load in the calculations, double-click any cell with blue text within the spreadsheet. All parameters shown in blue can be modified
with specific system or device parameters.
The Voltage Rail VP is the power supply nominal DC output expressed in volts. This value is taken at the input pin
of the PWM power device.
The modulation index is the amount of full-scale signal that can be output from the PWM amplifier. This value is
usually below 0.95 and is determined by both the modulator effectiveness and the ability of the PWM output power
stage to handle small input pulse widths.
The PWM output Rds_on is the nominal series resistance of the MOSFET device as specified by the manufacture.
As the temperature of the power device rises, so will this value.
The inductor resistance is the resistance value of the inductor at DC. This value is specified by the inductor manufacture and should be as low as possible. Typical values are between 0.05 and 0.15 ohms.
The capacitor's Equivalent Series Resistance (ESR) is specified by the component manufacturer. The capacitor
ESR is only used in the half-bridge configurations when a capacitor is required to block any DC offset.
The "Power Out" calculation is the total power output for the specified channel. This is not the power delivered to
the speaker load. The "Power Out" formula is: Power out = (Vrms)2/Rtotal, where Vrms is the root mean square of
the VP voltage rail multiplied by the modulation index. Rtotal is the sum of all the resistances along the amplifier
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output path. In addition to the speaker load, these include the resistance of all the MOSFETs and filter inductors.
Substituting gives:
Power out = (VP * Mod. Index/2.83)2 / (Spkr Load + PWM Rds_on + Ind. Resistance).
The 2.83 value in the denominator converts the DC value of VP to an RMS value. For half-bridge configurations,
the ESR of the capacitor is also added to the total resistance.
For full-bridge amplifier outputs, the current flow will actually incur the PWM Rds_on and the Inductor Resistance
twice as it flows from one PWM output, through the load, and into the other PWM output. Because of the additional
impedance, this configuration is a little less efficient than the others.
The parallel full-bridge amplifier is not only capable of supplying more current, it is more efficient than the fullbridge configuration. The reason is that since the two PWM outputs are in parallel, then the PWM Rds_on resistance of each output is also in parallel. This effectively cuts the total PWM Rds_on in half.
To calculate the actual power delivered to the speaker load, the "Total current" is calculated. The actual "Power at
the load" is then the square of the "Total current" multiplied times the speaker load. This power level is somewhat
lower than the "Power out" calculation due to the other resistive losses within the circuit. The 0.1% THD+N power
number is the typical full scale, non-clipping output power at the load. The 1% THD+N power value represents a
small amount of clipping when the output of the PWM modulator is brought above full scale and is approximately
the 0.1% power output multiplied by 1.15. To approximate the generally specified 10% THD+N power value given
by many amplifier manufacturers, multiply the 0.1% THD+N power output value by 1.75.
The "Power dissipation per channel" is the amount of power that will be dissipated within the MOSFETs for a single
channel. The calculation is simply the square of the "Total current" multiplied by the PWM Rds_on for that channel.
For bridge-tied channels, the effective PWM Rds_on is double that for a half-bridge channel. This increases the
overall power dissipation for a channel which is configured for full-bridge operation. If multiple channels are contained within a single power output stage module, then the total power dissipation for all channels must be summed
in order to calculate proper cooling requirements.
Assumptions
Voltage Rail Vp (volts)
Modulation Index
PWM Output Rds_on (ohms)
Capacitor ESR (ohms)
Inductor Resistance (ohms)
35
0.87
0.25
0.05
0.05
Calculations
Power out (W)
Total current (rms)
Total current (peak)
0.1% THD+N
1% THD+N (x1.15)
10% THD+N (x1.75)
Effciency
Power Dissipation (W) per
channel
4
Half-Bridged Tied
8
6
4
13.86 18.23 26.61
1.29
1.69
2.47
1.82
2.40
3.50
Full-Bridge Tied
8
6
4
53.85 70.16 100.67
2.50
3.26
4.68
3.54
4.61
6.61
Parallel Full-Bridge Tied
8
6
4
55.46 72.93 106.46
2.58
3.39
4.95
3.64
4.79
7.00
13.28
15.28
23.25
0.96
17.23
19.81
30.15
0.94
24.47
28.14
42.83
0.92
50.09 63.79 87.54
57.60 73.35 100.67
87.66 111.63 153.19
0.93
0.91
0.87
53.13 68.91 97.89
61.10 79.24 112.57
92.99 120.59 171.31
0.96
0.94
0.92
0.42
0.72
1.53
3.13
5.32
10.94
1.66
2.87
6.12
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3. REVISION HISTORY
Release
Rev 1
Date
May 2005
Changes
1st Preliminary Release
Table 1. Revision History
Contacting Cirrus Logic Support
For all product questions and inquiries contact a Cirrus Logic Sales Representative.
To find the one nearest to you go to www.cirrus.com
IMPORTANT NOTICE
Cirrus Logic, Inc. and its subsidiaries ("Cirrus") believe that the information contained in this document is accurate and reliable. However, the information is subject
to change without notice and is provided "AS IS" without warranty of any kind (express or implied). Customers are advised to obtain the latest version of relevant
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Microsoft, is a registered trademark of Microsoft Corporation.
AN268REV1
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``` | 3,280 | 13,492 | {"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} | 2.84375 | 3 | CC-MAIN-2017-26 | longest | en | 0.86864 |
https://leowiki.com/how-to-find-pmf-coin-toss-1656107849 | 1,696,149,768,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00134.warc.gz | 401,282,383 | 14,906 | Trang chủ » blog » Coin Flipping Probability Mass Function & Expectation
Coin Flipping Probability Mass Function & Expectation
$\begingroup$ The point is identical simpleton : The order of flip matters .
You are given four random variables $X_i, i = 1,2,3,4$, where each one stands for a flip of a coin, in ascending order of indices. You have to calculate the keep up quantity : $X = 8X_1 + 4X_2 + 2X_3 + X_4$, right ? This can have values between $0$ and $15$ .
For the foremost part : now, how is $X_i$ ? Well, it is $0$ or $1$ with probability half, so the expectation respect is just $E [ X ] = 8 \times 0.5 + 4 \times 0.5 + 2 \times 0.5+1 \times 0.5 = 7.5$, because the expectation of one toss is $0 \times 0.5 +1 \times 0.5 = 0.5$.
How about the mass officiate ? Well, each issue, from $0$ to $15$, has an equal probability of orgasm, because each one has a unique representation in binary, and the probability of heads and tails are the lapp. hence, it is merely $P ( one ) = \frac { 1 } { 16 }$, for $i=0,1, \ldots, 15$ ( there are sixteen possibilities in total, each occurs evenly credibly ) .
In the second question : The arithmetic mean of each $X_i$ nowadays is $0 \times \frac 13 + 1\times \frac 23 = \frac 23$, so then $E [ X ] = ( 8+4+2+1 ) \times\frac 23 = 10$ .
finally, here, every number has a unique binary star expansion, but this time the heads and tails do n’t have the same probability, so the mass affair changes, and it changes as follows : given the count of $1$ s in the expansion, so many tails are required for the issue to come, which happens with higher probability. so more the ones, more the probability .
so, we look at the binary star expansions :
1 ) The numbers having no $1$ sulfur is merely zero .
2 ) The numbers having precisely one $1$ are $2$, $4$, $8$, $1$ .
3 ) The numbers having precisely two $1$ s are $3$, $5$, $6$, $9$, $10$, $12$.
4 ) The numbers having precisely three ones are $7$, $11$, $13$, $14$ .
5 ) only $15$ has four ones .
therefore, we can write down the probabilities as follows :
1 ) For zero, the probability is $\frac { 1 } { 3^4 } = \frac { 1 } { 81 }$ .
2 ) For the numbers with only one $1$, the probability is $\frac { 1 } { 3^3 } \times \frac { 2 } { 3 } = \frac { 2 } { 81 }$ .
3 ) For the numbers having $2$ ones, the probability is $\frac { 1 } { 3^2 } \times \frac { 2^2 } { 3^2 } = \frac { 4 } { 81 }$.
4 ) For the numbers with three $1$ randomness, the probability is $\frac { 1 } { 3 } \times \frac { 2^3 } { 3^3 } = \frac { 8 } { 81 }$ .
5 ) For the numbers with all $1$ mho, the probability is $\frac { 2^4 } { 3^4 } = \frac { 16 } { 81 }$ .
hence, this is the probability distribution serve in the end : P ( x ) = \begin { cases } \frac { 1 } { 81 } & x=0 \\ \frac { 2 } { 81 } & x=1,2,4,8 \\ \frac { 4 } { 81 } & x= 3,5,6,9,10,12 \\ \frac { 8 } { 81 } & x = 7,11,13,14 \\ \frac { 16 } { 81 } & x = 15 \\ \end { cases }
reference : https://leowiki.com
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# In Rhombus Abcd; (I) If ∠A = 74° ; Find ∠B and ∠C. (Ii) If Ad = 7.5 Cm ; Find Bc and Cd. - Mathematics
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Sum
In rhombus ABCD;
(i) if ∠A = 74° ; find ∠B and ∠C.
(ii) if AD = 7.5 cm ; find BC and CD.
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#### Solution
AD || BC
∠A + ∠B = 180°
74° + ∠B = 180°
∠B =180° – 74°= 106°
opposite angles of Rhombus are equal.
∠A = ∠C = 74°
Sides of Rhombus are equal.
BC = CD = AD = 7.5 cm
(i) ∠B = 106° ; ∠C = 74°
(ii) BC = 7.5 cm and CD = 7.5 cm
Concept: Types of Quadrilaterals - Properties of Rhombus
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#### APPEARS IN
Selina Concise Mathematics Class 8 ICSE
Chapter 17 Special Types of Quadrilaterals
Exercise 17 | Q 3 | Page 198
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# JID: APM
## Applied Mathematical Modelling
journal homepage: www.elsevier.com/locate/apm
## An interval-valued hesitant fuzzy multigranulation rough set
over two universes model for steam turbine fault diagnosis
Chao Zhang a,b, Deyu Li a,b,, Yimin Mu c, Dong Song d
a
School of Computer and Information Technology, Shanxi University, Taiyuan 030006, Shanxi, China
b
Key Laboratory of Computational Intelligence and Chinese Information Processing of Ministry of Education, Taiyuan 030006, Shanxi,
China
c
College of Mechanical and Electrical Engineering, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, Jiangsu, China
d
Taiyuan Institute of China Coal Technology and Engineering Group, Taiyuan 030006, Shanxi, China
a r t i c l e i n f o a b s t r a c t
Article history: In the eld of mechanical engineering, steam turbine fault diagnosis is a dicult task
Received 17 September 2015 for mechanical engineers who are confronted with challenges in dealing with copious
Revised 13 August 2016
amounts of uncertain information. Different mechanical engineers may have their own
Accepted 19 October 2016
opinions about the system fault knowledge base that differs slightly from other mechani-
Available online xxx
cal engineers. Thus, to solve the problems presented by uncertain data analysis and group
Keywords: decision-making in steam turbine fault diagnosis, we propose a new rough set model that
Steam turbine fault diagnosis combines interval-valued hesitant fuzzy sets with multigranulation rough sets over two
Group decision-making universes, called an interval-valued hesitant fuzzy multigranulation rough set over two
Interval-valued hesitant fuzzy sets universes. In the multigranulation framework, both basic denitions and some important
Multigranulation rough sets over two properties of the proposed model are presented. Then, we develop a general approach to
universes steam turbine fault diagnosis by using the proposed model. Lastly, an illustrative example
Interval-valued hesitant fuzzy is provided to verify the established approach and demonstrate its validity and applicabil-
multigranulation rough sets over two
ity.
universes
1. Introduction
Fault diagnosis technology has grown in importance in multiple areas of engineering, and issues with fault diagnosis
technology have attracted much attention from researchers and practitioners. Until now, there are various data-based fault
diagnosis techniques that have been developed, including expert systems [1], neural networks [2], and fuzzy approaches
[35]. Taking advantages of human expertise and experiential knowledge, the expert system has been successfully utilized
in steam turbine fault diagnosis. However, the limitations of acquiring expert knowledge and maintaining the rules database
for steam turbines are revealed in expert system-based fault diagnosis. An additional potential weakness of this approach
is that the rules database may include conicting expert knowledge that may vary from case to case. This variation may
preclude a general mathematical formulation that can be utilized for fault diagnosis in different elds. Another typical
fault diagnosis technique, neural networks, addresses the shortcomings of expert systems by direct acquisition of fault
knowledge from a set of training samples and exhibition of highly nonlinear input-output relationships. However, neural
network-based fault diagnosis requires sucient and compatible training samples to ensure proper training of the model.
Corresponding author at: School of Computer and Information Technology, Shanxi University, No.92 Wucheng Road, Taiyuan 030 0 06, China.
E-mail addresses: zhch3276152@163.com (C. Zhang), lidysxu@163.com (D. Li), minminwell@163.com (Y. Mu), tymkysd@126.com (D. Song).
http://dx.doi.org/10.1016/j.apm.2016.10.048
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
http://dx.doi.org/10.1016/j.apm.2016.10.048
JID: APM
ARTICLE IN PRESS [m3Gsc;October 31, 2016;7:37]
## 2 C. Zhang et al. / Applied Mathematical Modelling 000 (2016) 112
Free from the limitations of these models, the fuzzy approach can be used to suciently model the complicated relationship
between faults and their features in steam turbine fault diagnosis. Additionally, many kinds of inherent uncertainties, such
as inaccuracy, incompleteness, and inconsistency, are rampant in expressions of fault information available to engineers.
Based on fuzzy set theory [6] and fuzzy logic [7], fuzzy approaches have been widely applied to fault diagnosis processes.
The classical fuzzy set is limited by imperfect and uncertain information induced by several sources of vagueness, so
several modications and extensions have been introduced, including interval-valued fuzzy sets [8], type-2 fuzzy sets [9],
intuitionistic fuzzy sets [10], and Pythagorean fuzzy sets [11]. These extensions overcome the limitations of classical fuzzy
sets in different ways but are based on similar rationales.
Recently, considering that decision-makers are usually hesitant and irresolute for one thing or another when establishing
a common membership degree, Torra and Narukawa [12] and Torra [13] introduced the concept of hesitant fuzzy sets,
opening a new door for research on decision-making under hesitant environments. However, for real-life decision-making
problems, the available information may not be sucient for experts to provide their preferences with crisp values. Thus, a
possible solution is to represent such preferences by interval values. As an extension form of hesitant fuzzy sets, Chen et al.
[14] presented the concept of interval-valued hesitant fuzzy sets, which permits the membership degrees of an element to
a given set are dened by several possible interval values. It is evident that interval-valued hesitant fuzzy sets can reect
individuals hesitancy more objectively than other widely developed fuzzy set approaches. Thereafter, many scholars have
studied interval-valued hesitant fuzzy sets from different points of view and obtained many meaningful results [1519].
Compared to classical fuzzy sets, since interval-valued hesitant fuzzy sets can better model insuciency and hesitancy in
available information, this approach can better handle the more uncertain information in fault diagnosis well, and provide
mechanical engineers with an improved way to include their understanding about the system fault knowledge base.
Considering the above facts, as one of the commonly used multi-attribute decision-making methods, rough set theory
[20] is an effective tool to acquire knowledge with its core concepts of lower and upper approximations. Moreover, rough
set theory has demonstrated advantages in selecting signicant attributes and explaining decisions according to attribute
reductions and distinguishable relations. Over the past few years, two kinds of important generalizations of rough set
theory, fuzzy rough sets [21] and multigranulation rough sets over two universes [22], have been developed in different
application contexts. To effectively solve the problems of interval-valued hesitant fuzzy data analysis in fault diagnosis
technology, we intend to integrate rough set theory with interval-valued hesitant fuzzy data analysis. We next itemize the
necessities of combining those two theories with interval-valued hesitant fuzzy sets in detail.
(1) Rough sets and fuzzy sets, as two primary theories utilized for handling uncertain information in various information
systems, are commonly viewed as correlative, but distinct and complementary. Determination of the best approach to gener-
alize a rough set model to the fuzzy case is required for the development of rough set theory. In order to deal with various
fuzzy information systems by utilizing rough set theory, Dubois and Prade [21] rst constructed the concept of fuzzy rough
sets. This work has been expanded by many recent studies of the fusion of hesitant fuzzy set theory with rough sets [2326].
(2) In realistic rough set-based decision-making problems, when considering concept approximations and rules acqui-
sitions in the background of multi-source information systems, in order to acquire knowledge eciently, a reasonable way
is to analyze these multi-source information systems directly rather than gathering each information systems as an entire
information system. In this situation, the single-granulation rough set model exposes limitations that the computational
times of knowledge discovery are long. Therefore, it is useful to describe a target concept through multiple binary relations
according to a users different requirements. From the perspective of granular computing [27], Qian et al. [28,29] introduced
multigranulation rough sets in optimistic and pessimistic styles. Based on multigranulation rough sets model, Sun and Ma
[22] further pointed out many kinds of decision-making information, such as the relationship between faults and their
features in steam turbine fault diagnosis, involves two different types of objects, each of which belongs to a different
universe of discourse, and the utilization of two universes and their interrelations may invalidate multigranulation rough
sets model. Thus, by utilizing rough sets over two universes model [3034], they proposed multigranulation rough sets
over two universes. In light of the above, multigranulation rough sets over two universes could not only describe the
decision-making information in different universes, but also synthesize multi-group expert preferences to form a nal
decision result by aggregating multiple binary relations. Thus, multigranulation rough sets over two universes can be seen
as a relatively superior information fusion strategy under conditions of several different granulation levels.
On the basis of the above analysis, it is meaningful to utilize multigranulation rough sets over two universes model
for multi-source information systems analysis. However, multigranulation rough sets over two universes can only handle
crisp information systems and have limitations in processing various fuzzy information systems. Considering that interval-
valued hesitant fuzzy sets can model insuciency and hesitancy in decision-making information effectively, we aim to
develop a new data analysis model in fault diagnosis techniques under the environment of interval-valued hesitant fuzzy
information. Thus, to expand the application domain of multigranulation rough sets over two universes, through combining
interval-valued hesitant fuzzy sets with multigranulation rough sets over two universes, it is necessary to construct an
interval-valued hesitant fuzzy multigranulation rough set over two universes model and further explore its application in
steam turbine fault diagnosis.
The paper is organized as follows. In Section 2, we briey introduce interval-valued hesitant fuzzy sets, interval-valued
hesitant fuzzy rough sets over two universes, and multigranulation rough sets over two universes. In Section 3, we
present interval-valued hesitant fuzzy multigranulation rough sets over two universes and discuss relevant properties.
Section 4 presents an analysis of the decision-making problems in steam turbine fault diagnosis. In Section 5, we give the
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
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application of the developed decision-making approach to a steam turbine fault diagnosis case and make some comparison
analysis. In Section 6, the paper is concluded and some directions for future work are suggested.
2. Preliminaries
Some preliminary concepts are given in this section to explain our proposal, including interval-valued hesitant fuzzy
sets, interval-valued hesitant fuzzy rough sets over two universes, and multigranulation rough sets over two universes.
These will be utilized in the subsequent analysis.
## 2.1. Interval-valued hesitant fuzzy sets
Interval-valued hesitant fuzzy sets (IVHFSs) were originally introduced by Chen et al. [14]. By replacing crisp numbers
with interval numbers in the environment of hesitant fuzzy sets, interval-valued hesitant fuzzy sets can be regarded as a
more powerful structure in reecting an experts hesitance in expressing preferences over objects. Prior to the introduction
of interval-valued hesitant fuzzy sets, we rst present the concept of hesitant fuzzy sets.
Hesitant fuzzy sets (HFSs) were introduced by Torra and Narukawa [12] and Torra [13]. HFSs permit the membership
degree of an element to a reference set expressed by several possible values between 0 and 1.
Denition 2.1 [13]. Let U be the universe of discourse, and a hesitant fuzzy set F on U is dened as a function hF (x) that
returns a subset of [0, 1], which can be expressed as the following mathematical symbol:
F = {x, hF (x )|x U },
where hF (x) is a set of some different nite values in [0, 1], describing the possible membership degrees of the element x
U to the set F. For convenience, hF (x) is called a hesitant fuzzy element. The set of all hesitant fuzzy elements is called
HFEs.
In what follows, we present the denition of interval-valued hesitant fuzzy sets introduced by Chen et al. [14].
Denition 2.2 [14]. Let U be the universe of discourse, and D[0, 1] be the set of all closed subintervals of [0, 1], an interval-
valued hesitant fuzzy set A on U can be expressed as the following mathematical symbol:
A = {x, hA (x )|x U },
where hA (x): U D[0, 1] denotes all possible interval-valued membership degrees of the element x U to the set A.
For convenience, hA (x) is called an interval-valued hesitant fuzzy element, which reads hA (x ) = { | hA (x )}, where =
[ L , U ] is an interval number, and L = inf and U = sup denote the lower and upper limits of , respectively. An
interval-valued hesitant fuzzy element is the basic unit of an interval-valued hesitant fuzzy set. The set of all interval-valued
hesitant fuzzy elements is called IVHFEs.
Suppose that U is the universe of discourse, then we denote the set of all interval-valued hesitant fuzzy sets on U as
IVHF(U). Therefore, A IVHF(U).
In what follows, we introduce some special interval-valued hesitant fuzzy sets.
(1) A is referred to as an empty interval-valued hesitant fuzzy set [24] if and only if hA (x ) = {[0, 0]} for all x in U, the
empty interval-valued hesitant fuzzy set is denoted by in this paper.
(2) A is referred to as a full interval-valued hesitant fuzzy set [24] if and only if hA (x ) = {[1, 1]} for all x in U, the full
interval-valued hesitant fuzzy set is denoted by U in this paper.
Example 2.1. Let U = {x1 , x2 , x3 } be a universe set, hA (x1 ) = {[0.1, 0.2], [0.3, 0.4]}, hA (x2 ) = {[0.4, 0.5], [0.5, 0.6]} and
hA (x3 ) = {[0.3, 0.5], [0.6, 0.7]} are three IVHFEs to a set A. Then, the interval-valued hesitant fuzzy set A can be expressed
as follows:
A = {x1 , {[0.1, 0.2], [0.3, 0.4]}, x2 , {[0.4, 0.5], [0.5, 0.6]}, x3 , {[0.3, 0.5], [0.6, 0.7]}}.
To compare the magnitude of different interval numbers and IVHFEs, we rst present the methods of comparing two
interval numbers introduced by Xu and Da [35].
Denition 2.3 [35]. Let a = [aL , aU ] and b = [bL , bU ] be two interval numbers with the following: (1) a = b if aL = bL and
aU = bU ; (2) a + b = [aL + bL , aU + bU ]. Let la = aU aL and lb = bU bL , then the degree of possibility of a b is dened as
bU aL
p(a b) = max 1 max ,0 ,0 .
la + lb
Similarly, the degree of possibility of a b is dened as
aU bL
p(a b) = max 1 max ,0 ,0 .
la + lb
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## 4 C. Zhang et al. / Applied Mathematical Modelling 000 (2016) 112
The above denition is proposed to compare and rank different interval numbers. In what follows, we further present
the score function for IVHFEs.
Denition 2.4 [14]. For an IVHFE hA (x), s(hA (x ) ) = l (h 1(x )) h (x ) , where l(hA (x)) is the number of interval values in an
A A
IVHFE hA (x) and s(hA (x)) is an interval value belonging to [0, 1]. For two IVHFEs, hA (x) and hB (x), if s(hA (x)) s(hB (x)), then
hA (x) hB (x).
It is clear that different IVHFEs have different number of interval values, and the interval values might be out of order.
Thus, Chen et al. [14] introduced the following assumptions.
(1) All elements in each IVHFE hA (x) are arranged in increasing order by Denition 2.3. Suppose that hA (k ) (x ) is the
kth largest interval number in the IVHFE hA (x). Then, hA (k ) (x ) is expressed as hA (k ) (x ) = [hA (k )L (x ), hA (k )U (x )], where
hA (k )L (x ) = inf hA (k ) (x ) and hA (k )U (x ) = sup hA (k ) (x ) denote the lower and upper limits of hA (k ) (x ), respectively.
(2) If for two IVHFEs hA (x) and hB (x), l(hA (x)) = l(hB (x)), then the two IVHFEs hA (x) and hB (x) should be of the same length
l = max{l (hA (x ) ), l (hB (x ) )}. If the number of elements in hA (x) is less than in hB (x), we should extend hA (x) by adding
the maximum element until hA (x) and hB (x) are the same length.
Based on the above assumptions, Zhang et al. [24] dened some operational laws for interval-valued hesitant fuzzy sets.
## (1) The complement of A is denoted by Ac such that x U, hAc ( x ) = hA ( x ) =
(k )U ( k )L
{[1 hA (x ), 1 hA (x )]|k = 1, 2, . . . l}.
(2) The intersection of A and B is denoted by A B such that x U, h AB ( x ) = hA ( x ) hB ( x ) =
{[hA (k)L (x ) hB (k)L (x ), hA (k)U (x ) hB (k)U (x )]|k = 1, 2, . . . l}.
(3) The union of A and B is denoted by A B such that x U, h AB ( x ) = hA ( x ) hB ( x ) =
{[hA (k)L (x ) hB (k)L (x ), hA (k)U (x ) hB (k)U (x )]|k = 1, 2, . . . l}.
In the above denition, the operations c , and are dened on interval-valued hesitant fuzzy sets, respectively.
Conversely, the operations , and are dened on corresponding interval-valued hesitant fuzzy elements, respectively.
We next present the properties of the above operations.
Theorem 2.1 [24]. Let U be the universe of discourse, and suppose that A, B and C are three interval-valued hesitant fuzzy sets.
Then, the following properties are true:
## (1) Double negation law: ( hA (x ) ) = hA (x );
(2) De Morgans laws: (hA (x ) hB (x )) = ( hA (x ) ) ( hB (x ) ) and (hA (x ) hB (x )) = ( hA (x ) ) ( hB (x ) );
(3) Commutativity: hA (x ) hB (x ) = hB (x ) hA (x ) and hA (x ) hB (x ) = hB (x ) hA (x );
(4) Associativity: hA (x ) (hB (x ) hC (x )) = (hA (x ) hB (x )) hC (x ) and hA (x ) (hB (x ) hC (x )) = (hA (x ) hB (x )) hC (x );
(5) Distributivity: hA (x ) (hB (x ) hC (x )) = (hA (x ) hB (x )) (hA (x ) hC (x )) and hA (x ) (hB (x ) hC (x )) =
(hA (x ) hB (x )) (hA (x ) hC (x )).
Example 2.2. Suppose that A and B are two interval-valued hesitant fuzzy sets. Let hA (x ) = {[0.1, 0.2], [0.3, 0.4]} and hB (x ) =
{[0.4, 0.5], [0.5, 0.6]} be two IVHFEs. Then, we obtain the complement, intersection and union as follows:
(1) hA (x ) = {[1 0.2, 1 0.1], [1 0.4, 1 0.3]} = {[0.6, 0.7], [0.8, 0.9]}; similarly, hB (x ) = {[0.4, 0.5], [0.5, 0.6]};
(2) hA (x ) hB (x ) = {[0.1 0.4, 0.2 0.5], [0.3 0.5, 0.4 0.6]} = {[0.1, 0.2], [0.3, 0.4]};
(3) hA (x ) hB (x ) = {[0.1 0.4, 0.2 0.5], [0.3 0.5, 0.4 0.6]} = {[0.4, 0.5], [0.5, 0.6]}.
Denition 2.6 [24]. Let U be the universe of discourse, A, B IVHF(U), if hA (x) hB (x) holds for each x U such that
hA (x ) hB (x ) hA (k )L (x ) hB (k )L (x ), hA (k )U (x ) hB (k )U (x ), k = 1, 2, . . . l, then A is referred to as an interval-valued hesitant
fuzzy subset of B, and is denoted by A B. Note that is reexive, transitive and antisymmetric on IVHF(U).
## 2.2. Interval-valued hesitant fuzzy rough sets over two universes
In this subsection, we rst present interval-valued hesitant fuzzy (IVHF) relations over two universes.
Denition 2.7 [24]. Let U, V be two universes of discourse. An interval-valued hesitant fuzzy relation R from U to V is
dened by:
R = {(x, y ), hR (x, y )|(x, y ) U V },
where hR (x, y): U V D[0, 1] is a set of interval values in D[0, 1]. The family of all IVHF relations on U V is denoted
by IVHFR(U V).
According to the IVHF relation over two universes, we present the denition of interval-valued hesitant fuzzy rough sets
over two universes as follows.
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
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Denition 2.8 [24]. Let U, V be two universes of discourse, and R IVHFR(U V), the pair (U, V, R) is called an interval-
valued hesitant fuzzy approximation space over two universes. For any A IVHF(V), the lower and upper approximations of
A are dened by:
R (A ) = x, hR(A ) (x ) |x U ;
R (A ) = x, hR(A ) (x ) |x U ,
where hR(A ) (x ) = yV {hRc (x, y ) hA (y )}; hR(A ) (x ) = yV {hR (x, y ) hA (y )}. The pair (R(A ), R(A )) is referred to as an interval-
valued hesitant fuzzy rough set over two universes of A with respect to (U, V, R), both R (A) and R(A ) are interval-valued
hesitant fuzzy sets, R and R are lower and upper IVHF rough approximation operators, respectively.
## 2.3. Multigranulation rough sets over two universes
According to two different approximation strategies, Sun and Ma [22] established two different multigranulation rough
sets over two universes including optimistic and pessimistic ones.
Denition 2.9 [30]. Let U, V be two universes of discourse. R is a family of binary compatibility relations from U to V in
terms of a binary mapping family Fi , where Fi : U 2V , u{v V|(u, v) Ri .}, Ri R, i = 1, 2, . . . , m. Then, we call the
ordered triple set (U, V, R) the multigranulation approximation space over two universes.
By Denition 2.9, the following denition gives the formal representation of optimistic and pessimistic multigranulation
rough sets over two universes.
Denition 2.10 [22]. Let F and G be two binary mappings from universe U to V. For any X V, the optimistic and pessimistic
lower and upper multigranulation approximations with respect to (U, V, R) are dened by:
aprOF +G (X ) = {x U |F (x ) X G(x ) X };
aprF +G (X ) = aprOF +G (X c )c ;
O
aprPF +G (X ) = {x U |F (x ) X G(x ) X };
aprF +G (X ) = aprPF +G (X c )c .
P
## The pair (aprO
F +G
(X ), aprOF +G (X )) is referred to as an optimistic multigranulation rough set over two universes. Similarly, we
call the pair (aprPF +G (X ), aprPF +G (X ) ) a pessimistic multigranulation rough set over two universes.
In the above denition, the word optimistic means that in multiple independent granular structures, at least one
granular structure must satisfy the inclusion condition between an equivalence class and a target concept. Conversely, the
word pessimistic means that all granular structures must satisfy the inclusion condition between an equivalence class and
a target concept.
## 3. IVHF multigranulation rough sets over two universes
In this section, based on constructive approach to interval-valued hesitant fuzzy rough sets over two universes, we
extend interval-valued hesitant fuzzy relations over two universes into the background of multigranulation rough sets.
## 3.1. Optimistic IVHF multigranulation rough sets over two universes
Denition 3.1. Let U, V be two universes of discourse and Ri IV HF R(U V )(i = 1, 2, . . . , m) be m interval-valued hesitant
fuzzy relations over U V, the pair (U, V, Ri ) is called an interval-valued hesitant fuzzy multigranulation approximation
space over two universes. For any A IVHF(V), the optimistic IVHF multigranulation lower and upper approximations over
two universes of A are dened by:
m
O
i=1 Ri ( A ) = x, hm (x ) |x U ;
O
i=1 Ri (A )
m O
R
i=1 i ( A ) = x, h O ( x ) |x U ,
i=1 Ri (A )
m
where hm O
(A )
m (x ) = i=1 yV hRi c (x, y ) hA (y ) ; hm O ( x ) = m
i=1 yV
hRi (x, y ) hA (y ) .
i=1 Ri i=1 Ri (A )
O m O
The pair ( m i=1 Ri (A ), i=1 Ri (A ) ) is an optimistic IVHF multigranulation rough set over two universes of A with respect
to (U, V, Ri ). Additionally, the IVHF multigranulation rough set over two universes will reduce to an IVHF rough set over two
universes if m = 1.
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
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Theorem 3.1. Let U, V be two universes of discourse and Ri IV HF R(U V )(i = 1, 2, . . . , m) be m interval-valued hesitant fuzzy
relations over U V. For any A, A IVHF(V), the optimistic IVHF multigranulation lower and upper approximations over two
universes satisfy the following properties:
m O O m O m O
(1) Ri ( Ac ) = ( m
i=1 i=1 Ri (A ) ) ,
c
i=1 Ri (A ) = ( i=1 Ri (A ) ) ;
c c
m m O m O
(2) A A m
O O
i=1 Ri (A ) i=1 Ri (A ), A A i=1 Ri (A ) i=1 Ri (A );
m O O O O
O m O
i=1 Ri (A A ) = i=1 Ri (A ) i=1 Ri (A ), i=1 Ri (A A ) = i=1 Ri (A ) i=1 Ri (A );
m m m m
(3)
m O O O O
O O
i=1 Ri (A A ) i=1 Ri (A ) i=1 Ri (A ), i=1 Ri (A A ) i=1 Ri (A ) i=1 Ri (A ).
m m m m m
(4)
Proof.
m O
i=1 Ri (A ) = {x, i=1 yV {hRi c (x, y ) hAc (y )}|x U } = {x, i=1 yV
(1) For all x U, we have c m m
{( hRi (x, y ) ) ( hA (y ))}|x U } = {x, i=1 yV { (hRi (x, y ) hA (y ))}|x U } = {x, (i=1 yV {(hRi (x, y )
m m
O m O m O
hA (y ))} )|x U } = ( m i=1 Ri (A ) ) .
c
i=1 Ri (A ) = ( i=1 Ri (A ) ) is similarly obtained.
c c
## (2) Since A A , by Denition 2.6, we have hA (y )hA (y ) hA (k )L (y ) hA(k )L (y ), hA (k )U (y ) hA(k )U (y ).
Thus, it follows that m {h (ck )L (x, y ) hA (k )L (y )} m
i=1 yV
{h (ck )L (x, y ) hA(k )L (y )} and m
i=1 yV
{h (ck )U (x, y )
i=1 yV
R i R i R i
hA (k )U (y )} m {hR (ck )U (x, y ) hA(k )U (y )}.
i=1 yV i
m m O m O
Therefore, we have A A m
O O
i=1 Ri (A ) i=1 Ri (A ). A A i=1 Ri (A ) i=1 Ri (A ) is obtained in an iden-
tical fashion.
m O
i=1 Ri (A A ) = {x, i=1 yV {hR c (x, y ) hAA (y )}|x U } = {x, i=1 yV {hR c (x, y )
(3) For all x U, we have m m
i i
(hA (y ) hA (y ))}|x U } = {x, [m {hR (ck )L (x, y ) (hA (k )L (y ) hA(k )L (y ))}, m
i=1 yV
{hR (ck )U (x, y ) (hA (k )U (y )
i=1 yV
i i
hA(k )U (y ))}]|x U } = {x, [m {(hR (ck )L (x, y ) hA (k )L (y )) (hR (ck )L (x, y ) hA(k )L (y ))}, m
i=1 yV
{(hR (ck )U (x, y )
i=1 yV
i i i
hA (k )U (y )) (hR (ck )U (x, y ) hA(k )U (y ))}]|x U } = {x, hm O (x )|x U } {x, hm O (x )|x U} =
i i=1 Ri (A ) i=1 Ri (A )
m O m O m O
m O m O
i=1 Ri ( A ) i=1 Ri (A ) Similarly, i=1 Ri (A A ) = i=1 Ri (A ) i=1 Ri (A ) is obtained.
m m m m O
Ri ( A A ) Ri (A ) and Ri ( A A )
O O O
(4) Based on above conclusions, it is easy to obtain i=1 i=1 Ri ( A ) i=1 i=1
m O m O
i=1 Ri (A ) i=1 Ri (A ).
In this theorem, (1) indicates the complement of optimistic interval-valued hesitant fuzzy multigranulation rough sets
over two universes; (2) indicates the monotone of optimistic interval-valued hesitant fuzzy multigranulation rough sets
over two universes with respect to a variety of interval-valued hesitant fuzzy targets; (3) and (4) indicate the multiplication
and addition of optimistic interval-valued hesitant fuzzy multigranulation rough sets over two universes.
Theorem 3.2. Let U, V be two universes of discourse and Ri , Ri IV HF R(U V )(i = 1, 2, . . . , m) be two interval-valued hesitant
fuzzy relations over U V. If Ri Ri , for any A IVHF(V), the following properties are true.
m m
Ri (A )
O O
(1) i=1 i=1 Ri (A ), for all A IVHF(V);
m O m O
(2) i=1 Ri (A ) i=1 Ri (A ), for all A IVHF(V).
Proof. Since Ri Ri , according to Denition 2.6, we have hR (ck )L (x, y ) h(ck )L (x, y ) and hR (ck )U (x, y ) h(ck )U (x, y )
i Ri i Ri
m O
for all (x, y) (U V). Therefore, it follows that i=1 i R ( A ) = { x, m
i=1 yV
{ h R c ( x, y ) h A ( y )}|x U } = i
{x, [m {hR (ck )L (x, y ) hA (k )L (y )}, m
i=1 yV
{hR (ck )U (x, y ) hA (k )U (y )}]|x U } {x, [m
i=1 yV
{hR(ck )L (x, y ) hA (k )L (y )},
i=1 yV
i i i
(k )U m m m
(x, y ) hA (k)U (y )}]|x U } = Ri (A )
O O O
m {hR c
i=1 yV i=1 Ri (A ) Hence, we have i=1 i=1 Ri ( A ). Similarly,
i
m O m O
i=1 Ri (A ) i=1 Ri (A ) is obtained.
In this theorem, (1) and (2) indicate the lower and upper approximations in optimistic interval-valued hesitant
fuzzy multigranulation rough sets over two universes are monotonic with respect to the monotonic forms of multiple
interval-valued hesitant fuzzy relations.
## 3.2. Pessimistic IVHF multigranulation rough sets over two universes
Denition 3.2. Let U, V be two universes of discourse and Ri IV HF R(U V )(i = 1, 2, . . . , m) be m interval-valued hesitant
fuzzy relations over U V, the pair (U, V, Ri ) is called an interval-valued hesitant fuzzy multigranulation approximation
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
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## C. Zhang et al. / Applied Mathematical Modelling 000 (2016) 112 7
space over two universes. For any A IVHF(V), the pessimistic IVHF multigranulation lower and upper approximations over
two universes of A are dened by:
m P
Ri ( A ) = x, hm Ri
P
(A )
( x ) |x U ;
i=1
i=1
m
P
Ri ( A ) = x, hm P
Ri ( A )
(x ) |x U ,
i=1
i=1
where hm ( x ) = m
P
i=1 yV
hR c (x, y ) hA (y ) ; hm P (x ) = m =1
yV hRi (x, y ) hA (y ) .
i=1 Ri (A )
i
i=1 i ( )
i
m P
R A
m P
The pair R
i=1 i ( A ), R
i=1 i ( A ) is a pessimistic IVHF multigranulation rough set over two universes of A with respect
to (U, V, Ri ).
Theorem 3.3. Let U, V be two universes of discourse and Ri IV HF R(U V )(i = 1, 2, . . . , m) be m interval-valued hesitant fuzzy
relations over U V. For any A, A IVHF(V), the pessimistic IVHF multigranulation lower and upper approximations over two
universes satisfy the following properties:
m P P m P m P
(1) Ri ( Ac ) = ( m
i=1 i=1 Ri (A ) ) ,
c
i=1 Ri (A ) = ( i=1 Ri (A ) ) ;
c c
P m P
(2) A A m
P P
i=1 Ri (A ) i=1 Ri (A ), A A i=1 Ri (A ) i=1 Ri (A );
m m
m P P P P
P m P
i=1 Ri (A A ) = i=1 Ri (A ) i=1 Ri (A ), i=1 Ri (A A ) = i=1 Ri (A ) i=1 Ri (A );
m m m m
(3)
m P P P P
P P
i=1 Ri (A A ) i=1 Ri (A ) i=1 Ri (A ), i=1 Ri (A A ) i=1 Ri (A ) i=1 Ri (A ).
m m m m m
(4)
In this theorem, (1) indicates the complement of pessimistic interval-valued hesitant fuzzy multigranulation rough sets
over two universes; (2) indicates the monotone of pessimistic interval-valued hesitant fuzzy multigranulation rough sets
over two universes with respect to a variety of interval-valued hesitant fuzzy targets; (3) and (4) indicate the multiplication
and addition of pessimistic interval-valued hesitant fuzzy multigranulation rough sets over two universes.
Theorem 3.4. Let U, V be two universes of discourse and Ri , Ri IV HF R(U V )(i = 1, 2, . . . , m) be two interval-valued hesitant
fuzzy relations over U V. If Ri Ri , for any A IVHF(V), the following properties are true.
m m
Ri (A )
P P
(1) i=1 i=1 Ri (A ), for all A IVHF(V);
m P m P
(2) i=1 Ri (A ) i=1 Ri (A ), for all A IVHF(V).
In this theorem, (1) and (2) indicate the lower and upper approximations in pessimistic interval-valued hesitant
fuzzy multigranulation rough sets over two universes are monotonic with respect to the monotonic forms of multiple
interval-valued hesitant fuzzy relations.
3.3. Relationships between optimistic and pessimistic IVHF multigranulation rough sets over two universes
Theorem 3.5. Let U, V be two universes of discourse and Ri IV HF R(U V )(i = 1, 2, . . . , m) be m interval-valued hesitant fuzzy
relations over U V. For any A IVHF(V), the optimistic and pessimistic IVHF multigranulation lower and upper approximations
over two universes satisfy the following properties:
m P m O
(1) i=1 Ri (A ) i=1 Ri (A );
m P m O
(2) i=1 Ri (A ) i=1 Ri (A ).
m O
i=1 Ri (A ) = {x, i=1 yV {hRi c (x, y ) hA (y )}|x U } {x, i=1 yV {hRi c (x, y ) hA (y )}|x U } =
Proof. For any x U, m m
m P m P m O
i=1 Ri (A ). i=1 Ri (A ) i=1 Ri (A ) is similarly obtained.
From Theorem 3.5, it is noted that the pessimistic interval-valued hesitant fuzzy multigranulation lower approximation
is included in the optimistic interval-valued hesitant fuzzy multigranulation lower approximation, while the optimistic
interval-valued hesitant fuzzy multigranulation upper approximation is included in the pessimistic interval-valued hesitant
fuzzy multigranulation upper approximation.
## 4. Steam turbine fault diagnosis approach
In this section, we present a novel method to the decision-making problem of steam turbine fault diagnosis with the aid
of IVHF multigranulation rough sets over two universes. Specically, some key points of the established decision-making
approach are summarized in the following subsections.
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
http://dx.doi.org/10.1016/j.apm.2016.10.048
JID: APM
ARTICLE IN PRESS [m3Gsc;October 31, 2016;7:37]
## 4.1. Application model
In the domain of electric power production and management, steam turbine generator units are a signicant piece of
equipment in the power industry. One of the common fault types of a steam turbine generator unit is vibration, which can
be triggered by many factors, such as mechanical structure, load, and vacuum degree. To deal with this type of mechanical
fault diagnosis, failures of steam turbine generator units have been specied based on the different amplitude ratios
of vibration signals in various frequency ranges. Moreover, many researchers have established the relationship between
the cause and fault phenomena of steam turbine generator units, and numerous studies have been conducted based on
intelligent computational and mathematical methods, such as expert systems, neural networks, and fuzzy approaches.
However, few studies have addressed the situation where the vibration signal amplitude ratio value is in the background of
hesitant and interval ambiguity. By utilizing the model of IVHF multigranulation rough sets over two universes, we can not
only express the vibration signal amplitude ratio value by using interval-valued hesitant fuzzy elements, but can also solve
group decision-making problems by providing optimistic and pessimistic strategies.
There are two kinds of objections related to steam turbine fault diagnosis, namely, the fault pattern sets and the
frequency range sets of the steam turbine generator unit. Therefore, we let U = {x1 , x2 , . . . , x j } be a set of fault patterns and
V = {y1 , y2 , . . . , yk } be a set of frequency ranges. Let Ri IV HF R(U V )(i = 1, 2, . . . , m) be m interval-valued hesitant fuzzy
relations over U V, which reects the steam turbine fault diagnosis knowledge base with interval-valued hesitant fuzzy
information given by m experts. We also let A IVHF(V) be the fault testing sample. Then, we obtain an interval-valued
hesitant fuzzy decision information system (U, V, Ri , A) of a typical steam turbine fault diagnosis procedure.
Next, we present a decision-making approach for the above-mentioned problem by using IVHF multigranulation rough
sets over two universes. First, according to Denitions 3.1 and 3.2, we separately compute lower and upper approxima-
tions of optimistic and pessimistic IVHF multigranulation rough sets over two universes of A. That is, we obtain the sets
m O m O m P m P
i=1 Ri (A ), i=1 Ri (A ), i=1 Ri (A ) and i=1 Ri (A ). Then, according to the operational laws presented in [24]:
## hA (x ) hB (x ) = {[hA (k )L (x ) + hB (k )L (x ) hA (k )L (x )hB (k )L (x ), hA (k )U (x ) + hB (k )U (x ) hA (k )U (x )hB (k )U (x )]},
m O m O m P m P
we further obtain the sets i=1 Ri ( A ) i=1 Ri (A ) and i=1 Ri ( A )
Ri (A ). Based on Denition 2.4, we can
i=1
m O m O
calculate the score function values of interval-valued hesitant fuzzy elements belonging to i=1 Ri (A ) i=1 Ri (A ) and
m P m P
i=1 Ri (A ) i=1 Ri (A ). Based on a previously established decision-making strategy [32], we then present the decision
rules for steam turbine fault diagnosis using the proposed rough set model. First, to facilitate the decision-making process,
we denote:
O
m O
m
T1 = imax Ri ( A ) ( xi ) Ri ( A ) ( xi ) ,
xi U
i=1 i=1
P
m P
m
T2 = jmax Ri ( A ) x j Ri ( A ) x j ,
x j U
i=1 i=1
O P
m O
m
m P
m
T3 = kmax Ri ( A ) ( xk ) Ri ( A ) ( xk ) Ri ( A ) ( xk ) Ri ( A ) ( xk ) .
xk U
i=1 i=1 i=1 i=1
From the viewpoint of decision-making, optimistic multigranulation rough sets are based on the seeking common
ground while reserving differences (SCRD) strategy, which implies that one reserves both common decisions and inconsis-
tent decisions at the same time. Thus, this opinion can be viewed as a risk-seeking decision strategy. In contrast, pessimistic
multigranulation rough sets are based on the seeking common ground while eliminating differences (SCED) strategy, this
strategy indicates that one reserves common decisions while deleting inconsistent decisions. Hence, this opinion can be
seen as a risk-averse decision strategy. According to the above decision rules, T1 is the optimistic decision-making criterion,
T2 is the pessimistic decision-making criterion, and T3 is the weighted decision-making criterion of T1 and T2 with the
weighted value 0.5. Based on the above denitions, the decision rules can be presented as follows:
## (1) If T1 T2 T3 = , then xi (i T1 T2 T3 ) is the determined fault pattern;
(2) If T1 T2 T3 = and T1 T2 = , then xi (i T1 T2 ) is the determined fault pattern. Otherwise, if T1 T2 T3 = and
T1 T2 = , then xi (i T3 ) is the determined fault pattern.
4.2. Algorithm for steam turbine fault diagnosis using IVHF multigranulation rough sets over two universes
We present an algorithm for steam turbine fault diagnosis by utilizing IVHF multigranulation rough sets over two
universes:
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
http://dx.doi.org/10.1016/j.apm.2016.10.048
JID: APM
ARTICLE IN PRESS [m3Gsc;October 31, 2016;7:37]
## C. Zhang et al. / Applied Mathematical Modelling 000 (2016) 112 9
Algorithm 1 Steam turbine fault diagnosis based on IVHF multigranulation rough sets over two universes.
Require: The relation between the universes U and V provided by an expert (U, V, Ri ) and a fault testing sample A.
Ensure: The determined fault pattern.
m O m O m P m P
1: Calculate i=1 Ri (A ), i=1 Ri (A ), i=1 Ri (A ) and i=1 Ri (A ), respectively;
m O m O m P m P
2: Calculate i=1 Ri (A ) i=1 Ri (A ) and i=1 Ri (A ) i=1 Ri (A ), respectively;
m O m O m P m P
3: Determine the score function values for the sets i=1 Ri (A ) i=1 Ri (A ) and i=1 Ri (A ) i=1 Ri (A ), respectively;
4: Compute T1 , T2 , T3 , T1 T2 T3 and T1 T2 , and conrm the determined fault pattern.
Table 1
Knowledge of system fault given by expert 1.
R1 y1 y2 y3 y4 y5
x1 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.85, 0.88], [0.85, 0.93]}
x2 {[0, 0]} {[0.27, 0.32], [0.28, 0.33]} {[0.08, 0.1], [0.09, 0.13]} {[0.54, 0.6], [0.56, 0.7]} {[0, 0]}
x3 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.3, 0.58], [0.4, 0.6]}
x4 {[0.09, 0.1], [0.1, 0.12]} {[0.78, 0.8], [0.8, 0.82]} {[0, 0]} {[0.08, 0.1], [0.1, 0.11]} {[0, 0]}
x5 {[0.09, 0.11], [0.11, 0.12]} {[0.09, 0.1], [0.1, 0.11]} {[0.08, 0.1], [0.1, 0.12]} {[0.09, 0.11], [0.11, 0.12]} {[0.18, 0.2], [0.19, 0.21]}
x6 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.18, 0.2], [0.2, 0.22]}
x7 {[0, 0]} {[0, 0]} {[0.08, 0.11], [0.1, 0.12]} {[0.86, 0.89], [0.88, 0.93]} {[0, 0]}
x8 {[0, 0]} {[0.27, 0.3], [0.29, 0.32]} {[0.08, 0.1], [0.09, 0.12]} {[0.54, 0.6], [0.57, 0.62]} {[0, 0]}
x9 {[0.85, 0.9], [0.89, 0.93]} {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x10 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
R1 y6 y7 y8 y9
x1 {[0.03, 0.05], [0.04, 0.06]} {[0.04, 0.06], [0.05, 0.07]} {[0, 0]} {[0, 0]}
x2 {[0, 0]} {[0, 0]} {[0, 0]} {[0.07, 0.12], [0.08, 0.14]}
x3 {[0.4, 0.6], [0.45, 0.62]} {[0.08, 0.12], [0.1, 0.13]} {[0, 0]} {[0, 0]}
x4 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x5 {[0.08, 0.12], [0.1, 0.13]} {[0.08, 0.12], [0.1, 0.13]} {[0.08, 0.1], [0.1, 0.12]} {[0.08, 0.1], [0.1, 0.12]}
x6 {[0.12, 0.15], [0.15, 0.17]} {[0.37, 0.4], [0.4, 0.45]} {[0, 0]} {[0.22, 0.25], [0.24, 0.28]}
x7 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x8 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x9 {[0, 0]} {[0, 0]} {[0.08, 0.1], [0.09, 0.12]} {[0, 0]}
x10 {[0.77, 0.8], [0.8, 0.83]} {[0.19, 0.2], [0.2, 0.23]} {[0, 0]} {[0, 0]}
5. Case study
In this section, we illustrate the eciency of our proposed algorithm by using a steam turbine fault diagnosis problem
with IVHFS information. To enhance the accuracy and reliability of the steam turbine fault diagnosis procedure, we aim to
solve the problem in the context of group decision-making. In group decision-making, each mechanical engineer might have
differing thoughts about the system fault knowledge base, but should also have the common goal to reach a nal consensus.
Moreover, the knowledge of system faults has been modeled and studied by many scholars utilizing sets such as interval-
valued intuitionistic fuzzy sets [3], vague sets [4], and single-valued neutrosophic sets [5]. To express the aforementioned
hesitant and interval ambiguity, based on existing knowledge of system faults and following detailed discussions about
several previous cases of steam turbine fault diagnosis with mechanical engineers, we obtained the required knowledge of
the system fault dataset with respect to this paper from a state-owned scientic and technological enterprise.
Let U = {x1 , x2 , . . . , x10 } be a set of fault patterns, where x j ( j = 1, 2, . . . , 10) represents unbalance, pneumatic force
couple, offset center, oil-membrane oscillation, radial impact friction of rotor, symbiosis looseness, damage of antithrust
bearing, surge, looseness of bearing block, and nonuniform bearing stiffness. We also let V = {y1 , y2 , . . . , yk } be a set of fre-
quency ranges, where yk (k = 1, 2, . . . , 9) stands for frequency ranges for different spectra: C1 (0.01 0.39 f ), C2 (0.4 0.49 f ),
C3 (0.5f), C4 (0.51 0.99 f ), C5 (f), C6 (2f), C7 (3 5 f ), C8 (odd times of f), and C9 (high frequency > 5f). The steam turbine fault
diagnosis knowledge base with IVHFS information is presented in Tables 1,2 and 3. The fault testing sample is also given as
follows. We aim to seek the determined fault pattern by utilizing the proposed model.
In steam turbine fault diagnosis, assume that we take a fault testing sample, which is represented by the following
IVHFS information:
A = {y1 , {[0.39, 0.4], [0.39, 0.41]}, y2 , {[0.07, 0.09], [0.08, 0.1]}, y3 , {[0, 0]}, y4 , {[0.06, 0.07], [0.06, 0.08]},
y5 , {[0, 0]}, y6 , {[0.12, 0.14], [0.13, 0.15]}, y7 , {[0, 0]}, y8 , {[0, 0]}, y9 , {[0.33, 0.36], [0.34, 0.37]}}.
Next, we calculate lower and upper approximations of optimistic and pessimistic IVHF multigran-
3 O
ulation rough sets over two universes of A. i=1 Ri (A ) = {x1 , {[0.07, 0.15], [0.12, 0.15]}, x2 , {[0.3, 0.45],
[0.4, 0.46]}, x3 , {[0.38, 0.6], [0.42, 0.6]},.x4 , {[0.18, 0.22], [0.2, 0.22]}, x5 , {[0.79, 0.82], [0.81, 0.82]}, x6 , {[0.55, 0.63], [0.6,
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
http://dx.doi.org/10.1016/j.apm.2016.10.048
JID: APM
ARTICLE IN PRESS [m3Gsc;October 31, 2016;7:37]
## 10 C. Zhang et al. / Applied Mathematical Modelling 000 (2016) 112
Table 2
Knowledge of system fault given by expert 2.
R2 y1 y2 y3 y4 y5
x1 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.85, 0.93]}
x2 {[0, 0]} {[0.28, 0.31]} {[0.09, 0.11], [0.1, 0.13]} {[0.55, 0.7]} {[0, 0]}
x3 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.3, 0.58]}
x4 {[0.09, 0.11], [0.1, 0.12]} {[0.78, 0.82]} {[0, 0]} {[0.08, 0.11]} {[0, 0]}
x5 {[0.09, 0.12]} {[0.09, 0.1], [0.1, 0.11]} {[0.08, 0.12]} {[0.09, 0.12]} {[0.18, 0.19], [0.18, 0.21]}
x6 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.18, 0.22]}
x7 {[0, 0]} {[0, 0]} {[0.08, 0.12]} {[0.86, 0.88], [0.86, 0.93]} {[0, 0]}
x8 {[0, 0]} {[0.27, 0.32]} {[0.08, 0.11], [0.09, 0.12]} {[0.54, 0.62]} {[0, 0]}
x9 {[0.85, 0.93]} {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x10 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
R2 y6 y7 y8 y9
x1 {[0.03, 0.06], [0.04, 0.07]} {[0.04, 0.07]} {[0, 0]} {[0, 0]}
x2 {[0, 0]} {[0, 0]} {[0, 0]} {[0.08, 0.13]}
x3 {[0.4, 0.58], [0.42, 0.62]} {[0.08, 0.1], [0.1, 0.13]} {[0, 0]} {[0, 0]}
x4 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x5 {[0.08, 0.13]} {[0.08, 0.12], [0.1, 0.13]} {[0.08, 0.12]} {[0.08, 0.09], [0.1, 0.12]}
x6 {[0.12, 0.16], [0.14, 0.17]} {[0.37, 0.45]} {[0, 0]} {[0.22, 0.26], [0.24, 0.28]}
x7 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x8 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x9 {[0, 0]} {[0, 0]} {[0.08, 0.12]} {[0, 0]}
x10 {[0.77, 0.79], [0.79, 0.83]} {[0.19, 0.23]} {[0, 0]} {[0, 0]}
Table 3
Knowledge of system fault given by expert 3.
R3 y1 y2 y3 y4 y5
x1 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.85, 0.9], [0.88, 0.93]}
x2 {[0, 0]} {[0.28, 0.3], [0.29, 0.32]} {[0.09, 0.12]} {[0.55, 0.65], [0.6, 0.7]} {[0, 0]}
x3 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.3, 0.55], [0.31, 0.58]}
x4 {[0.09, 0.11]} {[0.78, 0.81], [0.8, 0.82]} {[0, 0]} {[0.08, 0.1], [0.8, 0.11]} {[0, 0]}
x5 {[0.09, 0.1], [0.1, 0.12]} {[0.09, 0.11]} {[0.08, 0.09], [0.08, 0.12]} {[0.09, 0.1], [0.1, 0.12]} {[0.18, 0.21]}
x6 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0.18, 0.2], [0.18, 0.22]}
x7 {[0, 0]} {[0, 0]} {[0.08, 0.1], [0.1, 0.12]} {[0.86, 0.93]} {[0, 0]}
x8 {[0, 0]} {[0.27, 0.31], [0.29, 0.32]} {[0.08, 0.12]} {[0.54, 0.58], [0.55, 0.62]} {[0, 0]}
x9 {[0.85, 0.88], [0.88, 0.93]} {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x10 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
R3 y6 y7 y8 y9
x1 {[0.03, 0.06]} {[0.04, 0.05], [0.05, 0.07]} {[0, 0]} {[0, 0]}
x2 {[0, 0]} {[0, 0]} {[0, 0]} {[0.08, 0.1], [0.1, 0.13]}
x3 {[0.4, 0.62]} {[0.08, 0.13]} {[0, 0]} {[0, 0]}
x4 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x5 {[0.08, 0.1], [0.09, 0.13]} {[0.08, 0.13]} {[0.08, 0.1], [0.1, 0.12]} {[0.08, 0.12]}
x6 {[0.12, 0.17]} {[0.36, 0.44], [0.37, 0.45]} {[0, 0]} {[0.22, 0.28]}
x7 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x8 {[0, 0]} {[0, 0]} {[0, 0]} {[0, 0]}
x9 {[0, 0]} {[0, 0]} {[0.08, 0.1], [0.09, 0.12]} {[0, 0]}
x10 {[0.77, 0.83]} {[0.19, 0.21], [0.2, 0.23]} {[0, 0]} {[0, 0]}
0.64]}, x7 , {[0.07, 0.14], [0.12, 0.14]}, . .x8 , {[0.38, 0.46], [0.42, 0.46]}, x9 , {[0.39, 0.4], [0.39, 0.41]}, x10 , {[0.17, 0.23], [0.21,
3 O
0.23]}}; i=1 Ri (A ) = {x1 , {[0.03, 0.05], [0.03, 0.06]}, x2 , {[0.07, 0.12], [0.08, 0.14]}, x3 , {[0.12, 0.14], [0.13, 0.15]}, .x4 ,
{[0.09, 0.1], [0.09, 0.11]}, x5 , {[0.09, 0.12], [0.1, 0.13]}, x6 , {[0.22, 0.25], [0.22, 0.28]}, x7 , {[0.06, 0.07],
[0.06, 0.08]}, . .x8 , {[0.07, 0.09], [0.08, 0.1]}, x9 , {[0.39, 0.4], [0.39, 0.41]}, x10 , {[0.12, 0.14], [0.13, 0.15]}};
3 P
i=1 Ri (A ) = {x1 , {[0.07, 0.12], [0.07, 0.15]}, x2 , {[0.3, 0.4], [0.3, 0.45]}, x3 , {[0.38, 0.55], [0.38, 0.6]}, .x4 , {[0.18, 0.2],
[0.18, 0.22]}, x5 , {[0.79, 0.81], [0.79, 0.82]}, x6 , {[0.55, 0.6], [0.55, 0.63]}, . .x7 , {[0.07, 0.12], [0.07, 0.14]},
3 P
x8 , {[0.38, 0.43], [0.38, 0.46]}, x9 , {[0.39, 0.4], [0.39, 0.41]}, x10 , {[0.17, 0.2], [0.17, 0.23]}}; i=1 Ri (A ) =
{x1 , {[0.03, 0.06], [0.04, 0.07]}, x2 , {[0.08, 0.13], [0.1, 0.14]}, x3 , {[0.12, 0.14], [0.13, 0.15]}, .x4 , {[0.09, 0.11], [0.1, 0.12]},
x5 , {[0.09, 0.13], [0.11, 0.13]}, x6 , {[0.22, 0.28], [0.24, 0.28]}, . .x7 , {[0.06, 0.07], [0.06, 0.08]}, x8 , {[0.07, 0.09], [0.08,
0.1]}, x9 , {[0.39, 0.4], [0.39, 0.41]}, x10 , {[0.12, 0.14], [0.13, 0.15]}}.
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
http://dx.doi.org/10.1016/j.apm.2016.10.048
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ARTICLE IN PRESS [m3Gsc;October 31, 2016;7:37]
## C. Zhang et al. / Applied Mathematical Modelling 000 (2016) 112 11
m O m O m P m P 3 O
Then, we further obtain i=1 Ri ( A ) i=1 Ri ( A ) and i=1 Ri ( A ) i=1 Ri ( A ) as follows: i=1 Ri ( A )
3 O
i=1 Ri (A ) = {x1 , {[0.0979, 0.1925], [0.1464, 0.201]}, x2 , {[0.349, 0.516], [0.448, 0.5356]},.x3 , {[0.4544, 0.656],
[0.4954, 0.66]}, x4 , {[0.2538, 0.298], [0.272, 0.3058]}, x5 , {[0.8089, 0.8416], [0.829, 0.8434]}, x6 , {[0.649,
0.7225], [0.688, 0.7408]}, x7 , {[0.1258, 0.2002], [0.1728, 0.2088]}, x8 , {[0.4234, 0.5086], [0.4664, 0.514]}.,
3 P 3 P
x9 , {[0.6279, 0.64], [0.6279, 0.6519]}, x10 , {[0.2696, 0.3378], [0.3127, 0.3455]}}; i=1 Ri (A ) i=1 Ri (A ) =
{x1 , {[0.0979, 0.1728], [0.1072, 0.2095]}, x2 , {[0.356, 0.478], [0.37, 0.527]},x3 , {[0.4544, 0.613], [0.4606, 0.66]}, x4 ,
{[0.2538, 0.288], [0.262, 0.3136]}, x5 , {[0.8089, 0.8347], [0.8131, 0.8434]}, x6 , {[0.649, 0.712], [0.658, 0.7336]}, x7 ,
{[0.1258, 0.1816], [0.1258, 0.2088]}, x8 , {[0.4234, 0.4813], [0.4296, 0.514]}., x9 , {[0.6279, 0.64], [0.6279, 0.6519]}, x10 ,
{[0.2696, 0.312], [0.2779, 0.3455]}}.
Finally, according to Denition 2.4, we calculate the score function values of interval-valued hesitant fuzzy elements
O O P P
belonging to 3i=1 Ri (A ) 3i=1 Ri (A ) and 3i=1 Ri (A ) 3i=1 Ri (A ), respectively. The ranking results for optimistic and
pessimistic IVHF multigranulation rough sets on two universes of A are identical. That is, the ranking order of all faults is:
x5 > x6 > x9 > x3 > x8 > x2 > x10 > x4 > x7 > x1 . According to the diagnostic order of the above results, it is not dicult
to obtain T1 T2 T3 = {5}. Thus, from the arguments of the above results, it is easy to nd that the turbine vibration fault
is initiated by rotor radial impact friction.
To validate the effectiveness of the proposed model based on IVHF multigranulation rough sets over two universes,
a comparison analysis is conducted by utilizing the most commonly used aggregation operators for interval-valued hes-
itant fuzzy information. As established in previous work [14], we let h j ( j = 1, 2, . . . , n) be a collection of IVHFEs and
w = (1/n, 1/n, . . . , 1/n)T be the weight vector of of hj with equal weight. Then, we have the following special aggregation
operators.
(1) The interval-valued hesitant fuzzy averaging (IVHFA) operator IV HF A(h1 , h2 , . . . , hn ) = nj=1 ( n1 h j ) =
n 1/n n 1/n
{[1 j=1 (1 j ) , 1 j=1 (1 j ) ]|1 h1 , 2 h2 , . . . , n hn};
L U
1/n
(2) The interval-valued hesitant fuzzy geometric (IVHFG) operator IV HF G(h1 , h2 , . . . , hn ) = nj=1 h j =
n 1/n n 1/n
{[ j=1 ( j ) , j=1 ( j ) ]|1 h1 , 2 h2 , . . . , n hn}.
L U
Through utilizing the above two aggregation operators, we can aggregate the IVHF relations R1 , R2 and R3 in Tables 1,
2 and 3 to a single IVHF relation R. Then, we calculate the score function values of interval-valued hesitant fuzzy elements
belonging to R(A ) R(A ) within the background of IVHF rough sets over two universes. The ranking order of all faults for
the IVHFA and IVHFG operators is x5 > x6 > x9 > x3 > x8 > x2 > x10 > x4 > x7 > x1 , the ranking order is consistent with
that obtained by Algorithm 1.
By comparing the aggregation methods of the IVHFA and IVHFG operators and of IVHF multigranulation rough sets on
two universes model utilized in steam turbine fault diagnosis, the main fault of the testing sample is rotor radial impact
friction. Compared to the other aggregation operators, the IVHF multigranulation rough set over two universes model takes
full advantages of interval-valued hesitant fuzzy sets and multigranulation rough sets in steam turbine fault diagnosis
procedures. On one hand, the hesitancy membership function in interval-valued hesitant fuzzy sets provides mechanical en-
gineers with more exible vehicles by which to convey understanding about the system fault knowledge base. On the other
hand, the multigranulation rough set method can be viewed as a superior information fusion strategy that allows synthesis
of each engineers view to form a nal conclusion by providing optimistic and pessimistic decision-making strategies. If the
ranking order of all faults for the optimistic and pessimistic versions of IVHF multigranulation rough sets over two universes
are different from each other, the mechanical engineer could refer to results according to risk preference. Thus, the ranking
result of optimistic IVHF multigranulation rough sets over two universes is often adopted when the decision-maker is
more prone to pursue risk. Conversely, the pessimistic counterpart is adopted when the decision-maker intends to select
a conservative fault diagnosis strategy. In light of the above, the superiorities of IVHF multigranulation rough sets over two
universes could greatly decrease uncertainties and increase diagnostic accuracies of steam turbine fault diagnosis.
6. Conclusions
The focal point of interest in the paper is to propose a novel rough set approach, called an interval-valued hesitant
fuzzy multigranulation rough set over two universes, for solving steam turbine fault diagnosis problems. In the framework
of this study, basic denitions and some useful theorems of optimistic and pessimistic interval-valued hesitant fuzzy
multigranulation rough sets over two universes were explored. Then, we proposed a general approach to decision-making
related to steam turbine fault diagnosis. Lastly, case study outcomes show that the proposed fault diagnosis method could
effectively handle steam turbine fault diagnosis problems. There still exist some challenging issues in the theoretical and
practical researches of the proposed rough set model. First, it is meaningful to further study attribute reduction approaches,
uncertainty measures, and topological structures for interval-valued hesitant fuzzy multigranulation rough sets over two
universes. Second, how to apply the proposed approach to other complex and uncertain decision-making situations will
also be taken into account in the future.
over two universes model for steam turbine fault diagnosis, Applied Mathematical Modelling (2016),
http://dx.doi.org/10.1016/j.apm.2016.10.048
JID: APM
ARTICLE IN PRESS [m3Gsc;October 31, 2016;7:37]
## 12 C. Zhang et al. / Applied Mathematical Modelling 000 (2016) 112
Acknowledgments
The authors are very grateful to the subject editor and anonymous reviewers for their constructive and insightful com-
ments and suggestions, which have improved the quality of this paper. This work was supported by the National Natural
Science Foundation of China (Nos. 61272095, 61303107, 61432011, 61573231, 61672331 and U1435212), Shanxi Science and
Technology Infrastructure (No. 20150910010102), Shanxi Scholarship Council of China (No. 2013014), Project supported
by National Science and Technology (No. 2012BAH33B01) and Natural Science Foundation of Shanxi (No. 201601D021076).
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Syst. 31 (9) (2016) 921943.
[35] Z. Xu, Q. Da, The uncertain OWA operator, Int. J. Intell. Syst. 17 (6) (2002) 569575. | 21,149 | 60,309 | {"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-2019-26 | latest | en | 0.842681 |
http://www.enotes.com/homework-help/evaluate-integral-4-top-integal-sign-0-bottom-e-349438 | 1,469,315,611,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823802.12/warc/CC-MAIN-20160723071023-00235-ip-10-185-27-174.ec2.internal.warc.gz | 433,493,488 | 10,212 | # evaluate integral 4 on top integal sign 0 at bottom e^3xdxI am also having trouble inputing this into my TI-83 to solve any suggestions??
Asked on
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Posted on
Evaluate `int_0^4e^(3x)dx` :
Note that `inte^udu=e^u+C`
Let `u=3x,du=3` . When `x=0,u=0;x=4,u=12`
To make the integrand look like `e^(3x)3dx` we can mulitply by 3 and `1/3` to get:
`int_0^4e^(3x)(1/3)3xdx` or `1/3int_0^4e^(3x)3dx` .
Substituting for u, and changing the limits of integration we get:
`1/3int_0^12 e^udu=1/3[e^u|_0^12]=1/3[e^12-e^0]`
`~~1/3[162754.7914-1]~~54251.26`
-----------------------------------------------------------------
`int_0^4e^(3x)dx~~54251.26`
----------------------------------------------------------------
** In the graphing calculator graph the function `y=e^((3x))` (Note the use of parantheses). Then hit 2nd trace;7;0;4 to see the area and the result. Use the `e^x` key instead of the `e` key.
If the function will not graph make sure your Stat Plots are off (Go into stat plot anfd hit 4 for plots off -- or in the y= screen you can arrow up to the highlighted plots and hit enter to turn them off.
Make sure your window includes the x values from 0 to 4 (at least)
Those are the most common errors I can think of.
Sources:
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# Supply and demand for decontaminated blood
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
A recent report indicatesthat nearly 50 Americans contract HIV each year through blood transfusions.Although every pint of blood donated in the United States undergoes a batteryof nine different tests, existing screening methods can detect only the antibodiesproduced by the b0dyâ??s immune systemâ?"not foreign agents in the blood. Since it takes weeks or even months for these antibodies to build up in the blood, newly infected HIV donors can pass along the virus through blood that has,passed existing screening tests. Happily, researchers have developed a series ofnew tests aimed at detecting and removing infections from donated blood beforeit is used in transfusions. The obvious benefit t of these tests is the reduced incidence of infection through blood transfusions. The report indicates that the current price of decontaminated blood is \$80 per pint. However, if the newscreening methods are adopted, the demand and supply for decontaminatedblood will change to Qd = 175 â?" P and Qâ?? = 2P â?" 200. What price do youexpect to prevail if the new screening methods are adopted? How many units ofblood will be used in the United States? What is the level of consumer and pro-ducer surplus? Illustrate your ?ndings in a graph
https://brainmass.com/economics/general-equilibrium/supply-and-demand-for-decontaminated-blood-367518
#### Solution Preview
To find the equilibrium price, we set Qd = Qs:
2P - 200 = 175 - P
3P = 375
P = 125
Thus we know that the new price will be \$125 per pint.
To find the quantity demanded, we can insert this price ...
#### Solution Summary
Economic considerations of screening the blood supply for HIV.
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# STD and Sharpe Calculations
Discussion in 'Risk Management' started by dragonman, Jan 22, 2013.
1. ### dragonman
Hi, I have questions regarding calculating standard deviation and Sharpe ratio with respect to a stock portfolio:
1. Is a standard deviation of a stock portfolio generally expressed as % (i.e., by measuring the deviation of the stock portfolio's return from its average return)? For instance, if my portfolio returns were 2%, 3%, 4% and -5% at the recent 4 months, will the standard deviation be 4.08% or just 4.08?
2. Is a Sharpe ratio of a stock portfolio generally expressed as % as well? For instance, if the total return of the portfolio in the recent year was 10%, the risk free rate was 0.25%, and the portfolio's standard deviation was 4%, will the Sharpe ratio be 2.43% or just 2.43?
2. ### harric
1. 4.08%
2. 2.43
Be aware that typically you would use annualized standard deviation if you are using annualized returns in the calculation. Your standard deviation in this case is monthly.
3. ### dragonman
Thanks. If I want to annualize the 4.08% STD would I multiply it by 12 and make a square root to the result (so it would be 14.13%)?
Also, I want to make sure that the Sharpe ratio is usually calculated based on the annualized STD figure (and not the monthly STD figure). Please correct me if I'm wrong, thanks.
4. ### harric
You need to multiply monthly STD by SQRT(12) = 3.46
So your annualized Sharpe in this case would be 0.70
5. ### dragonman
I multiplied the monthly STD (4.08) by SQRT(12) and the result was 14.14%. Could you please elaborate how did you reach 3.46? Thanks.
6. ### harric
SQRT(12) = 3.46
7. ### dragonman
Ok, so we basically said the same thing. I referred to the final result (after multiplying 4.08 by 3.46) and you referred to the 3.46 itself. Thanks for the clarification.
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# (4) (B) (C) (D) 7. If velocity of the particle is given by , where denotes the position of the particle and initially particle was at , then which of the following are correct? (A) At . the position of the particle is at . (B) Particle's acceleration at . is . (C) Particle's acceleration is throughout the motion. (D) Particle will never go in negative direction from it's starting position.
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Question Text (4) (B) (C) (D) 7. If velocity of the particle is given by , where denotes the position of the particle and initially particle was at , then which of the following are correct? (A) At . the position of the particle is at . (B) Particle's acceleration at . is . (C) Particle's acceleration is throughout the motion. (D) Particle will never go in negative direction from it's starting position. Updated On Jul 21, 2022 Topic Modern Physics Subject Physics Class Class 11 Answer Type Video solution: 1 Upvotes 77 Avg. Video Duration 11 min | 392 | 1,647 | {"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} | 2.78125 | 3 | CC-MAIN-2024-30 | latest | en | 0.903472 |
https://mathematica.stackexchange.com/questions/146178/ndsolve-solution-goes-to-zero-for-large-domain/146187 | 1,568,810,327,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00181.warc.gz | 570,171,757 | 31,720 | ndsolve solution goes to zero for large domain
I am trying to solve numerically the diffusion equation with absorbing boundary conditions on a finite domain. The initial condition is an approximation of a delta function centered in x0 with standard deviation sigma.
I find that changing the size of the domain slightly has a dramatic (non-physical) effect on the solution evaluated at a finite time. Why is this? Why don't I get a warning from Mathematica? This code reproduces the problem:
sigma = 1/32;
T = 10;
x0 = 1;
L = 24;
sol1 = NDSolve[{D[pn[x, t], {t, 1}] == 1/8 D[pn[x, t], {x, 2}],
pn[0, t] == 0, pn[L, t] == 0,
pn[x, 0] == 1/Sqrt[2 Pi sigma^2] Exp[-(x - x0)^2/(2 sigma^2)]},
pn[x, t], {x, 0, L}, {t, 0, T}];
L = 26;
sol2 = NDSolve[{D[pn[x, t], {t, 1}] == 1/8 D[pn[x, t], {x, 2}],
pn[0, t] == 0, pn[L, t] == 0,
pn[x, 0] == 1/Sqrt[2 Pi sigma^2] Exp[-(x - x0)^2/(2 sigma^2)]},
pn[x, t], {x, 0, L}, {t, 0, T}];
t = T;
GraphicsRow[{Plot[Evaluate[pn[x, t] /. sol1], {x, 0, 24},
PlotRange -> All],
Plot[Evaluate[pn[x, t] /. sol2], {x, 0, 26}, PlotRange -> All]}]
On the left I plot the solution with domain size L=24, on the right I plot the solution with L=26. As you can see, the change is dramatic (notice the y axis values) and Mathematica produces no warning.
How can I solve this issue and be able to solve the equation numerically for L larger than 24?
• Tip; Setting the variable of integration t to a numeric value with t = T messes things up when you (or I) recompute the solutions. It's very inconvenient. You can avoid needing such a workaround by solving for pn instead of pn[x, t] in your NDSolve calls. Alternatively, you could use the code pn[x, t] /. sol /. t -> T. – Michael E2 May 16 '17 at 21:00
You have control the quality of the grid/mesh of the spatial variable. Either with "MaxStepSize":
NDSolve[{D[ppn[x, t], {t, 1}] == 1/8 D[ppn[x, t], {x, 2}],
ppn[0, t] == 0, ppn[L, t] == 0,
ppn[x, 0] ==
1/Sqrt[2 Pi sigma^2] Exp[-(x - x0)^2/(2 sigma^2)]}, ppn, {x, 0,
L}, {t, 0, T},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MaxStepSize" -> 1}}]
or with "MinPoints":
NDSolve[{D[ppn[x, t], {t, 1}] == 1/8 D[ppn[x, t], {x, 2}],
ppn[0, t] == 0, ppn[L, t] == 0,
ppn[x, 0] ==
1/Sqrt[2 Pi sigma^2] Exp[-(x - x0)^2/(2 sigma^2)]}, ppn, {x, 0,
L}, {t, 0, T},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MinPoints" -> Ceiling@L}}] | 889 | 2,421 | {"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} | 3.359375 | 3 | CC-MAIN-2019-39 | latest | en | 0.784438 |
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Mathematics - Next Generation (2017)
New York State Learning Standards
Feel free to use or edit a copy
includes Teacher and Student dashboards
### Measures 1 skill fromGrade 3Mathematics - Next Generation (2017)New York State Learning Standards
3.NF1
Track each student's skills and progress in your Mastery dashboards
• edit the questions
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### Our brand new solo games combine with your quiz, on the same screen
Correct quiz answers unlock more play!
6 questions
• Q1
Max and his friends made a flag for their school presentation. They divided the flag into equal sections and colored 3 of the sections gray. The flag is show below. What fraction of the flag is gray?
4/8
3/8
2/5
3/5
300s
3.NF1
• Q2
Sam was working on a math problem. He was asked to shade parts of the figure so that 5/8 of the figure should be shaded. What should Sam do to complete the math problem?
Sam should shade three more parts of the figure.
Sam should shade four more parts of the figure.
Sam should shade five more parts of the figure.
Sam should shade two more parts of the figure.
300s
• Q3
Jonathan divided a garden equally into 6 parts. He planted seeds in 5 of the parts. In what fraction of the garden did Jonathan plant seeds?
1/6
5/6
6/5
1/5
300s
• Q4
What is the fraction shown in this image?
4/4
6/8
None are correct
4/8
300s
• Q5
Which fraction is greater than?
3/5
2/3
300s
• Q6
Mr. George gave an fraction problem of 2/3 x 4/5. What is the first step he must do to solve this equation?
Mr. George must divide both sides until their equal.
Mr. George must subtract the bottom fractions in order to continue solving.
Mr. George must cross multiply from bottom to top.
Mr. George must add both equations first.
300s
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Introduction to rational and irrational numbers | Algebra I | Khan Academy
Introduction to rational and irrational numbers | Algebra I | Khan Academy
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Square Root of 10
The square root of 10 is expressed as √10 in the radical form and as (10)½ or (10)0.5 in the exponent form. The square root of 10 rounded up to 10 decimal places is 3.1622776602. It is the positive solution of the equation x2 = 10.
• Square Root of 10: 3.1622776601683795
• Square Root of 10 in exponential form: (10)½ or (10)0.5
• Square Root of 10 in radical form: √10
1 What Is the Square Root of 10? 2 Is Square Root of 10 Rational or Irrational? 3 How to Find the Square Root of 10? 4 Tips and Tricks 5 Challenging Questions 6 FAQs on Square Root of 10
## What Is the Square Root of 10?
The square root of any number n can be written as √n. This indicates that there is a number a such that:
• a × a = n
Now, it can also be written as:
• a2 = n
• a = √n
Here, a is called as the square root of n. Now, if n = 10, then a = √10 is the square root of 10.
• The square root of 10 is 3.16227.
## Is the Square Root of 10 Rational or Irrational?
The square root of 10 is an irrational number with never-ending digits. The square root of numbers which are perfect squares like 9, 16, 25, and 100 are integer numbers, but the square root of numbers which are not perfect squares are irrational with never-ending digits. The value of the square root of 10 in decimal form is 3.16227… The square root of any number has two values; one is positive and the other is negative.
• √10 = 3.16227 and √10 = – 3.16227
## How to Find the Square Root of 10?
The square root of 10 or any number can be calculated in many ways. Here are two methods: approximation (Hit and Trial) and long division method.
Let’s see how to find √10 by approximation method:
• √9 < √10 < √16
• 3 < √10 < 4
Multiply each by 10.
• 30 < 10 √10 < 40
• √900 < 10√10 < √1600
Let’s move closer to the inequality.
• √961 < 10√10 < √1024
• 31 < 10√10 < 32
Divide each by 10.
• 3.1 < √10 < 3.2
Take the average of both lower and upper limits.
• √10 approx. = (3.1+3.2)/2
• √10 approx. = 3.15
### Square Root of 10 By Long Division
The long division method helps us find more accurate values of the square roots of any number. Let’s see how to find the square root of 10 using the long division method.
• Step 1: Divide the number 10 by 3 (because 32 = 9 is a perfect square number less than 10)
• Step 2: Take the same divisor 3 as the quotient. Multiply the quotient and the divisor and subtract the result from 10.
• Step 3: Take the same quotient 3 and add with the divisor 3.
• Step 4: Add a decimal point after the quotient 3 and bring down two zeros and place it after 1 so that it becomes 100. We need to find a digit which, when placed at the end of 6 and multiplying the resultant number with the same digit, gives a number less than 100.
61 × 1 = 61
Subtract 61 from 100.
100 – 61 = 39
• Step 5: Bring down two zeros again and place it after 39, so that it becomes 3900. Take 1 and add it to 61. 61 + 1 = 62. We need to again find a number that when placed at the end of 62 and when the result is multiplied with the same number, we get a number less than 3900.
626 × 6 = 3756
• Write the same number after 1 in the quotient. Subtract 3756 from 3900.
3900 – 3756 = 144
• Step 6: Repeat the process until we get the remainder equal to zero.
The square root of 10 up to two places is obtained by the long division method.
Explore square roots using illustrations and interactive examples
Tips and Tricks:
• Use the hit and trial method while calculating the square root of any number.
• The square root of any number can be assumed as the number between the square roots of the two nearest perfect squares of that number.
For example, square root of 54 lies between the square roots of 49 and 64.
i.e. √49 < √54 < √64
7 < √54 < 8
• Square root of a number n can have two values ±√n
Challenging Questions:
• Find the square root of 1000 by approximation method.
• Find the square root of 10000 by long division method.
## Square Root of 10 Solved Examples
1. Example 1: Joseph, a salesman, wants to build a rectangular floor garage to store goods for his shop. He wants the floor to be of length 20 feet and width 10√10 feet. If he has to paint the floor at \$10 per square feet, how much money would he have to pay?
Solution
Length of the rectangle floor = 20 feet
Width of the rectangle floor = 10√10 feet
Area of the floor will be = Length × Width
= 20 × 10√10
= 200√10
= 200 × 3.162
= 632.4 feet2
Cost of painting the floor = 10\$ / feet2
Total Cost = 632.4 × 10
Total Cost = \$6324
Joseph has to pay \$6324.
2. Example 2
Help Joseph evaluating the value of √10 × √10 × √10
Solution
The value of √10 = 3.16227
Value of √10 × √10 × √10 = 31.6227
3. Example: If the area of a square is 10 in2. Find the length of the side of the square.
Solution:
Let ‘a’ be the length of the side of the square.
⇒ Area of the square = a2 = 10 in2
⇒ a = ±√10 in
Since length can’t be negative,
⇒ a = √10 = 3.162 in
## FAQs on the Square Root of 10
### What is the Value of the Square Root of 10?
The square root of 10 is 3.16227.
### Why is the Square Root of 10 an Irrational Number?
Upon prime factorizing 10 i.e. 21 × 51, 2 is in odd power. Therefore, the square root of 10 is irrational.
### What is the Square Root of -10?
The square root of -10 is an imaginary number. It can be written as √-10 = √-1 × √10 = i √10 = 3.162i
where i = √-1 and it is called the imaginary unit.
### What is the Value of 12 square root 10?
The square root of 10 is 3.162. Therefore, 12 √10 = 12 × 3.162 = 37.947.
### Evaluate 13 plus 12 square root 10
The given expression is 13 + 12 √10. We know that the square root of 10 is 3.162. Therefore, 13 + 12 √10 = 13 + 12 × 3.162 = 13 + 37.947 = 50.947
### What is the Square of the Square Root of 10?
The square of the square root of 10 is the number 10 itself i.e. (√10)2 = (10)2/2 = 10.
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# Final Exam Notes 3489
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Algorithmic Problem Solving
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Simulation Modeling H - Simulation Modeling with Builtin...
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1 Simulation Modeling with Built-in Excel Tools
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2 Announcements To download DecisionTools Suite to your computer, you have to go to the "Resources" area of the KCC Intranet, then look under "Configuring/Installing Software" and you'll find links to instructions for each of the recent versions of Windows. https://intra.krannert.purdue.edu/admin/kcc/Pages/Resources.aspx Computer Lab on Monday, October 24.
3 Why is simulation useful? Consider two investment alternatives: A: Invest \$10,000. Probability of a \$100,000 gain is 0.10 Probability of a \$10,000 loss is 0.90 B: Invest \$10,000 Probability of a \$500 gain is 1.0 Which would you choose? Why?
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4 Investments Analysis On Average, option A is twice as good as option B! Do we ever actually receive the average? Decisions made based only on the average can be very poor. Other examples Product development Insurance
5 Risk-Informed Decision Making Appropriate and inappropriate uses of averages. Managers manage risk. Simulation gives us a tool to help us evaluate risk. Risk: The uncertainty associated with an undesirable outcome. Risk is not the same as just being uncertain about something, and is not just the possibility of a bad outcome. Risk considers the likelihood of an undesirable outcome (e.g., the probability) as well as the magnitude of that outcome.
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6 “Flaw of Averages” (Sam Savage) Article by Sam Savage (http://www.stanford.edu/~savage/faculty/savage/ )
7 Simulation Model A simulation model is a computer model that imitates a real-life situation. Like other decision models, it has parameters (uncontroll-able inputs), decision variables (controllable inputs), and outputs (objective, consequences, etc.) Simulation model incorporates uncertainty in one or more parameters (uncontrollable inputs) Simulation Model Parameters Decision Variables Objective Consequences Determined by decision-maker Some of them are random inputs Probability and statistics Spreadsheet
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8 How Does Spreadsheet Simulation Model Work? We begin with input variables and then relate them with appropriate Excel formulas to output variables of interest. Simulation uses random numbers, which represent uncertain parameters, to drive the whole simulation process.
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Unformatted text preview: ; 0 ≤ i, j ≤ k. (6.63) Note that ∏i Qij represents the rate at which forward transitions go from i to j , and ∏i represents the rate at which forward transitions leave node i. Equation (6.63) takes advantage of the fact that ∏i is also the rate at which forward transitions enter node i, and thus the rate at which backward transitions leave node i. Using the conjecture that the backward time system is a Jackson network with routing probabilities {Q∗ ; 0 ≤ i, j ≤ k}, we can ij write down the backward transition rates in the same way as (6.59-6.61), ∗ qm ,m 0 = ∏0 Q∗j 0 = = for m 0 = m + e j µi Q∗0 i µi Q∗ ij (6.64) 0 (6.65) 0 (6.66) for m = m − e i , mi > 0, 1 ≤ i ≤ k for m = m − e i + e j , m − i > 0, 1 ≤ i, j ≤ k. If we substitute (6.63) into (6.64)-(6.66), we obtain ∗ qm ,m 0 = ∏j Qj 0 for m 0 = m + e j , = (µi /∏i )∏0 Q0i = (µi /∏i )∏j Qj i 0 1≤j≤k for m = m − e i , mi > 0, 0 for m = m − e i + e j , (6.67) 1≤i≤k (6.68) mi > 0, 1 ≤ i, j ≤ k. (6.69) This gives us our hypothesized backward transition rates in terms of the parameters of the original Jackson network. To use theorem 6.5, P must verify that there is a set of positive we P numbers, p(m ), satisfying m p(m ) = 1 and m ∫m pm < 1, and a set of non-negative ∗ numbers qm 0 ,m satisfying the following two sets of equations: ∗ p(m )qm ,m 0 = p(m 0 )qm 0 ,m X X ∗ qm ,m 0 = qm ,m 0 m for all m , m 0 (6.70) for all m . (6.71) m0 We verify (6.70) by substituting (6.59)-(6.61) on the left side of (6.70) and (6.67)-(6.69) on the right side. Recalling that ρi is defined as ∏i /µi , and cancelling out common terms on each side, we have p(m ) = p(m 0 )/ρj 0 p(m ) = p(m )ρi 0 p(m ) = p(m )ρi /ρj for m 0 = m + e j (6.72) 0 for m = m − e i , mi > 0 0 for m = m − e i + e j , (6.73) mi > 0. (6.74) Looking at the case m 0 = m − e i , and using this equation repeatedly to get from state (0, 0, . . . , 0) up to an arbitrary m, we obtain p(m ) = p(0, 0, . . . , 0) k Y i=1 ρmi . i (6.75) 6.7. JACKSON NETWORKS 263 It is easy to verify that (6.75) satisfies (6.72)-(6.74) for all possible transitions. Summing over all m to solve for p(0, 0, . . . , 0), we get X X X Xm 1= p(m ) = p(0, 0, . . . , 0) ρm1 ρm2 . . . ρk k 1 2 m1 ,m2 ,... ,mk m1 m2 −1 = p(0, 0 . . . , 0)(1 − ρ1 ) mk −1 (1 − ρ2 ) . . . (1 − ρk )−1 . Thus, p(0, 0, . . . , 0) = (1 − ρ1 )(1 − ρ2 ) . . . (1 − ρk ), and substituting this in (6.75), we get p(m ) = k Y i=1 k i Yh pi (mi ) = (1 − ρi )ρmi . i (6.76) i=1 where pi (m) = (1 − ρi )ρm is the steady state distribution of a single M/M/1 queue. Now i that we have found the steady state distribution implied by our assumption about the backward process being a Jackson network, our remaining task is to verify (6.71) P P ∗ To verify (6.71), i.e., m 0 qm ,m 0 = m 0 qm ,m 0 , first consider the right side. Using (6.64) to sum over all m 0 = m + e j , then (6.65) to sum over m 0 = m − e i (for i such that mi > 0), and finally (6.66) to sum over m 0 = m − e i + e j , (again for i such that mi > 0), we get X m0 ∗ qm ,m 0 = k X ∏0 Q∗j + 0 j =1 X µi Q∗0 + i i:mj >0 X i:mi >0 Using the fact Q∗ is a stochastic matrix, then, X X ∗ qm ,m 0 = ∏0 + µi . m0 µi k X Q∗ . ij (6.77) j =1 (6.78) i:mi >0 The left hand side of (6.71) can be summed in the same way to get the result on the right side of (6.78), but we can see that P must be the result by simply observing that ∏0 is this the rate of exogenous arrivals and i:mi>0 µi is P overall rate of P the service completions in state m . Note that this also verifies that ∫m = m 0 qm ,m 0 ≥ ∏0 + i µi , and since ∫m is P bounded, m ∫m p(m ) < 1. Since all the conditions of Theorem 6.5 are satisfied, p(m ), as given in (6.76), gives the steady state probabilities for the Jackson network. This also verifies that the backward process is a Jackson network, and hence the exogenous departures are Poisson and independent. Although the exogenous arrivals and departures in a Jackson network are Poisson, the endogenous processes of customers travelling from one node to another are typically not Poisson if there are feedback paths in the network. Also, although (6.76) shows that the numbers of customers at the different nodes are independent random variables at any given time in steady state, it is not generally true that the number of customers at one node at one time is independent of the number of customers at another node at another time. There are many generalizations of the reversibility arguments used above, and many network situations in which the nodes have independent states at a common time. We discuss just two of them here and refer to Kelly, [13], for a complete treatment. 264 CHAPTER 6. MARKOV PROCESSES WITH COUNTABLE STATE SPACES For the first generalization, assume that the service time at each node depends on the number of customers at that node, i.e., µi is replaced by µi,mi . Note that this includes the M/M/m type of situation in which each node has several independent exp...
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## This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.
Ask a homework question - tutors are online | 1,880 | 5,418 | {"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-2017-17 | longest | en | 0.811129 |
http://openstudy.com/updates/503eb515e4b00813010c0474 | 1,448,579,802,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447860.26/warc/CC-MAIN-20151124205407-00322-ip-10-71-132-137.ec2.internal.warc.gz | 174,510,991 | 27,794 | ## haganmc 3 years ago the Figure shows four charges at the corners of a square of side L. Assume q and Q are positive. What is the magnitude of the net Force on q?
1. haganmc
|dw:1346286892631:dw|
2. bhaweshwebmaster
|dw:1346298352562:dw|
3. haganmc
the top right is also positive
4. haganmc
|dw:1346298573822:dw|
5. haganmc
|dw:1346298617109:dw|
6. bhaweshwebmaster
|dw:1346299022875:dw| This is the real figure.. and the resultant force is along the diagonal but repulsive..
7. haganmc
the answer is $(2-\sqrt{2}) \frac{ KQq }{ L ^{2} }$.. i'm still confused on how to get it
8. bhaweshwebmaster
See that F is the force by 2 -Qs on q... each of which is equal to F=Q∗q /(4Πϵ ∗L^2) and are attractive.. And each of these F act on q at 90 degree.. so their resultant is F ' = F*√2 towards the inside of the square .. along the diagonal.. And, the force on q by 4Q is 4Q∗q/(4Πϵ *2d^2) = 2F.. along the diagonal but outside the square.. So the forces 2F and F√2 are opposite .. Hence the resultant force will be 2F−F√2 = (2-√2) F Now put the value of F=Q∗q /(4Πϵ ∗L^2) to get the actual force..resultant. 4Πϵ can be written into a single constant K.
9. haganmc
alright thanks!
10. amallari
Hi, did you have to draw vectors for this question? I can't seem to get it right. | 445 | 1,290 | {"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} | 3.59375 | 4 | CC-MAIN-2015-48 | longest | en | 0.8876 |
https://www.mathworks.com/matlabcentral/cody/problems/44369-circle-pentagon-overlap/solutions/1301393 | 1,566,724,421,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027323246.35/warc/CC-MAIN-20190825084751-20190825110751-00201.warc.gz | 881,705,543 | 15,922 | Cody
# Problem 44369. Circle/Pentagon Overlap
Solution 1301393
Submitted on 19 Oct 2017 by Joel Cottrell
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
p = [0,5; 4.76,1.55; 2.94,-4.05; -2.94,-4.05; -4.76,1.55]; cp = [0,0]; r = 5; y_correct = 5; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
2 Pass
p = [0,5; 4.76,1.55; 2.94,-4.05; -2.94,-4.05; -4.76,1.55]; cp = [0,0]; r = 4; y_correct = 0; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
3 Pass
p = [0,5; 4.76,1.55; 2.94,-4.05; -2.94,-4.05; -4.76,1.55]; cp = [0,0]; r = 15; y_correct = 5; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
4 Pass
p = [0,5; 4.76,1.55; 2.94,-4.05; -2.94,-4.05; -4.76,1.55]; cp = [2,0]; r = 5; y_correct = 2; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
5 Pass
p = [0,5; 4.76,1.55; 2.94,-4.05; -2.94,-4.05; -4.76,1.55]; cp = [2,0.75]; r = 5; y_correct = 3; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
6 Pass
p = [0,5; 4.76,1.55; 2.94,-4.05; -2.94,-4.05; -4.76,1.55]; cp = [7.5,0]; r = 5; y_correct = 1; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
7 Pass
p = [0,5; 4.76,1.55; 2.94,-4.05; -2.94,-4.05; -4.76,1.55]; cp = [0,-5]; r = 9; y_correct = 4; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
8 Pass
p = [23.66,11.42; 24.37,5.58; 19.05,3.10; 15.04,7.40; 17.89,12.54]; cp = [20,8]; r = 5; y_correct = 5; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
9 Pass
p = [23.66,11.42; 24.37,5.58; 19.05,3.10; 15.04,7.40; 17.89,12.54]; cp = [19,8]; r = 5; y_correct = 3; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
10 Pass
p = [23.66,11.42; 24.37,5.58; 19.05,3.10; 15.04,7.40; 17.89,12.54]; cp = [19.5,10]; r = 5; y_correct = 2; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
11 Pass
p = [23.66,11.42; 24.37,5.58; 19.05,3.10; 15.04,7.40; 17.89,12.54]; cp = [19.5,10]; r = 6.6; y_correct = 4; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
12 Pass
p = [23.66,11.42; 24.37,5.58; 19.05,3.10; 15.04,7.40; 17.89,12.54]; cp = [19.5,10]; r = 7; y_correct = 5; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
13 Pass
p = [30.94,36.26; 35.61,27.09; 28.34,19.82; 19.17,24.49; 20.78,34.65]; cp = [26.97,28.45]; r = 8.75; y_correct = 5; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct))
14 Pass
p = [30.94,36.26; 35.61,27.09; 28.34,19.82; 19.17,24.49; 20.78,34.65]; cp = [26.97,28.45]; r = 8; y_correct = 0; assert(isequal(circle_pentagon_overlap(p,cp,r),y_correct)) | 1,255 | 2,649 | {"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} | 3.75 | 4 | CC-MAIN-2019-35 | longest | en | 0.58308 |
https://in.mathworks.com/matlabcentral/answers/311468-c0-and-x-are-scalars-c-vector-and-p-scalar-if-c-is-then-p-c0-if-c-is-a-scalar-then?s_tid=prof_contriblnk | 1,685,398,260,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00224.warc.gz | 368,994,553 | 32,176 | # c0 and x are scalars, c - vector, and p - scalar. If c is [ ], then p = c0. If c is a scalar, then p = c0 + c*x . Else, p =
3 views (last 30 days)
Subramanian Mahadevan on 9 Nov 2016
Answered: Govind Mishra on 14 Mar 2018
function [p] = poly_val(c0,c,x)
N = length(c);
n=1:1:N;
if (N<=1)
if(isempty(c))
p=c0;
else
p= c0+(c*x);
end
end
if(N>1)
p = c0+(sum(c(n).*(power(x,n))));
end
end
##### 2 CommentsShow 1 older commentHide 1 older comment
Subramanian Mahadevan on 10 Nov 2016
My question is to create a function which performs in the above mentioned way.
Thorsten on 14 Nov 2016
function y = mypolyval(c0,c,x)
if isempty(c)
y = c0;
else
y = c0 + power(x, 1:numel(c))*c(:);
end
The case y = c0 + c*x is already covered by the else. And you can use matrix multiplication of a row and a column vector r*c instead of sum(r.*c').
##### 2 CommentsShow 1 older commentHide 1 older comment
Raunil Raj on 6 Mar 2018
really! I went through the same problem. I however don't understand as to how c(:) can convert any row vector or column vector into a column vector. Can someone please explain?
Jorge Briceño on 29 Jan 2018
Here is my solution:
function p = poly_val(c0,c,x)
format long
n=(1:1:length(c));
c=c(:)' & This part converts any array/matrix into a colunm vector and transpose...
% it afterwards, since you are working with row vector properties.
if isempty(c)
p=c0;
elseif isscalar(c)
p=c0+sum(c.*x);
else
p=c0+sum((c.*(x.^n)));
end
end
KSSV on 10 Nov 2016
Edited: KSSV on 10 Nov 2016
function [p] = poly_val(c0,c,x)
N = length(c); % length of c
if N == 0 % if c is empty
p = c0 ;
elseif N == 1 % if c is a scalar
p = c0+c*x ;
else % if c is a vector
p = c0+(sum(c.*(power(x,N))));
end
KSSV on 14 Nov 2016
N is the number of elements in C.
Gabir Yusuf on 8 Aug 2017
if true
function p = poly_val(c0,c,x)
n=length(c);
if sum(size(c))==0
p = c0;
elseif isscalar(c)
p = c0 + c*x;
else
y=1:n;
z=x.^y;
if size(c)==[1 n]
p=sum(c.*z)+c0;
else
c=c';
p=sum(c.*z)+c0;
end
end
end
This is such a long code
Anshuman Panda on 19 Aug 2017
function p=poly_val(c0,c,x) a=length(c); if a==0 p=c0; else if a==1 p=c0+c*x; else p=c0 + power(x , 1:a)*c(:); end end end
Darío Pascual on 12 Mar 2018
function p=poly_val(c0,c,x)
N = length(c);
n=1:1:N;
d=size(c);
if(isempty(c))
p=c0;
end
if N==1
p= c0+(c*x);
end
if N>1
if d(1)==1
p = c0+(sum(c(n).*(power(x,n))));
else
c=c'
p = c0+(sum(c(n).*(power(x,n))));
end
end
Govind Mishra on 14 Mar 2018
function [p] = poly_val(c0,c,x)
if(iscolumn(c)) c=transpose(c); end
N = length(c); n=1:1:N; if (N<=1) if(isempty(c)) p=c0; else p= c0+(c*x); end end if(N>1) p = c0+(sum(c(n).*(power(x,n)))); end end | 980 | 2,634 | {"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} | 3.75 | 4 | CC-MAIN-2023-23 | latest | en | 0.683404 |
http://web2.0calc.com/questions/what-is-two-plus-five | 1,508,516,496,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00046.warc.gz | 373,086,659 | 6,078 | +0
# what is two plus five
0
136
5
what is two plus five
Guest May 31, 2017
Sort:
#1
+328
0
((7*(10^0)-2+5)*7)/10+2-5+3
or more simply, seven
ZZZZZZ May 31, 2017
#2
+1
Let's make it more complicated.
f(x) = (((-((x)(x-(((7*(10^0)-2+5)*7)/10+2-5+3)-1)(x-(((7*(10^0)-2+5)*7)/10+2-5+3)))/((((x)(x-(((7*(10^0)-2+5)*7)/10+2-5+3)-1)(x-(((7*(10^0)-2+5)*7)/10+2-5+3))))^2)^0.5))+1)/2
Or maybe it isn't
Guest May 31, 2017
#3
+80
0
mathguardian23 May 31, 2017
#4
+328
0
or sqrt(-72)
ZZZZZZ May 31, 2017
#5
0
sqrt((-7)^2)*
Can't do the square root of a negative number pal
Guest Jun 1, 2017
### 26 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 365 | 855 | {"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} | 3.625 | 4 | CC-MAIN-2017-43 | longest | en | 0.733208 |
https://oeis.org/A228236 | 1,695,379,501,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00305.warc.gz | 499,081,818 | 4,242 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A228236 Numbers k for which a sum k+bitcount(k) can be also obtained as a sum k2 +bitcount(k2) for some other k2<>k . Here bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k. 4
3, 4, 11, 12, 14, 15, 16, 17, 19, 20, 27, 28, 29, 31, 32, 34, 35, 36, 43, 44, 46, 47, 48, 49, 51, 52, 59, 60, 62, 65, 67, 68, 75, 76, 78, 79, 80, 81, 83, 84, 91, 92, 93, 95, 96, 98, 99, 100, 107, 108, 110, 111, 112, 113, 115, 116, 123, 124, 125, 126, 127, 128 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS In other words, numbers k such that A228085(A092391(k)) > 1. LINKS Antti Karttunen, Table of n, a(n) for n = 1..10000 EXAMPLE 0 is not in this sequence because the sum 0+A000120(0)=0 cannot be obtained with any other value of k than k=0. 1 is not in this sequence because the sum 1+A000120(1)=2 cannot be obtained with any other value of k than k=1. 2 is not in this sequence because the sum 2+A000120(2)=3 cannot be obtained with any other value of k than k=2. 3 IS in this sequence because the sum 3+A000120(3)=5 can also be obtained with value k=4, as also 4+A000120(4)=5, and thus also 4 is in this sequence. PROG (Scheme, with Antti Karttunen's IntSeq-library) (define A228236 (MATCHING-POS 1 0 (lambda (k) (> (A228085 (A092391 k)) 1)))) CROSSREFS Complement: A228090. Subsets: A228091, A228237. Cf. also A092391, A228085. Sequence in context: A195589 A339578 A244005 * A344346 A047457 A226632 Adjacent sequences: A228233 A228234 A228235 * A228237 A228238 A228239 KEYWORD nonn AUTHOR Antti Karttunen, Aug 17 2013 STATUS approved
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Last modified September 22 05:50 EDT 2023. Contains 365519 sequences. (Running on oeis4.) | 738 | 2,080 | {"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-2023-40 | latest | en | 0.778502 |
http://en.wikipedia.org/wiki/Quantum_Fourier_transform | 1,432,789,611,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929230.43/warc/CC-MAIN-20150521113209-00044-ip-10-180-206-219.ec2.internal.warc.gz | 83,150,282 | 15,726 | # Quantum Fourier transform
In quantum computing, the quantum Fourier transform is a linear transformation on quantum bits, and is the quantum analogue of the discrete Fourier transform. The quantum Fourier transform is a part of many quantum algorithms, notably Shor's algorithm for factoring and computing the discrete logarithm, the quantum phase estimation algorithm for estimating the eigenvalues of a unitary operator, and algorithms for the hidden subgroup problem.
The quantum Fourier transform can be performed efficiently on a quantum computer, with a particular decomposition into a product of simpler unitary matrices. Using a simple decomposition, the discrete Fourier transform can be implemented as a quantum circuit consisting of only $O(n^2)$ Hadamard gates and controlled phase shift gates, where $n$ is the number of qubits.[1] This can be compared with the classical discrete Fourier transform, which takes $O(n2^n)$ gates (where $n$ is the number of bits), which is exponentially more than $O(n^2)$. However, the quantum Fourier transform acts on a quantum state, whereas the classical Fourier transform acts on a vector, so not every task that uses the classical Fourier transform can take advantage of this exponential speedup.
The best quantum Fourier transform algorithms known today require only $O(n \log n)$ gates to achieve an efficient approximation.[2]
## Definition
The quantum Fourier transform is the classical discrete Fourier transform applied to the vector of amplitudes of a quantum state. The classical (unitary) Fourier transform acts on a vector in $\mathbb{C}^N$, (x0, ..., xN−1) and maps it to the vector (y0, ..., yN−1) according to the formula:
$y_k = \frac{1}{\sqrt{N}} \sum_{j=0}^{N-1} x_j \omega^{jk}$
where $\omega = e^{\frac{2 \pi i}{N}}$ is a primitive Nth root of unity.
Similarly, the quantum Fourier transform acts on a quantum state $\sum_{i=0}^{N-1} x_i |i\rangle$ and maps it to a quantum state $\sum_{i=0}^{N-1} y_i |i\rangle$ according to the formula:
$y_k = \frac{1}{\sqrt{N}} \sum_{j=0}^{N-1} x_j \omega^{jk}.$
This can also be expressed as the map
$|j\rangle \mapsto \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} \omega^{jk} |k\rangle.$
Equivalently, the quantum Fourier transform can be viewed as a unitary matrix acting on quantum state vectors, where the unitary matrix $F_N$ is given by
$F_N = \frac{1}{\sqrt{N}} \begin{bmatrix} 1&1&1&1&\cdots &1 \\ 1&\omega&\omega^2&\omega^3&\cdots&\omega^{N-1} \\ 1&\omega^2&\omega^4&\omega^6&\cdots&\omega^{2(N-1)}\\ 1&\omega^3&\omega^6&\omega^9&\cdots&\omega^{3(N-1)}\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\ 1&\omega^{N-1}&\omega^{2(N-1)}&\omega^{3(N-1)}&\cdots&\omega^{(N-1)(N-1)} \end{bmatrix}.$
## Properties
### Unitarity
Most of the properties of the quantum Fourier transform follow from the fact that it is a unitary transformation. This can be checked by performing matrix multiplication and ensuring that the relation $FF^{\dagger}=F^{\dagger}F=I$ holds, where $F^\dagger$ is the Hermitian adjoint of $F$. Alternately, one can check that vectors of norm 1 get mapped to vectors of norm 1.
From the unitary property it follows that the inverse of the quantum Fourier transform is the Hermitian adjoint of the Fourier matrix, thus $F^{-1}=F^{\dagger}$. Since there is an efficient quantum circuit implementing the quantum Fourier transform, the circuit can be run in reverse to perform the inverse quantum Fourier transform. Thus both transforms can be efficiently performed on a quantum computer.
## Circuit implementation
Quantum circuit representation of the quantum Fourier transform
The quantum Fourier transform can be approximately implemented for any N; however, the implementation for the case where N is a power of 2 is much simpler. Suppose N = 2n. We have the orthonormal basis consisting of the vectors
$|0\rangle, \ldots , |2^n - 1\rangle.$
The basis states enumerate all the possible states of the qubits:
$| x \rangle = | x_1 x_2 \ldots x_n \rangle = | x_1 \rangle \otimes | x_2 \rangle \otimes \cdots \otimes | x_n \rangle$
where, with tensor product notation $\otimes$, $|x_j\rangle$ indicates that qubit $j$ is in state $x_j$, with $x_j$ either 0 or 1. By convention, the basis state index $x$ orders the possible states of the qubits lexicographically, i.e., by converting from binary to decimal in this way:
$x = x_1 2^{n-1} + x_2 2^{n-2} +\cdots + x_n 2^0.\quad$
It is also useful to borrow fractional binary notation:
$[0. x_1 \ldots x_m] = \sum_{k = 1}^m x_k 2^{-k}.$
For instance, $[0.x_1] = \frac{x_1}{2}$ and $[0.x_1 x_2] = \frac{x_1}{2}+\frac{x_2}{2^2}.$
With this notation, the action of the quantum Fourier transform can be expressed as:
$|x_1 x_2 \ldots x_n \rangle \mapsto \frac{1}{\sqrt{N}} \ \left(|0\rangle + e^{2 \pi i \, [0.x_n] }|1\rangle\right) \otimes \left(|0\rangle + e^{2 \pi i \, [0.x_{n-1} x_n] }|1\rangle\right) \otimes \cdots \otimes \left(|0\rangle + e^{2 \pi i \, [0.x_1 x_2 \ldots x_n] }|1\rangle\right),$
where the output qubit 1 is in a superposition of state $|0\rangle$ and $e^{2 \pi i \, [0.x_n] }|1\rangle$, and so on for the other qubits.
In other words, the discrete Fourier transform, an operation on n-qubits, can be factored into the tensor product of n single-qubit operations, suggesting it is easily represented as a quantum circuit. In fact, each of those single-qubit operations can be implemented efficiently using a Hadamard gate and controlled phase gates. The first term requires one Hadamard gate, the next one requires a Hadamard gate and a controlled phase gate, and each following term requires an additional controlled phase gate. Summing up the number of gates gives $1 + 2 + \cdots + n = n(n+1)/2 = O(n^2)$ gates, which is polynomial in the number of qubits.
## Example
Consider the quantum Fourier transform on 3 qubits. It is the following transformation:
$|j\rangle \mapsto \frac{1}{\sqrt{2^3}} \sum_{k=0}^{2^3-1} \omega^{jk} |k\rangle,$
where $\omega$ is a primitive eighth root of unity satisfying $\omega^8=\left(e^{\frac{2\pi i}{8}}\right)^8=1$ (since $N = 2^3 = 8$).
The matrix representing this transformation on 3 qubits is
$F_{2^3} = \frac{1}{\sqrt{2^3}} \begin{bmatrix} 1&1&1&1&1&1&1&1 \\ 1&\omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6&\omega^7 \\ 1&\omega^2&\omega^4&\omega^6&\omega^8&\omega^{10}&\omega^{12}&\omega^{14} \\ 1&\omega^3&\omega^6&\omega^9&\omega^{12}&\omega^{15}&\omega^{18}&\omega^{21} \\ 1&\omega^4&\omega^8&\omega^{12}&\omega^{16}&\omega^{20}&\omega^{24}&\omega^{28} \\ 1&\omega^5&\omega^{10}&\omega^{15}&\omega^{20}&\omega^{25}&\omega^{30}&\omega^{35} \\ 1&\omega^6&\omega^{12}&\omega^{18}&\omega^{24}&\omega^{30}&\omega^{36}&\omega^{42} \\ 1&\omega^7&\omega^{14}&\omega^{21}&\omega^{28}&\omega^{35}&\omega^{42}&\omega^{49} \\ \end{bmatrix} = \frac{1}{\sqrt{2^3}} \begin{bmatrix} 1&1&1&1&1&1&1&1 \\ 1&\omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6&\omega^7 \\ 1&\omega^2&\omega^4&\omega^6&1&\omega^2&\omega^4&\omega^6 \\ 1&\omega^3&\omega^6&\omega&\omega^4&\omega^7&\omega^2&\omega^5 \\ 1&\omega^4&1&\omega^4&1&\omega^4&1&\omega^4 \\ 1&\omega^5&\omega^2&\omega^7&\omega^4&\omega&\omega^6&\omega^3 \\ 1&\omega^6&\omega^4&\omega^2&1&\omega^6&\omega^4&\omega^2 \\ 1&\omega^7&\omega^6&\omega^5&\omega^4&\omega^3&\omega^2&\omega \\ \end{bmatrix}.$
The 3-qubit quantum Fourier transform is the following operation:
$|x_1, x_2, x_3 \rangle \mapsto \frac{1}{\sqrt{2^3}} \ \left(|0\rangle + e^{2 \pi i \, [0.x_3] }|1\rangle\right) \otimes \left(|0\rangle + e^{2 \pi i \, [0.x_2 x_3] }|1\rangle\right) \otimes \left(|0\rangle + e^{2 \pi i \, [0.x_1 x_2 x_3] }|1\rangle\right).$
This quantum circuit implements the quantum Fourier transform on the quantum state $|x_1,x_2,x_3\rangle$.
The quantum gates used in the circuit above are the Hadamard gate and the controlled phase gate $R_\theta$.
As calculated above, the number of gates used is $n(n+1)/2$ which is equal to 6, for n = 3.
## References
1. ^ Michael Nielsen and Isaac Chuang (2000). Quantum Computation and Quantum Information. Cambridge: Cambridge University Press. ISBN 0-521-63503-9. OCLC 174527496.
2. ^ L. Hales, S. Hallgren, An improved quantum Fourier transform algorithm and applications, Proceedings of the 41st Annual Symposium on Foundations of Computer Science, p. 515, November 12–14, 2000
• K. R. Parthasarathy, Lectures on Quantum Computation and Quantum Error Correcting Codes (Indian Statistical Institute, Delhi Center, June 2001)
• John Preskill, Lecture Notes for Physics 229: Quantum Information and Computation (CIT, September 1998) | 2,799 | 8,571 | {"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": 44, "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-2015-22 | latest | en | 0.821881 |
http://www.solutioninn.com/youve-just-taken-a-job-at-a-investment-banking-firm | 1,503,263,745,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106990.33/warc/CC-MAIN-20170820204359-20170820224359-00169.warc.gz | 679,887,842 | 7,615 | # Question: You ve just taken a job at a investment banking firm
You’ve just taken a job at a investment banking firm and been given the job of calculating the appropriate nominal interest rate for a number of different Treasury bonds with different maturity dates. The real risk- free interest rate that you have been told to use is 2.5%, and this rate is expected to continue on into the future without any change. Inflation is expected to be constant over the future at a rate of 2.0%. Since these are bonds that are issued by the U. S. Treasury, they do not have any default risk or any liquidity risk ( that is, there is no liquidity- risk premium). The maturity- risk premium is dependent upon how many years the bond has to maturity. The maturity- risk premiums are as follows:
Given this information, what should the nominal rate of interest on Treasury bonds maturing in 0– 1 year, 1– 2 years, 2– 3 years, and 3– 4 years be?
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https://www.calctool.org/machines-and-mechanisms/pulley | 1,702,047,948,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100762.64/warc/CC-MAIN-20231208144732-20231208174732-00896.warc.gz | 762,758,362 | 60,081 | # Pulley Calculator
Created by Davide Borchia
Last updated: Aug 16, 2022
You can find belt and pulley systems in your car's engine, the luggage belt at the airport, and in many other places: learn how to calculate pulleys' speed, diameter, and much more with our comprehensive and helpful tool!
Learn:
• What is a belt and pulley system;
• How to calculate the RPM of a pulley and its diameter;
• How to calculate the pulley ratio;
• How to calculate the torque in a belt and pulley system;
• The formulas to calculate the belt speed, tension, and length.
## What is a pulley?
A pulley is a simple machine: a mechanical device with a basic construction (i.e., no gears, no heat cycles, and so on) that changes the direction and/or the magnitude of a force.
In the most basic configuration, a single pulley is a wheel rotating around an axle, only able to change the direction of a force (without giving you any mechanical advantage). This is pretty straightforward to understand — the story changes when the number of pulleys increases.
## Adding pulleys: what is a belt and pulley system
When you add a pulley, your simple machine can change the magnitude of the applied force too. You can see a similar example of this in our gear ratio calculator. Pulleys connected by a rope (an open stretch of material in which you can identify a beginning and an end) allow for the multiplication and rotation of the tension of the rope, with a final mechanical advantage ideally not dependent on the size of the pulleys. If you connect the set of pulleys through a closed loop of material, a belt, however, the size of the pulleys matter.
But what is a belt-pulley system? We are talking of a system of two to more pulleys connected by a flat, smooth belt. The system is not dissimilar from the chain transmission of bikes, and it's actually more widespread than you think: joined to the astounding math of Möbius strips it allows your car to move (and to require less maintenance).
When you build a simple belt-pulley system, you have to consider only a few important parameters:
• The radiuses of the pulleys;
• The rotational speeds of the pulleys; and
• The distance between the pulleys.
Many other parameters and quantities can be calculated starting from these.
If you want to move your belt and pulley system, you need to apply a driving force to it: it may be an engine, your legs (in a bike), the wind, or anything else. The pulley that is moved in the first place is called the driving pulley; the other ones are, regardless of their position and role, driven pulleys.
## Pulley calculations: pulley RPM and diameter calcualtions
We begin our analysis of the belt and pulley calculator with quantities independent from the belt. The diameters and rotations per minute of two pulleys connected by a belt are related by a simple formula:
$n_1\cdot d_1 = n_2\cdot d_2$
Here you can find:
• The diameters of the pulleys, $d_1$ and $d_2$;
• The RPM of both pulleys: $n_1$ and $n_2$.
We can also calculate the pulley ratio, though it has not much importance as for gears:
$\text{ratio}=\frac{d_{\text{Driven}}}{d_{\text{Driving}}}$
The diameters of the pulleys in a belt and pulley system are the quantities defining the mechanical advantage of your machine.
#### Pulley torque calculations
When you know a pulley's diameter and RPM, you can also define its torque. The torque depends, obviously, on the input power on the driving pulley. In rotational systems. the power is equal to the product of torque and rotational speed:
$P=\tau\cdot\omega$
Where:
• $\tau$ is the torque on the pulley; and
• $\omega$ its rotational speed.
The rotational speed, or angular velocity, is usually defined in degrees per second or radians per second. To convert it to rotations per minute, use this formula:
$\omega_{\text{rad/s}} = \frac{\omega_{\text{RPM}\cdot 2\pi}}{60\ \text{s}}$
If you know the input power in watts, the formula for a pulley's torque is:
$P = \tau_1 \cdot \frac{n_1\cdot 2\pi}{60\ \text{s}}$
Where $n_1$ and $\tau_1$ are quantities related to the driving pulley.
The same formula applies when you calculate the torque of the driven pulley. Since, ideally, the power is transmitted without dissipation, and you can calculate the pulley rotational speed, we can calculate the torque in a belt and pulley system with a generic equality:
$\tau_1 \cdot n_1 = \tau_2 \cdot n_2$
## Belt and pulley calculator: belt speed, length and tension formulas
Let's finally introduce the belt in our pulley calculator. Some of the calculations require the knowledge of an additional parameter: the distance between the pulley's centers, $D$.
#### Belt speed calculator
To calculate the speed of a belt in a belt and pulley system, just... think! In an ideal case, there is no slippage (which means that the points of contact between the belt and pulleys don't move). The speed of the belt, then, is equal to the tangential speed of both pulleys, which is (unsurprisingly) the same:
$v_{\text{belt}} = \frac{n_1\cdot\pi\cdot d_1}{60\ \text{s}}= \frac{n_2\cdot\pi\cdot d_2}{60\ \text{s}}$
This equality is easily understandable if you return to the previous section, where you'd see that the product of diameter and RPM is a constant in a belt and pulley system.
#### Calculation for the tension in a pulley and belt system
To calculate the tension in a pulley system driven by a belt, we use the following formula:
$T_{\text{belt}} = \frac{P}{v_{\text{belt}}}$
Where $P$ is the power driving the pulleys.
#### Lenght of the belt in a belt and pulley system
The length of the belt is a fundamental quantity: when you translate your machine from the drawing board to the physical world, you want the belt to be as taut as possible. If your belt is too long, the stretch "leaving" the driving pulley would be slack, and the tension on that side would be lost.
To calculate the length of a belt in a machine composed by belts and pulleys, this formula comes in handy:
$l_{\text{belt}} = \frac{d_1\!\cdot\! \pi}{2}\!+\!\frac{d_2\!\cdot\! \pi}{2}\! +\! 2\!\cdot\! D \!+\! \frac{(d_1\!-\!d_2)^2}{4 \!\cdot\!D}$
Mind that this is an approximation of the exact formula!
Davide Borchia
Transmitting power
W
Pulley centers
ft
Driver pulley
Diameter
ft
Angular velocity
RPM
Drive torque
Nm
Driven pulley
Diameter
ft
Angular velocity
RPM
Driven torque
Nm
Belt
Belt length
ft
Belt velocity
ft/s
Belt tension
N
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Week 13 results that the Lions need to keep their playoff hopes alive
Ironically enough, I feel like the most unrealistic result in Week 13 is the Lions winning their game, but they did almost beat the Viking earlier this year, so let's assume they win to keep this wild playoff ride going by picking up the win. If you really believe in the Lions playoff hopes and you decided to do a Lions playoff parlay for Week 13, you'd win \$15,517 on a \$100 bet if all eight of these teams won.
Week 14 results that the Lions need to keep their playoff hopes alive
Two weeks ago, you would have been laughed at for suggesting the Jets might be the Saints, but that doesn't sound so unrealistic anymore. Also, with Russell Wilson struggling, a Texans upset over the Seahawks doesn't' seem completely crazy. Once again, the Lions winning seems like the most improbable part of this list.
Week 15 results that the Lions need to keep their playoff hopes alive
• Lions beat Cardinals
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• Buccaneers beat Saints
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Week 16 results that the Lions need to keep their playoff hopes alive
• Lions beat Falcons
• Rams beat Vikings
• Giants beat Eagles
• Buccaneers beat Panthers
• Cowboys beat Washington
• Seahawks beat Bears
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Week 17 results that the Lions need to keep their playoff hopes alive
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Week 18 results that the Lions need to keep their playoff hopes alive
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• Vikings beat Bears
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# Exam 1 Review - Stat 145: Exam 1 Review Questions Note:...
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Stat 145: Exam 1 Review Questions Note: This is not intended to be a preview of the actual exam. Rather, it is meant to give you an idea of the types of questions that will be asked. There are concepts in these review problems that will not appear on the actual exam, just as there will be concepts on the actual exam that are not covered in these review problems. 1. In an introductory statistics class with 136 students, the professor recorded how much money each student had in their possession during the first class of the semester. The histogram below shows the data collected: (a) Approximately how many students had less than \$20 in their possession? (b) Describe the shape, center, and spread of this distribution. (c) If the outlier on the right side of the histogram is removed, will the standard deviation increase, decrease, or stay the same? Explain. 2. Bottles of a popular cola contain, on average, 300 milliliters (ml) of cola. The standard deviation is 3 ml. Assume that the distribution of the contents of the bottles is
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## Exam 1 Review - Stat 145: Exam 1 Review Questions Note:...
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We have ECEV = DE something else because the electrons and holes recombining are always close to the band edges for energy conservation.
DE something else refers to the unavoidable condition, that "something else" has to provide the energy needed for generation, or must take away the energy released during recombination.
3. Now we look at the entropy. Recombination reduces the entropy of the system (full bands are more orderly than bands with a few wildly moving holes and electrons. The "something else", that takes energy out of the system may in addition take some entropy out of it, too. However, no easy law can be formulated.
The first two points determine if a recombination/generation event - which we from now on are going to call an R/G-event - can take place at all, i.e. if it is allowed; the third point comes in - in principle - when we discuss the probability of an allowed R/G-event to take place. This insight, however, will only be used in an indirect way in what follows.
The major quantities are the recombination rate R and the generation rate G.
The recombination rate R is the more important one of the two. It is related to the carrier density ne,h by
R = dne,h
dt
It is always directly given by the rate at which the carrier density decreases and it does not matter which carrier type we are looking at because dne/dt = dnh/dt as long as the carriers disappear in pairs by recombination.
Note that the equilibrium condition of constant carrier concentration does not mean that dne,h/dt = 0, but only that + dne,h/dt = – dne,h/dt since n is an average quantity. It is not unlike the drift velocity of electrons which can be zero despite large individual velocities of the individual electrons.
Recombination and Generation in Direct Semiconductors
If we first look at recombination in direct semiconductors, we need holes and electrons at the same position in the band diagram ; i.e. in k-space. However, that does not imply that they are at the same position in real space. For a recombination event they have to find each other; i.e. we also need them to be at about the same position in real space.
The recombination rate R thus must be proportional to the two concentrations, ne and nh, because the probability of finding a partner scales with the concentration. We thus can write down the recombination rate R as
R = r · ne · nh = r · Neffh · Neffe · exp – ECEFe
kT
· exp – EFhEV
kT
With r = proportionality constant. We also only assume only local equilibrium as evidenced by the use of quasi Fermi energies.
We can rewrite this equation as follows
R = r · Neffh · Neffe · exp – EC + EFeEFh + EV
kT
= r · Neffh · Neffe · exp – ECEV
kT
· exp – EFeEFh
kT
Using our old relation for the intrinsic carrier concentration ni
ni2 = Neffh · Neffe · exp – EC – EV
kT
we finally obtain
R = r · ni2 · exp – EFeEFh
kT
Note again that we have not invoked total equilibrium, but only local equilibrium in the bands - we use the Quasi Fermi energies EFe,h. That is essential; after all it is recombination and generation that restore equilibrium between the bands and the SRH theory only makes sense for non-equilibrium.
If we were to consider total thermal equilibrium, we know that the generation rate G must be identical to the recombination rate R; both Quasi Fermi energies are identical (= EF) and R = r · ni2 applies.
Note that we did not assume intrinsic conditions; the Fermi energy can have any value, i.e. the semiconductor may be doped.
In essence, what we see is that the recombination rate in non-equilibrium depends very much on the actual carrier concentration.
So far it was easy and straigth-forward. Now comes an important point.
In contrast to the recombination rate R, the generation rate G does not depend (very much) on the carrier concentration; it is just a reflection on the thermal energy contained in the system and therefore pretty much constant. In other words, under most conditions we have .
G = Gtherm ¹ G(n) » constant
We may, from the above consideration, equate G under all conditions with the recombination rate for equilibrium, i.e.
G = Gtherm = r · ni2
In non-equilibrium, which will be the normal case for devices under operation, the difference (R – G) is no longer zero, but has some value
U = R – G
Since R is mostly (but not always) larger than G under non-equilibrium conditions, U is the net rate of recombination (or, on special occasions, the net generation rate).
Using the expressions derived so far, we obtain .
U = R – exp – Gtherm = r · (ne · nh – ni2) = r · ni2 · æ ç è ö ÷ ø
This equation tells us for example, how fast a non-equilibrium carrier concentration will decay, i.e. how fast full equilibrium will be reached, or, if we keep the non-equilibrium concentration fixed for some reason, what kind of recombination current we must expect.
This is so, because U, the difference between recombination and generation, times the charge is nothing but a net current flowing from the conduction band to the valence band (for positive U).
Determining the Proportionality Constant r
We still need to determine the proportionality constant r.
This is not so easy, but we can make a few steps in the right direction. We assume in a purely classical way that an electron (or hole) moves with some average velocity v through the lattice, and whenever it encounters a hole (or electron), it recombines.
The problem is the word "encounters". If the particles were to be small spheres with a diameter dp, "encountering" would mean that parts of such a sphere would be found in the cylinder with diameter dp formed by another moving sphere because that would cause a physical contact.
Our particles are not spheres, but for the purpose of scattering theory we treat them as such, except that the diameter of the cylinder that characterizes its "scattering size" is called scattering cross section s and has a numerical value that need not be identical to the particle size - but it also will not be wildly different.
One electron now covers a volume v · s per second and all N e electrons (a number, not a density) probe per second the volume
Vprobed = N e · v · s
Any time an electron encounters a hole in the volume it probes, it recombines. The recombination R rate then is simply the number of encounters per second.
How many holes are "hit" per second? In other words, how many are to be found in the volume probed? That is easy: The number Nhof holes encountered in the Volume probed by electrons and thus the recombination rate is
N h = n h · Vprobed = n h · N e · v · s = R
With nh = density of holes in the sample. You many wonder if this is correct, considering that the holes move around, too, but simply realize that the density of holes is nevertheless constant.
The formula is a bit unsatisfying, because it contains the volume density of holes, but the absolute number of electron.
That is easily remedied, however, if we express Ne, the number of electrons, by their density ne via Ne = ne · V with V = sample volume. This gives us
R
V
= Recombinations per
s and cm –3
= n e · n h · v · s
In other words, if we use the density of the electron and holes, we obtain a recombination rate density, i.e. recombination events per s and cm–3 - as it should be. As always, we are going to be a bit sloppy about keeping densities and numbers apart. But there is no real problem: Just look at the dimensions you get, and you know what it is
A comparison with the formula from above yields
r = v · s
This leaves us with finding the proper value for s. Whereas this is difficult (in fact, the equation above is more useful for determining s from measurements of R than to calculate r), we are still much better off than with r alone:
Whereas we had no idea about a rough value for r, we do know something about v (it is the group velocity of the carriers considered), and we know at least the rough order of magnitude for s: We would expect it to be in the general range of atomic dimensions (give or take an order of magnitude).
You might wonder now, why we assume that any "meeting" of the elctrons and holes lead to recombination, given that we have to preserve momentum, too. You are right, but remember:
We are treating direct semiconductors here! Since we only consider the mobile electrons and holes, we only consider the ones at the band edges - and those have the same k-vector in the reduced band diagram!
Useful Approximations and the Lifetime t
We now consider non-equilibrium, but describe it in terms of deviations from equilibrium. Then it is sensible to rewrite the carrier densities (or numbers, take whatever you like) in terms of the equilibrium density ne,h(equ) plus/minus some delta:
ne,h = ne,h(equ) + Dne,h
This is one of the decisive "tricks" to get on with the basic equations, because it allows to specify particular cases (e.g. Dne,h << ne,h(equ) or whatever), and then resort to approximations. We will encounter this "trick" fairly often.
We obtain after some shuffling of the terms form the equation for the net recombination rate U
U = æ è ö ø
So far everything is still correct. But now we consider a first special, if rather general case:
We assume that Dne = Dnh = Dn , i.e. that only additional electron - hole pairs were created in non-equilibrium. We than may simplify the equation to
U = æ è ö ø
We can simplify even more. For the extrinsic case where one carrier density - lets say for example nh - is far larger than ne or Dn (i.e. we have a p-doped semiconductor), we may neglect the terms Dn · ne(equ) and Dn2 and obtain
U » r · nh · Dn
U was the difference between the recombination and the generation rate. We are now looking at an approximation where only some Dn in the concentration of the minority carriers is noticeably different from equilibrium conditions (where we always have U = 0).
We thus may write .
U = R(equ) + R(D) – G(equ) = R(D)
With R(D) denoting the additional recombination due to the excess minorities. Remembering the basic definition of R we see that now we have
U = d(Dne)
dt
= r · nh · Dne = v · s · nh · Dne
This is a differential equation for Dne(t), it has the simple solution
Dne(t) = Dne(t = 0) · exp – t
t
The quantity demanded by the general solution is, of course, the life time of the minority carriers. We now have a formula for this prime parameter, it comes out to be
t = 1
v · s · nh
= 1
v · s · nmaj
The last equality generalizes for both types of carriers - it is always the density of the majority carriers that determine the lifetime of the minority carriers. This is clear enough considering the "hit and recombine"scenario that we postulated at he beginning
Substituting r · nh with 1/t in the equation for U yields
U = Dn
t
In other words: The recombination rate in excess of the recombination rate in equilibrium is simply given by the excess concentration of minority carriers divided by their life time.
In yet other words:
The net current flowing from the band containing the minority carriers to the other band is given by U (times the elementary charge, of course), because U gives the net amount of carriers "flowing" from here to there! And that is the definition of a current!
This result not only justifies our earlier approach, it gives us the minority carrier life time in more basic quantities including (at least parts) of its temperature dependence via the thermal velocity v and the majority carrier concentration nh - the T-dependence of which we know.
Since 1/nh is more or less proportional to the resistivity, we expect t to increase linearly with the resistivity which it does as illustrated before, at least for resistivities that are not too low.
A rough order of magnitude estimate gives indeed a good value for many direct semiconductors:
s » 10 –15 cm2 Þ t » 10 –9 s = 1 ns
v = 107 cm/s
nh = 1017 cm-3
Recombination and Generation in Indirect Semiconductors
In indirect semiconductors, direct recombination is theoretically impossible or, being more realistic, very improbable.
In general, a recombination event needs a third partner to allow conservation of energy and crystal momentum.
Under most (but not all) circumstances, this third partner is a lattice defect, most commonly an impurity atom, with an energy state "deep" in the band gap, i.e. not close to the band edges.
Recombination then is determined by these "deep states" or deep levels, and is no longer an intrinsic or just doping dependent property.
How the recombination and generation depends on the properties of deep levels is the subject of the proper Shockley-Read-Hall theory (what we did so far was just a warming-up exercise). It is a lengthy theory with long formulas; here we will just give an outline of the important results. More topics will be covered in an advanced module.
First we look at the situation in a band diagram that shows the relevant energy levels plus the mid-band energy level EMB, which will come in handy later on.
Besides the energy level of the "deep level" defect, we now need four transition rates instead of just one recombination rate:
• Red, the rate with which electrons from the conductance band transit from the conduction band to the deep level, or more simply put, occupy the deep level with the energy EDL - in short: the rate with which they are going down to the deep level. It is important to realize that this identical to the rate with which holes, i.e. unoccupied states at the deep level are going up to the conduction band.
• Reu, the rate with which electrons occupying the deep level state go up to the conduction band (or holes in the conduction band are going down to the deep level).
• Rhu, the rate with which holes from the valence band go up to the deep level (or the electrons in the deep level go down to the valence band), and finally
• Rhd, the rate with which holes from the deep level go down to the valence band (or electrons up to the deep level).
The equilibrium concentration of electrons (and holes) on the deep level is, as always, given by the Fermi distribution. We have
nDL = NDL · f(EDL, T) = density of negatively charged deep levels with one electron sitting on it, and
n0DL = NDL · [1 – f(EDL, T)] = density of deep levels with no electron sitting at it. This is identical to the density of holes sitting "on" the deep level. NDL, of course, is the density of deep level states, e.g. the density of impurity atoms. It's written with capital N (otherwise used for absolute numbers) to avoid confusion with the carrier densities.
To make life easier, we assumed that the deep level is normally neutral, i.e. does not contain an unalterable fixed charge, and can only accommodate one electron or hole, respectively.
We may now write down formulas for the transition rates in direct analogy to the consideration of the recombination rate in direct semiconductors as given above. For Red we have
Red = r · ne · n0DL = v · se · ne · NDL · [1 – f(EDL, T)]
With se = scattering cross section (also called capture cross section) of the deep level for electrons.
For the other transition rate Reu we have to think a little harder. Since the electron is trapped at the deep level and does not move around in space, we look at the alternative description with holes in the conduction band going down and obtain
Reu = r' · ( Neffe – ne) · nDL
With r' = some proportionality constant, principally different from r, and Neffene = density of holes in the conduction band.
That last statement needs perhaps a little thought. Just consider how many free places, corresponding to holes, you have in the conduction band!
Since ne is much smaller than Neffe, we may approximate this equation by
Reu » r' · Neffe · NDL · f(EDL, T)
We have not invoked some cross section and thermal velocity here, because the electron now is localized and doesn't move around. We also used a different proportionality constant r' because the situation is not fully symmetric to the reverse process. It is conventional, to call the quantity e e = r' · Neffe the emission probability for electrons of the deep level.
The emission probability contains the information about the generation of carriers from the deep level; in this it is comparable to the generation rate from the valence band for the simple recombination theory considered above.
Now, if we assume that the transitions from conductance band electrons to the deep level and their re-emission to the conduction band are in local equilibrium (which does not necessarily entail total equilibrium), we have
Reu = Red, and from this we get - after some shuffling of the terms - for the emission probability e e in local equilibrium:
e e = v · se · ne · [1 – f(EDL, T)]
f(EDL, T)
Again, as in the case of the generation rate G for direct semiconductors, we may assume that the emission probability e e is pretty much constant and this is a crucial point for what follows.
Since we want to find quantities like life times as a function of the density and energy level of the deep level, it is useful to use the mid-band energy level as a reference, and to rewrite the equation for e e in terms of this mid-band level EMB via the relations
1 – f(EDL, T)
f(EDL, T)
= exp – EF – EDL
kT
EMB = ECEV
2
ni = Neffe · exp – ECEMB
kT
These equations may need a little thought. The first one came up before in a similar way, the second simply defines mid band-gap, and the last one uses the fact that the Fermi energy for intrinsic semiconductors is in mid band gap (at least in a good approximation).
Using these equations, we first rewrite the formula for the density of electrons in the conduction band and obtain
ne = Neff e · exp – ECEF
kT
= Neffe · exp – ECEMB
kT
· exp – EMBEF
kT
= ni · exp EFEMB
kT
Putting everything together, we get for the emission probability
e e = v · se · ni · exp EDLEMB
kT
This is the best we can do to describe the traffic of electrons between the deep level and the conduction band.
Next, we do the matching calculation for the transitions rates with the valence band, Rhu and Rhd.
Except, we won't do it. Too boring - everything is quite similar. As the final result for the emission probability for the holes, eh, we obtain exactly what we should expect anyway:
eh = v · sh · ni · exp EMBEDL
kT
We captured the electron and hole traffic betwen a deep level and the conduction or valence band, respectively, with these equations- always for local equilibrium of the deep level with the respective band. Now we consider the intraband generation and recombination rate, G and R.
This is exactly the same thing as the money traffic form one major bank to another one via an intermediate bank. Each bank can deposit and withdraw money from all three accounts, while the total amount of all the money must be kept constant. If it would be your money, you sure like hell would want to and be able to keep track of it. So let's do it with electrons and holes, too.
With G we still denote the rate of electron - hole pair generation taking place directly between the bands; by thermal or other energies, e.g. by illumination. It is thus the rate with which electrons and holes are put directly into the conduction or valence band, no matter what goes on between the deep level and the bands.
We may for some added clarity, decompose G into Gperfect, the generation always going on even in a hypothetical perfect semiconductor, and Gne for whatever is added in non-equilibrium (e.g. the generation by light). We have G = Gperfect + Gne.
After all, before we put in "our" deep levels or switched on the light, the hypothetically perfect crystal already must have had some generation and recombination, too (and Rperfect = Gperfect must obtain). However, we can expect that Rperfect is rather small in a perfect indirect semiconductor, which makes Gperfect rather small, too.
Now comes a special point. If we introduce deep levels into a perfect indirect semiconductor, G stays unchanged for the old reasons, but Rperfect essentially reduces to zero - i.e. all recombination takes place via deep levels.
The rate of change of the electron and hole concentration in their bands is then the sum total of all processes withdrawing and depositing electrons or holes, i.e.
dne
dt
= Gperfect + Gne – Rperfect + Reu – Red = Gne – (Red – Reu)
dnh
dt
= Gperfect + Gne – Rperfect + Rhd – Rhu = Gne – (Rhu – Rhd)
Note that Gperfect – Rperfect = 0 by definition.
Local equilibrium between the bands and the deep level, still not necessarily implying total equilibrium, now demands that both dne/ dt and dnh/dt must be zero.
That means that the density of electrons in the conductance band and the density of the holes in the valence band do not change with time anymore. However, that does not mean that they have their global equilibrium value, only that we have a so-called steady state (in global non-equilibrium) which, on the time scales considered, appears to keep things at a constant value.
As an example, a piece of semiconductor under constant illumination conditions will achieve a steady state in global non-equilibrium conditions. The carrier concentrations in the bands will be constant, but not at their equilibrium values if light generates electron hole pairs all the time.
This gives us the simple equations
Red – Reu = Rhu – Rhd
Essentially, this says that the total electron or hole traffic or current (= difference of the partial rates (times elementary charge)) from the conduction or valence band, respectively, to the deep level are identical and equal to the extra band-to-band generation current produced in non-equilibrium for the given material and situation.
But steady state also implies that their must be an additional recombination exactly equal to Gne and that is of course exactly what the terms Red – Reu or Rhu – Rhd denote: They are identical to the additional recombination rates needed for balancing the additional generation Gne, or simply
We thus have
Red – Reu = R – Rperfect = R – Gperfect := UDL
The quantity UDL is exactly analogous to the difference (R – G) defined for direct semiconductors.
UDL is also the difference between the recombination to a deep level and the emission from it. For the example considered so far (additional generation via illumination) it must be positive, there is more recombination than generation
However, our treatment is completely general; UDL can have any value - if it is negative, we would have more generation via deep levels than recombination.
Of course, UDL makes only sense for global non-equilibrium conditions, because for global equilibrium UDL must be zero!
All we have to do now is to express the Re 's with the formluas from above. Inserting the equations for the various R's, the emission probabilities, and setting se = sh for simplicities sake, we get, after some shuffling of the terms, the final equation
UDL =
v · se · NDL · (ne · nh – ni2)
ne + nh + 2ni · cosh EDLEMB
kT
The cosh (= hyperbolic cosine) comes from the sum of the two exponential functions. Its value is 1 for EDL = EMB; it increases symmetrically for deviations of EDL from the mid-level energy EMB.
A chain hanging down from two posts has exactly a cosh(x) shape - that's the way to memorize the general shape of a cosh curve. If you want to look more closely at the cosh function, activate the link.
The equation for R is quite similar to the one we had for direct semiconductors, as far as the denominator is concerned. We will explore a little more what it implies.
For global equilibrium, the mass action law ne · nh = ni2 applies, and UDL = 0. In other words, there is no net recombination, i.e. recombination in excess of what is always going on.
Without deep levels UDL = 0! The recombination rate then is fixed and simply Rperfect .
The recombination rate - everything else being constant - is directly proportional to the density of the deep levels and their scattering cross section (or capture cross section as it is called in this case).
Since the recombination rate is highest for deep levels exactly in mid-band (look at the cosh function), defects with levels near mid-band are more efficient in recombining carriers than those with levels farther off the mid-band position.
As before, lets look at some special case. Again, we write the carrier densities as ne,h = ne,h(equ) + Dn assuming equal D`s for electron and holes.
This gives us
U v · se ·NDL · [ne(equ) + Dn] · [nh(equ) + Dn] – ni2
ne(equ) + nh(equ) + 2 Dn + 2ni · cosh[(EDLEMB)/kT]
Looking at a p-doped semiconductor and only considering the large densities nh as in the example before, we obtain
U = v · sh · NDL · nh (equ)
nh (equ) + 2ni · cosh[(EDLEMB)/kT]
Again, as before, U is equal to the change in minority carrier concentration dne/dt which gives us
dDne
dt
= v · sh · NDL · nh(equ)
nh(equ) + 2ni · cosh[(EDLEMB)/kT]
Since ni is also much smaller than nh(equ), we may neglect the whole cosh term, too - as long as cosh[(EDL EMB)/kT] is not large, i.e. for deep levels around mid-band.
The solution of the differential equation now becomes trivial and we have
Dne(t) = Dne (t = 0) · exp – t
t
with t = minority life time or better recombination life time in indirect semiconductors defined by
t = 1
v · sh · NDL
This is the same equation as before except that the concentration of the majority carriers (holes in the valence band for the example) now is replaced by the concentration of (mid-band) deep levels.
That this formula is a useful approximation is shown in the two illustrations below:
Dependence of the life time on the deep level position relative to the mid level
- it is fairly constant (and small) as long as the deep level is approximately in mid band.
Dependence of life time on deep level concentration - it is linear as predicted.
The picture on the right illustrates a sad fact hidden in all these equations: it doesn't take much dirt (or contamination, to use the proper word) to considerably degrade the life time. Interstitial gold atoms obviously are felt at 10 –14 cm3, or at concentrations far below ppb. | 6,606 | 27,350 | {"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-2018-34 | latest | en | 0.925772 |
http://cboard.cprogramming.com/c-programming/55714-can%27t-wrap-my-mind-around.html | 1,455,007,170,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701156627.12/warc/CC-MAIN-20160205193916-00149-ip-10-236-182-209.ec2.internal.warc.gz | 34,256,894 | 14,647 | # can't wrap my mind around it...
This is a discussion on can't wrap my mind around it... within the C Programming forums, part of the General Programming Boards category; Hello All, Basically in this app, I have a function. (Is that descriptive or what?) What I need to do ...
1. ## can't wrap my mind around it...
Hello All,
Basically in this app, I have a function. (Is that descriptive or what?)
What I need to do is to pass this function all possible values or a certain variable until it returns a specific string. For instance, the values should look like:
aaaa
aaab
aaac
--//--
aaaz
aaba
aabb
aabc
--//--
zzzy
zzzz
aaaaa (note: goes one char longer, does it again.)
This process is called "brute forcing" by some people. And I need to try all those values up to 56 chars long. And the only way I can think of is to have 56 nested for() loops, obviously very ugly.
SOOOO... anyone have any ideas how i might contruct a better algorithm, such that I don't have eight-zillion nested for()'s???
2. It's called "permutation"
I'm sure a board search will show up something along those lines
3. Just imagine incrementing a base-26 number by one, where the digits are represented using 'a' through 'z'.
If you can't get your head around that, do it for base-10 numbers first.
001, 002, ... 009, 010, 011, ...
Hint: 'a'+1 = 'b'
gg
4. Well, I have a var:
Code:
```char c[26];
strcpy(c,"abcde...xyz");```
and thus i can go:
Code:
```for (i=0;i<26;i++) {
myvar[current-char]=c[i];
dostuff(myvar);
}```
but still, thats 56 of them
and I'm looking into permutations.
5. Originally Posted by crepincdotcom
This process is called "brute forcing" by some people. And I need to try all those values up to 56 chars long.
Let's say the function is only one instruction.
Further, let's assume that your processor executes 10 Gigainstructions per second.
The brute-forcing would then take very long time
26^56 / 10^10 = 1.79 * 10^69 second or 5.49 * 10^61 years.
Your algorithm would take zillions times the age of the universe to complete.
6. yes I know
I don't plan to have it ever get past about 10-13 chars, but I'd just like to be able to thats all.
By the way I have a beowulf cluster that this will be running on too. Sure, not 56 a cray, but it works ok.
Now, on a different note.
I searched all over the forums. I found lots of nice cpp examples... but object oriented stuff scares me. I did find one in c though. (below). It works great. EXCEPT.... it gives you all the permutations of a given string. So, if I put in abc, I get all the different types out... the same length. So abcdefghij.... I can't get a 4 char string out of it.
Can anyone tell me how I might hack this to make it work? To tell the truth, I'm not even sure I know how this version works completly... but I'll try.
Code:
```#include <stdio.h>
#include <string.h>
typedef enum { FALSE, TRUE } BOOL;
void swap(char *a, char *b)
{
char t = *a;
*a = *b; *b = t;
}
void rev(char *begin, char *end)
{
if(*end == '\0') end--;
while(begin < end)
{ swap(begin, end); begin++; end--; }
}
BOOL NextPerm(char *first, char *last)
{
char *a, *b, *c;
if(first == last) return FALSE;
a = first; a++;
if(a == last) return FALSE;
a = last; a--;
for(;/*ever*/;)
{
b = a--;
if(*a < *b)
{
c = last;
while(!(*a < *--c));
swap(a, c); rev(b, last);
return TRUE;
}
if(a == first)
{ rev(first, last); return FALSE; }
}
}
void InsertionSort(char *a, int l, int r)
{
char v;
int i, j;
for(i = l + 1; i <= r; i++)
{
j = i - 1; v = a[i];
while(j >= l && (v < a[j]))
{ a[j+1] = a[j]; j--; }
a[j+1] = v;
}
}
int main(int argc, char *argv[])
{
char *begin, *end;
char *chars="abcdefghijklmnopqrstuvwxyz";
InsertionSort(chars, 0, strlen(chars) - 1);
while(NextPerm(chars, chars + strlen(chars)))
{
printf("%s\n", chars);
}
printf("%s\n", chars);
return 0;
}```
Thanks,
7. I don't think that this is suitable for what you want to do. The reason being, that this gives you permutations. For example, putting in abc, you would get (in some order):
abc
acb
bac
bca
cab
cba
Not only does this not include ones like simply ab or ac as you suggested, it also does not produce permutations using one or more of each character like aaa.
I think code plug might be on to something though...
Davey
8. I understand what codeplug says, but doesnt that still leave me with 56 nested for()'s, for each char in the string?
9. Problem: Given a decimal integer in the form of a C-string, develop an algorithm for incrementing that integer by 1. You may not convert the C-string to any other data type and the update to the C-string must be done in-place (no copies).
Example: Gven this string, "0099", the resulting string should be "0100".
Do that and you'll see how to solve the other problem.
gg
10. Originally Posted by crepincdotcom
I understand what codeplug says, but doesnt that still leave me with 56 nested for()'s, for each char in the string?
Can't look at it from another point of view, eh? Well, that happens sometimes. Here's one slow solution:
Code:
```#include <stdio.h>
#include <string.h>
typedef enum {FALSE, TRUE} BOOL;
#define MAXLENGTH 8
#define STARTLENGTH 4
int main(int argc, char **argv)
{
char c[MAXLENGTH];
BOOL bRun;
int length, i;
memset((void*)c, 'a', sizeof(c));
length = STARTLENGTH;
bRun = TRUE;
do
{
for (i = length-1; i >= 0; i--)
fputc(c[i], stdout);
fputc('\n', stdout);
for (i = 0; i < length; i++)
if (++c[i] > 'z')
{
c[i] = 'a';
if (i == length-1)
{
if (++length > sizeof(c))
bRun = FALSE;
break;
}
}
else
break;
} while (bRun);
return 0;
}``` | 1,585 | 5,557 | {"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-2016-07 | latest | en | 0.920925 |
https://www.aqua-calc.com/calculate/food-weight-to-volume/substance/red-blank-enchilada-blank-sauce-coma-and-blank-upc-column--blank-884395000100 | 1,566,386,764,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315936.22/warc/CC-MAIN-20190821110541-20190821132541-00092.warc.gz | 727,038,997 | 8,084 | # Volume of RED ENCHILADA SAUCE, UPC: 884395000100
## food weight to volume conversions
### calculate volume of generic and branded foods per weight
#### Volume, i.e. how many spoons, cups,gallons or liters in 100 gram of REDENCHILADA SAUCE, UPC: 884395000100
centimeter³ 98.58 milliliter 98.58 foot³ 0 US cup 0.42 Imperial gallon 0.02 US dessertspoon 13.33 inch³ 6.02 US fluid ounce 3.33 liter 0.1 US gallon 0.03 meter³ 9.86 × 10-5 US pint 0.21 metric cup 0.39 US quart 0.1 metric dessertspoon 9.86 US tablespoon 6.67 metric tablespoon 6.57 US teaspoon 20 metric teaspoon 19.72
#### Weight
gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000
Nutrient (find foodsrich in nutrients) Unit Value /100 g BasicAdvancedAll Proximates Energy kcal 42 Protein g 1.67 Total lipid (fat) g 0.83 Carbohydrate,bydifference g 6.67 Fiber,totaldietary g 0 Sugars, total g 1.67 Minerals Calcium, Ca mg 0 Iron, Fe mg 0.6 Sodium, Na mg 733 Vitamins Vitamin C,totalascorbic acid mg 0 Vitamin A, IU IU 833 Lipids Fatty acids,totalsaturated g 0 Cholesterol mg 0
#### See how many calories in0.1 kg (0.22 lbs) of REDENCHILADA SAUCE, UPC:884395000100
From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 42 175.73 Total 42 175.73
• 253.60517 grams [g] of RED ENCHILADA SAUCE, UPC: 884395000100 fill 1 metric cup
• 8.46575 ounces [oz] of RED ENCHILADA SAUCE, UPC: 884395000100 fill 1 US cup
• RED ENCHILADA SAUCE, UPC: 884395000100 weigh(s) 253.61 gram per (metric cup) or 8.47 ounce per (US cup), and contain(s) 42 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• Ingredients: WATER, TOMATO PASTE, CHILI POWDER, CONTAINS LESS THAN 2%: SALT, MODIFIED CORNSTARCH, SUGAR, SOYBEAN OIL, ONION POWDER, PAPRIKA, GARLIC POWDER, CITRIC ACID, CHICKEN BROTH POWDER, CUMIN, OREGANO, WHITE PEPPER, DISODIUM INOSINATE AND DISODIUM GUANYLATE, NATURAL FLAVOR.
• For instance, compute how many cups or spoons a pound or kilogram of “RED ENCHILADA SAUCE, UPC: 884395000100” fills. Volume of the selected food item is calculated based on the food density and its given weight. Visit our food calculations forum for more details.
• Reference (ID: 19760)
• USDA National Nutrient Database for Standard Reference; National Agricultural Library; United States Department of Agriculture (USDA); 1400 Independence Ave., S.W.; Washington, DC 20250 USA.
#### Foods, Nutrients and Calories
TOMATO ALFREDO, UPC: 739276000499 weigh(s) 262.06 gram per (metric cup) or 8.75 ounce per (US cup), and contain(s) 81 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
THE ESSENTIAL BAKING COMPANY, SUNNY SEEDED WHITE, GLUTEN FREE BREAD, UPC: 813305011998 contain(s) 333 calories per 100 grams or ≈3.527 ounces [ price ]
Foods high in Copper, Cu and Recommended Dietary Allowances (RDAs) for Copper
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Moonlight Sand weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Porphyry, solid weighs 2 547 kg/m³ (159.00402 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Rapeseed oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
A milligram per cubic millimeter (mg/mm³) is a derived metric SI (System International) measurement unit of density used to measure volume in cubic millimeters in order to estimate weight or mass in milligrams
Dynamic viscosity is a measure of how resistive a fluid is to flow. Viscosity describes how thick a fluid is
Convert long ton per US gallon to stone per US tablespoon or convert between all units of density measurement
#### Calculators
Capsule capacity calculator: weight, volume and number of capsules | 1,210 | 4,081 | {"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-2019-35 | latest | en | 0.408541 |
https://brainmass.com/math/trigonometry/trigonometry-questions-102767 | 1,675,125,176,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00488.warc.gz | 154,791,622 | 75,435 | Explore BrainMass
# Trigonometry Questions
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
1. Consider the graph of y = tan x.
(a) How does it show that the tangent of 90 degrees is undefined?
(b) What are other undefined x values?
(c) What is the value of the tangent of angles that are close to 90 degrees (say 89.9 degrees and 90.01 degrees)?
(d) How does the graph show this?
2. A nautical mile depends on latitude. It is defined as length of a minute of arc of the earth's radius. The formula is N(P) = 6066 - 31 cos 2P, where P is the latitude in degrees.
(a) Using the Library and other course resources, find the exact latitude (to 4 decimal places) of where you live, used to live, work, or used to work (include the zip code).
(b) Using the latitude found in part a and the formula N(P), find the length of a nautical mile to the nearest foot at that location.
(c) Next, use the formula N(P) to find the latitude where the nautical mile is 6051 feet.
(d) Name two cities in the Northern Hemisphere and two in the Southern that are close to the latitude found in part c.
3. When graphed using polar coordinates, the center of a regular nonagon is at the origin and one vertex is at (6, 0 degrees) or (6, 0 radians). Find the polar coordinates of the other vertices in both degrees and radians.
https://brainmass.com/math/trigonometry/trigonometry-questions-102767
#### Solution Preview
Consider the graph of y = tan x.
The answers are next to the question, but read the longer explanation below.
(a) How does it show that the tangent of 90 degrees is undefined? There is a vertical asymptote at x = 90 degrees (a vertical line that the graph approaches, but never intersects).
(b) What are other undefined x values? 90 degrees (pi/2 radians) plus multiples of 180 degrees.
(c) What is the value of the tangent of angles that are close to 90 degrees (say 89.9 degrees and 90.01 degrees)? They are very large positive numbers, or very small negative numbers (you can find the exact values of tan(89.9) and tan(90.01) on your calculator if you'd like, but the closer you get to 90, the larger or smaller the number will become).
(d) How does the graph show this? The graph increases steeply (y takes on larger and larger values) as it gets closer to 90 degrees from the left, and decreases steeply (y takes on smaller and smaller negative values) as it gets closer to 90 degrees from the right.
The graph of y = tanx looks like this:
The graph is in green, and the ...
\$2.49 | 646 | 2,643 | {"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} | 3.859375 | 4 | CC-MAIN-2023-06 | longest | en | 0.87921 |
https://math.libretexts.org/Bookshelves/PreAlgebra/Prealgebra_1e_(OpenStax) | 1,718,411,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00385.warc.gz | 333,783,989 | 31,329 | # Prealgebra 1e (OpenStax)
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Prealgebra is designed to meet scope and sequence requirements for a one-semester prealgebra course. The text introduces the fundamental concepts of algebra while addressing the needs of students with diverse backgrounds and learning styles. Each topic builds upon previously developed material to demonstrate the cohesiveness and structure of mathematics.
This page titled Prealgebra 1e (OpenStax) is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 791 | 2,309 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-26 | latest | en | 0.336246 |
https://metanumbers.com/1021128 | 1,643,452,111,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304883.8/warc/CC-MAIN-20220129092458-20220129122458-00338.warc.gz | 436,471,987 | 7,528 | # 1021128 (number)
1,021,128 (one million twenty-one thousand one hundred twenty-eight) is an even seven-digits composite number following 1021127 and preceding 1021129. In scientific notation, it is written as 1.021128 × 106. The sum of its digits is 15. It has a total of 6 prime factors and 32 positive divisors. There are 336,960 positive integers (up to 1021128) that are relatively prime to 1021128.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 15
• Digital Root 6
## Name
Short name 1 million 21 thousand 128 one million twenty-one thousand one hundred twenty-eight
## Notation
Scientific notation 1.021128 × 106 1.021128 × 106
## Prime Factorization of 1021128
Prime Factorization 23 × 3 × 157 × 271
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 255282 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,021,128 is 23 × 3 × 157 × 271. Since it has a total of 6 prime factors, 1,021,128 is a composite number.
## Divisors of 1021128
32 divisors
Even divisors 24 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 2.57856e+06 Sum of all the positive divisors of n s(n) 1.55743e+06 Sum of the proper positive divisors of n A(n) 80580 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1010.51 Returns the nth root of the product of n divisors H(n) 12.6722 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,021,128 can be divided by 32 positive divisors (out of which 24 are even, and 8 are odd). The sum of these divisors (counting 1,021,128) is 2,578,560, the average is 80,580.
## Other Arithmetic Functions (n = 1021128)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 336960 Total number of positive integers not greater than n that are coprime to n λ(n) 7020 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 79869 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 336,960 positive integers (less than 1,021,128) that are coprime with 1,021,128. And there are approximately 79,869 prime numbers less than or equal to 1,021,128.
## Divisibility of 1021128
m n mod m 2 3 4 5 6 7 8 9 0 0 0 3 0 3 0 6
The number 1,021,128 is divisible by 2, 3, 4, 6 and 8.
• Arithmetic
• Abundant
• Polite
## Base conversion (1021128)
Base System Value
2 Binary 11111001010011001000
3 Ternary 1220212201120
4 Quaternary 3321103020
5 Quinary 230134003
6 Senary 33515240
8 Octal 3712310
10 Decimal 1021128
12 Duodecimal 412b20
20 Vigesimal 67cg8
36 Base36 lvwo
## Basic calculations (n = 1021128)
### Multiplication
n×y
n×2 2042256 3063384 4084512 5105640
### Division
n÷y
n÷2 510564 340376 255282 204226
### Exponentiation
ny
n2 1042702392384 1064732608530289152 1087228279083317101203456 1110199238163789424917682618368
### Nth Root
y√n
2√n 1010.51 100.699 31.7885 15.9153
## 1021128 as geometric shapes
### Circle
Diameter 2.04226e+06 6.41594e+06 3.27575e+12
### Sphere
Volume 4.45994e+18 1.3103e+13 6.41594e+06
### Square
Length = n
Perimeter 4.08451e+06 1.0427e+12 1.44409e+06
### Cube
Length = n
Surface area 6.25621e+12 1.06473e+18 1.76865e+06
### Equilateral Triangle
Length = n
Perimeter 3.06338e+06 4.51503e+11 884323
### Triangular Pyramid
Length = n
Surface area 1.80601e+12 1.2548e+17 833748
## Cryptographic Hash Functions
md5 50b3174fdb1ff67ce3437049cf972967 a0c930be6596863c63e9af33959eb25f2aaefb9a e36f0b6888761bf5e8f438235ff4640b0717c7fe61da8eaa1d3f20f620aa8e8f c1fa557345675380bf81b58323b092a8275f39e0e5d056708b949ae6b8e36d645e61c556a67b225678584df86ea6c36e9beae65851ce5f4ed7b2070205a6f223 bb1766c3ac3cf478611f87a182adc5b79bf328df | 1,523 | 4,259 | {"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} | 3.609375 | 4 | CC-MAIN-2022-05 | latest | en | 0.806826 |
https://quantumcomputing.stackexchange.com/questions/16890/are-spin-glass-problems-np-complete/16902 | 1,627,642,999,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.52/warc/CC-MAIN-20210730091645-20210730121645-00184.warc.gz | 488,058,882 | 43,038 | # Are spin-glass problems NP (-complete)?
It is well known that finding ground states for spin-glass systems (Ising, XY...) is NP-hard (at least as hard as the hardest NP-problems) so that they can be efficiently used to solve other NP problems like the Traveling Salesman Problem.
My question is: is the problem NP-complete? This seems to be what is claimed here. To my understanding, this would mean that apart from NP-hard, the problem is NP itself. But I don't know an obvious algorithm to check if a given solution is the true ground state in polynomial time? And actually, I think that a similar argument can be made for the traveling salesman problem.
• I suggest to add the tag "spin-glasses" btw, as I think that it would be useful for future reference (but lack the reputation) – Wouter Apr 7 at 5:23
## Background
Computational problems come in a variety of types, for example:
• decision problem: given input $$x$$, output "YES" if $$x$$ belongs to a fixed set $$L$$ and output "NO" otherwise,
• function problem: given input $$x$$, output a $$y$$ such that $$(x, y)$$ belongs to a fixed relation $$R$$,
• optimization problem: given objective $$f$$ and constraints $$g$$, output $$x$$ that maximizes $$f$$ while satisfying $$g$$,
• sampling problem: given a probability distribution $$P$$, output a sample $$x\sim P$$.
Together with the computational model and resource bounds, the type of relevant problems is a key ingredient in a definition of a computational complexity class.
## Traveling Salesman Problem
The class NP is defined as a class of decision problems. The Traveling Salesman Problem formulated as the task of finding the shortest loop traversing all nodes in a graph exactly once is a function problem and thus - as you suspected in your question - does not belong to NP. However, TSP has a decision version which for a given graph and a threshold length $$w$$ asks whether there is a loop traversing all nodes in the graph exactly once and whose total length is less than $$w$$. This problem belongs to NP. The certificate is the path and it is easy to see that it can be checked in polynomial time on a deterministic Turing machine.
## Spin glasses
Similarly, the problem of finding the ground state of a spin glass system is a function problem and hence not in NP. The problem has a decision variant which for a given spin glass system and an energy threshold $$E$$ asks whether there is an energy eigenstate with energy below $$E$$. Under certain assumptions about the Hamiltonian, this problem belongs to NP.
For example, in the absence of a transversal field, the Ising model is diagonal in a product basis and thus has a product eigenstate for every eigenvalue. Each product state admits a description of size polynomial in the number of sites. We can use such a state as a certificate that can be checked in polynomial time to prove that the model has a state with energy below the threshold $$E$$.
## Terminology
The expression "TSP is NP-complete" is an imprecise shortcut for "the decision version of TSP is NP-complete". Leaving "decision version" implicit is generally acceptable because only decision problems may be meaningfully declared to be NP-complete.
That said, note that it does make sense to say that a computational problem $$P$$ which is not a decision problem is NP-hard. See Wikipedia for an explanation.
• Thank you! Is it still true that the optimization version of the spin glass problem gives the optimization of traveling salesman with only polynomial overhead if none of them are NP? – Wouter Apr 7 at 7:49
• Another question: isn't the notion also meaningful for function problems? (e.g. prime factorization, permanents) – Wouter Apr 7 at 8:24
• Re 1: Yes. The distinction of decision and function problems does not affect our ability to efficiently map from graph problems (such as TSP and Max-Cut) to the Ising model. Re 2: Yes, it is meaningful (even though imprecise) to talk about NP-completeness of such problems, because it is easy to convert them into corresponding decision problems when greater precision is needed. This happens e.g. when we begin to wonder what a suitable certificate looks like (as you did in your question). You may also be interested in the FNP complexity class. – Adam Zalcman Apr 7 at 17:01
• Thank you! Why are you referring to the standard versions of the problems as ´function problems´ rather than ´opnimization problems´? To me, a function problem would be something like: "Given a fixed loop of the TSP, how long is it?" or, "What's the energy of this fixed state in the spin-glass?" which are not hard at all – Wouter Apr 8 at 1:23
• Usual formulations of the TSP and ground state problems are indeed optimization problems. However, optimization problems can be thought of as function problems with extra structure. I called them "function problems" since I didn't need to involve any of the special structure that optimization problems have. Your other examples are also function problems. – Adam Zalcman Apr 8 at 2:09
Adam gave an excellent answer explaining that the "decision version" of the Ising model could be considered NP-complete, but the term doesn't make sense without that extra context.
I'll answer some of the other parts which I think were not yet addressed:
"But I don't know an obvious algorithm to check if a given solution is the true ground state in polynomial time?"
If you are given a solution to an arbitrary Ising model, for example: 001100110...01101 where 0 means spin-up and 1 means spin-down, there is no way in general to know that it is the true ground state, without checking all $$2^n$$ possible solutions, which means it can't be done in polynomial time with respect to the number of spins $$n$$. If you have a very specific Ising model, for example the one where there's no linear terms and all couplings between the spins are negative in strength, then you can find the ground state in polynomial time, but there's no way to verify that a solution is the true ground state in polynomial time in general.
Finally, since you talked about spin-glass systems in general (not just the Ising model, as you specifically mentioned the XY model which unlike the Ising model is not diagonal), I'll mention that the 2-local Hamiltonian problem is QMA-complete, which is the analog of NP-completeness, for when you have access to a quantum computer. Any Hamiltonian can be decomposed into a sum of products of spin-matrices, which describes a k-local spin-glass. It was proven that finding the ground state energy of the k-local Hamiltonian accurately is QMA-complete long before the more strong result in that paper I linked earlier, which shows that 2-local Hamiltonians are enough (you don't even need k-local interactions).
In fact, the 2D Ising model with transverse fields is universal, which was proved in this Science paper 5 years ago.
• Thanks for your answer. To be sure I understand you correctly: finding ground states cannot be verified in polynomial time on a quantum computer (as opposed to modeling time-evolution for the system) as I read once before. But you're saying that verification on a quantum computer (of the optimization version of the spin glass problem) can be done polynomially? – Wouter Apr 8 at 3:24
• And this is irrespective of if the spin model is quantum or classical in nature? I think that the classical ones can be harder sometimes... maybe you would be interested also in taking a look at my question here physics.stackexchange.com/questions/621460/… – Wouter Apr 8 at 3:24
• @Wouter I can't find where I said that anything can be done polynomially except for the submodular Ising problem which is a very specific example. – user1271772 Apr 8 at 3:36
• I see that in the first sentence of my first comment, I wrote ´verified´when I meant ´computed´. But you were saying that all quadratic Hamiltonians become NP-like (QMA) problems (can be checked in polynomial time) on a quantum computer ? – Wouter Apr 8 at 3:45
• "the 2-local Hamiltonian problem is QMA-complete, which is the analog of NP-completeness, for when you have access to a quantum computer" – Wouter Apr 8 at 3:49
I've found this Venn Diagram to be incredibly helpful in understanding the differences between P, NP, NP-Complete, and NP-Hard.
You're right that Traveling Salesman would not be considered NP-Complete. It's NP-Hard, but not in NP. The only way to verify a solution to Traveling Salesman (that we know of today) is to compare it to the optimal solution... and to get that optimal solution, you'd need to solve the problem. There are a whole bunch of NP-Hard problems that are too hard to be NP, including spin-glass. | 1,968 | 8,686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-31 | latest | en | 0.922663 |
http://increaseverticaljumptips.com/how-to-increase-vertical-jump/vertical-jump-kinematic-analysis/ | 1,550,342,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480905.29/warc/CC-MAIN-20190216170210-20190216192210-00534.warc.gz | 127,008,121 | 9,519 | Vertical Jump Kinematic Analysis
By | January 15, 2017
Projectile motion part 1 Onedimensional motion Physics Khan Academy
Welcome back. I'm not going to do a bunch ofprojectile motion problems, and this is because I thinkyou learn more just seeing someone do it, and thinkingout loud, than all the formulas. I have a strange notion that Imight have done more harm than good by confusing you with a lotof what I did in the last
couple of tutorials, so hopefullyI can undo any damage if I have done any, or even betterhopefully, you did learn from those, and we'lljust add to the learning. Let's start with ageneral problem. Let's say that I'm at the topof a cliff, and I jump instead of throwing something,I just jump off the cliff. We won't worry about my motionfrom side to side, but just assume that I gostraight down.
We could even think that someonejust dropped me off of the top of the cliff. I know these are getting kindof morbid, but let's just assume that nothingbad happens to me. Let's say that at the topof the cliff, my initial velocity velocity initial isgoing to be 0, because I'm stationary before the persondrops me or before I jump. At the bottom of the cliffmy velocity is
100 meters per second. My question is, what is theheight of this cliffé I think this is a good timeto actually introduce the direction notion ofvelocity, to show you this scalar quantity. Let's assume up is positive,and down is negative. My velocity is actually 100meters per second down I
could have assumedthe opposite. The final velocity is 100 metersper second down, and since we're saying that downis negative, and gravity is always pulling you down, we'regoing to say that our acceleration is equal togravity, which is equal to minus 10 meters persecond squared. I just wrote that ahead oftimes, because when we're
dealing with anything ofthrowing or jumping or anything on this planet, wecould just use this constant the actual number is 9.81, butI want to be able to do this without a calculator, so I'lljust stick with minus 10 meters per second squared. It's pulling me down, so that'swhy the minus is there. My question is: I know myinitial velocity, I know my final velocity, right before Ihit the ground or right when I
PASCO Capstone Tutorial Analysis
Elizabeth: Hello and welcome. I'm Elizabeth. Chong: I'm Chong. E: We're here today to use the PASCO Capstone software, and talk about the feature of tutorial analysis. I've already started an experiment in Capstone, and I have two graph displays and I've imported a movie. The movie's been prerecorded, but Chong'sgoing to explain what the movie's about. C: The movie is going to have a ballisticcart on a PASCO PAStrack,
which is bigger than the one that's shownhere, moving, and a ball shooting up out of it. We're going to track the ball. E: Let's see what the movie looks like. I'llgo ahead and play it here. There it is. C: I notice, Elizabeth, that the movie's veryslow. E: It does appear very slow, almost as ifit were slow motion. Why is thaté
C: That's because the tutorial was shot usinga high speed camera. The other thing I notice is the cart caughtthe ball. E: The cart did catch the ball. That's whywe have this device. This is meant to show the independence ofmotion in the X and Y direction. If we are able to perhaps use our tutorial analysisto show that, that would be a great use for this tool. C: Let's look at how do we go about correctingfor the camera being a high speed camera. E: Great idea. We can actually do that.
We can open up the properties in here, andwe can go into the Movie Playback. C: In this case, the tutorial was filmed in 210frames per second. E: I'll enter 210 here instead of 29. Now let's play that back and see if that looksdifferent. C: That looks more like real life. E: That looks exactly like real life, exactly. C: Perfect. E: Now as Chong said, 210 frames per second.
We're going to end up clicking the mouse on the ball to track it. If we work at 210 frames per second, that's going to be a lot of clicks just to follow that little tiny ball. C: Do we have a tool to address thaté E: We do in fact have a tool. We can addressthat. Let's tell it to advance on each click, let'ssay seven frames. We'll go into the properties and we'll dothe same thing. Actually, before we do that, we need to enterinto tutorial analysis mode. I'll do that.
In tutorial analysis mode now, what I can dois back in the properties in the Overlay menu, I can tell it instead of advancing every oneframe, for it to advance seven. Each mouse click will now be equal to sevenframes. C: Now that we corrected for time, is thereany way that we can correct for distanceé If we wanted our data to be actually usablefor real life. E: That's a great idea. Real world data, yes.We have here a caliper. It's draggable. You'll notice that when Iclick on either end of it, I can drag it. The handy thing is, why don't you tell them aboutthe length of the real track.
Simi Motion Inverse Dynamics
Hello and welcome to another Simi Motion tutorialtutorial. Today we are going to show you how to createinverse dynamics. With inverse dynamics you can calculate jointangles as well as forces like muscle torques. To create the inverse dynamic, you need movementtutorials with force plate data and a static trial. A static trial is a short tutorial of personstanding in a relaxed state, with all the used markers of the markersetalready applied. To demonstrate the creation of an inversedynamics, we already have such a project with alreadytracked and assigned markers of the lower body.
If you also have come to this point with yourproject, you can start the inverse dynamics. Find Project and select Inverse Dynamics. For the body model, the model for static markerscan be used for full body acquisitions, as well as onlyparts of the body. All body parts which can be calculated withthe available markers will be calculated. For static, select the 3Dcalculation of yourstatic trial. If you do not have it right here in your openproject, you can also import it from another one. Next, select the 3Dcalculation from yourmovement trial for dynamic 3Ddata.
The forces for both legs have to be chosenfrom the force plate data you acquired together with your movement tutorials. For the data to be calculated accordingly,you also need to enter height, weight and sex of the person you tested. It is also important to select the right valuefor the marker size. Now you have also some options to finetunethe results, like interpolation and filtering as well assetting the force threshold for your force data. On the right side of the window you can chooseall the data rows you want to have calculated.
When you have entered everything, click Calculateto finish and close the window. Now you have calculated the inverse dynamics. To test if the data is calculated correctly,you can check with the 3Dview. Via right click and properties you can selectthe XModel and display the calculated inverse dynamicsas skeleton. If the skeleton looks good, the inverse dynamicscalculation was a success. Now you know, how to calculate inverse dynamicsand check if it was correct. | 1,647 | 7,149 | {"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} | 3.765625 | 4 | CC-MAIN-2019-09 | latest | en | 0.95703 |
http://metamath.tirix.org/mpests/lvecvscan | 1,726,189,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00437.warc.gz | 20,081,545 | 3,397 | # Metamath Proof Explorer
## Theorem lvecvscan
Description: Cancellation law for scalar multiplication. ( hvmulcan analog.) (Contributed by NM, 2-Jul-2014)
Ref Expression
Hypotheses lvecmulcan.v $⊢ V = Base W$
lvecmulcan.s
lvecmulcan.f $⊢ F = Scalar W$
lvecmulcan.k $⊢ K = Base F$
lvecmulcan.o
lvecmulcan.w $⊢ φ → W ∈ LVec$
lvecmulcan.a $⊢ φ → A ∈ K$
lvecmulcan.x $⊢ φ → X ∈ V$
lvecmulcan.y $⊢ φ → Y ∈ V$
lvecmulcan.n
Assertion lvecvscan
### Proof
Step Hyp Ref Expression
1 lvecmulcan.v $⊢ V = Base W$
2 lvecmulcan.s
3 lvecmulcan.f $⊢ F = Scalar W$
4 lvecmulcan.k $⊢ K = Base F$
5 lvecmulcan.o
6 lvecmulcan.w $⊢ φ → W ∈ LVec$
7 lvecmulcan.a $⊢ φ → A ∈ K$
8 lvecmulcan.x $⊢ φ → X ∈ V$
9 lvecmulcan.y $⊢ φ → Y ∈ V$
10 lvecmulcan.n
11 df-ne
12 biorf
13 11 12 sylbi
14 10 13 syl
15 lveclmod $⊢ W ∈ LVec → W ∈ LMod$
16 6 15 syl $⊢ φ → W ∈ LMod$
17 eqid $⊢ 0 W = 0 W$
18 eqid $⊢ - W = - W$
19 1 17 18 lmodsubeq0 $⊢ W ∈ LMod ∧ X ∈ V ∧ Y ∈ V → X - W Y = 0 W ↔ X = Y$
20 16 8 9 19 syl3anc $⊢ φ → X - W Y = 0 W ↔ X = Y$
21 1 2 3 4 18 16 7 8 9 lmodsubdi
22 21 eqeq1d
23 1 18 lmodvsubcl $⊢ W ∈ LMod ∧ X ∈ V ∧ Y ∈ V → X - W Y ∈ V$
24 16 8 9 23 syl3anc $⊢ φ → X - W Y ∈ V$
25 1 2 3 4 5 17 6 7 24 lvecvs0or
26 1 3 2 4 lmodvscl
27 16 7 8 26 syl3anc
28 1 3 2 4 lmodvscl
29 16 7 9 28 syl3anc
30 1 17 18 lmodsubeq0
31 16 27 29 30 syl3anc
32 22 25 31 3bitr3d
33 14 20 32 3bitr3rd | 691 | 1,368 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "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 | 0.174541 |
https://www.physicsforums.com/threads/verify-subset-is-a-subspace.323518/ | 1,521,667,219,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647692.51/warc/CC-MAIN-20180321195830-20180321215830-00487.warc.gz | 849,131,746 | 16,435 | # Verify Subset is a Subspace
1. Jul 5, 2009
Okay then. I just read the section of Axler on subspaces. It says that if U is a subset of V, then to check that U is a subspace of V we need only check that U satisfies the following:
$$0\in U$$
$$u,v\in U\text{ implies }u+v\,\in\,U$$
closed under scalar multiplication
$$a\in\mathbf{F}\text{ and }u\in U\text{ implies }au\in U$$
Now I am supposed to use these axioms to verify that for $b\in\mathbf{F}$, then
$${(x_1,x_2,x_3,x_4)\in\mathbf{F}^4:x_3=5x_4+b}$$
is a subspace of F4 iff b=0.
I am not exactly sure how to actually apply those 3 axioms to this problem?
How does one test that $u,v\in U\text{ implies }u+v\,\in\,U$?
(x1,x2,x3,x4)+(y1,y2,y3,y4)=(x1+y1,x2+y2,5x4+b+5y4+b,x4+y4) ?
Last edited: Jul 5, 2009
2. Jul 6, 2009
### Office_Shredder
Staff Emeritus
$$(x_1,x_2,x_3,x_4) + (y_1,y_2,y_3,y_4) = (x_1 + y_1, x_2 + y_2, x_3 + y_3, x_4 + y_4)$$
by definition. So there are two parts to this problem
1) Show if b=0 then this is a subspace
2) Show if this is a subspace then b=0
To work towards 1), if
$$x_3 = 5x_4$$ and $$y_3 = 5y_4$$
show the RHS satisfies the necessary relationship. Rinse and repeat for the other stuff
To do part 2, you need to find a contradiction. So look at the three things required for the subset to be a subspace, and see if all of them hold
3. Jul 6, 2009
Okay. So just dealing with the x3+y3 part of the addition
x3+y3=5x4+5y4+2b.. I am sorry, but I still do not see why b must equal 0 what part of the definition is being violated if it is not?
4. Jul 6, 2009
### snipez90
Check the additive identity first. What is the additive identity? Now what happens if b is not 0?
5. Jul 6, 2009
Okay, I see now. I forgot about the other properties for a moment.
Can someone help me out wit the formalism of this? If this were a HW problem, my 'proof' would need to follow a certain format. Should I name the proposed subspace and then continue. Like this.
Proposition:
$$U={(x_1,x_2,x_3,x_4)\in\mathbf{F}^4:x_3=5x_4+b)\text{ is a subspace of }\mathbf{F}^4\text{ iff }b=0$$
Proof:
Now I am not sure how to formally state that since a subspace must include the additive identity then b must equal 0. How do you math savvy types do this?
6. Jul 6, 2009
### Dick
You don't have to be that math savvy. You want to solve the equations x3=0, x4=0 and x3=5*x4+b. For what values of b is that possible?
7. Jul 6, 2009 | 857 | 2,419 | {"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} | 4 | 4 | CC-MAIN-2018-13 | longest | en | 0.840586 |
https://gmatclub.com/forum/m20-q32-96714.html?fl=similar | 1,508,326,363,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00785.warc.gz | 695,299,136 | 40,625 | It is currently 18 Oct 2017, 04:32
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# m20 q32
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
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Intern
Joined: 10 Jun 2010
Posts: 9
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m20 q32 [#permalink]
### Show Tags
04 Jul 2010, 05:47
A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?
a. at 2:00 pm
b. at 2:30 pm
c. at 3:00 pm
d. at 3:30 pm
e. at 4:00 pm
[Reveal] Spoiler:
When only the valve is open, the pool is filling at a rate of $$\frac{1}{4}$$ an hour. When both the valve and the drain are open, the pool is filling at a rate of $$\frac{1}{4} - \frac{1}{5}= \frac{1}{20}$$ an hour. Let $$x$$ denote the time when only the valve was open. Then both the valve and the drain were open for $$11-1-x = 10-x$$ hours. Now we can compose the equation $$\frac{1}{4}x+\frac{1}{20}(10-x)=1$$ which reduces to $$\frac{x}{5}=0.5$$ from where $$x=2.5$$ . Thus, the drain was opened at $$1:00 + 2:30 = 3:30$$ pm.
The correct answer is D.
Can someone please explain/ simplify the wording of the problem for me please?
The explanation provided is straightforward enough, but I could not set the correct equation up because I could get my head round the 3rd statement
Does the pool fill at 11pm the following day, or was it full at 11pm the previous day?
To derive the solution, does it even matter if the pool was full the previous day or the it filled up the following day?
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Manager
Joined: 16 Apr 2010
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Re: m20 q32 [#permalink]
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04 Jul 2010, 14:14
Hi,
You have to assume that the pool was empty. Otherwise you cannot solve the problem.
The valve that fills the pool was opened at 1:00 pm in the afternoon just at my lunch time .
If the pool was filled by 11:00 pm and not earlier, when was the drain opened? Here 11:00pm is 10 hours after my lunch time and ie 1 hour just before midnight.
I hope I got your questions right and answered them accordingly.
regards,
Jack
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Re: m20 q32 [#permalink]
### Show Tags
04 Jul 2010, 21:50
suddenly it all makes sense
thanks a lot jakolik!
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Re: m20 q32 [#permalink] 04 Jul 2010, 21:50
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# m20 q32
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Moderator: Bunuel
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 992 | 3,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2017-43 | latest | en | 0.940427 |
https://www.numbersaplenty.com/607426559 | 1,716,569,371,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00270.warc.gz | 799,598,192 | 3,088 | Search a number
607426559 = 911666769
BaseRepresentation
bin100100001101001…
…001011111111111
31120022222110120222
4210031021133333
52221000122214
6140135131555
721024016625
oct4415113777
91508873528
10607426559
11291970623
1214b513bbb
1398ac8ccc
145a95b515
15384d838e
hex243497ff
607426559 has 4 divisors (see below), whose sum is σ = 608094240. Its totient is φ = 606758880.
The previous prime is 607426553. The next prime is 607426571. The reversal of 607426559 is 955624706.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-607426559 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (607426553) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 332474 + ... + 334295.
It is an arithmetic number, because the mean of its divisors is an integer number (152023560).
Almost surely, 2607426559 is an apocalyptic number.
607426559 is a deficient number, since it is larger than the sum of its proper divisors (667681).
607426559 is an equidigital number, since it uses as much as digits as its factorization.
607426559 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 667680.
The product of its (nonzero) digits is 453600, while the sum is 44.
The square root of 607426559 is about 24646.0252170609. The cubic root of 607426559 is about 846.8982959129.
The spelling of 607426559 in words is "six hundred seven million, four hundred twenty-six thousand, five hundred fifty-nine".
Divisors: 1 911 666769 607426559 | 514 | 1,739 | {"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} | 2.984375 | 3 | CC-MAIN-2024-22 | latest | en | 0.847709 |
https://de.scribd.com/document/399955158/agung1-2 | 1,586,107,914,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371606067.71/warc/CC-MAIN-20200405150416-20200405180916-00445.warc.gz | 416,095,164 | 57,744 | Sie sind auf Seite 1von 17
Tugas Analisa Struktur 4
BAB I
TRUSS 2D
1.1 Soal
Agung saputro (30201504026) 1
Tugas Analisa Struktur 4
1.3 Persiapan Data
a. Data Propertis :
No E A
1 2.1 e 7 0.004
2 2.1 e 7 0.001
b. Data Koordinat :
No x Y
1 0 0
2 4.5 0
3 9 0
4 13.5 0
5 18
6 13.5 2.5
7 9 5
8 4.5 2.5
c. Data Batang :
No J1 J2 Prop
1 1 2 1
2 8 2 1
3 1 8 2
4 2 3 1
5 8 3 2
6 8 7 2
7 7 3 1
8 3 4 1
9 6 3 2
10 7 6 2
11 4 5 1
Agung saputro (30201504026) 2
Tugas Analisa Struktur 4
12 6 4 1
13 6 5 2
d. Data Tumpuan :
1 1 1 1
8 5 0 1
No Fx Fy
8 3.5 -5
7 3.5 -5
5 0 -5
2 0 -6
3 0 -6
4 0 -6
1.4 Input Data
%****rangka batang 2D---TEKNIK SIPIL UNISSULA SEMARANG----
%****nama file : soal1_agung oleh :agung p 18/4/2018----
%****struktur : rangka batang soal 1
clear all
clc
%****General Data-------
type='t2d'
nfile='agung saputro'
%****Property---i,E,A--------------
prop=[ 1 2.1e7 0.004
2 2.1e7 0.001];
%****Coordinates-----------
%****--- Joint X Y ---
coord=[ 1 0 0
2 4.5 0
3 9 0
4 13.5 0
5 18 0
6 13.5 2.5
7 9 5
8 4.5 2.5];
%****Element data------------------
Agung saputro (30201504026) 3
Tugas Analisa Struktur 4
%****--- Element J1 J2 No.prop---
element=[ 1 1 2 1
2 8 2 1
3 1 8 2
4 2 3 1
5 8 3 2
6 8 7 2
7 7 3 1
8 3 4 1
9 6 3 2
10 7 6 2
11 4 5 1
12 6 4 1
13 6 5 2];
%****Nodal Restraint----------
%****---- Joint JR1 JR2 --
Support=[ 1 1 1
5 0 1];
%****---Joint FX FY --
JL=[ 8 3.5 -5
7 3.5 -5
5 0 -5
2 0 -6
3 0 -6
4 0 -6];
AML=[ 1 0 0 0 0];
%****Call function for analysis trs2d 2d--------------------
[dof,index,coord,element]=TRUSS2D_sdata(prop,element,coord,type);
[S,Sm,SmS,Cx,Cy,RT,L,A,Joint,Xj,Xk,Yj,Yk]=...
TRUSS2D_stiff(prop,element,coord,index,nfile);
[IR,IF,Support]=TRUSS2D_ldata(Support,dof);
[DF,AR,AM]=TRUSS2D_analysis_result...
(element,dof,index,IF,IR,S,Sm,JL,AML,RT,Support,type,nfile);
[joint_disp,support_reaction,beam_endforces]=TRUSS2D_print_result...
(JL,AML,dof,Support,element,IF,IR,DF,AR,AM,type,nfile);
%****end data------------------------------------------------
TRUSS2D_xplot
Agung saputro (30201504026) 4
Tugas Analisa Struktur 4
STIFFNESS MATRIX METHOD : PROGRAM TRUSS2D
File Name : Ari p_2.txt
Joint Fx Fy
7 0.000 -5.000
6 0.000 -6.000
5 3.500 -5.000
4 0.000 -6.000
3 3.500 -5.000
2 0.000 -6.000
Member Fx1 Fy1 Fx2 Fy2
1 0.000 0.000 0.000 0.000
Support Restraints :
Joint Rx Ry
1 1 1
8 0 1
Joint Displacements :
Joint Dx Dy
1 0.0000e+00 0.0000e+00
2 1.8254e-03 -3.2925e-02
3 9.5070e-03 -3.2746e-02
4 3.6509e-03 -3.6561e-02
5 4.9606e-03 -3.5491e-02
6 5.3826e-03 -3.3767e-02
7 -1.1762e-03 -3.3588e-02
8 7.1143e-03 0.0000e+00
Support Reactions :
Joint Rx Ry
1 -7.0000e+00 1.5042e+01
8 0.0000e+00 1.7958e+01
Member End Forces (axial/normal):
Member axial force : + tension - compression
1 34.075
2 -30.973
3 6.000
4 34.075
5 -13.327
6 -21.649
7 32.325
8 -11.325
9 17.972
10 -25.653
Agung saputro (30201504026) 5
Tugas Analisa Struktur 4
11 32.325
12 6.000
13 -36.978
1.6 Cek Titik Keseimbangan
Support Reactions :
Joint Rx Ry
1 -7.0000e+00 1.5042e+01
8 0.0000e+00 1.7958e+01
∑Fx = 0
Rx = Px
Px = (F1*2)
= (-3.5*2)
=-7
Rx = - 7
-Px +Rx = 0
- 7+7 = 0 (oke)
∑Fy = 0
Py + Ry = 0
Py = - ((P1*3) + (P2*3))
= -((5*3) + (6*3))
= - ((15) + (18))
= -33
Ry = (15 +18) = 33
Py +Ry = 0
-33 + 33 = 0 (oke)
Agung saputro (30201504026) 6
Tugas Analisa Struktur 4
Agung saputro (30201504026) 7
Tugas Analisa Struktur 4
-20,65 -24,66
17,00
-37,92
-20,65 -24,66
-9,27
5,95 5,95
-13,27
-37,92
-29,92
-9,27
5,95 17,00 5,95
-13,27
-29,92
33,16 33,16
33,14 33,14
36,65 36,65
36,66 36,66
Axial Force Diagram, Comb: CASE1, Units: kN-m
LinPro 2.7 | Enes Siljak | eness@bosnia.ba | www.line.co.ba
Agung saputro (30201504026) 8
Tugas Analisa Struktur 4
SOAL 2
TRUSS 2D
Agung saputro (30201504026) 9
Tugas Analisa Struktur 4
Gambar 2.2 Penomoran Batang dan Join
Persiapan Data
Data Propertis :
No E A
1 2.1 e 7 0.002
Agung saputro (30201504026) 10
Tugas Analisa Struktur 4
Data Koordinat :
No x y
1 0 0
2 2 0
3 2 2
4 0 2
5 1 1
Data Batang :
No J1 J2 Prop
1 1 2 1
2 2 3 1
3 3 4 1
4 1 4 1
5 1 5 1
6 2 5 1
7 3 5 1
8 4 5 1
Data Tumpuan :
No JR1 JR2 JR3
1 1 1 0
2 1 1 0
No Fx Fy
3 0 -10
4 2 -10
Agung saputro (30201504026) 11
Tugas Analisa Struktur 4
Input Data
%****RANGKA BATANG 2D---TEKNIK SIPIL UNISSULA SEMARANG----
%****nama file : agung 2----oleh :agung saputro 30201504026----
%****struktur : rangka batang fleming hal 80 -- kip.in
clear all
clc
%****General Data---
type='t2d'
nfile= agung_TrussNo2'
%****Property---i E A -------
prop = [ 1 2.1e7 0.002];
%****Coordinates----------------
%****--- Joint X Y -------
coord=[ 1 0 0
2 2 0
3 2 2
4 0 2
5 1 1];
%****Element data---------------
%****--- Element J1 J2 prop---
element=[1 1 2 1
2 2 3 1
3 3 4 1
4 1 4 1
5 1 5 1
6 2 5 1
7 3 5 1
8 4 5 1];
%****Nodal Restraint-----------
%****---- Joint JR1 JR2 ---
Support=[ 1 1 1
2 1 1 ];
%****---Joint FX FY -------
JL=[ 3 0 -10
4 2 -10];
%****---Elem. Fx1 Fy1 Fx2 Fy2 ---
AML=[ 1 0 0 0 0 ];
%****Call function for analysis t2d-----------------------
[dof,index,coord,element]=TRUSS2D_sdata(prop,element,coord,type);
[S,Sm,SmS,Cx,Cy,RT,L,A,Joint,Xj,Xk,Yj,Yk]=...
TRUSS2D_stiff(prop,element,coord,index,nfile);
[IR,IF,Support]=TRUSS2D_ldata(Support,dof);
[DF,AR,AM]=TRUSS2D_analysis_result...
(element,dof,index,IF,IR,S,Sm,JL,AML,RT,Support,type,nfile);
[joint_disp,support_reaction,beam_endforces]=TRUSS2D_print_result...
(JL,AML,dof,Support,element,IF,IR,DF,AR,AM,type,nfile);
%****end data------------------------------------------------
TRUSS2D_xplot
%
Agung saputro (30201504026) 12
Tugas Analisa Struktur 4
Output Data
STIFFNESS MATRIX METHOD : PROGRAM TRUSS2D
File Name : agung_TrussNo2_2.txt
Joint Fx Fy
3 0.000 -10.000
4 2.000 -10.000
Member Fx1 Fy1 Fx2 Fy2
1 0.000 0.000 0.000 0.000
Support Restraints :
Joint Rx Ry
1 1 1
2 1 1
Joint Displacements :
Joint Dx Dy
1 0.0000e+00 0.0000e+00
2 0.0000e+00 0.0000e+00
3 2.1625e-04 -4.0829e-04
4 1.4836e-04 -3.1306e-04
5 6.7344e-05 -1.6336e-04
Support Reactions :
Joint Rx Ry
1 1.4258e+00 8.0000e+00
2 -3.4258e+00 1.2000e+01
Member End Forces (axial/normal):
Member axial force : + tension - compression
1 0.000
2 -8.574
3 1.426
4 -6.574
5 -2.016
6 -4.845
7 -2.016
8 -4.845
<<< end of file >>>
Agung saputro (30201504026) 13
Tugas Analisa Struktur 4
Cek Titik Keseimbangan
Support Reactions :
Joint Rx Ry
1 1.4258e+00 8.0000e+00
2 -3.4258e+00 1.2000e+01
∑Fx =0
Px + Rx = 0
Px =F
=2
Rx = (Rx1 + Rx2)
= (1,4258 + (-3,4258))
= -2
Px + Rx = 0
2 + -2 = 0
= 0 (oke)
∑Fy =0
Py + Ry = 0
Py = - ( P*2 )
= - (10*2)
= - 20
Ry = ( Ry1 + Ry2 )
= ( 8 + 12 )
= 20
Py +Ry = 0
- 20+ 20 = 0 (oke)
Agung saputro (30201504026) 14
Tugas Analisa Struktur 4
Gambar Deformasi
Gambar 2.3 Deformasi
Freebody Diagram
Member End Forces (axial/normal):
Member axial force : + tension - compression
1 0.000
2 -8.574
3 1.426
4 -6.574
5 -2.016
Agung saputro (30201504026) 15
Tugas Analisa Struktur 4
6 -4.845
7 -2.016
8 -4.845
Reaksi
Agung saputro (30201504026) 16
Tugas Analisa Struktur 4 | 3,598 | 7,099 | {"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} | 2.984375 | 3 | CC-MAIN-2020-16 | latest | en | 0.12324 |
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