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Global Constraint Catalog: Csymmetric_cardinality
<< 5.397. symmetric_alldifferent_loop5.399. symmetric_gcc >>
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}
by W. Kocjan.
\mathrm{๐๐ข๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}\left(\mathrm{๐
๐ฐ๐๐},\mathrm{๐
๐ฐ๐ป๐}\right)
\mathrm{๐
๐ฐ๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}-\mathrm{๐๐๐},\mathrm{๐๐๐}-\mathrm{๐๐๐๐},๐-\mathrm{๐๐๐},๐-\mathrm{๐๐๐}\right)
\mathrm{๐
๐ฐ๐ป๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}-\mathrm{๐๐๐},\mathrm{๐๐๐}-\mathrm{๐๐๐๐},๐-\mathrm{๐๐๐},๐-\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐},\left[\mathrm{๐๐๐๐๐},\mathrm{๐๐๐},๐,๐\right]\right)
|\mathrm{๐
๐ฐ๐๐}|\ge 1
\mathrm{๐
๐ฐ๐๐}.\mathrm{๐๐๐๐๐}\ge 1
\mathrm{๐
๐ฐ๐๐}.\mathrm{๐๐๐๐๐}\le |\mathrm{๐
๐ฐ๐๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐},\mathrm{๐๐๐๐๐}\right)
\mathrm{๐
๐ฐ๐๐}.๐\ge 0
\mathrm{๐
๐ฐ๐๐}.๐\le \mathrm{๐
๐ฐ๐๐}.๐
\mathrm{๐
๐ฐ๐๐}.๐\le |\mathrm{๐
๐ฐ๐ป๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐ป๐},\left[\mathrm{๐๐๐๐๐},\mathrm{๐๐๐},๐,๐\right]\right)
|\mathrm{๐
๐ฐ๐ป๐}|\ge 1
\mathrm{๐
๐ฐ๐ป๐}.\mathrm{๐๐๐๐๐}\ge 1
\mathrm{๐
๐ฐ๐ป๐}.\mathrm{๐๐๐๐๐}\le |\mathrm{๐
๐ฐ๐ป๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐ป๐},\mathrm{๐๐๐๐๐}\right)
\mathrm{๐
๐ฐ๐ป๐}.๐\ge 0
\mathrm{๐
๐ฐ๐ป๐}.๐\le \mathrm{๐
๐ฐ๐ป๐}.๐
\mathrm{๐
๐ฐ๐ป๐}.๐\le |\mathrm{๐
๐ฐ๐๐}|
Put in relation two sets: for each element of one set gives the corresponding elements of the other set to which it is associated. In addition, it constraints the number of elements associated with each element to be in a given interval.
\left(\begin{array}{c}โฉ\begin{array}{cccc}\mathrm{๐๐๐๐๐}-1\hfill & \mathrm{๐๐๐}-\left\{3\right\}\hfill & ๐-0\hfill & ๐-1,\hfill \\ \mathrm{๐๐๐๐๐}-2\hfill & \mathrm{๐๐๐}-\left\{1\right\}\hfill & ๐-1\hfill & ๐-2,\hfill \\ \mathrm{๐๐๐๐๐}-3\hfill & \mathrm{๐๐๐}-\left\{1,2\right\}\hfill & ๐-1\hfill & ๐-2,\hfill \\ \mathrm{๐๐๐๐๐}-4\hfill & \mathrm{๐๐๐}-\left\{1,3\right\}\hfill & ๐-2\hfill & ๐-3\hfill \end{array}โช,\hfill \\ โฉ\begin{array}{cccc}\mathrm{๐๐๐๐๐}-1\hfill & \mathrm{๐๐๐}-\left\{2,3,4\right\}\hfill & ๐-3\hfill & ๐-4,\hfill \\ \mathrm{๐๐๐๐๐}-2\hfill & \mathrm{๐๐๐}-\left\{3\right\}\hfill & ๐-1\hfill & ๐-1,\hfill \\ \mathrm{๐๐๐๐๐}-3\hfill & \mathrm{๐๐๐}-\left\{1,4\right\}\hfill & ๐-1\hfill & ๐-2,\hfill \\ \mathrm{๐๐๐๐๐}-4\hfill & \mathrm{๐๐๐}-\varnothing \hfill & ๐-0\hfill & ๐-1\hfill \end{array}โช\hfill \end{array}\right)
\mathrm{๐๐ข๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}
3\in \mathrm{๐
๐ฐ๐๐}\left[1\right].\mathrm{๐๐๐}โ1\in \mathrm{๐
๐ฐ๐ป๐}\left[3\right].\mathrm{๐๐๐}
1\in \mathrm{๐
๐ฐ๐๐}\left[2\right].\mathrm{๐๐๐}โ2\in \mathrm{๐
๐ฐ๐ป๐}\left[1\right].\mathrm{๐๐๐}
1\in \mathrm{๐
๐ฐ๐๐}\left[3\right].\mathrm{๐๐๐}โ3\in \mathrm{๐
๐ฐ๐ป๐}\left[1\right].\mathrm{๐๐๐}
2\in \mathrm{๐
๐ฐ๐๐}\left[3\right].\mathrm{๐๐๐}โ3\in \mathrm{๐
๐ฐ๐ป๐}\left[2\right].\mathrm{๐๐๐}
1\in \mathrm{๐
๐ฐ๐๐}\left[4\right].\mathrm{๐๐๐}โ4\in \mathrm{๐
๐ฐ๐ป๐}\left[1\right].\mathrm{๐๐๐}
3\in \mathrm{๐
๐ฐ๐๐}\left[4\right].\mathrm{๐๐๐}โ4\in \mathrm{๐
๐ฐ๐ป๐}\left[3\right].\mathrm{๐๐๐}
The number of elements of
\mathrm{๐
๐ฐ๐๐}\left[1\right].\mathrm{๐๐๐}=\left\{3\right\}
belongs to interval
\left[0,1\right]
\mathrm{๐
๐ฐ๐๐}\left[2\right].\mathrm{๐๐๐}=\left\{1\right\}
\left[1,2\right]
\mathrm{๐
๐ฐ๐๐}\left[3\right].\mathrm{๐๐๐}=\left\{1,2\right\}
\left[1,2\right]
\mathrm{๐
๐ฐ๐๐}\left[4\right].\mathrm{๐๐๐}=\left\{1,3\right\}
\left[2,3\right]
\mathrm{๐
๐ฐ๐ป๐}\left[1\right].\mathrm{๐๐๐}=\left\{2,3,4\right\}
\left[3,4\right]
\mathrm{๐
๐ฐ๐ป๐}\left[2\right].\mathrm{๐๐๐}=\left\{3\right\}
\left[1,1\right]
\mathrm{๐
๐ฐ๐ป๐}\left[3\right].\mathrm{๐๐๐}=\left\{1,4\right\}
\left[1,2\right]
\mathrm{๐
๐ฐ๐ป๐}\left[4\right].\mathrm{๐๐๐}=\varnothing
\left[0,1\right]
|\mathrm{๐
๐ฐ๐๐}|>1
|\mathrm{๐
๐ฐ๐ป๐}|>1
\mathrm{๐
๐ฐ๐๐}
\mathrm{๐
๐ฐ๐ป๐}
The most simple example of applying
\mathrm{๐๐ข๐๐๐๐๐๐๐}_\mathrm{๐๐๐}
is a variant of personnel assignment problem, where one person can be assigned to perform between
m
\left(n\le m\right)
jobs, and every job requires between
p
q
\left(p\le q\right)
persons. In addition every job requires different kind of skills. The previous problem can be modelled as follows:
For each person we create an item of the
\mathrm{๐
๐ฐ๐๐}
For each job we create an item of the
\mathrm{๐
๐ฐ๐ป๐}
There is an arc between a person and the particular job if this person is qualified to perform it.
\mathrm{๐๐ข๐๐๐๐๐๐๐}_\mathrm{๐๐๐}
constraint generalises the
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}
constraint by allowing a variable to take more than one value.
A first flow-based arc-consistency algorithm for the
\mathrm{๐๐ข๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}
constraint is described in [KocjanKreuger04]. A second arc-consistency filtering algorithm exploiting matching theory [DulmageMendelsohn58] is described in [Cymer12], [CymerPhD13].
\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐ข๐๐๐๐๐๐๐}_\mathrm{๐๐๐}
\mathrm{๐๐๐ก๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}
\mathrm{๐๐}_\mathrm{๐๐๐}
combinatorial object: relation.
constraint type: decomposition, timetabling constraint.
filtering: flow, bipartite matching.
\mathrm{๐
๐ฐ๐๐}
\mathrm{๐
๐ฐ๐ป๐}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐},\mathrm{๐๐๐๐}\right)
โข
\mathrm{๐๐}_\mathrm{๐๐๐}
\left(\mathrm{๐๐๐๐}.\mathrm{๐๐๐๐๐},\mathrm{๐๐๐๐}.\mathrm{๐๐๐}\right)โ
\mathrm{๐๐}_\mathrm{๐๐๐}
\left(\mathrm{๐๐๐๐}.\mathrm{๐๐๐๐๐},\mathrm{๐๐๐๐}.\mathrm{๐๐๐}\right)
โข\mathrm{๐๐๐๐}.๐\le \mathrm{๐๐๐๐}_\mathrm{๐๐๐}\left(\mathrm{๐๐๐๐}.\mathrm{๐๐๐}\right)
โข\mathrm{๐๐๐๐}.๐\ge \mathrm{๐๐๐๐}_\mathrm{๐๐๐}\left(\mathrm{๐๐๐๐}.\mathrm{๐๐๐}\right)
โข\mathrm{๐๐๐๐}.๐\le \mathrm{๐๐๐๐}_\mathrm{๐๐๐}\left(\mathrm{๐๐๐๐}.\mathrm{๐๐๐}\right)
โข\mathrm{๐๐๐๐}.๐\ge \mathrm{๐๐๐๐}_\mathrm{๐๐๐}\left(\mathrm{๐๐๐๐}.\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐}
=|\mathrm{๐
๐ฐ๐๐}|*|\mathrm{๐
๐ฐ๐ป๐}|
The graph model used for the
\mathrm{๐๐ข๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}
is similar to the one used in the
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐}
or in the
\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
constraints: we use an equivalence in the arc constraint and ask all arc constraints to hold.
\mathrm{๐๐๐๐}
graph property, all the arcs of the final graph are stressed in bold.
\mathrm{๐๐ข๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
\mathrm{๐
๐ฐ๐๐}
\mathrm{๐
๐ฐ๐ป๐}
, the number of arcs of the initial graph is equal to
|\mathrm{๐
๐ฐ๐๐}|ยท|\mathrm{๐
๐ฐ๐ป๐}|
. Therefore the maximum number of arcs of the final graph is also equal to
|\mathrm{๐
๐ฐ๐๐}|ยท|\mathrm{๐
๐ฐ๐ป๐}|
and we can rewrite
\mathrm{๐๐๐๐}=|\mathrm{๐
๐ฐ๐๐}|ยท|\mathrm{๐
๐ฐ๐ป๐}|
\mathrm{๐๐๐๐}\ge |\mathrm{๐
๐ฐ๐๐}|ยท|\mathrm{๐
๐ฐ๐ป๐}|
. So we can simplify
\underline{\overline{\mathrm{๐๐๐๐}}}
\overline{\mathrm{๐๐๐๐}}
|
Limiting Reagents | Brilliant Math & Science Wiki
Aditya Virani, Satyabrata Dash, and Jimin Khim contributed
A limiting reagent is a reactant that is totally consumed in a chemical reaction. The limiting reagent, as its name implies, limits the amount of product produced during the reaction. For example, let's take a look at the following reaction in which hydrogen and oxygen react to form water:
2\text{H}_2(g)+\text{O}_2(g)\rightarrow2\text{H}_2\text{O}(l).
When this reaction occurs, exactly two hydrogen molecules and an oxygen molecule react to produce two water molecules. Suppose there are 10 moles of hydrogen and 7 moles of oxygen. Since the molar ratio of hydrogen and oxygen molecules is not 2:1, one must be left over after the reaction has fully occurred, which would be oxygen in this case. Thus, all 10 moles of hydrogen will react with 5 moles of oxygen to produce 10 moles of water, leaving 2 moles of oxygen, as shown in the equation below:
\begin{aligned} &2\text{H}_2(g)&&+&&\text{O}_2(g)&&\rightarrow&&2\text{H}_2\text{O}(l)\\ &\ \ \ \ 10&& &&\ \ \ \ 7&& &&\ \ \ \ \ 0\\ &\ -10&& &&\ -5&& &&+10\\ \hline &\ \ \ \ \ 0&& &&\ \ \ \ 2&& &&\ \ \ \ 10. \end{aligned}
In this example, hydrogen is the limiting reagent and oxygen is the excess reagent. The amount of product formed is limited by the amount of hydrogen. In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents.
The following is the chemical equation of the combustion of methane. 64 grams of methane reacts with 96 grams of oxygen. Identify the limiting reagent, and calculate the amount (in grams) of carbon dioxide produced from this reaction. Use the following atomic weights:
\text{H}=1, \text{C}=12, \text{O}=16.
\text{CH}_4+2\text{O}_2\rightarrow\text{CO}_2+2\text{H}_2\text{O}.
First, we find the number of moles of reactants we have. Since one mole of methane is
12+4\times1=16
grams and one mole of oxygen is
2\times16=32
grams, we have
64\div16=4
moles of methane and
96\div32=3
moles of oxygen. Since one methane molecule is to react with two oxygen molecules, oxygen is the limiting reagent and methane is the excess reagent. 1.5 moles of methane will react with 3 moles of oxygen to produce 1.5 moles of carbon dioxide and 3 moles of water, as shown in the equation below:
\begin{aligned} &\text{CH}_4&&+&&2\text{O}_2&&\rightarrow&&\text{CO}_2&&+&&2\text{H}_2\text{O}\\ &\ \ 4&& &&\ \ 3&& &&\ \ 0&& &&\ \ \ 0\\ &-1.5&& &&-3&& &&+1.5&& &&+3\\ \hline &\ 2.5&& &&\ \ 0&& &&\ 1.5&& &&\ \ \ 3. \end{aligned}
Therefore, the limiting reagent is oxygen, and the amount of produced carbon dioxide is
1.5\times(12+2\times16)=66
_\square
Cite as: Limiting Reagents. Brilliant.org. Retrieved from https://brilliant.org/wiki/limiting-reagents/
|
Anscombe's quartet - Wikipedia
Four data sets with the same descriptive statistics, yet very different distributions
Anscombe's quartet comprises four data sets that have nearly identical simple descriptive statistics, yet have very different distributions and appear very different when graphed. Each dataset consists of eleven (x,y) points. They were constructed in 1973 by the statistician Francis Anscombe to demonstrate both the importance of graphing data when analyzing it, and the effect of outliers and other influential observations on statistical properties. He described the article as being intended to counter the impression among statisticians that "numerical calculations are exact, but graphs are rough."[1]
Sample variance of x : s2
x 11 exact
Sample variance of y : s2
y 4.125 ยฑ0.003
{\displaystyle R^{2}}
0.67 to 2 decimal places
The first scatter plot (top left) appears to be a simple linear relationship, corresponding to two variables correlated where y could be modelled as gaussian with mean linearly dependent on x.
The second graph (top right); while a relationship between the two variables is obvious, it is not linear, and the Pearson correlation coefficient is not relevant. A more general regression and the corresponding coefficient of determination would be more appropriate.
In the third graph (bottom left), the modelled relationship is linear, but should have a different regression line (a robust regression would have been called for). The calculated regression is offset by the one outlier which exerts enough influence to lower the correlation coefficient from 1 to 0.816.
Finally, the fourth graph (bottom right) shows an example when one high-leverage point is enough to produce a high correlation coefficient, even though the other data points do not indicate any relationship between the variables.
It is not known how Anscombe created his datasets.[7] Since its publication, several methods to generate similar data sets with identical statistics and dissimilar graphics have been developed.[7][8] One of these, the Datasaurus Dozen, consists of points tracing out the outline of a dinosaur, plus twelve other data sets that have the same summary statistics.[9][10][11] Datasaurus Dozen was created by Justin Matejka and George Fitzmaurice. The process is described in their paper โSame stats, different graphs: generating datasets with varied appearance and identical statistics through simulated annealingโ.
The Datasaurus Dozen proves us as much as Anscombe Quartet why visualizing our data is important as summary statistics can be the same, while data distributions can be very different.
^ a b Anscombe, F. J. (1973). "Graphs in Statistical Analysis". American Statistician. 27 (1): 17โ21. doi:10.1080/00031305.1973.10478966. JSTOR 2682899.
^ Elert, Glenn (2021). "Linear Regression". The Physics Hypertextbook.
^ Janert, Philipp K. (2010). Data Analysis with Open Source Tools. O'Reilly Media. pp. 65โ66. ISBN 978-0-596-80235-6.
^ Chatterjee, Samprit; Hadi, Ali S. (2006). Regression Analysis by Example. John Wiley and Sons. p. 91. ISBN 0-471-74696-7.
^ Saville, David J.; Wood, Graham R. (1991). Statistical Methods: The geometric approach. Springer. p. 418. ISBN 0-387-97517-9.
^ Tufte, Edward R. (2001). The Visual Display of Quantitative Information (2nd ed.). Cheshire, CT: Graphics Press. ISBN 0-9613921-4-2.
^ a b Chatterjee, Sangit; Firat, Aykut (2007). "Generating Data with Identical Statistics but Dissimilar Graphics: A follow up to the Anscombe dataset". The American Statistician. 61 (3): 248โ254. doi:10.1198/000313007X220057. JSTOR 27643902. S2CID 121163371.
^ Matejka, Justin; Fitzmaurice, George (2017). "Same Stats, Different Graphs: Generating Datasets with Varied Appearance and Identical Statistics through Simulated Annealing". Proceedings of the 2017 CHI Conference on Human Factors in Computing Systems: 1290โ1294. doi:10.1145/3025453.3025912. S2CID 9247543.
^ Murray, Lori L.; Wilson, John G. (April 2021). "Generating data sets for teaching the importance of regression analysis". Decision Sciences Journal of Innovative Education. 19 (2): 157โ166. doi:10.1111/dsji.12233. ISSN 1540-4595. S2CID 233609149.
^ Andrienko, Natalia; Andrienko, Gennady; Fuchs, Georg; Slingsby, Aidan; Turkay, Cagatay; Wrobel, Stefan (2020), "Visual Analytics for Investigating and Processing Data", Visual Analytics for Data Scientists, Cham: Springer International Publishing, pp. 151โ180, doi:10.1007/978-3-030-56146-8_5, ISBN 978-3-030-56145-1, S2CID 226648414, retrieved 2021-04-20
^ Matejka, Justin; Fitzmaurice, George (2017). "Same Stats, Different Graphs: Generating Datasets with Varied Appearance and Identical Statistics through Simulated Annealing". Autodesk Research. Archived from the original on 2020-10-04. Retrieved 2021-04-20.
Animated examples from Autodesk called the "Datasaurus Dozen".
Documentation for the datasets in R.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Anscombe%27s_quartet&oldid=1080690040"
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Compute H-infinity optimal controller - MATLAB hinfsyn - MathWorks รญโขลรชยตยญ
\left[\begin{array}{c}z\\ y\end{array}\right]=\left[\begin{array}{cc}{P}_{11}& {P}_{12}\\ {P}_{21}& {P}_{22}\end{array}\right]\left[\begin{array}{c}w\\ u\end{array}\right],
{H}_{\infty }
{H}_{\infty }
\mathrm{รยณ}
\mathrm{รยณ}รขยย1.5
G\left(s\right)=\frac{s-1}{s+1},{W}_{1}=\frac{0.1\left(s+100\right)}{100s+1},{W}_{2}=0.1,no\phantom{\rule{0.2777777777777778em}{0ex}}{W}_{3}.
{W}_{1}
\begin{array}{c}dx=Ax+{B}_{1}w+{B}_{2}u\\ z={C}_{1}x+{D}_{11}w+{D}_{12}u\\ y={C}_{2}x+{D}_{21}w+{D}_{22}u.\end{array}
\left[\begin{array}{cc}Aรขยยj\mathrm{รย}& {B}_{2}\\ {C}_{1}& {D}_{12}\end{array}\right]
has full column rank for all frequencies รโฐ. By default, hinfsyn automatically adds extra disturbances and errors to the plant to ensure that the restriction on P12 and P21 is met. This process is called regularization. If you are certain your plant meets the conditions, you can turn off regularization using the Regularize option of hinfsynOptions.
Target performance level, specified as a positive scalar. hinfsyn attempts to compute a controller such that the Hโ of the closed-loop system does not exceed gamTry. If this performance level is achievable, then the returned controller has gamma รขโฐยค gamTry. If gamTry is not achievable, hinfsyn returns an empty controller.
In general, the solution to the infinity-norm optimal control problem is nonunique. The controller returned by hinfsyn is only one particular solution, K. For the default Riccati-based method, info.AS contains the all-solution controller parameterization KAS. All solutions with closed-loop performance of รยณ or less are parameterized by a free stable contraction map Q, which is constrained by
{รขยยQรขยย}_{\infty }<\mathrm{รยณ}
{รขยย{T}_{{y}_{1}{u}_{1}}รขยย}_{\infty }รขยย\underset{\mathrm{รย}}{\mathrm{sup}}{\mathrm{รย}}_{\mathrm{max}}\left({T}_{{y}_{1}{u}_{1}}\left(j\mathrm{รย}\right)\right)<\mathrm{รยณ}.
{T}_{{y}_{1}{u}_{1}}
\begin{array}{c}d{x}_{e}=A{x}_{e}+{B}_{1}{w}_{e}+{B}_{2}u+{L}_{x}e\\ u={K}_{u}{x}_{e}+{L}_{u}e\\ {w}_{e}={K}_{w}{x}_{e}.\end{array}
e=yรขยย{C}_{2}{x}_{e}รขยย{D}_{21}{w}_{e}รขยย{D}_{22}u.
\text{Entropy = }\frac{{\mathrm{รยณ}}^{2}}{2\mathrm{รย}}{รขยยซ}_{รขยย\infty }^{\infty }\mathrm{ln}|\mathrm{det}Iรขยย{\mathrm{รยณ}}^{รขยย2}{T}_{{y}_{1}{u}_{1}}\left(j\mathrm{รย}{\right)}^{รขยยฒ}{T}_{{y}_{1}{u}_{1}}\left(j\mathrm{รย}\right)|\left[\frac{{s}_{o}{}^{2}}{{s}_{0}{}^{2}+{\mathrm{รย}}^{2}}\right]d\mathrm{รย}
{T}_{{y}_{1}{u}_{1}}
For all methods, the function uses a standard รยณ-iteration technique to determine the optimal value of the performance level รยณ. รยณ-iteration is a bisection algorithm that starts with high and low estimates of รยณ and iterates on รยณ values to approach the optimal Hโ control design.
At each value of รยณ, the algorithm tests a รยณ value to determine whether a solution exists. In the Riccati-based method, the algorithm computes the smallest performance level for which the stabilizing Riccati solutions X = Xโ/รยณ and Y = Yโ/รยณ exist. For any รยณ greater than that performance level and in the range gamRange, the algorithm evaluates the central controller formulas (K formulas) and checks the closed-loop stability of CL = lft(P,K). This step is equivalent to verifying the conditions:
min(eig(X)) รขโฐยฅ 0
min(eig(Y)) รขโฐยฅ 0
A รยณ that meets these conditions passes. The stopping criterion for the bisection algorithm requires the relative difference between the last รยณ value that failed and the last รยณ value that passed be less than 0.01. (You can change this criterion using hinfsynOptions.) hinfsyn returns the controller corresponding to the smallest tested รยณ value that passes. For discrete-time controllers, the algorithm performs additional computations to construct the feedthrough matrix DK.
Use the Display option of hinfsynOptions to make hinfsyn display values showing which of the conditions are satisfied for each รยณ value tested.
\left[\begin{array}{cc}Aรขยยj\mathrm{รย}I& {B}_{2}\\ {C}_{1}& {D}_{12}\end{array}\right]
has full column rank for all รโฐ รขหล R.
\left[\begin{array}{cc}Aรขยยj\mathrm{รย}I& {B}_{1}\\ {C}_{2}& {D}_{21}\end{array}\right]
has full row rank for all รโฐ รขหล R.
When these rank conditions do not hold, the controller may have undesirable properties. If D12 and D21 are not full rank, then the Hโ controller K might have large high-frequency gain. If either of the latter two rank conditions does not hold at some frequency รโฐ, the controller might have very lightly damped poles near that frequency.
{รขยยQรขยย}_{\infty }<1
{u}_{2}\left(t\right)={K}_{FI}\left[\begin{array}{c}x\left(t\right)\\ {u}_{1}\left(t\right)\end{array}\right]
{รขยยQรขยย}_{\infty }<1
{รขยยQรขยย}_{\infty }<\mathrm{รยณ}
, where รยณ is info.gamma. This new constraint ensures that the all-solutions controller KAS has a finite limit as gamTry รขโ โ โ.
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Reflected Gray To Ordinary - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Combinatorics : Iterator : Mixed Radix : Reflected Gray To Ordinary
ReflectedGrayToOrdinary(a,m)
ReflectedGrayToOrdinary converts a mixed-radix reflected Gray code tuple to the ordinary mixed-radix tuple of the same rank.
The a parameter is the reflected Gray code tuple. It is a list or one-dimensional rtable of nonnegative integers. The first element is the low-order element.
\mathrm{with}โก\left(\mathrm{Iterator}:-\mathrm{MixedRadix}\right):
Compare, by rank, the mixed-radix Gray codes with the ordinary mixed-radix tuples.
\mathrm{radices}โ[4,3,2]:
Mโ\mathrm{Iterator}:-\mathrm{MixedRadixTuples}โก\left(\mathrm{radices}\right):
Gโ\mathrm{Iterator}:-\mathrm{MixedRadixGrayCode}โก\left(\mathrm{radices}\right):
\mathbf{for}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}g\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}G\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}aโ\mathrm{ReflectedGrayToOrdinary}โก\left(g,\mathrm{radices}\right);\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}\mathrm{printf}โก\left("%2d : %d : %d : %2d\n",\mathrm{Rank}โก\left(G\right),g,a,\mathrm{Rank}โก\left(M,a\right)\right)\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathbf{end}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}:
1 : 0 0 0 : 0 0 0 : 1
10 : 1 2 0 : 1 2 0 : 10
Knuth, Donald Ervin. The Art of Computer Programming, volume 4, fascicle 2; generating all tuples and permutations, sec. 7.2.1.1, generating all n-tuples, p. 19, eq. 51.
The Iterator[MixedRadix][ReflectedGrayToOrdinary] command was introduced in Maple 2016.
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Global Constraint Catalog: Calldifferent_except_0
<< 5.15. alldifferent_cst5.17. alldifferent_interval >>
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\right)
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐},\mathrm{๐๐๐}\right)
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
to take distinct values, except those variables that are assigned value 0.
\left(โฉ5,0,1,9,0,3โช\right)
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
constraint holds since all the values (that are different from 0) 5, 1, 9 and 3 are distinct.
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
{V}_{1}\in \left[0,4\right]
{V}_{2}\in \left[1,2\right]
{V}_{3}\in \left[1,2\right]
{V}_{4}\in \left[0,1\right]
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
\left(โฉ{V}_{1},{V}_{2},{V}_{3},{V}_{4}โช\right)
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|>2
\mathrm{๐๐๐๐๐๐๐}
\left(2,\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐},0\right)
\mathrm{๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}.\mathrm{๐๐๐}\right)>1
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}.\mathrm{๐๐๐}
that are both different from 0 can be swapped; a value of
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
Quite often it appears that, for some modelling reason, you create a joker value. You do not want that normal constraints hold for variables that take this joker value. For this purpose we modify the binary arc constraint in order to discard the vertices for which the corresponding variables are assigned value 0. This will be effectively the case since all the corresponding arcs constraints will not hold.
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
constraint is described in [Cymer12]. The algorithm is based on the following ideas:
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
constraint to var-perfect matchingsA var-perfect matching is a maximum matching covering all vertices representing variables. in a bipartite graph derived from the domain of the variables of the constraint in the following way: to each variable of the
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
constraint corresponds a variable and a joker vertices, while to each potential value corresponds a value vertex; there is an edge between a variable vertex and a value vertex if and only if that value belongs to the domain of the corresponding variable; there is an edge between a variable vertex and its corresponding value vertex.
Second, Dulmage-Mendelsohn decomposition [DulmageMendelsohn58] is used to characterise all edges that do not belong to any var-perfect matching, and therefore prune the corresponding variables.
n
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
0..n
\mathrm{๐ ๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐ข}
characteristic of a constraint: joker value, all different, sort based reformulation, automaton, automaton with array of counters.
constraint type: value constraint, relaxation.
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐ถ๐ฟ๐ผ๐๐๐ธ}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}\right)
โข\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}\ne 0
โข\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}=\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}
\mathrm{๐๐๐}_\mathrm{๐๐๐๐}
\le 1
The graph model is the same as the one used for the
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}
constraint, except that we discard all variables that are assigned value 0.
\mathrm{๐๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
holds since all the strongly connected components have at most one vertex: a value different from 0 is used at most once.
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
{\mathrm{๐
๐ฐ๐}}_{i}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
{๐}_{i}
{\mathrm{๐
๐ฐ๐}}_{i}
{๐}_{i}
{\mathrm{๐
๐ฐ๐}}_{i}\ne 0โ{๐}_{i}
. The automaton counts the number of occurrences of each value different from 0 and finally imposes that each non-zero value is taken at most one time.
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
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Conservation of energy - Wikiversity
Conservation of energy (the First Law of Thermodynamics) is a simple law stating that although energy may change form, it cannot disappear altogether, nor can it be created.
More recently, this law has been modified to take into account the equivalence of mass and energy (
{\displaystyle E=mc^{2}}
) discovered by Einstein. This new law is called the law of conservation of mass and energy. It states that although mass may convert to energy, and vice versa, neither may disappear without compensation in the other quantity. However, the original form of the law is adequate in most everyday situations.
A concrete example of energy conservation is found with falling objects. Near the surface of the Earth, the gravitational potential energy of an object is given by
{\displaystyle mgh}
, where m is the object's mass, g is the acceleration due to gravity at that place, and h is the object's height from where the object started to where it is now (measure with downward values negative). When an object falls, its potential energy is converted into kinetic energy (given by
{\displaystyle (1/2)mv^{2}}
). In math:
{\displaystyle -mgh={\frac {1}{2}}mv^{2}}
where m is mass, v is velocity. There is a negative sign in front of the potential energy because when the object is gaining kinetic energy, it will be losing potential energy.
Another, more useful way of interpreting conservation of energy is by using the following formula
{\displaystyle K_{1}+U_{1}+Wother=K_{2}+U_{2}}
Learn more about Conservation of energy
Retrieved from "https://en.wikiversity.org/w/index.php?title=Conservation_of_energy&oldid=2013154"
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Error propagation โ Web Education in Chemistry
In analytical chemistry, it is important to work as accurately and precisely as possible. Therefore, almost all analytical, volumetric glassware shows the error that is made when using the glassware, such that you can calculate the size of the error in the experiment. An example is given in the picture below, which shows a close-up of a 100 mL volumetric flask. The error that you make when using this flask is ยฑ0.1 mL.
Question: is this a random or systematic error?
In the remainder of this section, we will learn what this actually means and how it influences a final experimental result.
More on volumetric glassware
The error displayed on volumetric glassware is the random error resulting from the production process. In the case of the volumetric flask above, this would mean that a collection of identical flasks together has an error of ยฑ0.1 mL (in other words: the standard deviation is 0.1 mL). However, individual flasks from the collection may have an error of +0.05 mL or -0.07 mL (Question: are these systematic or random errors?). For accurate results, you should constantly use different glassware such that errors cancel out. A second option is to calibrate the glassware: determine the volume by weighing. The error after calibration should be much smaller than the error shown on the glassware. Moreover, this error has now become random instead of systematic! Since this requires a lot of work each time you want to use volumetric glassware, we will from now on assume that errors shown on volumetric glassware are random errors. For example, each time when using the depicted volumetric flask properly, the volume will be 100 mL with an error of ยฑ0.1 mL.
As a general rule, the last reported figure of a result is the first with uncertainty. Assume that we have measured the weight of an object: 80 kg. To indicate that we are not sure of the last digit,we can write 80 ยฑ 1 kg. If we would have used a better scale to weigh the object, we might have found 80.00 ยฑ 0.01 kg. Question: is the second result more precise or more accurate than the first?
We can also display the error in a relative way. For instance, 80 ยฑ 1 kg is identical to 80 ยฑ 1.25%. The order of magnitude of the result should be as clear as possible. Therefore, the preferred notation of for instance 0.0174 ยฑ 0.0002 is (1.74 ยฑ 0.02)ยท10-2.
When pipetting a volume with a certain pipette, the error in the final volume will be identical to the error shown on the pipette. But what happens to the error of the final volume when pipetting twice with the same pipette? Is it the same error as when using the pipette only once? And is there an error difference between using the same pipette twice or two times a different pipette? This becomes even more difficult when weighing a certain amount of salt and dissolving it in water to a certain volume. The error in weighing is shown on the scale and the error in volume on the volumetric flask, but what is the error in the density of this solution,
\rho = m/V
Error propagation is able to answer all these questions. Basically, it it is a set of mathematical rules that explain how certain errors affect the final answer. Here, we will cover the most important and most used error propagation rules, including some practical examples.
The rule for error propagation with addition and subtraction is as follows. If
z = ax+by
and
being constants and
x
y
z
variables, the absolute error in
z
\Delta x
\Delta y
\Delta z
x
y
z
Example 1: Suppose you pipette subsequently 10.00 ยฑ 0.023 mL and 25.00 ยฑ 0.050 mL in a beaker with uncalibrated pipettes. What is the error in the total volume of 35 mL?
Solution: In this example,
x
= 10.00 mL,
y
z
\Delta x
= 0.023 mL and
\Delta y
= 0.050 mL. The total error can now be calculated via:
Note that in this example, both
and
b
are 1, because we use the two pipettes only once. The final answer is that you have pipetted 35.00 ยฑ 0.055 mL.
Example 2: You pipette three times 10.00 ยฑ 0.023 mL in a beaker with the same, uncalibrated pipette. What is the error in the total volume of 30 mL?
x
\Delta x
a
= 3. The total error can now be calculated:
So, you have pipetted 30.00 ยฑ 0.069 mL.
In the first example, two different pipettes were used. The errors made when pipetting with these pipettes are thus independent (they have nothing to do with each other). In the second example, the same pipette is used three times. Therefore, the errors in this example are dependent. The error is even exactly the same each time you pipette, because you are using the same pipette! That means that pipetting three times with the same pipette will lead to a bigger total error compared to pipetting three times with three different pipettes (indeed, errors cancel out to a certain extent in the latter case). This is exactly the reason that we are not allowed to add the errors in example 2 as we have done in example 1. Moreover, note that the repetition factor
a
is also squared (Example 2).
The rule for error propagation with multiplication and division is: suppose that
z=a\cdot x\cdot y
z = a \cdot x / y
, again with
a
x
y
z
variables. The relative error in
z
\Delta x
\Delta y
\Delta z
x
y
z
Why do we use a relative error here and not an absolute error? The answer lies in the fact that, in the case of multiplication and division,
x
y
often represent different (physical) quantities, whereas with addition and subtraction
x
y
represent the same quantities (otherwise, we could not add them to begin with). You can for instance add two masses or subtract two volumes, but the addition of a mass and a volume is meaningless (e.g. what does '10 g + 3 mL' mean?). Division of mass and volume is not meaningless: it provides the density of a specific sample. The error in density cannot be calculated by simply adding the errors in mass and volume, because they are different quantities. That is why the total error is calculated with relative errors, which are unitless. The absolute error in
z
can be calculated by multiplying the relative error, found with the rule above, with
z
Example 3: You pipette 9.987 ยฑ 0.004 mL of a salt solution in an Erlenmeyer flask and you determine the mass of the solution: 11.2481 ยฑ 0.0001 g. What is the error in the density of the solution that can be calculated from these data?
Solution: In this case,
x
= 11.2481 g,
y
= 9.987 mL,
\Delta x
= 0.0001 g and
\Delta y
= 0.004 mL. The density
z
can be calculated via
z = x/y
\rho = m/V
), thus
z = 11.2481 / 9.987 = 1.12627
gยทmL-1. The relative error equals:
This means that the error in the final answer is 0.04% of the final answer itself. The absolute error can therefore be calculated via multiplication with
z
\Delta z = 4.0\cdot10^{-4} \times 1.12627 = 4.5\cdot10^{-4}
. Using the right amount of significant figures, the final answer is that the density of the salt solution is equal to 1.1263 ยฑ 0.0005 gยทmL-1.
Example 4: Technically, the solution given above is not fully complete. The mass of a sample is always obtained by 'taring' the balance (i.e. setting the mass of the empty flask to 0). The obtained mass is therefore the difference between two masses:
m=m_{1}-m_{0}
. The error on such a balance, as also used during the practicals, is a random error. The total error when weighing can thus be obtained by using the error propagation rule for addition and subtraction. This total error should then be used to calculate the error in the density.
The weighing error is given by:
This does not influence the final result of example 3 (verify this!). Also verify that the full equation for calculting the relative error in the density is given by:
We have seen that a mass is always obtained as a difference between two masses: the error given on the balance can thus not be directly used in error propagation calculations (see Example 4). The same holds for a volume added via a burette: this is also the difference between an initial and a final volume and therefore the error propagation rule for addition and subtraction should be used. This becomes even worse when (correctly!) doing a titration: the blank (a difference between two volumes itself) should be subtracted from the result of an experiment (again a difference in two volumes). This results in a difference between two differences:
(V_{1}-V_{0})-(V_{1, blank} - V_{0, blank})
. Luckily, the total error in the volume can be calculated easily:
In the practical manual, you can find a table that lists the error propagation rules, including those for mathematical operations not covered here, such as logarithms and exponents.
Continue with the questions on this subject.
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f(x)=2^x-3
\lim\limits_ { x \rightarrow \infty } ( 2 ^ { x } - 3 )
\lim\limits_ { x \rightarrow - \infty } ( 2 ^ { x } - 3 )
Sketch your graph without a calculator. Then use a calculator to verify whether or not it is correct, identifying and correcting any errors.
The limits are asking what happens to the curve on the far right and the far left.
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Censored Data and Survival Analysis - Fizzy
Censored Data and Survival Analysis
Censorships in data is a condition in which the value of a measurement or observation is only partially observed. Censored data is one kind of missing data, but is different from the common meaning of missing value in machine learning. We usually observe censored data in a time-based dataset. In such datasets, the event is been cut off beyond a certain time boundary. We can apply survival analysis to overcome the censorship in the data.
Usually, there are two main variables exist, duration and event indicator. In this context, duration indicates the length of the status and event indicator tells whether such event occurred.
For example, in the medical profession, we don't always see patients' death event occur -- the current time, or other events, censor us from seeing those events. But it does not mean they will not happen in the future.
Given this situation, we still want to know even that not all patients have died, how can we use the data we have currently. Or how can we measure the population life expectancy when most of the population is alive.
There are several censored types in the data. The most common one is right-censoring, which only the future data is not observable. Others like left-censoring means the data is not collected from day one of the experiment.
Below is an example that only right-censoring occurs, i.e. everyone starts at time 0.
where the censoring time is at 50. Blue lines stand for the observations are still alive up to the censoring time, but some of them actually died after that. Red lines stand for the observations died before time 50, which means those death events are observed in the dataset.
Survival analysis was first developed by actuaries and medical professionals to predict survival rates based on censored data. Survival analysis can not only focus on medical industy, but many others.
There are several statistical approaches used to investigate the time it takes for an event of interest to occur. For example:
Customer churn: duration is tenure, the event is churn;
Machinery failure: duration is working time, the event is failure;
Visitor conversion: duration is visiting time, the event is purchase.
In R, the may package used is survival. In Python, the most common package to use us called lifelines.
Kaplan-Meier Estimator is a non-parametric statistic used to estimate the survival function from lifetime data. The Kaplan-Meier Estimate defined as:
\hat{S}(t) = \prod_{t_i \lt t} \frac{n_i - d_i}{n_i}
d_i
are the number of death events at time
t
n_i
is the number of subjects at risk of death just prior to time
t
Here is an short example using lifelines package:
# load KaplanMeierFitter
# define time and event columns
(source: lifelines)
This is an full example of using the Kaplan-Meier, results available in Jupyter notebook: survival_analysis/example_dd.ipynb
The Kaplan-Meier Estimator is an univariate model. It is not so helpful when many of the variables can affect the event differently. Further, the Kaplan-Meier Estimator can only incorporate on categorical variables.
To include multiple covariates in the model, we need to use some regression models in survival analysis. There are a few popular models in survival regression: Coxโs model, accelerated failure models, and Aalenโs additive model.
The Cox Proportional Hazards (CoxPH) model is the most common approach of examining the joint effects of multiple features on the survival time. The hazard function of Cox model is defined as:
h_{i}(t)=h_{0}(t) e^{\beta_{1} x_{i 1}+\cdots+\beta_{p} x_{i p}}
h_{0}(t)
is the baseline hazard,
x_{i 1},...,x_{i p}
are feature vectors, and
\beta_{1},...,\beta{p}
are coefficients. The only time component is in the baseline hazard,
h_{0}(t)
. In the above product, the partial hazard is a time-invariant scalar factor that only increases or decreases the baseline hazard. Thus a changes in covariates will only increase or decrease the baseline hazard.
The major assumption of Cox model is that the ratio of the hazard event for any two observations remains constant over time:
\frac{h_{i}(t)}{h_{j}(t)} = \frac{h_{0}(t) e^{\eta_{i}}}{h_{0}(t) e^{\eta_{j}}} = \frac{e^{\eta_{i}}}{e^{\eta_{j}}}
i
j
are any two observations.
The Cox model is a semi-parametric model which mean it can take both numerical and categorical data. But categorical data requires to be preprocessed with one-hot encoding. For more information on how to use One-Hot encoding, check this post: Feature Engineering: Label Encoding & One-Hot Encoding.
Here we use a numerical dataset in the lifelines package:
# import Cox model and load data
rossi_dataset = load_rossi()
# call CoxPHFitter
cph.fit(rossi_dataset, duration_col='week', event_col='arrest', show_progress=True)
# print statistical summary
cph.print_summary() # access the results using cph.summary
We metioned there is an assumption for Cox model. It can be tested by check_assumptions() method in lifelines package:
cph.check_assumptions()
Further, Cox model uses concordance-index as a way to measure the goodness of fit. Concordance-index (between 0 to 1) is a ranking statistic rather than an accuracy score for the prediction of actual results, and is defined as the ratio of the concordant pairs to the total comparable pairs:
0.5 is the expected result from random predictions,
1.0 is perfect concordance and,
0.0 is perfect anti-concordance (multiply predictions with -1 to get 1.0)
This is an full example of using the CoxPH model, results available in Jupyter notebook: survival_analysis/example_CoxPHFitter_with_rossi.ipynb
Survival Analysis in Machine Learning
There are several works about using survival analysis in machine learning and deep learning. Please check the packages for more information.
DeepSurv: adaptively select covariates in Cox model.
TFDeepSurv: TensorFlow version of DeepSurv with some improvements.
LASSO for Cox: use LASSO for covariates selection in Cox model.
More examples about survival analysis and further topics are available at: https://github.com/huangyuzhang/cookbook/tree/master/survival_analysis/
Davidson-Pilon, C., Kalderstam, J., Zivich, P., Kuhn, B., Fiore-Gartland, A., Moneda, L., . . .Rendeiro, A. F. (2019, August).Camdavidsonpilon/lifelines: v0.22.3 (late).Retrieved from https://doi.org/10.5281/zenodo.3364087 doi: 10.5281/zenodo.3364087
Fox, J. (2002). Cox proportional-hazards regression for survival data. An R and S-PLUS companion to applied regression,2002.
Simon, S. (2018).The Proportional Hazard Assumption in Cox Regression. The Anal-ysis Factor.
Steck, H., Krishnapuram, B., Dehing-oberije, C., Lambin, P., & Raykar, V. C. (2008). Onranking in survival analysis: Bounds on the concordance index. InAdvances in neuralinformation processing systems(pp. 1209โ1216).
Ture, M., Tokatli, F., & Kurt, I. (2009). Using kaplanโmeier analysis together with decisiontree methods (c&rt, chaid, quest, c4. 5 and id3) in determining recurrence-free survivalof breast cancer patients.Expert Systems with Applications,36(2), 2017โ2026.
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Mean Value Theorem | Brilliant Math & Science Wiki
Beakal Tiliksew, Yao Liu, Hobart Pao, and
The mean value theorem (MVT), also known as Lagrange's mean value theorem (LMVT), provides a formal framework for a fairly intuitive statement relating change in a function to the behavior of its derivative. The theorem states that the derivative of a continuous and differentiable function must attain the function's average rate of change (in a given interval). For instance, if a car travels 100 miles in 2 hours, then it must have had the exact speed of 50 mph at some point in time.
Suppose that a function
f
[a,b],
(a,b).
Then, there is a number
c
a<c<b
f'(c)=\frac{f(b)-f(a)}{b-a}.
Simple-sounding as it is, the mean value theorem actually lies at the heart of the proof of the fundamental theorem of calculus, and is itself based ultimately on properties of the real numbers. There is a slight generalization known as Cauchy's mean value theorem; for a generalization to higher derivatives, see Taylor's theorem.
The statement seems reasonable upon inspection of an example or two. Below,
f'(c)
is the slope of the tangent lines in the interval
(a,b)
\frac{f(b)-f(a)}{b-a}
is the slope of the secant line joining the two endpoints
\big(a, \, f(a)\big)
\big(b, \, f(b)\big)
Note that the mean value theorem does not restrict
c
to only one value, nor does it tell us where
c
is (other than inside the interval).
f
is not required to be differentiable at the end points. Two examples suffice to illustrate it:
f(x)=\sqrt{1-x^2}
[-1, 1]
f(x)=\arcsin x
[-1, 1]
The mean value theorem is best understood by first studying the restricted case known as Rolle's theorem.
f
[a, b]
(a, \, b)
f(a) = f(b)
. Then, there is a number
c
a<c<b
f'(c) = 0
In other words, if a function has the same value at two points, then it must "level" somewhere between those points. By considering whether the function is increasing or decreasing immediately after the first point, it becomes clear that neither option can continue indefinitely if the function is to return to the same value; therefore, there must be a local maximum or minimum before the next point occurs.
Rolle's theorem quickly turns into the mean value theorem by simply skewing the graph of the function.
Let all be as in the theorem statement above.
Define a new function
h
as the difference between
f
and the line passing through points
\big(a, f(a)\big)
\big(b, f(b)\big)
. This line has equation
y-f(a)=\dfrac{f(b)-f(a)}{b-a} (x-a) \implies y= f(a)+\dfrac{f(b)-f(a)}{b-a} (x-a).
has equation
h(x)=f(x)- f(a)-\dfrac{f(b)-f(a)}{b-a} (x-a).
Next, Rolle's theorem is useful. The function
h
satisfies the conditions of the theorem:
h
[a,b]
because it is the sum of
and a first-degree polynomial, both of which are continuous.
The function is differentiable on
(a,b)
f
and the first-degree polynomial are differentiable. In fact, we can compute it directly:
h'(x)= f'(x)-\frac{f(b)-f(a)}{b-a} .
h(a)=h(b).
By Rolle's theorem, there exists a value
c
(a,b)
h'(c)=0
h'(c)=f'(c)-\dfrac{f(b)-f(a)}{b-a} =0 \implies f'(c)=\dfrac{f(b)-f(a)}{b-a}.\ _\square
f(x)
5 \leq f'(x) \leq 14
x
and
be the maximum and minimum values, respectively, that
f(11) - f(3)
a+b
Determine all the numbers
c
that satisfy the conclusion of the mean value theorem for
f(x)=x^{3}+2x^{2}+x
[-2,2]
f(x)
is a polynomial which is continuous and differentiable over any interval, so the mean value theorem applies. Indeed, we can compute that
f'(x)=3x^2+4x+1
, and the average rate of change is
\frac{f(b)-f(a)}{b-a} = \frac{f(2)-f(-2)}{2-(-2)} = \frac{18 - (-2)}{4} = 5.
We want to find the values of
c
f'(c)=5:
\begin{aligned} 3c^{2}+4c+1&=5\\ 3c^{2}+4c-4&=0\\ (c+2)(3c-2)&=0\\ c&=-2, ~ \frac{2}{3}. \end{aligned}
c=-2
is one of the endpoints, so the mean value theorem actually guarantees another
c
in the interior, and indeed we have
c=\frac{2}{3}
_\square
A car starts from rest and drives a distance of
10 \text{ km}
30 \text{ min}
. Use the mean value theorem to show that the car attains a speed of
20 \text{ km/hr}
at some point(s) during the interval.
The mean value theorem says that the average speed of the car (the slope of the secant line) is equal to the instantaneous speed (slope of the tangent line) at some point(s) in the interval.
\frac{\Delta y}{\Delta x}=\frac{10 \text{ km}-0}{0.5 \text{ hr}-0}=20 \text{ km/hr}.
From the mean value theorem, we can find
c
f'(c)=\frac{dx}{dt}=20 \text{ km/hr}. \ _\square
f(x)
is a differentiable function for all
x
f'(x)\leq 7
x
f(2)=-4
f(5)?
f(x)
is differentiable on all intervals, we can choose any two points. So from the mean value theorem, we have
\begin{aligned} \frac{f(5)-f(2)}{5-2}=f'(c)&\leq 7\\ \frac{f(5)-(-4)}{3}&\leq 7\\ f(5)&\leq 17. \end{aligned}
So the maximum possible value of
f(5)
17
_\square
f(x)
is an arbitrary quadratic polynomial:
f(x)=Kx^{2}+Lx+M \quad (K\neq 0),
show that the point
c
whose existence is guaranteed by the mean value theorem is the mid-point of the interval
[a,b]
From the mean value theorem, we have
\begin{aligned} f'(c)&=\frac { f(b)-f(a) }{ b-a } \\ 2Kc+L &=\frac { \left(K{ b }^{ 2 }+Lb+M\right)-\left(K{ a }^{ 2 }+La+M\right) }{ b-a } \\ &=\frac { K(b-a)(b+a)+L(b-a) }{ b-a } \\ &=K(b+a)+L\\ 2c&=b+a\\\\ \Rightarrow c&=\frac { b+a }{ 2 }. \ _\square \end{aligned}
Does there exist a function
f(x)
f(0)=-1
f(2)=4
f'(x)\leq 2
x?
If such a function exists, then from the mean value theorem there is a number
c
0<c<2
f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{5}{2}.
But this is impossible because of the assumption
f'(x)\leq 2
. Therefore, such a function does not exist.
_\square
f(x)
[7,15]
f(7) = 21
f'(x) \leq 14
7\leq x \leq 15
f(15)?
f(x) = \sqrt{x-1}
x
[1,\ 38]
The name of the mean value theorem may require a little explanation. For a function
f(x)
[a, b]
, not necessarily continuous, one may define its mean value, or the "average", by the formula
\text{Mean} (f) = \dfrac{1}{b-a} \displaystyle \int_{a}^{b} f(x) \, dx,
provided, of course, that the (Riemann) integral exists. The reason is simple: the integral gives the area under the curve
y=f(x)
, and dividing by the length of the interval yields the "average" height of the function between
and
b
. In picture, we must have parts of the curve
y=f(x)
go above the mean value, and parts go below. If
f(x)
is continuous, then it must cross the mean value at some point. That is the "original" mean value theorem.
f(x)
[a, b]
c
and
b
\int_{a}^{b} f(x) \, dx = f(c) (b-a) .
What does this have to do with the (actual) mean value theorem, other than the semblance of the indeterminate
c
? Well, if
F(x)
f(x
), the fundamental theorem of calculus (Newton-Leibniz formula) makes the left-hand side equal to
F(b) - F(a)
\frac{F(b)-F(a)}{b-a} = f(c) = F'(c).
This is precisely the mean value theorem, but for
F(x)
However, that does not count as a real proof of the mean value theorem. First of all, the fundamental theorem of calculus actually relies on the mean value theorem in its proof. Secondly, here
F(x)
is indeed continuous on
[a,b]
(a,b)
, but its derivative furthermore is required to be continuous for the Riemann integral above to exist. One can construct (though somewhat ad hoc) functions that are differentiable, but its derivative fails to be continuous at one point, in a way that makes it not (Riemann) integrable.
We can also use the mean value theorem to prove certain inequalities.
Use the mean value theorem to prove that
\ln(x+1) < x
x > 0.
Suppose that our function was
f(t) = \ln(t + 1) - t
. Note that
is just a dummy variable as we will be using
x
in our interval of choice. Now, given that
f(t)
is defined, continuous, and differentiable over the interval
[0, x]
, by the mean value theorem, we see that
\begin{aligned} f'(c) &= \frac{f(x) - f(0)}{x - 0}\ \text{ for } c \in(0, x)\\ \frac{1}{c + 1} - 1 &= \frac{\ln(x+1) - x}{x}. \end{aligned}
By observation, we know that
c + 1 > 1,
\frac{1}{c + 1} < 1
\frac{1}{c + 1} - 1 < 0
\frac{\ln(x+1) - x}{x} < 0 \implies \ln(x+1) < x
x > 0.\ _\square
Cite as: Mean Value Theorem. Brilliant.org. Retrieved from https://brilliant.org/wiki/mean-value-theorem/
|
Fractional Part Function | Brilliant Math & Science Wiki
Patrick Corn, Christopher Williams, Hamza A, and
\lfloor x \rfloor
is defined to be the greatest integer less than or equal to the real number
x
. The fractional part function
\{ x \}
is defined to be the difference between these two:
x
be a real number. Then the fractional part of
x
\{x\}= x -\lfloor x \rfloor.
y=\{x\}.
For nonnegative real numbers, the fractional part is just the "part of the number after the decimal," e.g.
\{3.64 \} = 3.64 - \lfloor 3.64 \rfloor = 3.64 - 3 = 0.64.
But for negative real numbers, this is no longer the case:
\{-3.64 \} = -3.64 - \lfloor -3.64 \rfloor = -3.64 - (-4) = 0.36.
Note that in both cases
\{ x \}
The following are some examples of how fractional part functions work:
\{ 1 \} = 1 - 1 = 0.
\left\{ \sqrt{2} \right\} = \sqrt{2}-1 = 0.4142\ldots.
\{ \pi \} = \pi - 3 = 0.14159\ldots.
\left\{ -\frac{17}5 \right\} = -\frac{17}5 - (-4) = \frac35.
Properties of the Fractional Part
Fractional Parts and Integral Calculus
The following are some properties of the fractional part:
0 \le \{ x \} < 1
0 = \{ x \}
x
\{x \} + \{-x\} = \begin{cases} 0 & \text{if } x \text{ is an integer} \\ 1 & \text{otherwise.} \end{cases}
and
b
b > 0
\big\{ \frac{a}{b} \big\} = \frac{r}{b}
r
is the remainder from dividing
a
b
\{ 9 \} + \{-9\} = 0+0 = 0
\{9.01\} + \{-9.01\} = 0.01+0.99 = 1
. The fractional part is always nonnegative.
47 = 13 \cdot 3 + 8
\left\{\frac{47}{13} \right\} = \frac{8}{13}
For problems involving the floor function and the fractional part function, it often helps (for ease of notation) to write
x = n+r ,
n = \lfloor x \rfloor
r = \{ x\}
. Then
0 \le r < 1
x
be a positive real number such that
x^2+\{x\}^2 = 27.
x
x = n+r
as suggested. Now note that
x^2 \le x^2+\{x\}^2 < x^2+1
26 < x^2 \le 27
n = 5
\begin{aligned} (5+r)^2+r^2 &= 27 \\ 25+10r+2r^2 &=27 \\ 2r^2+10r-2 &= 0, \end{aligned}
r = \frac{-5+\sqrt{29}}2
x = 5 + r = \frac{5+\sqrt{29}}2
_\square
\left \{\frac 1x \right \} = \{ x \}
Find the number of solution(s) of
x
[1,6]
such that the equation above is satisfied.
\{ x \}
x
Find the smallest real number
m
such that for all positive real numbers
x
\{ x \} + \left\{ \frac1{x} \right\} < m.
x < 1
x = \frac1y, y \ge 1,
\{ x \} + \big\{ \frac1{x} \big\} = \{ y \} + \big\{ \frac1{y} \big\}.
So we may assume
x \ge 1
\big\{ \frac1{x} \big\} = \frac1{x}
x = n+r
n
a positive integer, the left side becomes
r + \frac1{n+r}.
r
, this is maximized when the denominator is minimized, i.e.
n = 1
r + \frac1{1+r}
0 \le r < 1
, its derivative being
1-\frac1{(1+r)^2}
r \to 1^-
, the sum goes to
1+\frac1{1+1} = \frac{3}{2}
\frac{3}{2}
_\square
(
Exercise: If
x
is allowed to be negative, the answer is
m=2.)
Find the number of real
x
\{x\} \lfloor x \rfloor- 2\lfloor x \rfloor = \{x\} -1
\{\cdot \}
\lfloor \cdot \rfloor
There are many interesting integrals involving the fractional part function. A good way to evaluate definite integrals of this type is to break up the interval of integration into intervals on which the greatest integer function is constant; then the original integral is a sum of integrals which are easier to evaluate.
\int_0^1 \left\{ \frac1{x} \right\} \, dx.
\left(\frac1{n+1},\frac1n \right]
\left\lfloor \frac1{x} \right\rfloor = n
. So the integral on that interval becomes
\begin{aligned} \int_{1/(n+1)}^{1/n} \left( \frac1{x}-n \right) \, dx &= \ln \frac1n-\ln \frac1{n+1} -n\left(\frac1n-\frac1{n+1}\right) \\ &= \ln(n+1)-\ln n - \frac1{n+1}. \end{aligned}
\sum_{n=1}^{\infty} \int_{1/(n+1)}^{1/n} \left\{ \frac1{x} \right\} \, dx = \sum_{n=1}^\infty \left(\ln(n+1)-\ln n - \frac1{n+1}\right)
k^\text{th}
partial sum telescopes to
\ln(k+1)-\left( \frac12 + \cdots + \frac1{k+1} \right)
. The limit is
1-\gamma
\gamma
is the famous Euler-Mascheroni constant.
_\square
\int_1^{\infty}\frac{\{x\}}{x^3}\, dx=a-\frac{\pi^b}{c}
The above integral is true for positive integers
a, b,
c
a+b+c?
\{x\}
x
\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ 3 } } dx=-\frac { H }{ U } -Mฮณ+\frac { { M }_{ 1 } }{ U_1 } \ln { (Sฯ) } -B\ln { A }
A
is the GlaisherโKinkelin constant, all other variables are positive integers, and all the fractions mentioned are coprime.
H+U+M+{ M }_{ 1 }+{ U }_{ 1 }+S+B.
\{x\}
x.
This is a part of "Who's up to the challenge?"
\displaystyle \int_{0}^{1}{\left(\left\{ \frac{1}{x}\right\} \{ 2x \}\right) \frac{dx}{x} }
The above integral is equal to
A+\ln{B}-C\gamma,
A,B,C
A+B+C.
\{ \}
denotes the fractional part and
\gamma
\int _{ 0 }^{ 1 }{ \left\{ \dfrac { 1 }{ { x }^{1/6 } } \right\} \,dx } =\dfrac { A }{ B } -\dfrac { { \pi }^{ C } }{ D }
The equation above holds true for positive integers
A,B,C,
D
A
B
coprime.
A+B+C+D
\{ \cdot \}
Cite as: Fractional Part Function. Brilliant.org. Retrieved from https://brilliant.org/wiki/factional-part-function/
|
NDF_CLEN
The routine returns the length of the specified character component of an NDF (i.e. the number of characters in the LABEL, TITLE or UNITS component).
CALL NDF_CLEN( INDF, COMP, LENGTH, STATUS )
\ast
\ast
Name of the character component whose length is required: โLABELโ, โTITLEโ or โUNITSโ.
Length of the component in characters.
The length of an NDF character component is determined by the length of the VALUE string assigned to it by a previous call to NDF_CPUT (note that this could include trailing blanks).
If the specified component is in an undefined state, then a length of zero will be returned.
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Fluids, Popular Questions: ICSE Class 9 PHYSICS, Concise Physics 9 - Meritnation
Manpreet & 2 others asked a question
ร
Me asked a question
Please make me underatand the law of flotation
Please solve the following question expert sirs
{m}^{-3}
ร{10}^{3}
{m}^{-3}
Please solve question 5.
Varisha Anwer asked a question
Please solve 5 ,6 and 7 on a page
Please solve question no 3 ,4 and 5
Expert please answer it urgent.
Ruchi Sengupta asked a question
Can you name a colour that doesn't have "e" in it.
Please solve it quickly??????
prasannamvsnk asked a question
A cubical tank carrying a non viscous liquid moves down a smooth inclined plane
Question number 10 th solution with reason
describe a simple experiment to demonstrate that a liquid enclosed in a vessel exerts pressure in all directions
Please solve it experts...
At what depth below the surface of water will the pressure be equal to 1 atmosphere. the atmospheric pressure is 10 raised to power 5 Pascal .the density of water is 10 raise to power 3 kg metre per second and gravity equals to 9.8 M per second square
Ananya Bountra asked a question
How is the reading of a barometer affected when it is taken to a deep mine?
Srinivas Katti asked a question
Why is the value of g not taken here?? Experts pls help me..
Calculate the area of an object which experiences a pressure of 1000 Pa by a force of 100 N
Q. What is a siphon system ?
Please solve 8,9,10,11and 12 on a page
Antxxfd asked a question
Please solve1 question
Please solve question no 9 and 10
What is SI unit of relative density? Expert please answer it urgent
Q no. 1, 2, 3, 5:
A 1. A man first swims in seawater and then in river water.
(i) Compare the weight of seawater and river water displaced by him.
(ii) Where doeshe find it easier to swim and Why?
Q 2. State the law of floatation.
Q. 3. Will a body weight more in air or in vacuum when weighed with a spring balance? Give a reason for your answer.
Q 5. Shile floating, is the weight of the body greater than, equat to or lesser than the upthrust?
A man can lift 60Kg load in air .How much load can he lift in water Relative density of load is 5.6
Siddhant Sp Singh asked a question
We even need to state the assumption made in part 1,
A solid weights 100 gf in air and 88 gf when completely immersed in water. Calculate: (i) the upthrust (ii) the volume of the solid, (iii) the realtive density of the solid.
Q16 all parts
Junnu asked a question
What isacceleration
Please solve @ it experts._-
Please solve question no 13 and 14 on a page
Agr dunya hilana chahate ho to pehle hilana band kro....
Please solve 13,14,15 and 16 on a page
What is the action of a siphon system ?
Please solve these questions am unable to solve them
Two cylindrical vessels fitted with Pistons A and B of area of cross section 8 cm square and 320 cm square respectively are joined at their bottom by a tube and they are completely filled with water when a mass of 4 Kg is placed on Piston A, find a(l) the pressure on Piston A (ll)the pressure on Piston B and (III)the thrust on piston B. please answer it. Urgent
please solve the following question experts
a solid weighs 1.5 kgf in air and 0.9 Kgf in a liquid of density 1.2 into 10 raise to the power 3 kg per metre cube calculate relative density of solid. Experts please answer it urgent
Rishika Prasad asked a question
Que 2 plzz send the answer
why the pressure of the atmosphere does not break the glass window
Umakant asked a question
3. A rec\mathrm{tan}gular solid block has sides 4ร10ร20 cm and a density of 7.5 g c{m}^{-1}.if it rests on a horizontal flat surface calculate the minimum and maximum pressure it can exert\left(in SI units\right) . Also calculate the thrust in both cases. \left(Ans: 3000 Pa and 15000 Pa , 60 N\right)
Please solve question no 1,2,3,4 and 5
Please sum 2
Please solve @it experts. .
A balloon is descending with uniform acceleration a (a<g) . Weight of the balloon with basket and its content is W. What wight w of the ballast should be released so that the balloon will begin to rise with the same uniform acceleration a? Neglect air resistance.
Please solve the first question
Kuber Bhatt asked a question
A sphere of iron and another of wood of the same radius are held under water. Compare the upthrust on the two spheres.
Relative density of silver is 10.8. What is the density of silver in SI unit?
What is law of floatation , explain in simple words
8.9ร{10}^{3} kg {m}^{-3}
Keishav asked a question
what force is applied on a piston of area of cross section 2 cm square to obtain of force 150 Newton on the piston of area of cross section 12 centimetre square in a hydraulic machine. please answer it. Urgent
I want the solution for it
weather forecasting plastic balloon of volume 15 metre cube contain hydrogen of density is 0.09 kg per metre cube the volume of an equipment carried by the balloon is negligible compared to its own value the mass of empty balloon alone is 7.15 kg the balloon is floating in air of density 1.3 kg per metre cube calculate 1) the mass of a hydrogen in the balloon 2) the mass of hydrogen and balloon 3) the total mass of hydrogen balloon and equipment if the mass of equipment is X kg 4) the mass of air displaced by balloon and 5) the mass of equipment using the law of floatation
What are values of g and G ??? Please tell.
Q14 ols solve fast
@please solve the following question...experts....
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Hour angle - Wikipedia
Coordinates used in the oul' equatorial coordinate system
Find sources: "Hour angle" โ news ยท newspapers ยท books ยท scholar ยท JSTOR (April 2018) (Learn how and when to remove this template message)
The hour angle is indicated by an orange arrow on the celestial equator plane. The arrow ends at the oul' hour circle of an orange dot indicatin' the apparent place of an astronomical object on the oul' celestial sphere.
In astronomy and celestial navigation, the oul' hour angle is the bleedin' angle between two planes: one containin' Earth's axis and the bleedin' zenith (the meridian plane), and the other containin' Earth's axis and a holy given point of interest (the hour circle).[1]
It may be given in degrees, time, or rotations dependin' on the bleedin' application. The angle may be expressed as negative east of the oul' meridian plane and positive west of the oul' meridian plane, or as positive westward from 0ยฐ to 360ยฐ. Bejaysus here's a quare one right here now. The angle may be measured in degrees or in time, with 24h = 360ยฐ exactly. In celestial navigation, the feckin' convention is to measure in degrees westward from the prime meridian (Greenwich hour angle, GHA), from the oul' local meridian (local hour angle, LHA) or from the oul' first point of Aries (sidereal hour angle, SHA).
The hour angle is paired with the oul' declination to fully specify the feckin' location of a point on the oul' celestial sphere in the oul' equatorial coordinate system.[2]
1 Relation with right ascension
2 Solar hour angle
3 Sidereal hour angle
Relation with right ascension[edit]
As seen from above the bleedin' Earth's north pole, a star's local hour angle (LHA) for an observer near New York (red dot). Sure this is it. Also depicted are the star's right ascension and Greenwich hour angle (GHA), the feckin' local mean sidereal time (LMST) and Greenwich mean sidereal time (GMST). The symbol ส identifies the oul' vernal equinox direction.
Assumin' in this example the day of the feckin' year is the oul' March equinox so the sun lies in the oul' direction of the oul' grey arrow then this star will rise about midnight, the cute hoor. Just after the feckin' observer reaches the feckin' green arrow dawn comes and overwhelms with light the bleedin' visibility of the bleedin' star about six hours before it sets on the feckin' western horizon. Me head is hurtin' with all this raidin'. The Right Ascension of the star is about 18h
The local hour angle (LHA) of an object in the oul' observer's sky is
{\displaystyle {\text{LHA}}_{\text{object}}={\text{LST}}-\alpha _{\text{object}}}
{\displaystyle {\text{LHA}}_{\text{object}}={\text{GST}}+\lambda _{\text{observer}}-\alpha _{\text{object}}}
where LHAobject is the bleedin' local hour angle of the feckin' object, LST is the local sidereal time,
{\displaystyle \alpha _{\text{object}}}
is the feckin' object's right ascension, GST is Greenwich sidereal time and
{\displaystyle \lambda _{\text{observer}}}
is the observer's longitude (positive east from the prime meridian).[3] These angles can be measured in time (24 hours to a circle) or in degrees (360 degrees to a circle)โone or the bleedin' other, not both.
Negative hour angles (โ180ยฐ < LHAobject < 0ยฐ) indicate the feckin' object is approachin' the feckin' meridian, positive hour angles (0ยฐ < LHAobject < 180ยฐ) indicate the feckin' object is movin' away from the oul' meridian; an hour angle of zero means the oul' object is on the oul' meridian.
Solar hour angle[edit]
Observin' the oul' Sun from Earth, the bleedin' solar hour angle is an expression of time, expressed in angular measurement, usually degrees, from solar noon. Chrisht Almighty. At solar noon the hour angle is zero degrees, with the feckin' time before solar noon expressed as negative degrees, and the oul' local time after solar noon expressed as positive degrees. Sure this is it. For example, at 10:30 AM local apparent time the feckin' hour angle is โ22.5ยฐ (15ยฐ per hour times 1.5 hours before noon).[4]
The cosine of the oul' hour angle (cos(h)) is used to calculate the oul' solar zenith angle. At solar noon, h = 0.000 so cos(h) = 1, and before and after solar noon the bleedin' cos(ยฑ h) term = the bleedin' same value for mornin' (negative hour angle) or afternoon (positive hour angle), so that the Sun is at the bleedin' same altitude in the oul' sky at 11:00AM and 1:00PM solar time.[5]
Sidereal hour angle[edit]
The sidereal hour angle (SHA) of a body on the oul' celestial sphere is its angular distance west of the oul' vernal equinox generally measured in degrees, like. The SHA of a star varies by less than a bleedin' minute of arc per year, due to precession, while the bleedin' SHA of a planet varies significantly from night to night. SHA is often used in celestial navigation and navigational astronomy, and values are published in astronomical almanacs.[citation needed]
^ :10.1002/9780470209738.ch2. ISBN 9780470209738.
^ Schowengerdt, R, like. A, the cute hoor. (2007). "Optical radiation models". Arra' would ye listen to this. Remote Sensin'. pp. 45โ88. Arra' would ye listen to this shite? doi:10.1016/B978-012369407-2/50005-X. ISBN 9780123694072.
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Kinematics in Two Dimensions: An Introduction | Physics | Course Hero
Figure 1. Walkers and drivers in a city like New York are rarely able to travel in straight lines to reach their destinations. Instead, they must follow roads and sidewalks, making two-dimensional, zigzagged paths. (credit: Margaret W. Carruthers)
Suppose you want to walk from one point to another in a city with uniform square blocks, as pictured in Figure 2.
Figure 2. A pedestrian walks a two-dimensional path between two points in a city. In this scene, all blocks are square and are the same size.
An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line path form a right triangle, and so the Pythagorean theorem, a2 + b2 = c2, can be used to find the straight-line distance.
The Pythagorean theorem relates the length of the legs of a right triangle, labeled a and b, with the hypotenuse, labeled c. The relationship is given by: a2+ b2 = c2. This can be rewritten, solving for c:
c\text{ = }\sqrt{{a}^{2}\text{+ }{b}^{2}}
The hypotenuse of the triangle is the straight-line path, and so in this case its length in units of city blocks is
\sqrt{(\text{(9 blocks)}^{2} + \text{(5 blocks)}^{2}}= 10.3 \text{ blocks}
, considerably shorter than the 14 blocks you walked. (Note that we are using three significant figures in the answer. Although it appears that โ9โ and โ5โ have only one significant digit, they are discrete numbers. In this case โ9 blocksโ is the same as โ9.0 or 9.00 blocks.โ We have decided to use three significant figures in the answer in order to show the result more precisely.)
Figure 4. The straight-line path followed by a helicopter between the two points is shorter than the 14 blocks walked by the pedestrian. All blocks are square and the same size.
The fact that the straight-line distance (10.3 blocks) in Figure 4 is less than the total distance walked (14 blocks) is one example of a general characteristic of vectors. (Recall that vectors are quantities that have both magnitude and direction.)
As for one-dimensional kinematics, we use arrows to represent vectors. The length of the arrow is proportional to the vectorโs magnitude. The arrowโs length is indicated by hash marks in Figure 2 and Figure 4. The arrow points in the same direction as the vector. For two-dimensional motion, the path of an object can be represented with three vectors: one vector shows the straight-line path between the initial and final points of the motion, one vector shows the horizontal component of the motion, and one vector shows the vertical component of the motion. The horizontal and vertical components of the motion add together to give the straight-line path. For example, observe the three vectors in Figure 4. The first represents a 9-block displacement east. The second represents a 5-block displacement north. These vectors are added to give the third vector, with a 10.3-block total displacement. The third vector is the straight-line path between the two points. Note that in this example, the vectors that we are adding are perpendicular to each other and thus form a right triangle. This means that we can use the Pythagorean theorem to calculate the magnitude of the total displacement. (Note that we cannot use the Pythagorean theorem to add vectors that are not perpendicular. We will develop techniques for adding vectors having any direction, not just those perpendicular to one another, in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods.)
The person taking the path shown in Figure 4 walks east and then north (two perpendicular directions). How far he or she walks east is only affected by his or her motion eastward. Similarly, how far he or she walks north is only affected by his or her motion northward.
Figure 5. This shows the motions of two identical ballsโone falls from rest, the other has an initial horizontal velocity. Each subsequent position is an equal time interval. Arrows represent horizontal and vertical velocities at each position. The ball on the right has an initial horizontal velocity, while the ball on the left has no horizontal velocity. Despite the difference in horizontal velocities, the vertical velocities and positions are identical for both balls. This shows that the vertical and horizontal motions are independent.
College Physics. Authored by: OpenStax College. Located at: http://cnx.org/contents/[email protected]:[email protected]/Kinematics-in-Two-Dimensions-A. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected]
PhET Interactive Simulations. Provided by: University of Colorado Boulder . Located at: http://phet.colorado.edu. License: Public Domain: No Known Copyright
3.1 Kinematics in Two Dimensions An Introduction - College Physics OpenStax 2.pdf
3.1 Kinematics in Two Dimensions_ An Introduction.pdf
Chp 3 Sections 1 and 2 Two Dimensions, Vector Math Open Stax.doc
Ch 3.1 Kinematics in Two Dimensions An IntroductionCh 3.2 Vector Addition and Subtraction Graphical
i. Kinematics in Two Dimensions and Projectile Motion.pdf
Lab 2 - Kinematics in Two Dimensions.docx
Kinematics in two dimensions.pptx
PHYSICS 1 โข Union County College
Kinematics+in+Two+Dimensions+summary+notes.pdf
Topic_04 Chapter_03 Giancoli (Kinematics in two dimensions) with Australian notation.pdf
ECU 48 โข Edith Cowan University
Chapter 4 - Kinematics in Two Dimensions - Projectile Motion
CN0401 Kinematics in two dimensions.doc
PHYSICS 1 โข St. Francis High
05-Kinematics in Two Dimensions
PHYSICS Physics 1E โข McMaster University
Ch 03 Kinematics in Two Dimensions
PHY 136 โข University of Toronto, Mississauga
Chapter 3 - Kinematics in Two Dimensions and Vectors Exercises
CHM 1046L โข Miami Dade College, Miami
5.Kinematics of particle_Motion in Two Dimensions_Projectile.pdf
PHYSICS MECHANICS โข University of Botswana-Gaborone
PHYS 1301 Powerpoint 12 Kinematics in Two Dimensions - Motion under Constant Acceleration.pptx
PHYSICS 1301 โข C.E. King High School
Chapter 03 - Kinematics in Two Dimensions; Vectors
Kinematics in Two Dimensions.docx
Unit 4 - Kinematics in Two Dimensions - Projectile Motion.pdf
Kinematics-in-two-dimensions-and-vectors-1jql12x
NONE none โข Shades Valley High Sch
Projectile_Motion_Physics_Problems_Kinematics_In_Two_Dimensions
Chapter 4 - Kinematics in Two Dimensions - Solutions.pdf
PHYSICS 4A โข University of the Fraser Valley
Kinematics in Two Dimensions report
PHYS 112 โข Suffolk University
Physics I Syllabus FALL 2020.docx
PHYSIC PHYS-2425 โข Tarrant County College, Northeast
physics-3-physics-cutnell-6-ch-03-kinematics-in-two-dimensions-819
PHYSIC 1B03 โข McMaster University
Ch3-Kinematics in Two Dimensions.pdf
EGN MISC โข Florida Atlantic University
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Modulate signal using OFDM method - MATLAB - MathWorks รญโขลรชยตยญ
NumGuardBandCarriers
InsertDCNull
PilotInputPort
PilotCarrierIndices
CyclicPrefixLength
NumSymbols
NumTransmitAntennnas
Specific to comm.OFDMModulator
Create and Modify OFDM Modulator
Create OFDM Modulator from OFDM Demodulator
Visualize Time-Frequency Resource Assignments for OFDM Modulator
Create OFDM Modulator and Specify Pilots
Create OFDM Modulator with Varying Cyclic Prefix Lengths
Determine OFDM Modulator Data Dimensions
Create OFDM Modulated Data
Modulate signal using OFDM method
The OFDMModulator object modulates a signal using the orthogonal frequency division modulation method. The output is a baseband representation of the modulated signal.
To modulate a signal using OFDM:
Create the comm.OFDMModulator object and set its properties.
hMod = comm.OFDMModulator
hMod = comm.OFDMModulator(Name,Value)
hMod = comm.OFDMModulator(hDemod)
hMod = comm.OFDMModulator creates an OFDM modulator System objectโข.
hMod = comm.OFDMModulator(Name,Value) specifies Properties using one of more name-value pair arguments. Enclose each property name in quotes. For example, comm.OFDMModulator('NumSymbols',8) specifies eight OFDM symbols in the time-frequency grid.
hMod = comm.OFDMModulator(hDemod) sets the OFDM modulator system object properties based on the specified OFDM demodulator system object comm.OFDMDemodulator.
FFTLength โ Number of FFT points
Number of Fast Fourier Transform (FFT) points, specified as a positive integer. The length of the FFT, NFFT, must be greater than or equal to 8 and is equivalent to the number of subcarriers.
NumGuardBandCarriers โ Number of subcarriers to the left and right guard bands
[6;5] (default) | two-element column vector of integers
Number of subcarriers allocated to the left and right guard bands, specified as a two-element column vector of integers. The number of subcarriers must fall within [0, รขลล NFFT/2รขลโน รขหโ 1]. This vector has the form [NleftG, NrightG], where NleftG and NrightG specify the left and right guard bands, respectively.
InsertDCNull โ Option to insert DC null
Option to insert DC null, specified as a numeric or logical 0 (false) or 1 (true). The DC subcarrier is the center of the frequency band and has the index value:
(FFTLength / 2) + 1 when FFTLength is even
(FFTLength + 1) / 2 when FFTLength is odd
PilotInputPort โ Option to specify pilot input
Option to specify pilot input, specified as a numeric or logical 0 (false) or 1 (true). If this property is 1 (true), you can assign individual subcarriers for pilot transmission. If this property is 0 (false), pilot information is assumed to be embedded in the input data.
PilotCarrierIndices โ Pilot subcarrier indices
Pilot subcarrier indices, specified as a column vector. If the PilotCarrierIndices property is set to 1 (true), you can specify the indices of the pilot subcarriers. You can assign the indices to the same or different subcarriers for each symbol. Similarly, the pilot carrier indices can differ across multiple transmit antennas. Depending on the desired level of control for index assignments, the dimensions of the property vary. Valid pilot indices fall in the range
\left[{N}_{\text{leftG}}+1,\text{รขยย}{N}_{\text{FFT}}/2\right]รขยยช\left[{N}_{\text{FFT}}/2+2,\text{รขยย}{N}_{\text{FFT}}รขยย{N}_{\text{rightG}}\right],
where the index value cannot exceed the number of subcarriers. When the pilot indices are the same for every symbol and transmit antenna, the property has dimensions Npilot-by-1. When the pilot indices vary across symbols, the property has dimensions Npilot-by-Nsym. If you transmit only one symbol but multiple transmit antennas, the property has dimensions Npilot-by-1-by-Nt., where Nt. is the number of transmit antennas. If the indices vary across the number of symbols and transmit antennas, the property has dimensions Npilot-by-Nsym-by-Nt. If the number of transmit antennas is greater than one, ensure that the indices per symbol must be mutually distinct across antennas to minimize interference.
To enable this property, set the PilotInputPort property to 1 (true).
CyclicPrefixLength โ Length of cyclic prefix
16 (default) | positive integer | row vector
Windowing โ Option to apply raised cosine window between OFDM symbols
Option to apply raised cosine window between OFDM symbols, specified as true or false. Windowing is the process in which the OFDM symbol is multiplied by a raised cosine window before transmission to more quickly reduce the power of out-of-band subcarriers. Windowing reduces spectral regrowth.
WindowLength โ Length of raised cosine window
Length of raised cosine window, specified as a positive scalar. This value must be less than or equal to the minimum cyclic prefix length. For example, in a configuration of four symbols with cyclic prefix lengths 12, 14, 16, and 18, the window length must be less than or equal to 12.
To enable this property, set the Windowing property to 1 (true).
NumSymbols โ Number of OFDM symbols
Number of OFDM symbols in the time-frequency grid, specified as a positive integer.
NumTransmitAntennnas โ Number of transmit antennas
Number of transmit antennas, used to transmit the OFDM modulated signal, specified as a positive integer.
waveform = hMod(insignal)
waveform = hMod(data,pilot)
waveform = hMod(insignal) applies OFDM modulation the specified baseband signal and returns the modulated OFDM baseband signal.
waveform = hMod(data,pilot) assigns the pilot signal, pilot, into the frequency subcarriers specified by the PilotCarrierIndices property value of the hMod system object. To enable this syntax set the PilotCarrierIndices property to true.
insignal โ Input baseband signal
Input baseband signal, specified as a matrix or 3-D array of numeric values. The input baseband signal must be of size Nf-by-Nsym-by-Nt. where Nf is the number of frequency subcarriers excluding guard bands and DC null.
Input data, specified as a matrix or 3-D array. The input must be a numeric of size Nd-by-Nsym-by-Nt. where Nd is the number of data subcarriers in each symbol. For more information on how Nd is calculated, see the PilotCarrierIndices property.
pilot โ Pilot signal
Pilot signal, specified as a 3-D array of numeric values. The pilot signal must be of size Npilot-by-Nsym-by-Nt.
waveform โ OFDM Modulated baseband signal
OFDM Modulated baseband signal, returned as a 2-D array. If the CyclicPrefixLength property is a scalar, the output waveform is of size ((NFFT+CPlen)รขยลฝNsym)-by-Nt. Otherwise, the size is (NFFTรขยลฝNsym+รขหโ(CPlen))-by-Nt.
info Provide dimensioning information for OFDM modulator
showResourceMapping Show the subcarrier mapping of the OFDM symbols created by the OFDM modulator System object
Create and display an OFDM modulator System objectโข with default property values.
hMod =
comm.OFDMModulator with properties:
NumGuardBandCarriers: [2x1 double]
InsertDCNull: false
PilotInputPort: false
CyclicPrefixLength: 16
Windowing: false
NumSymbols: 1
Modify the number of subcarriers and symbols.
hMod.FFTLength = 128;
hMod.NumSymbols = 2;
Verify that the number of subcarriers and the number of symbols changed.
disp(hMod)
Use the showResourceMapping object function to show the mapping of data, pilot, and null subcarriers in the time-frequency space.
showResourceMapping(hMod)
Create an OFDM demodulator System objectโข with default property values. Then, specify pilot indices for a single symbol and two transmit antennas.
Setting the PilotCarrierIndices property of the demodulator affects the number of transmit antennas in the OFDM modulator when you use the demodulator in the creation of the modulator. The number of receive antennas in the demodulator is uncorrelated with the number of transmit antennas.
ofdmDemod = comm.OFDMDemodulator;
ofdmDemod.PilotOutputPort = true;
ofdmDemod.PilotCarrierIndices = ...
cat(3,[12; 26; 40; 54],[13; 27; 41; 55]);
Use the OFDM demodulator to construct the OFDM modulator.
ofdmMod = comm.OFDMModulator(ofdmDemod);
Display the properties of the OFDM modulator and demodulator, verifying that the applicable properties match.
disp(ofdmMod)
PilotInputPort: true
PilotCarrierIndices: [4x1x2 double]
disp(ofdmDemod)
comm.OFDMDemodulator with properties:
RemoveDCCarrier: false
PilotOutputPort: true
The showResourceMapping method displays the time-frequency resource mapping for each transmit antenna.
Construct an OFDM modulator.
mod = comm.OFDMModulator;
Apply the showResourceMapping method.
showResourceMapping(mod)
Insert a DC null.
mod.InsertDCNull = true;
Show the resource mapping after adding the DC null.
Create an OFDM modulator and specify the subcarrier indices for the pilot signals. Specify the indices for each symbol and transmit antenna. When the number of transmit antennas is greater than one, set different pilot indices for each symbol between antennas.
Create an OFDM modulator System object, specifying two symbols and inserting a DC null.
mod = comm.OFDMModulator('FFTLength',128,'NumSymbols',2,...
'InsertDCNull',true);
Enable the pilot input port so you can specify the pilot indices.
mod.PilotInputPort = true;
Specify the same pilot indices for both symbols.
mod.PilotCarrierIndices = [12; 56; 89; 100];
Visualize the placement of the pilot signals and nulls in the OFDM time-frequency grid by using the showResourceMapping object function.
Specify different indices for the second symbol by concatenating a second column of pilot indices to the PilotCarrierIndices property.
mod.PilotCarrierIndices = cat(2,mod.PilotCarrierIndices, ...
[17; 61; 94; 105]);
Verify that the pilot subcarrier indices differ between the two symbols.
Increase the number of transmit antennas to two.
mod.NumTransmitAntennas = 2;
Specify the pilot indices for each of the two transmit antennas. To provide indices for multiple antennas while minimizing interference among the antennas, set the PilotCarrierIndices property as a 3-D array such that the indices for each symbol differ among antennas.
mod.PilotCarrierIndices = cat(3,[20; 50; 70; 110], [15; 60; 75; 105]);
Display the resource mapping for the two transmit antennas. The gray lines denote the insertion of custom nulls. The nulls are created by the object to minimize interference among the pilot symbols from different antennas.
Specify the length of the cyclic prefix for each OFDM symbol.
Create an OFDM modulator, specifying five symbols, four left and three right guard-band subcarriers, and the cyclic prefix length for each OFDM symbol.
mod = comm.OFDMModulator('NumGuardBandCarriers',[4;3],...
'NumSymbols',5,...
'CyclicPrefixLength',[12 10 14 11 13]);
Display the properties of the OFDM modulator, verifying that the cyclic prefix length changes across symbols.
disp(mod)
CyclicPrefixLength: [12 10 14 11 13]
Get the OFDM modulator data dimensions by using the info object function.
Construct an OFDM modulator System objectโข with user-specified pilot indices, an inserted DC null, and specify two transmit antennas.
hMod = comm.OFDMModulator('NumGuardBandCarriers',[4;3], ...
'PilotCarrierIndices',cat(3,[12; 26; 40; 54], ...
[11; 25; 39; 53]), ...
Use the info object function to get the modulator input data, pilot input data, and output data sizes.
info(hMod)
DataInputSize: [48 1 2]
OutputSize: [80 2]
Generate OFDM modulated symbols for use in link-level simulations.
Construct an OFDM modulator with an inserted DC null, seven guard-band subcarriers, and two symbols having different pilot indices for each symbol.
mod = comm.OFDMModulator( ...
'NumGuardBandCarriers',[4;3], ...
'PilotCarrierIndices',[12 11; 26 27; 40 39; 54 55], ...
Determine input data, pilot, and output data dimensions.
modDim = info(mod);
Generate random data symbols for the OFDM modulator. The structure variable, modDim, determines the number of data symbols.
dataIn = complex( ...
randn(modDim.DataInputSize),randn(modDim.DataInputSize));
Create a pilot signal that has the correct dimensions.
pilotIn = complex( ...
rand(modDim.PilotInputSize),rand(modDim.PilotInputSize));
Apply OFDM modulation to the data and pilot signals.
modData = mod(dataIn,pilotIn);
Use the OFDM modulator object to create the corresponding OFDM demodulator.
demod = comm.OFDMDemodulator(mod);
Demodulate the OFDM signal and output the data and pilot signals.
[dataOut, pilotOut] = demod(modData);
Verify that, within a tight tolerance, the input data and pilot symbols match the output data and pilot symbols.
isSame = (max(abs([dataIn(:) - dataOut(:); ...
pilotIn(:) - pilotOut(:)])) < 1e-10)
isSame = logical
x\left(t\right)=\underset{k=0}{\overset{Nรขยย1}{รขยย}}{X}_{k}{e}^{j2\mathrm{รย}k\mathrm{รย}ft},\text{รขยย}0รขยยคtรขยยคT,
\frac{1}{T}{รขยยซ}_{0}^{T}{\left({e}^{j2\mathrm{รย}m\mathrm{รย}ft}\right)}^{*}\left({e}^{j2\mathrm{รย}n\mathrm{รย}ft}\right)\text{รขยย}dt=\frac{1}{T}{รขยยซ}_{0}^{T}{e}^{j2\mathrm{รย}\left(mรขยยn\right)\mathrm{รย}ft}\text{รขยย}dt=0\text{รขยย}\text{for}\text{รขยย}mรขย n.
w\left(t\right)=\left\{\begin{array}{l}1,\text{รขยย}\text{}0รขยยค|t|<\frac{Tรขยย{T}_{W}}{2}\\ \frac{1}{2}\left\{1+\mathrm{cos}\left[\frac{\mathrm{รย}}{{T}_{W}}\left(|t|รขยย\frac{Tรขยย{T}_{W}}{2}\right)\right]\right\},\text{}\frac{Tรขยย{T}_{W}}{2}รขยยค|t|รขยยค\frac{T+{T}_{W}}{2}\\ 0,\text{}\text{otherwise}\end{array}
[1] Dahlman, Erik, Stefan Parkvall, and Johan Skรยถld. 4G LTE/LTE-Advanced for Mobile Broadband. Amsterdam: Elsevier, Acad. Press, 2011.
[2] Andrews, J. G., A. Ghosh, and R. Muhamed. Fundamentals of WiMAX. Upper Saddle River, NJ: Prentice Hall, 2007.
[3] Agilent Technologies, Inc., "OFDM Raised Cosine Windowing", https://rfmw.em.keysight.com/wireless/helpfiles/n7617a/ofdm_raised_cosine_windowing.htm.
[4] Montreuil, L., R. Prodan, and T. Kolze. "OFDM TX Symbol Shaping 802.3bn", https://www.ieee802.org/3/bn/public/jan13/montreuil_01a_0113.pdf. Broadcom, 2013.
[5] "IEEE Standard 802.16โข-2009." New York: IEEE, 2009.
qammod | ofdmmod
comm.OFDMDemodulator | comm.QPSKModulator
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Ideal mechanical rotational inertia - MATLAB - MathWorks Deutschland
The Inertia block represents an ideal mechanical translational inertia that is described with the following equation:
T=J\frac{d\omega }{dt}
T is inertia torque.
J is inertia.
By default, the block has one mechanical translational conserving port. The block positive direction is from its port to the reference point. This means that the inertia torque is positive if inertia is accelerated in the positive direction.
In some applications, it is customary to display inertia in series with other elements in the block diagram layout. To support this use case, the Number of graphical ports parameter lets you display a second port on the opposite side of the block icon. The two-port variant is purely graphical: the two ports have the same angular velocity, so the block functions exactly the same whether it has one or two ports. The block icon changes depending on the value of the Number of graphical ports parameter.
I โ Port that connects inertia to the circuit
Mechanical rotational conserving port that connects the inertia to the physical network.
J โ Second graphical port
Second mechanical rotational conserving port that lets you connect the inertia in series with other elements in the block diagram. This port is rigidly connected to port I, therefore the difference between the one-port and two-port block representations is purely graphical.
Inertia โ Constant inertia
Inertia value. The inertia is constant during simulation.
1 โ The block has one conserving port that connects it to the mechanical rotational circuit. When the block has one port, attach it to a connection line between two other blocks.
2 โ Selecting this option exposes the second port, which lets you connect the block in series with other blocks in the circuit. The two ports are rigidly connected and have the same angular velocity, therefore the block functions exactly the same as if it had one port.
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PermanentIndices - Maple Help
Home : Support : Online Help : Programming : Procedures and Functions : Cache Package : PermanentIndices
return a sequence of the permanent indices
PermanentIndices( cache )
cache table or procedure: the object whose indices (keys) are to be returned
The PermanentIndices command returns the indices of the permanent entries of the given cache table. The cache table can be given directly as cache, or cache can refer to a procedure that has, or can have, a cache remember table. If such a procedure is given and it has a cache remember table, the keys of the permanent indices from that table are returned. If the procedure does not have a table, NULL is returned.
PermanentIndices returns the indices in the same format as indices, that is, a sequence of lists where the contents of each list is an index of a permanent entry from the table.
The PermanentEntries command can be used to get the indices of the permanent entries.
\mathrm{c1}โ\mathrm{Cache}โก\left(\right)
\textcolor[rgb]{0,0,1}{\mathrm{c1}}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathrm{Cache}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{512}\right)
\mathrm{Cache}:-\mathrm{AddPermanent}โก\left(\mathrm{c1},[x],y\right)
\mathrm{Cache}:-\mathrm{AddPermanent}โก\left(\mathrm{c1},[y],z\right)
\mathrm{Cache}:-\mathrm{PermanentEntries}โก\left(\mathrm{c1}\right)
[\textcolor[rgb]{0,0,1}{z}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{y}]
\mathrm{Cache}:-\mathrm{PermanentIndices}โก\left(\mathrm{c1}\right)
[\textcolor[rgb]{0,0,1}{y}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{x}]
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathbf{proc}}\left(\textcolor[rgb]{0,0,1}{x}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{option}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathrm{cache}}\textcolor[rgb]{0,0,1}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{^}\textcolor[rgb]{0,0,1}{2}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{end proc}}
\mathrm{Cache}:-\mathrm{AddPermanent}โก\left(p,[2],2\right)
\mathrm{Cache}:-\mathrm{AddPermanent}โก\left(p,[3],3\right)
\mathrm{Cache}:-\mathrm{PermanentEntries}โก\left(p\right)
[\textcolor[rgb]{0,0,1}{2}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{3}]
\mathrm{Cache}:-\mathrm{PermanentIndices}โก\left(p\right)
[\textcolor[rgb]{0,0,1}{2}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{3}]
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Understanding of Convolutional Neural Network (CNN) โ Deep Learning - Freakblogger - Freak for blogging
In neural networks, Convolutional neural network (ConvNets or CNNs) is one of the main categories to do images recognition, images classifications.
Artificial Intelligence Jan 30, 2021 1 777 Add to Reading List
In neural networks, Convolutional neural network (ConvNets or CNNs) is one of the main categories to do images recognition, images classifications. Objects detections, recognition faces etc., are some of the areas where CNNs are widely used.CNNs use image recognition and classification in order to detect objects, recognize faces, etc.They are made up of neurons with learnable weights and biases. Each specific neuron receives numerous inputs and then takes a weighted sum over them, where it passes it through an activation function and responds back with an output.
CNNs are primarily used to classify images, cluster them by similarities, and then perform object recognition. Many algorithms using CNNs can identify faces, street signs, animals, etc.
Visualization of an image by a computer
Letโs say we have a color image in JPG form and its size is 480 x 480. The representative array will be 480 x 480 x 3, it will see h x w x d( h = Height, w = Width, d = Dimension ). Each of these numbers is given a value from 0 to 255 which describes the pixel intensity at that point. RGB intensity values of the image are visualized by the computer for processing.
A Classic CNN:
Contents of a classic Convolutional Neural Network: -
1.Convolutional Layer.
2.Activation operation following each convolutional layer.
3.Pooling layer especially Max Pooling layer and also others based on the requirement.
4.Finally Fully Connected Layer.
Convolution is the first layer to extract features from an input image. Convolution preserves the relationship between pixels by learning image features using small squares of input data. It is a mathematical operation that takes two inputs such as image matrix and a filter or kernel.
Consider a 5 x 5 whose image pixel values are 0, 1 and filter matrix 3 x 3 as shown in below
Then the convolution of 5 x 5 image matrix multiplies with 3 x 3 filter matrix which is called โFeature Mapโ as output shown in below
Figure : 3 x 3 Output matrix
Convolution of an image with different filters can perform operations such as edge detection, blur and sharpen by applying filters. The below example shows various convolution image after applying different types of filters (Kernels).
Figure : Some common filters
Data or imaged is convolved using filters or kernels. Filters are small units that we apply across the data through a sliding window. The depth of the image is the same as the input, for a color image that RGB value of depth is 4, a filter of depth 4 would also be applied to it. This process involves taking the element-wise product of filters in the image and then summing those specific values for every sliding action. The output of a convolution that has a 3d filter with color would be a 2d matrix.
Now, the best way to explain a convolutional layer is to imagine a flashlight that is shining over the top left of the image. In order to understand how this works, imagine as if a flashlight shines its light and covers a 5 x 5 area. And now, letโs imagine this flashlight sliding across all the areas of the input image. This flashlight is called a filter(or sometimes referred to as a neuron or a kernel) and the region that it is shining over is called the receptive field. This filter is also an array of numbers (the numbers are called weights or parameters).
In several cases, we incorporate techniques, including padding and strided convolutions, that affect the size of the output. As motivation, note that since kernels generally have width and height greater than
1
, after applying many successive convolutions, we tend to wind up with outputs that are considerably smaller than our input. If we start with a pixel image, layers of convolutions reduce the image to pixels, slicing off of the image and with it obliterating any interesting information on the boundaries of the original image. Padding is the most popular tool for handling this issue.
In other cases, we may want to reduce the dimensionality drastically, e.g., if we find the original input resolution to be unwieldy. Strided convolutions are a popular technique that can help in these instances.
As described above, one tricky issue when applying convolutional layers is that we tend to lose pixels on the perimeter of our image. Since we typically use small kernels, for any given convolution, we might only lose a few pixels, but this can add up as we apply many successive convolutional layers. One straightforward solution to this problem is to add extra pixels of filler around the boundary of our input image, thus increasing the effective size of the image. Typically, we set the values of the extra pixels to zero. In Fig. 6.3.1, we pad a input, increasing its size to . The corresponding output then increases to a matrix. The shaded portions are the first output element as well as the input and kernel tensor elements used for the output computation: .
Sometimes filter does not fit perfectly fit the input image. We have two options:
Pad the picture with zeros (zero-padding) so that it fits
Drop the part of the image where the filter did not fit. This is called valid padding which keeps only valid part of the image.
ReLU stands for Rectified Linear Unit for a non-linear operation. The output is ฦ(x) = max(0,x).
Why ReLU is important : ReLUโs purpose is to introduce non-linearity in our ConvNet. Since, the real world data would want our ConvNet to learn would be non-negative linear values.
Figure 7 : ReLU operation
There are other non linear functions such as tanh or sigmoid that can also be used instead of ReLU. Most of the data scientists use ReLU since performance wise ReLU is better than the other two.
Stride is the number of pixels shifts over the input matrix. When the stride is 1 then we move the filters to 1 pixel at a time. When the stride is 2 then we move the filters to 2 pixels at a time and so on. The below figure shows convolution would work with a stride of 2.
Pooling layers section would reduce the number of parameters when the images are too large. Spatial pooling also called subsampling or downsampling which reduces the dimensionality of each map but retains important information. Spatial pooling can be of different types:
Max pooling takes the largest element from the rectified feature map. Taking the largest element could also take the average pooling. Sum of all elements in the feature map call as sum pooling.
The layer we call as FC layer, we flattened our matrix into vector and feed it into a fully connected layer like a neural network.
In the above diagram, the feature map matrix will be converted as vector (x1, x2, x3, โฆ). With the fully connected layers, we combined these features together to create a model. Finally, we have an activation function such as softmax or sigmoid to classify the outputs as cat, dog, car, truck etc.,
https://medium.com/@RaghavPrabhu/understanding-of-convolutional-neural-network-cnn-deep-learning
https://towardsdatascience.com/basics-of-the-classic-cnn-a3dce1225add
https://medium.com/datadriveninvestor/introduction-to-how-cnns-work-77e0e4cde99b
Understanding of Convolutional Neural Network (CNN)- From Fully-Connected...
it is an easy explanation of CNN
How can you manage Diabetes during Summer Season
|
A plane flying with an airspeed of
540
mph in the direction
\text{N }45^\circ\text{ E}
45ยฐ
east of north) encounters a wind blowing at
47
\text{S }78^\circ\text{ E}
What are the true speed and direction of the plane?
\left\langle540\cos\left(45^\circ\right)+47\cos\left(-12^\circ\right),540\sin\left(45^\circ\right)+47\sin\left(-12^\circ\right)\right\rangle
How long will it take the plane to travel
1000
miles in the resultant direction?
Solve (magnitude of vector in part (a))
t=1000
If the pilot intended to fly
1000
miles in the direction
\text{N }45^\circ\text{ E}
, how far from his intended destination is he given the situation in part (b)?
intended distance vector:
\left\langle1000\cos(45^\circ),1000\sin(45^\circ)\right\rangle
actual distance vector:
\left\langle(540\cos(45^\circ)+47\cos(-12^\circ))t,540\sin(45^\circ)+47\sin(-12^\circ))t\right\rangle
Use the time you calculated in part (b).
Determine the magnitude of the difference of the vectors in steps 1 and 2.
|
This problem is a checkpoint for solving logarithmic equations. Solve each of the following equations.
\log_2(x^2+2x)=3
\log_5(\sqrt{x})+\log_5(3\sqrt{x})=2
\log_2(x)=\frac{1}{2}\log_2(9)+\log_2(x-4)
17^{\log_{17}(xโ6)}=3
Ideally at this point you are comfortable working with these types of problems and can complete them correctly. If you feel that you need more confidence when completing these types of problems, then review the Solving Logarithmic Equations Checkpoint materials available at cpm.org and try the practice problems provided. From this point on, you will be expected to do problems like these correctly and with confidence.
Click the following link: Solving Logarithmic Equations
|
Global Constraint Catalog: Camong_seq
<< 5.27. among_modulo5.29. among_var >>
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐}\left(\mathrm{๐ป๐พ๐},\mathrm{๐๐ฟ},\mathrm{๐๐ด๐},\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐},\mathrm{๐
๐ฐ๐ป๐๐ด๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐ป๐พ๐}
\mathrm{๐๐๐}
\mathrm{๐๐ฟ}
\mathrm{๐๐๐}
\mathrm{๐๐ด๐}
\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐}\right)
\mathrm{๐ป๐พ๐}\ge 0
\mathrm{๐ป๐พ๐}\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
\mathrm{๐๐ฟ}\ge \mathrm{๐ป๐พ๐}
\mathrm{๐๐ด๐}>0
\mathrm{๐๐ด๐}\ge \mathrm{๐ป๐พ๐}
\mathrm{๐๐ด๐}\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐},\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐ป๐๐ด๐},\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐ป๐๐ด๐},\mathrm{๐๐๐}\right)
Constrains all sequences of
\mathrm{๐๐ด๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
to take at least
\mathrm{๐ป๐พ๐}
values in
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\mathrm{๐๐ฟ}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\left(1,2,4,โฉ9,2,4,5,5,7,2โช,โฉ0,2,4,6,8โช\right)
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐}
constraint holds since the different sequences of 4 consecutive variables contains respectively 2, 2, 1 and 1 even numbers.
\mathrm{๐ป๐พ๐}<\mathrm{๐๐ด๐}
\mathrm{๐๐ฟ}>0
\mathrm{๐๐ด๐}>1
\mathrm{๐๐ด๐}<|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|>1
|\mathrm{๐
๐ฐ๐ป๐๐ด๐}|>0
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|>|\mathrm{๐
๐ฐ๐ป๐๐ด๐}|
\mathrm{๐ป๐พ๐}>0\vee \mathrm{๐๐ฟ}<\mathrm{๐๐ด๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\mathrm{๐ป๐พ๐}
\ge 0
\mathrm{๐๐ฟ}
\le \mathrm{๐๐ด๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐ฟ}=0
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐ด๐}=1
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐}
constraint occurs in many timetabling problems. As a typical example taken from [HoevePesantRousseauSabharwal06], consider for instance a nurse-rostering problem where each nurse can work at most 2 night shifts during every period of 7 consecutive days.
Beldiceanu and Carlsson [BeldiceanuCarlsson01] have proposed a first incomplete filtering algorithm for the
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐}
constraint. Later on, W.-J. van Hoeve et al. proposed two filtering algorithms [HoevePesantRousseauSabharwal06] establishing arc-consistency as well as an incomplete filtering algorithm based on dynamic programming concepts. In 2007 Brand et al. came up with a reformulation [BrandNarodytskaQuimperStuckeyWalsh07] that provides a complete filtering algorithm. One year later, Maher et al. use a reformulation in term of a linear program [MaherNarodytskaQuimperWalsh08] where (1) each coefficient is an integer in
\left\{-1,0,1\right\}
, (2) each column has a block of consecutive 1's or
-1
's. From this reformulation they derive a flow model that leads to an algorithm that achieves a complete filtering in
O\left({n}^{2}\right)
along a branch of the search tree.
sequence in Gecode, sequence in JaCoP.
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}
(single set of values replaced by individual values).
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐ }_\mathrm{๐๐}
\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐ }_\mathrm{๐๐}
constraint type: system of constraints, decomposition, sliding sequence constraint.
filtering: arc-consistency, linear programming, flow.
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐ด๐๐ป}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐ด๐}
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐ }_\mathrm{๐๐}
\left(\mathrm{๐ป๐พ๐},\mathrm{๐๐ฟ},\mathrm{๐๐๐๐๐๐๐๐๐๐},\mathrm{๐
๐ฐ๐ป๐๐ด๐}\right)
\mathrm{๐๐๐๐}
=|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-\mathrm{๐๐ด๐}+1
A constraint on sliding sequences of consecutive variables. Each vertex of the graph corresponds to a variable. Since they link
\mathrm{๐๐ด๐}
variables, the arcs of the graph correspond to hyperarcs. In order to link
\mathrm{๐๐ด๐}
consecutive variables we use the arc generator
\mathrm{๐๐ด๐๐ป}
. The constraint associated with an arc corresponds to the
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐ }_\mathrm{๐๐}
constraint defined at another entry of this catalogue.
\mathrm{๐๐ด๐๐ป}
arc generator with an arity of
\mathrm{๐๐ด๐}
on the items of the
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-\mathrm{๐๐ด๐}+1
\mathrm{๐๐๐๐}
=
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-\mathrm{๐๐ด๐}+1
\mathrm{๐๐๐๐}
\ge
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-\mathrm{๐๐ด๐}+1
\underline{\overline{\mathrm{๐๐๐๐}}}
\overline{\mathrm{๐๐๐๐}}
|
\boxed{\begin{array}{c|r:r:r:r:r} x & 0.1 & 0.01& 0.001& 0.0001& 0.00001\\ \hline \lfloor x \rfloor & 0 & 0 & 0 & 0 & 0 \end{array}}
By looking at the table above, is it true that as
x
approaches 0, then
\lfloor x \rfloor
approaches 0 as well?
\displaystyle \lim_{x\to0} \lfloor x\rfloor = 0
correct?
\lfloor \cdot \rfloor
It is correct It is not correct
\Large \lim_{x\to0} \dfrac{x}{x!} = \, ?
Note: Treat
x! = \Gamma(x+1)
by ุญุณู ุงูุนุทูุฉ
\lim_{x\to\infty} \dfrac{\sin x}x = \, ?
\displaystyle\lim_{x\rightarrow 3}
\dfrac{x^2-9}{x-3}
\large \lim_{x \to 1} \dfrac{|x-1|}{x-1} = \, ?
1 -1 0 Does not exist
|
Global Constraint Catalog: Csymmetric
<< 5.393. sum_squares_ctr5.395. symmetric_alldifferent >>
\mathrm{๐๐ข๐๐๐๐๐๐๐}\left(\mathrm{๐ฝ๐พ๐ณ๐ด๐}\right)
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐ก}-\mathrm{๐๐๐},\mathrm{๐๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฝ๐พ๐ณ๐ด๐},\left[\mathrm{๐๐๐๐๐ก},\mathrm{๐๐๐๐}\right]\right)
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐๐ก}\ge 1
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐๐ก}\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฝ๐พ๐ณ๐ด๐},\mathrm{๐๐๐๐๐ก}\right)
G
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
G
i
j
j
i
\left(\begin{array}{c}โฉ\begin{array}{cc}\mathrm{๐๐๐๐๐ก}-1\hfill & \mathrm{๐๐๐๐}-\left\{1,2,3\right\},\hfill \\ \mathrm{๐๐๐๐๐ก}-2\hfill & \mathrm{๐๐๐๐}-\left\{1,3\right\},\hfill \\ \mathrm{๐๐๐๐๐ก}-3\hfill & \mathrm{๐๐๐๐}-\left\{1,2\right\},\hfill \\ \mathrm{๐๐๐๐๐ก}-4\hfill & \mathrm{๐๐๐๐}-\left\{5,6\right\},\hfill \\ \mathrm{๐๐๐๐๐ก}-5\hfill & \mathrm{๐๐๐๐}-\left\{4\right\},\hfill \\ \mathrm{๐๐๐๐๐ก}-6\hfill & \mathrm{๐๐๐๐}-\left\{4\right\}\hfill \end{array}โช\hfill \end{array}\right)
\mathrm{๐๐ข๐๐๐๐๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
collection depicts a symmetric graph.
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|>2
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
The filtering algorithm for the
\mathrm{๐๐ข๐๐๐๐๐๐๐}
constraint is given in [Dooms06]. It removes (respectively imposes) the arcs
\left(i,j\right)
for which the arc
\left(j,i\right)
is not present (respectively is present). It has an overall complexity of
O\left(n+m\right)
n
m
respectively denote the number of vertices and the number of arcs of the initial graph.
\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐}_\mathrm{๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐ถ๐ฟ๐ผ๐๐๐ธ}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐}\mathtt{2}\right)
\mathrm{๐๐}_\mathrm{๐๐๐}
\left(\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐ก},\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐}\right)
\mathrm{๐๐๐ผ๐ผ๐ด๐๐๐ธ๐ฒ}
\mathrm{๐๐๐๐}
\mathrm{๐๐ข๐๐๐๐๐๐๐}
|
Compute aerodynamic forces and moments using aerodynamic coefficients, dynamic pressure, center of gravity, center of pressure, and velocity - Simulink - MathWorks Benelux
{C}_{sโb}=\left[\begin{array}{ccc}\mathrm{cos}\left(\alpha \right)& 0& \mathrm{sin}\left(\alpha \right)\\ 0& 1& 0\\ -\mathrm{sin}\left(\alpha \right)& 0& \mathrm{cos}\left(\alpha \right)\end{array}\right]
{C}_{wโs}=\left[\begin{array}{ccc}\mathrm{cos}\left(\beta \right)& \mathrm{sin}\left(\beta \right)& 0\\ -\mathrm{sin}\left(\beta \right)& \mathrm{cos}\left(\beta \right)& 0\\ 0& 0& 1\end{array}\right]
{C}_{wโb}=\left[\begin{array}{ccc}\mathrm{cos}\left(\alpha \right)\mathrm{cos}\left(\beta \right)& \mathrm{sin}\left(\beta \right)& \mathrm{sin}\left(\alpha \right)\mathrm{cos}\left(\beta \right)\\ -\mathrm{cos}\left(\alpha \right)\mathrm{sin}\left(\beta \right)& \mathrm{cos}\left(\beta \right)& -\mathrm{sin}\left(\alpha \right)\mathrm{sin}\left(\beta \right)\\ -\mathrm{sin}\left(\alpha \right)& 0& \mathrm{cos}\left(\alpha \right)\end{array}\right]
{F}_{A}^{w}\equiv \left[\begin{array}{c}-D\\ -C\\ -L\end{array}\right]={C}_{wโb}\cdot \left[\begin{array}{c}{X}_{A}\\ {Y}_{A}\\ {Z}_{A}\end{array}\right]\equiv {C}_{wโb}\cdot {F}_{A}^{b}
|
v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}
{v}_{\text{tot}}=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}
{v}_{\text{tot}}=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}
{v}_{\text{tot}}=\sqrt{({1.20}\text{ m/s})^{2} + ({0.750}\text{ m/s})^{2}}
\begin{array}{c}{{v}_{w}}=\sqrt{{{v}_{wx}}^{2}+{{v}_{wy}}^{2}} \\=\sqrt{({-13.0}\text{ m/s})^{2} + ({-9.29}\text{ m/s}^{2})}\end{array}
{{v}_{y}}^{2}={0}^{2}-2\left(9\text{.}\text{80}{\text{m/s}}^{2}\right)\left(-1\text{.}\text{50}\text{m}-0 m\right)=\text{29}\text{.}4{\text{m}}^{2}{\text{/s}}^{2}
v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}
v=\sqrt{({260\text{ m/s})}^{2}+({-5.42\text{ m/s}})^{2}}
v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}
{H}_{\text{average}}=\text{14}\text{.}\text{9}\frac{\text{km/s}}{\text{Mly}}
LAS-GenPhysics2_Q4_MELC_20-Week-7.pdf
PH 2 โข Iloilo Science and Technology University - formerly Western Visayas College of Science and Technolog
|
The adiabatic approximation | Less Than Epsilon
In time independent perturbation theory, we saw that a time dependent perturbation
H'(t)
can cause transitions
|\psi_a \rangle \rightarrow |\psi_b \rangle
For this to occur, frequency
\omega
of the perturbation had to match the spacing between those levels.
\omega_0 = \frac{E_b - E_a}{\hbar} \quad \text{(transition frequency)}
\omega \ll \frac{E_b - E_a}{\hbar}
This is a very slow perturbation, which is what we call the adiabatic limit. We expect system initially in eigen state
|\psi_a \rangle
will stay in that state (no transitions), but eigenstate itself will slowly change in time.
|\Psi(t) \rangle \sim |\Psi_a (t) \rangle
Example. Harmonic oscillator with time dependence
K(t) = K_0 \cos(\omega t)
Suppose system starts off in ground state at time
t=0
K(0) = K_0
\Psi(x, 0) = \psi_0(x) = \left(\frac{m \Omega}{\pi \hbar}\right)^{\frac{1}{4}} e^{-m \Omega_{0} x^{2} / 2 \hbar}
\Omega_{0}=\sqrt{\frac{k_{0}}{m}}
If spring constant changes very slowly,
(\omega \ll \Omega_0)
we expect the only effect is that
\Omega_0
becomes time dependent
\Omega_0 \rightarrow \Omega(t) = \sqrt{K(t)/m}
\Psi(x, t)=\left(\frac{m \Omega(t)}{\pi \hbar}\right)^{\frac{1}{4}} e^{-m \Omega(t) x^{2} / 2 \hbar}
If the time dependence is slow, the solution of the time dependent Schrodinger equation
i \hbar \frac{d|\Psi(t)\rangle}{d t}=H(t)|\Psi(t)\rangle
should be well approximated by a succession of eigenvalue problems.
H(t)\left|\Psi_{n}(t)\right\rangle= E_{n}(t)\left|\Psi_{n}(t)\right\rangle
Solving the time independent Schrodinger equation for each
t
|\Psi(t)\rangle \sim\left|\Psi_{n}(t)\right\rangle
|\Psi(0) \rangle=\left|\Psi_{n}(0)\right\rangle
H(t)
varies slowly in time with respect to level spacing, the system prepared in the
n
th eigenstate
|\Psi_n(0)\rangle
H(0)
will remain in the
n
th instantaneous eigenstate
|\Psi(t)
H(t)
, picking up only a phase factor.
\boxed{|\Psi(t) \rangle = e^{i\theta_n(t)}e^{i\gamma_n(t)}|\psi_n(t)\rangle}
\theta_n(t)
is the dynamical phase, with
E_n(t)
the instantaneous eigenenergies.
\theta_n(t) = -\frac{1}{\hbar} \int_{0}^{t} d t^{\prime} E_{n}\left(t^{\prime}\right)
\gamma_n(t)
is the geometrical phase.
\gamma_n(t) = i \int_0^t dt' \left\langle\psi_{n}\left(t^{\prime}\right)\left|\frac{\partial}{\partial t^{\prime}}\right|\psi_{n}\left(t^{\prime}\right)\right\rangle
Proof of the adiabatic theorem
|psi_n (\theta)\rangle
E_n(t)
via the time independent schrodinger equation,
H(t) |\psi_n \rangle = E_n(t) |\psi_n(t)\rangle
and treat
as a parameter and solve the time independent Schrodinger equation for each
t
t
\{\psi_n(t)\}
forms a complete orthonormal basis.
Expand solution
|Psi(t)\rangle
of full time dependent Schrodinger equation into that basis, for each
t
\Psi_{n}(t)=\psi_{n} e^{-i E_{n} t / h}
If the Hamiltonian changes with time, then the eigenfunctions and eigenvalues are time-dependant:
H(t) \psi_{n}(t)=E_{n}(t) \psi_{n}(t)
But, they are at least orthonormal and complete:
\left\langle \psi_n(t) | \psi_{m}(t)\right\rangle=\delta_{m n}
Hence, we can take our most general state, the solution to the time dependant Schrodinger equation
i \hbar \frac{\partial}{\partial t} \Psi(t)=H(t) \Psi(t)
and expand into the instantaneous basis:
\Psi(t)=\sum_{n} c_{n}(t) \psi_{n}(t) e^{i \theta_{n}(t)}
\theta_{n}(t) \equiv-\frac{1}{\hbar} \int_{0}^{t} E_{n}\left(t^{\prime}\right) d t^{\prime}
Substitute the instantaneous basis into the time dependent Schrodinger equation:
i \hbar \sum_{n}\left[\dot{c}_{n} \psi_{n}+c_{n} \dot{\psi}_{n}+i c_{n} \psi_{n} \dot{\theta}_{n}\right] e^{i \theta_{n}}=\sum_{n} c_{n}\left(H \psi_{n}\right) e^{i \theta_{n}}
\sum_{n} \dot{c}_{n} \psi_{n} e^{i \theta_{n}}=-\sum_{n} c_{n} \dot{\psi}_{n} e^{i \theta_{n}}
The inner product and orthonormality gives:
\sum_{n} \dot{c}_{n} \delta_{m n} e^{i \theta_{n}}=-\sum_{n} c_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle e^{i \theta_{n}}
\dot{c}_{m}(t)=-\sum_{n} c_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle e^{i\left(\theta_{n}-\theta_{m}\right)}
Now, take the time derivative of Schrodinger equation
\frac{\partial}{\partial t} H(t)\left|\psi_{n}(t)\right\rangle=\frac{\partial}{\partial l}\left(E_{n}(t)\left|\psi_{n}(t)\right\rangle\right)
\dot{H} \psi_{n}+H \dot{\psi}_{n}=\dot{E}_{n} \psi_{n}+E_{n} \dot{\psi}_{n}
Taking the inner product
\langle \psi_m (t) | \quad m\neq n
\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle+\left\langle\psi_{m}|H| \dot{\psi}_{n}\right\rangle=\dot{E}_{n} \delta_{m n}+E_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle
\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle=\left(E_{n}-E_{m}\right)\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle
\dot{c}_{m}(t)=-c_{m}\left\langle\psi_{m} | \dot{\psi}_{m}\right\rangle-\sum_{n \neq m} c_{n} \frac{\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle}{E_{n}-E_{m}} e^{(-i / \hbar)\int_{0}^{t}\left[E_{n}\left(t^{\prime}\right)-E_{u t}\left(t^{\prime}\right)\right] d t^{\prime}}
Now invoke the adiabatic approximation. Assume
\dot{H}
is extremely small, and drop the second term. We now have:
\dot{c}_{m}(t)=-c_{m}\left\langle\psi_{n} | \dot{\psi}_{m}\right\rangle
c_{m}(t)=c_{m}(0) e^{i \gamma_{m}(t)}
\gamma_{n}(t) \equiv i \int_{0}^{t}\left\langle\psi_{m}\left(t^{\prime}\right) | \frac{\partial}{\partial t^{\prime}} |\psi_{m}\left(t^{\prime}\right)\right\rangle d t^{\prime}
And so, the final answer is
\Psi_{n}(t)=e^{i \theta_{n}(t)} e^{i \gamma_{n}(t)} \psi_{n}(t)
Which is the statement of the adiabatic theorem.
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Global Constraint Catalog: Ctwo_orth_are_in_contact
<< 5.407. two_layer_edge_crossing5.409. two_orth_column >>
[Roach84], used for defining
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐ ๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐}\left(\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1},\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}\right)
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐},\mathrm{๐๐๐ฃ}-\mathrm{๐๐๐๐},\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}
|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}|>0
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐}
\left(2,\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด},\left[\mathrm{๐๐๐},\mathrm{๐๐๐ฃ},\mathrm{๐๐๐}\right]\right)
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}.\mathrm{๐๐๐ฃ}>0
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}.\mathrm{๐๐๐}\le \mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}.\mathrm{๐๐๐}
|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}|=|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}|
\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐๐ฃ}_\mathrm{๐๐๐}
\left(\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}\right)
\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐๐ฃ}_\mathrm{๐๐๐}
\left(\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}\right)
Enforce the following conditions on two orthotopes
{O}_{1}
{O}_{2}
i
, except one dimension, the projections of
{O}_{1}
{O}_{2}
i
have a non-empty intersection.
i
, the distance between the projections of
{O}_{1}
{O}_{2}
is equal to 0.
\left(\begin{array}{c}โฉ\mathrm{๐๐๐}-1\mathrm{๐๐๐ฃ}-3\mathrm{๐๐๐}-4,\mathrm{๐๐๐}-5\mathrm{๐๐๐ฃ}-2\mathrm{๐๐๐}-7โช,\hfill \\ โฉ\mathrm{๐๐๐}-3\mathrm{๐๐๐ฃ}-2\mathrm{๐๐๐}-5,\mathrm{๐๐๐}-2\mathrm{๐๐๐ฃ}-3\mathrm{๐๐๐}-5โช\hfill \end{array}\right)
Figure 5.408.1 shows the two rectangles of the example. The
\mathrm{๐๐ ๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐}
constraint holds since the two rectangles are in contact: the contact is depicted by a pink line-segment.
Figure 5.408.1. The two rectangles that are in contact of the Example slot where the contact is shown in pink
|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}|>1
\left(\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1},\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}\right)
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}
\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐ ๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
geometry: geometrical constraint, touch, contact, non-overlapping, orthotope.
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
\left(=\right)โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}\right)
โข\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}>\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}
โข\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}>\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}
\mathrm{๐๐๐๐}
=|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}|-1
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
\left(=\right)โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}\right)
\mathrm{๐๐๐ก}\left(\begin{array}{c}0,\begin{array}{c}\mathrm{๐๐๐ก}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐},\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}\right)-\hfill \\ \mathrm{๐๐๐}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐},\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}\right)\hfill \end{array}\hfill \end{array}\right)=0
\mathrm{๐๐๐๐}
=|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}|
\mathrm{๐๐๐๐}
graph property, the unique arc of the final graph is stressed in bold. It corresponds to the fact that the projection onto dimension 1 of the two rectangles of the example overlap.
\mathrm{๐๐ ๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐}
Consider the second graph constraint. Since we use the arc generator
\mathrm{๐๐
๐๐ท๐๐ถ๐}\left(=\right)
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}
, and because of the restriction
|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}|=|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}|
, the maximum number of arcs of the corresponding final graph is equal to
|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}|
\mathrm{๐๐๐๐}=|\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}|
\mathrm{๐๐๐๐}\ge |\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}|
\underline{\overline{\mathrm{๐๐๐๐}}}
\overline{\mathrm{๐๐๐๐}}
\mathrm{๐๐ ๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐พ๐๐ธ}{\mathtt{1}}_{i}
\mathrm{๐๐ธ๐}{\mathtt{1}}_{i}
\mathrm{๐ด๐ฝ๐ณ}{\mathtt{1}}_{i}
\mathrm{๐๐๐}
\mathrm{๐๐๐ฃ}
\mathrm{๐๐๐}
{i}^{th}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{1}
\mathrm{๐พ๐๐ธ}{\mathtt{2}}_{i}
\mathrm{๐๐ธ๐}{\mathtt{2}}_{i}
\mathrm{๐ด๐ฝ๐ณ}{\mathtt{2}}_{i}
\mathrm{๐๐๐}
\mathrm{๐๐๐ฃ}
\mathrm{๐๐๐}
{i}^{th}
\mathrm{๐พ๐๐๐ท๐พ๐๐พ๐ฟ๐ด}\mathtt{2}
collection. To each sextuple
\left(\mathrm{๐พ๐๐ธ}{\mathtt{1}}_{i},\mathrm{๐๐ธ๐}{\mathtt{1}}_{i},\mathrm{๐ด๐ฝ๐ณ}{\mathtt{1}}_{i},\mathrm{๐พ๐๐ธ}{\mathtt{2}}_{i},\mathrm{๐๐ธ๐}{\mathtt{2}}_{i},\mathrm{๐ด๐ฝ๐ณ}{\mathtt{2}}_{i}\right)
{S}_{i}
\left\{0,1,2\right\}
\left(\left(\mathrm{๐๐ธ๐}{\mathtt{1}}_{i}>0\right)\wedge \left(\mathrm{๐๐ธ๐}{\mathtt{2}}_{i}>0\right)\wedge \left(\mathrm{๐ด๐ฝ๐ณ}{\mathtt{1}}_{i}>\mathrm{๐พ๐๐ธ}{\mathtt{2}}_{i}\right)\wedge \left(\mathrm{๐ด๐ฝ๐ณ}{\mathtt{2}}_{i}>\mathrm{๐พ๐๐ธ}{\mathtt{1}}_{i}\right)\right)โ{S}_{i}=0
\left(\left(\mathrm{๐๐ธ๐}{\mathtt{1}}_{i}>0\right)\wedge \left(\mathrm{๐๐ธ๐}{\mathtt{2}}_{i}>0\right)\wedge \left(\mathrm{๐ด๐ฝ๐ณ}{\mathtt{1}}_{i}=\mathrm{๐พ๐๐ธ}{\mathtt{2}}_{i}\vee \mathrm{๐ด๐ฝ๐ณ}{\mathtt{2}}_{i}=\mathrm{๐พ๐๐ธ}{\mathtt{1}}_{i}\right)\right)โ{S}_{i}=1
\mathrm{๐๐ ๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐ ๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐}
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P and Q are points on the sides AB and AC of a ABC such that BP=CQ=x, PA= 6 cm,AQ= - Maths - Circles - 10762453 | Meritnation.com
P and Q are points on the sides AB and AC of a ABC such that BP=CQ=x, PA= 6 cm,AQ= 20 cm, BC=25 cm. If PAQ and quadrilateral BPQC have equal areas, then the value of x
\mathrm{We} \mathrm{know} \mathrm{that},\phantom{\rule{0ex}{0ex}}\mathrm{ar}\left(\mathrm{APQ}\right)+\mathrm{ar}\left(\mathrm{BPQC}\right)=\mathrm{ar}\left(\mathrm{ABC}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Therefore}, 2\left(\mathrm{APQ}\right)=\mathrm{ABC} \left[\because \mathrm{ar}\left(\mathrm{APQ}\right)=\mathrm{ar}\left(\mathrm{BPQC}\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{Applying} \mathrm{the} \mathrm{sine} \mathrm{formula}, \mathrm{this} \mathrm{is} \mathrm{equivalent} \mathrm{to} \phantom{\rule{0ex}{0ex}}2\left[\frac{1}{2}ร20ร6ร\mathrm{sinA}\right]=\frac{1}{2}\left(\mathrm{x}+20\right)\left(\mathrm{x}+6\right)\mathrm{sinA}\phantom{\rule{0ex}{0ex}}โ240\mathrm{sinA}=\left(\mathrm{x}+20\right)\left(\mathrm{x}+6\right)\mathrm{sinA}\phantom{\rule{0ex}{0ex}}โ240={\mathrm{x}}^{2}+26\mathrm{x}+120\phantom{\rule{0ex}{0ex}}โ{\mathrm{x}}^{2}+26\mathrm{x}-120=0\phantom{\rule{0ex}{0ex}}โ{\mathrm{x}}^{2}+30\mathrm{x}-4\mathrm{x}-120=0\phantom{\rule{0ex}{0ex}}โ\mathrm{x}\left(\mathrm{x}+30\right)-4\left(\mathrm{x}+30\right)=0\phantom{\rule{0ex}{0ex}}โ\left(\mathrm{x}+30\right)\left(\mathrm{x}-4\right)=0\phantom{\rule{0ex}{0ex}}\mathrm{And} \mathrm{since} \mathrm{x}>0 \mathrm{the} \mathrm{only} \mathrm{solution} \mathrm{is} \mathrm{x}=4.
Sarthak Ray answered this
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A Taylor series approximation uses a Taylor series to represent a number as a polynomial that has a very similar value to the number in a neighborhood around a specified
x
f(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots.
If only concerned about the neighborhood very close to the origin, the
n=2
approximation represents the sine wave sufficiently, and no higher orders are direly needed.[1]
a
to be a number that makes
f(a)
easy to compute.
x
f(x)
the number being approximated.
Using the first three terms of the Taylor series expansion of
f(x) = \sqrt[3]{x}
x = 8
\sqrt[3]{8.1}:
f(x) = \sqrt[3]{x} \approx 2 + \frac{(x - 8)}{12} - \frac{(x - 8)^2}{288} .
The first three terms shown will be sufficient to provide a good approximation for
\sqrt[3]{x}
. Evaluating this sum at
x = 8.1
gives an approximation for
\sqrt[3]{8.1}:
\begin{aligned} f(8.1) = \sqrt[3]{8.1} &\approx 2 + \frac{(8.1 - 8)}{12} - \frac{(8.1 - 8)^2}{288} \\ &=\color{#3D99F6}{2.008298}\color{#D61F06}{61111}\ldots \\ \\ \sqrt[3]{8.1} &={ \color{#3D99F6}{2.008298}\color{#D61F06}{85025}\dots}. \end{aligned}
With just three terms, the formula above was able to approximate
\sqrt[3]{8.1}
to six decimal places of accuracy.
_\square
Using the quadratic Taylor polynomial for
f(x) = \frac{1}{x^2},
\frac{1}{4.41}.
P_2(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2.
f(x) = \frac{1}{x^2},\quad f'(x) = \frac{-2}{x^3},\quad f''(x) = \frac{6}{x^4}.
But what about
and
x?
a
so that the values of the derivatives are easy to calculate. Rewriting the approximated value as
4.41 = (2+0.1)^2
a = 2
x = 2.1.
\begin{aligned} P_2(2.1) &= f(2)+\frac {f'(2)}{1!} (2.1-2)+ \frac{f''(2)}{2!} (2.1-2)^2\\ &= \frac14 +\frac {\hspace{3mm} \frac{-2}{8}\hspace{3mm} }{1!} (2.1-2)+ \frac{\hspace{3mm} \frac{6}{16}\hspace{3mm} }{2!} (2.1-2)^2 \\ &= \frac14 + \frac {-1}{4}(0.1) + \frac{3}{16}(0.01)\\ &= 0.25 - 0.025 + 0.001875 \\ &= 0.226875. \end{aligned}
\frac{1}{4.41} = 0.226757...,
so the approximation is only off by about 0.05%.
_\square
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Global Constraint Catalog: Cused_by
<< 5.411. two_orth_include5.413. used_by_interval >>
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1},\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}\right)
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}|\ge |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1},\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2},\mathrm{๐๐๐}\right)
All the values of the variables of collection
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
are used by the variables of collection
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
\left(โฉ1,9,1,5,2,1โช,โฉ1,1,2,5โช\right)
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
constraint holds since, for each value occurring within the collection
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}=โฉ1,1,2,5โช
, its number of occurrences within
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}=โฉ1,9,1,5,2,1โช
is greater than or equal to its number of occurrences within
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
Value 1 occurs 3 times within
โฉ1,9,1,5,2,1โช
and 2 times within
โฉ1,1,2,5โช
โฉ1,9,1,5,2,1โช
โฉ1,1,2,5โช
โฉ1,9,1,5,2,1โช
โฉ1,1,2,5โช
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
{U}_{1}\in \left\{1,5\right\}
{U}_{2}\in \left[1,2\right]
{U}_{3}\in \left[1,2\right]
{V}_{1}\in \left[0,2\right]
{V}_{2}\in \left[2,4\right]
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
\left(โฉ{U}_{1},{U}_{2},{U}_{3}โช,โฉ{V}_{1},{V}_{2}โช\right)
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}|>1
\mathrm{๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}.\mathrm{๐๐๐}\right)>1
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|>1
\mathrm{๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}.\mathrm{๐๐๐}\right)>1
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}.\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}\left(\mathrm{๐๐๐๐๐}\right)
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}\left(\mathrm{๐๐๐๐๐}\right)
As described in [BeldiceanuKatrielThiel04a] we can pad
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
with dummy variables such that its cardinality will be equal to that cardinality of
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
. The domain of a dummy variable contains all of the values. Then, we have a
\mathrm{๐๐๐๐}
constraint between the two sets. Direct arc-consistency and bound-consistency algorithms based on a flow model are also proposed in [BeldiceanuKatrielThiel04a], [BeldiceanuKatrielThiel05], [Katriel05]. The leftmost part of Figure 3.7.30 illustrates this flow model.
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
constraint to var-perfect matchingsA var-perfect matching is a maximum matching covering all vertices corresponding to the variables of
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
. in a bipartite intersection graph derived from the domain of the variables of the constraint in the following way. To each variable of the
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
\left(โฉ\mathrm{๐๐๐}-{U}_{1}\mathrm{๐๐๐}-{U}_{2},\cdots ,\mathrm{๐๐๐}-{U}_{|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}|}โช,โฉ\mathrm{๐๐๐}-{V}_{1}\mathrm{๐๐๐}-{V}_{2},\cdots ,\mathrm{๐๐๐}-{V}_{|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|}โช\right)
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|
reified constraints of the form:
{\sum }_{1\le j\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}|}\left({V}_{i}={U}_{j}\right)\ge {\sum }_{1\le j\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|}\left({V}_{i}={V}_{j}\right)
\left(i\in \left[1,|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|\right]\right)
\mathrm{๐๐๐}_\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐}
๐_\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}_\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}/\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}\mathrm{mod}\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}\in \mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐ข}_\mathrm{๐๐๐}
๐_\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
combinatorial object: multiset.
filtering: flow, bipartite matching, arc-consistency, bound-consistency, DFS-bottleneck.
modelling: inclusion.
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}\right)
\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}=\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}
โข\text{for}\text{all}\text{connected}\text{components:}
\mathrm{๐๐๐๐๐๐๐}
\ge
\mathrm{๐๐๐๐๐}
โข
\mathrm{๐๐๐๐๐}
=|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|
\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐}
graph properties, the source and sink vertices of the final graph are stressed with a double circle. Since there is a constraint on each connected component of the final graph we also show the different connected components. Each of them corresponds to an equivalence class according to the arc constraint. Note that the vertex corresponding to the variable assigned to value 9 was removed from the final graph since there is no arc for which the associated equality constraint holds. The
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
For each connected component of the final graph the number of sources is greater than or equal to the number of sinks.
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
Since the initial graph contains only sources and sinks, and since sources of the initial graph cannot become sinks of the final graph, we have that the maximum number of sinks of the final graph is equal to
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}2|
\mathrm{๐๐๐๐๐}=|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}2|
\mathrm{๐๐๐๐๐}\ge |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}2|
\underline{\overline{\mathrm{๐๐๐๐๐}}}
\overline{\mathrm{๐๐๐๐๐}}
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}
{S}_{i}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
{S}_{i+|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}|}
\mathrm{๐๐๐๐}_\mathrm{๐๐ข}
|
Matt DeCross, ๅฑ่ฑช ๅผต, Eli Ross, and
The small-angle approximation is the term for the following estimates of the basic trigonometric functions, valid when
\theta \approx 0:
\sin \theta \approx \theta, \qquad \cos \theta \approx 1 - \frac{\theta^2}{2} \approx 1, \qquad \tan \theta \approx \theta.
These estimates are widely used throughout mathematics and the physical sciences to simplify equations and make problems analytically tractable. For instance, solving a differential equation that looks like
\ddot{\theta} + \sin (\theta) = 0
is much trickier than solving its small-angle approximation
\ddot{\theta} + \theta= 0,
and the solutions to the latter are much more useful than the solutions to the former:
\sin(x)
and its small-angle approximation near
x=0
\cos(x)
x=0
Derivations of the Approximation
The small-angle approximations can be derived geometrically without the use of calculus. Consider the below diagram of a right triangle with one side tangent to a circle:
A right triangle with two sides formed from the radii of a circle and the third side tangent to the circle.
As long as the angle
\theta
is sufficiently small, the length of
s
(
the arc subtended by
\theta)
is very close to that of
s^{\prime}
, the third side of the triangle. The small-angle approximation thus corresponds to
s\approx s^{\prime}
in this diagram.
From the law of sines,
\frac{s^{\prime}}{\sin \theta} = \frac{r}{\sin 90^{\circ}} = r \implies s^{\prime} = r\sin \theta.
From the formula for arc length,
s = r \theta
. Comparing the expression for
s
to the expression for
s^{\prime}
, the small-angle approximation
s\approx s^{\prime}
must correspond to
\sin \theta \approx \theta.
Using the double angle identity
\cos 2 x = 1-2\sin^2 x
x = \frac{\theta}{2}
and substituting in
\sin \theta \approx \theta
, one obtains the small-angle approximation for cosine:
\cos \theta = 1 -2\sin^2 \frac{\theta}{2} \approx 1 - 2\left(\frac{\theta}{2}\right)^2 = 1 - \frac{\theta^2}{2} \approx 1.
The most straightforward way of deriving the small-angle approximations uses the calculus technique of Taylor series approximation for both sine and cosine:
\begin{aligned} \sin \theta &= \sum_{n=0}^{\infty} \frac{(-1)^n \theta^{2n+1}}{(2n+1)!} = \theta - \frac{\theta^3}{3!} + \mathcal{O} (\theta^5) \\ \cos \theta &= \sum_{n=0}^{\infty} \frac{(-1)^n \theta^{2n}}{(2n)!} = 1 - \frac{\theta^2}{2!} + \mathcal{O} (\theta^4). \end{aligned}
The small-angle approximations correspond to the low-order approximations of these Taylor series, as can be seen from the expansions above.
Percent errors for each of the small-angle approximations
\sin(x) \approx x
\cos (x) \approx 1
\tan (x) \approx x
. For very small angles
(x<0.1),
the approximation is excellent and the error is very small.
What is the small-angle approximation to
f(x) = \sec x \tan x?
The Taylor series for this function is
f(x) = x + \frac{5}{6} x^3 + \frac{61}{120} x^5 + \mathcal{O} \big(x^7\big).
The lowest-order small-angle approximation is therefore
f(x) \approx x
. This is consistent with substituting in the small-angle approximations for the sine and cosine functions, which gives
\sec x \approx 1, \tan x \approx x \implies \sec x \tan x \approx x.\ _\square
Find the small-angle approximation to
f(x) = \frac{\sec x \tan x - \csc^2 x}{\cot^2 x}.
Naively substituting
\sin (x) \approx x
\cos x \approx 1
, one finds the approximation
f(x) = \frac{x - \frac{1}{x^2}}{\frac{1}{x^2}} = x^3 - 1.
Using the improved approximation for the cosine function,
\cos x \approx 1 - \frac{x^2}{2}
f(x) = \frac{\frac{x}{\left(1-\frac{x^2}{2}\right)^2} - \frac{1}{x^2}}{\frac{\left(1-\frac{x^2}{2}\right)^2}{x^2}} = -1-x^2+x^3 -\frac{3}{4} x^4 \mathcal{O} \big(x^5\big).
If we instead Taylor expand the given function
f(x)
directly, we obtain
f(x) \approx -1 - x^2 + x^3 - \frac23 x^4 + \mathcal{O} \big(x^5\big),
which agrees with the improved approximation to the first three orders.
Plotting the three approximations, we see that the naive approximation does quite poorly, while the improved approximation is much better. The improved approximation for cosine added a lower-order term that was missed by the naive approximation, illustrating the dangers involved in naively approximating when dividing by functions. For the best accuracy, the Taylor series of
f(x)
itself should be used as opposed to the series for sine and cosine separately:
Small-angle approximations to
f(x)
based on different truncations of the Taylor series for cosine. The refined truncation is necessary to obtain any useful precision for small angles.
-\frac{\theta^3}{3}
-\frac{\theta^3}{6}
\frac{\theta^3}{6}
\frac{\theta^2}{2}
What is the lowest-order correction term to the small-angle approximation for the sine function
\sin \theta\approx \theta?
At what angle in radians to the nearest thousandth
\big(
[0,2\pi)\big)
does the relative error between
\tan \theta
and its lowest-order small-angle approximation exceed 5%?
The small-angle approximation is used ubiquitously throughout fields of physics including mechanics, waves and optics, electromagnetism, astronomy, and more. Below, a few well-known examples are explored to illustrate why the small-angle approximation is useful in physics.
Small oscillations of a simple pendulum are best modeled using the small-angle approximation.
The small oscillations of a simple pendulum are a basic example in mechanics where the small-angle approximation is absolutely essential to making any useful analytic progress. From the rotational form of Newton's second law, the torque
\tau
on a pendulum of mass
m
from gravity as it oscillates about a pivot point on a string of length
\ell
\tau = I \alpha \implies -\ell mg \sin \theta = m\ell^2 \ddot{\theta} \implies \ddot{\theta} + \frac{g}{\ell} \sin \theta = 0,
\theta
is the angle between the string and the vertical.
The solutions to this equation of motion can be found in terms of functions called elliptic integrals, which are difficult to work with by hand. However, employing the small-angle approximation,
\ddot{\theta} + \frac{g}{\ell} \sin \theta = 0 \implies \ddot{\theta} + \frac{g}{\ell} \theta = 0.
The new differential equation is easily solvable. The solutions go as
\theta(t) = A\cos \left(\sqrt{\frac{g}{\ell}} t\right)+B\sin\left(\sqrt{\frac{g}{\ell}} t\right)
A
B
depending on initial conditions, successfully reproducing the oscillatory behavior of the pendulum.
Angular Distance in Astronomy
The size or distance between celestial bodies in astronomy is typically written in terms of the angular diameter or apparent size, i.e. the angle
\theta
between the two bodies as seen from Earth. If the separation between two faraway points is
d
and the midpoint between the points is at distance
D
from Earth, then this angle obeys the relationship
\tan \frac{\theta}{2} = \frac{d}{2D}.
The diagram corresponding to this formula is below:
Angular separation between two faraway stars as seen from Earth
Using the small-angle approximation, the angular distance can be rewritten as
\theta = \frac{d}{D}.
The approximation is useful because typically the angular distance is the easiest to measure in astronomy and the difference between angles is so small that the angle itself is more useful than the sine.
There are 60 arcminutes in a degree, and the sun has an angular diameter of approximately
32
arcminutes. Knowing that the sun is about
8
light-minutes away from Earth, estimate the diameter of the sun.
8
light-minutes into the earth-sun distance
D:
D = ct = (3 \times 10^{8} \text{ m}/\text{s})(8\times 60 \text{ s}) = 1.44 \times 10^{11} \text{ m}.
Using the formula for angular distance in the small-angle approximation,
d \approx \theta D = (32 \text{ arcmin})\left(\frac{1^{\circ}}{60 \text{ arcmin}}\right) \left(\frac{2\pi \text{ radians}}{360^{\circ}}\right) \left(1.44 \times 10^{11} \text{ m}\right) = 1.34 \times 10^9 \text{ m},
which is accurate to less than
5 \%
_\square
8.4 \times 10^5
3.2 \times 10^6
3.7 \times 10^8
1.3 \times 10^9
The angular diameter of the moon as seen from Earth is about
30 \text{ arcmin}
when the moon is at a distance of
3.7 \times 10^8 \text{ m}
from Earth. Which of the following is the best approximation for the diameter of the moon in meters?
In single-slit diffraction, light passing through a barrier with a slit larger than one wavelength of the light has an intensity profile measured behind the barrier which exhibits a characteristic pattern of peaks and troughs. The condition for a minimum in this intensity distribution is
d \sin \theta = m \lambda,
d
is the slit width,
\theta
is the angle to the point of measurement from the center of the slit,
\lambda
is the wavelength of the light, and
m
is a nonzero integer.
This condition can be rewritten in terms of the vertical distance from the center of the measurement screen,
y
, as shown in the above diagram. Suppose the measurement screen is a distance of
D
from the barrier. Typically,
D
is taken to be much greater than
d
, and the small-angle approximation for
\theta
can be used. Then the formula for intensity minima becomes
y = \frac{m \lambda D}{d},
a convenient expression in terms of the wavelength of the light, width of the slit, and distance from the barrier to the screen.
\frac{\lambda a}{d}
\frac{2\lambda d}{a}
\frac{2\lambda a}{d}
\frac{\lambda d}{a}
In a single-slit diffraction experiment with slit width
a
, distance
d
between barrier and measurement screen, and light of wavelength
\lambda
, what is the width of the central maximum peak?
Cite as: Small-Angle Approximation. Brilliant.org. Retrieved from https://brilliant.org/wiki/small-angle-approximation/
|
Matrix Factorization - Fizzy
\begin{array}{c|cc} & m_1 & m_2 & m_3 & m_4 \\ \hline u_1 & & w_{12}^{um} & &\\ u_2 & w_{21}^{um} & & &\\ u_3 & & & w_{32}^{um} & \end{array}
How do we compute matrix factorizations?
\min _{W, Z}\left\|W Z^{T}-X\right\|_{\mathrm{Fro}}^{2}
||\cdot||_{Fro}
denotes the so-called Frobenius norm, i.e.:
\|X\|_{\mathrm{Fro}} :=\sqrt{\sum_{i=1}^{s} \sum_{j=1}^{n}\left|x_{i j}\right|^{2}}
Then we can add regularization to the model for
W
Z
\min _{W, Z}\left\|W Z^{T}-X\right\|_{\mathrm{Fro}}^{2}+R_{1}(W)+R_{2}(Z)
Here we need to choose suitable regularization functions
R_1
R_2
W\cdot Z^{T}
is equal to a value with more than one solution, thus this optimization problem is not a convex problem. We could use coordinate descent also called alternating minimization to solve this problem.
Coordinate Descent / Alternating Minimization
keep one fixed. ๅ
ถๅบๆฌๆ่ทฏๆฏไธๆฌกไผๅไธไธชๅๆฐ๏ผๅๆ ๏ผ๏ผ่ฝฎๆตๅพช็ฏ๏ผๅฐๅคๆไผๅ้ฎ้ขๅ่งฃไธบ็ฎๅ็ๅญ้ฎ้ข
W^{k+1}=\arg \min _{W}\left\|W\left(Z^{k}\right)^{T}-X\right\|^{2}
Z^{k+1}=\arg \min _{Z}\left\|W^{k+1} Z^{T}-X\right\|^{2}
\Rightarrow \left\{ \begin{array}{lr} W^{k+1}=X Z^{k}\left(\left(Z^{k}\right)^{T} Z^{k}\right)^{-1} & \\ Z^{k+1}=\left(\left(\left(W^{k+1}\right)^{T} W^{k+1}\right)^{-1}\left(W^{k+1}\right)^{T} X\right)^{T} & \end{array} \right.
We can obtain other kinds of matrix factorizations. A very popular one is the singular value decomposition:
ๅฅๅผๅผๅ่งฃ (Singular Value Decomposition)๏ผๅฎ่ฝๅฐไปปๆ็ฉ้ตๅ่งฃไธบไธคไธชๆญฃไบค้ตๅไธไธชๅฏน่ง้ต๏ผๅนถๆญ็คบๅบ็ฉ้ต็่ฎธๅคๆง่ดจใ
Every matrix
X \in \mathbb{R}^{s \times n}
X=U \Sigma V^{T}
U \in \mathbb{R}^{s \times s}
is unitary matrix, i.e.
U^{T} U=I
V \in \mathbb{R}^{n \times n}
V^{T} V=I
\Sigma= \left( \begin{array}{lllllll}{ \sigma_{1}} & {0} & {\ldots} & {0} & {0} & {\ldots} & {0} \\ {0} & \sigma_{2} & {\ldots} & {0} & {0} & {\ldots} & {0} \\ {\vdots} & {\vdots} & {\ddots} & {0} & {0} & {\ldots} & {0} \\ {0} & {0} & {\ldots} & \sigma_{s} & {0} & {\ldots} & {0} \\ \end{array} \right) \in \mathbb{R}^{s \times n}=\text { diagonal } \quad \text{for s less than n}
\sigma_1
is the largest,
\sigma_s
\Sigma= \left( \begin{array}{llll}{ \sigma_{1}} & {0} & {\ldots} & {0} \\ {0} & {\sigma_{2}} & {\ldots} & {0} \\ {0} & {0} & {\ddots} & {0} \\ {0} & {0} & {\ldots} & {\sigma_{n}} \\ {0} & {0} & {\ldots} & {0} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {0} & {0} & {\ldots} & {0} \end{array} \right) \in \mathbb{R}^{s \times n}=\text { diagonal } \quad \text{for} s>n
Now for unitary matrices, we have:
\|U x\|^{2}=\langle U x, U x\rangle=\left\langle x, {U}^{T} U x\right\rangle=\langle x, x\rangle=\|x\|^{2}
i.e., rotations do not affect the Euclidean norm (Frobenius norm)
\|X V w\|^{2}=\left\|U \Sigma V^{T} V w\right\|^{2}=\langle U \Sigma w, U \Sigma w\rangle==\langle\Sigma w, \Sigma w\rangle=\sum_{i=1}^{S} \sigma_{i}^{2}\left|w_{i}\right|^{2}
Principal Component Analysis is another matrix factorization algorithm which is similar to SVD.
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Mean predictive measure of association for surrogate splits in regression tree - MATLAB - MathWorks รญโขลรชยตยญ
ma = 3รโ3
{\mathrm{รยป}}_{jk}=\frac{\text{min}\left({P}_{L},{P}_{R}\right)รขยย\left(1รขยย{P}_{{L}_{j}{L}_{k}}รขยย{P}_{{R}_{j}{R}_{k}}\right)}{\text{min}\left({P}_{L},{P}_{R}\right)}.
{P}_{{L}_{j}{L}_{k}}
{P}_{{R}_{j}{R}_{k}}
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Global Constraint Catalog: Cin_relation
<< 5.180. in_intervals5.182. in_same_partition >>
Constraint explicitly defined by tuples of values.
\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐},\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}\right)
\mathrm{๐๐๐๐}
\mathrm{๐๐ก๐๐๐๐๐๐๐}
\mathrm{๐๐ก๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐ก๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐ก๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐ก๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐ก๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐}\right)
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐๐}
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}-\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐๐},\mathrm{๐๐๐}\right)
|\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐๐}|\ge 1
|\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}|\ge 1
|\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}|=|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐},\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐},\mathrm{๐๐๐๐๐}\right)
Enforce the tuple of variables
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
to take its value out of a set of tuples of values
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}
. The value of a tuple of variables
โฉ{V}_{1},{V}_{2},\cdots ,{V}_{n}โช
is a tuple of values
โฉ{U}_{1},{U}_{2},\cdots ,{U}_{n}โช
{V}_{1}={U}_{1}\wedge {V}_{2}={U}_{2}\wedge \cdots \wedge {V}_{n}={U}_{n}
\left(\begin{array}{c}โฉ5,3,3โช,\hfill \\ โฉ\mathrm{๐๐๐๐๐}-โฉ5,2,3โช,\mathrm{๐๐๐๐๐}-โฉ5,2,6โช,\mathrm{๐๐๐๐๐}-โฉ5,3,3โชโช\hfill \end{array}\right)
\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
โฉ5,3,3โช
corresponds to the third item of the collection of tuples
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}
|\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐๐}|>1
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}.\mathrm{๐๐๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}.\mathrm{๐๐๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}.\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}
Quite often some constraints cannot be easily expressed, neither by a formula, nor by a regular pattern. In this case one has to define the constraint by specifying in extension the combinations of allowed values.
\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐ก๐๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
in JaCoP (http://www.jacop.eu/). Within SICStus Prolog the constraint can be applied to more than a single tuple of variables and is called
\mathrm{๐๐๐๐๐}
. Within [BourdaisGalinierPesant03] this constraint is called
\mathrm{๐๐ก๐๐๐๐๐๐๐}
\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐}
in MiniZinc (http://www.minizinc.org/).
feasPairAC in Choco, infeasPairAC in Choco, relationPairAC in Choco, feasTupleAC in Choco, infeasTupleAC in Choco, relationTupleAC in Choco, extensional in Gecode, extensionalsupportVA in JaCoP, extensionalsupportMDD in JaCoP, extensionalsupportSTR in JaCoP, table in MiniZinc, case in SICStus, relation in SICStus, table in SICStus.
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐ฒ๐พ๐๐}
\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐}
characteristic of a constraint: tuple, derived collection.
constraint type: data constraint, extension.
\mathrm{๐๐๐}\left(\begin{array}{c}\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐๐}-\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐๐}\right),\hfill \\ \mathrm{๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\right)\right]\hfill \end{array}\right)
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐๐}
\mathrm{๐๐๐ฟ๐ป๐ด๐}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐},\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}\right)
\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐}
\left(\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}.\mathrm{๐๐๐},\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}.\mathrm{๐๐๐๐๐}\right)
\mathrm{๐๐๐๐}
\ge 1
\mathrm{๐๐๐๐}
\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
|
Global Constraint Catalog: Kconfiguration_problem
<< 3.7.50. Conditional constraint3.7.52. Connected component >>
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
A constraint that was used for modelling configuration problems. Within the context of configuration problems [Subbarayan05], it is crucial to identify all variable-value pairs which do not participate to any solution. This stems from the fact that one wants typically to avoid proposing invalid choices to the user of such configuration systems.
Note also that open constraints are also useful in the context of configuration problems.
|
quadrilateral ABCD is circumscribed to a circle with centre o if angle AOB = 115, then angle cod is - Maths - Circles - 7084218 | Meritnation.com
quadrilateral ABCD is circumscribed to a circle with centre o.if angle AOB = 115, then angle cod is?
\text{We know that, Opposite sides subtend supplementary angles at the centre of inscribed circle.}\phantom{\rule{0ex}{0ex}}\text{From The figure,}\phantom{\rule{0ex}{0ex}}\angle AOB+\angle COD={180}^{0}\phantom{\rule{0ex}{0ex}}\Rightarrow {115}^{0}+\angle COD={180}^{0}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle COD={180}^{0}-{115}^{0}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle COD={65}^{0}\phantom{\rule{0ex}{0ex}}Hence,\quad \quad \angle COD\quad is\quad {65}^{0}.
Saachi Awate answered this
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Linear Dependence of Vectors and Matrix Rank - Fizzy
Consider the row vectors below:
\mathbf{a}=\left[ \begin{array}{lll}{1} & {2} & {3}\end{array}\right] \quad \mathbf{d}=\left[ \begin{array}{lll}{2} & {4} & {6}\end{array}\right]
\mathbf{b}=\left[\begin{array}{lll}{4} & {5} & {6}\end{array}\right] \quad \mathbf{e}=\left[ \begin{array}{lll}{0} & {1} & {0}\end{array}\right]
\mathbf{c}=\left[ \begin{array}{lll}{5} & {7} & {9}\end{array}\right] \quad \mathbf{f}=\left[ \begin{array}{lll}{0} & {0} & {1}\end{array}\right]
You can think of an
r \times c
matrix as a set of r row vectors, each having c elements; or you can think of it as a set of c column vectors, each having r elements.
the maximum number of linearly independent column vectors in the matrix
\mathbb{R}^{r \times c}
r < c
, then the maximum rank of the matrix is r.
r > c
, then the maximum rank of the matrix is c.
Extended / Augmented Matrix
\begin{aligned} x+y+2 z &=3 \\ x+y+z &=1 \\ 2 x+2 y+2 z &=2 \end{aligned}
A=\left[ \begin{array}{lll}{1} & {1} & {2} \\ {1} & {1} & {1} \\ {2} & {2} & {2}\end{array}\right]
(A | B)=\left[ \begin{array}{lll|l}{1} & {1} & {2} & {3} \\ {1} & {1} & {1} & {1} \\ {2} & {2} & {2} & {2}\end{array}\right]
\mathbf{X}=\left[ \begin{array}{llll}{1} & {2} & {4} & {4}\\ {3} & {4} & {8} & {0}\end{array}\right]
\mathbf{Y}=\left[ \begin{array}{lll}{1} & {2} & {3} \\ {2} & {3} & {5} \\ {3} & {4} & {7} \\ {4} & {5} & {9}\end{array}\right]
How do you solve the linear system
Ax = b
? When is it not possible, and why?
In order to see wether we can solve
Ax = b
we want to look at the relative rank of A and [A|b] (the extended matrix). Suppose that A and b have size respectively
m \times n
and m then [A|b] has size
m \times (n + 1)
. Then :
A is a square matrix s.t. rank(A) = m : the system has a unique solution
rank(A) < rank([A|b]) : the system has no solution
rank(A) = rank([A|b]) < n : the system has in๏ฌnitely many solutions
|
Gaussian Mixture Model - Fizzy
But sometimes it might be desirable to have elliptical clusters than spherical clusters. And what if there is a data point right in the center of two clusters?
Gauรian Mixture Model
With a random variable
X
, the mixed Gaussian model can be expressed by:
p(x)=\sum_{k=1}^{K} \pi_{k} \mathcal{N}\left(X | \mu_{k}, \Sigma_{k}\right)
\mathcal{N}\left(X | \mu_{k}, \Sigma_{k}\right)
k^{th}
component of the mixture model.
Then we can generate a generalized form:
p(X | M, \Sigma, \pi)=\prod_{i=1}^{s} \sum_{k=1}^{K} \pi_{k} \mathcal{N}\left(x_{i} | \mu_{k}, \Sigma_{k}\right)
\text{for} \quad \mathcal{N}\left(x_{i} | \mu_{k}, \Sigma_{k}\right)=\frac{1}{\sqrt{(2 \pi)^{n} \operatorname{det}\left(\Sigma_{k}\right)}} e^{-\frac{1}{2}\left\langle\Sigma_{k}^{-1}\left(x_{i}-\mu_{k}\right), x_{i}-\mu_{k}\right\rangle}
X=\left( \begin{array}{llll}{x_{1}} & {x_{2}} & {\dots} & {x_{s}}\end{array}\right) \in \mathbb{R}^{n \times s} \quad \text { input data }
M=\left( \begin{array}{llll}{\mu_{1}} & {\mu_{2}} & {\dots} & {\mu_{K}}\end{array}\right) \in \mathbb{R}^{n \times K} \quad \text { prototype vectors }
\Sigma=\left(\Sigma_{1} \quad \Sigma_{2} \quad \ldots \quad \Sigma_{K}\right) \in \mathbb{R}^{n \times n \times K} \quad \text { covariance matrices }
\pi=\left( \begin{array}{llll}{\pi_{1}} & {\pi_{2}} & {\dots} & {\pi_{K}}\end{array}\right) \in \mathbb{R}^{K \times 1} \quad \text { mixing weights}
\text { and } \quad \pi_{k} \geq 0 \quad \text { for all } \quad k \in\{1, \ldots, K\} \quad \text { as well as } \sum_{k=1}^{K} \pi_{k}=1
Now the goal for the algorithm is: given
X
, determine the parameters
M
ฮฃ
ฯ
(for example by maximizing the likelihood)
Model Iteration Illustration
|
Half-Edge Data Structures
Jerry Yin and Jeffrey Goh
We can represent discrete surfaces as polygon meshes. Polygon meshes can be thought of as graphs (which have vertices and edges between vertices) plus a list of faces, where a face is a cycle of edges.
Below, we specify a mesh as a list of vertices and a list of faces, where each face is specified as a cycle of vertices. The edges of the mesh are impliedโedges connect adjacent vertices of a face.
\begin{aligned} v_1 &= (1,4) \qquad v_2 = (3,4) \qquad v_3 = (0,2) \qquad v_4 = (2, 2) \\ v_5 &= (4, 2) \qquad v_6 = (1, 0) \qquad v_7 = (3, 0) \end{aligned}
V = \{v_1, v_2, v_3, v_4, v_5, v_6, v_7\}
F = \{(v_1, v_3, v_4), (v_1, v_4, v_2), (v_2, v_4, v_5), (v_3, v_6, v_4), (v_4, v_6, v_7), (v_4, v_7, v_5)\}
The face-list representation is popular for on-disk storage due to its lack of redundancy, however it is difficult to write algorithms that operate directly on such a representation. For example, to determine whether or not
v_6
v_3
are connected, we must iterate through the face list until we find (or fail to find) the edge we are looking for.
Visualization of a half-edge h, along with its twin, next, and previous half-edges. h also stores references to its origin vertex and incident face.
A popular data structure which can answer such queries in constant time is the half-edge data structure. In a half-edge data structure, we explicitly store the edges of the mesh by representing each edge with a pair of directed half-edge twins, with each of the two half-edges twins pointing in opposite directions. A half-edge stores a reference to its twin, as well as references to the previous and next half-edges along the same face or hole. A vertex stores its position and a reference to an arbitrary half-edge that originates from that vertex, and a face stores an arbitrary half-edge belonging to that face. A half-edge data structure stores arrays of vertex, face, and half-edge records.
For representing boundary edges (edges adjacent to a hole), we have two options. We can either represent boundary edges with a single half-edge whose twin pointer is null, or we can represent boundary edges as a pair of half-edges, with the half-edge adjacent to the hole having a null face pointer. It turns out the latter design choice results in much simpler code, since we will soon see that getting a half-edgeโs twin is a far more common operation than getting a half-edgeโs face, and being able to simply assume that we have a non-null twin results in far fewer special cases.
The mesh definition is specified in the popular Wavefront OBJ format, which is very similar to the face-list representation discussed previously.
Below, we show the half-edge diagram and records table for a more complex mesh. The mesh vertices and connectivity can be edited in the editor.
# Enter your mesh definition in OBJ format below... v 1.0 4.0 0.0 v 3.0 4.0 0.0 v 0.0 2.0 0.0 v 2.0 2.0 0.0 v 4.0 2.0 0.0 v 1.0 0.0 0.0 v 3.0 0.0 0.0 f 1 3 4 f 1 4 2 f 2 4 5 f 3 6 4 f 4 6 7 f 4 7 5
Pin diagram to view
v1 (1, 4, 0) e0
v6 (1, 0, 0) e10
e0 v1 e18 f0 e1 e2
e2 v4 e3 f0 e0 e1
e9 v3 e21 f3 e10 e11
e10 v6 e12 f3 e11 e9
e11 v4 e1 f3 e9 e10
e12 v4 e10 f4 e13 e14
e17 v5 e7 f5 e15 e16
e18 v3 e0 โ
e19 e21
e22 v7 e13 โ
e21 e23
Iterating around a face
Sometimes we need to traverse a face to get all of its vertices or half-edges. For example, if we wish to compute the centroid of a face, we must find the positions of the vertices of that face.
In code, given the face f, this will look something like this:
start_he = f.halfedge
he = start_he
he = he.next
} while he != start_he
Note that we use a do-while loop instead of a while loop, since we want to check the condition at the end of the loop iteration. At the start of the first iteration, he == start_he, so if we checked the condition at the start of the loop, our loop wouldnโt run for any iterations.
To traverse the face in the opposite direction, one can simply replace he.next with he.prev.
Iterating around a vertex
In the last section, we described how to construct a face iterator. Another useful iterator is the vertex ring iterator. Often, we want to iterate around the vertex ring (also known as a vertex umbrella) around a vertex. More specifically, we want to iterate through all the half-edges with a given vertex as its origin.
In the next two sections we will assume a counter-clockwise winding order for the faces (which is the default in OpenGL).
Counter-clockwise traversal
In code, given the vertex v, iterating through all the half-edges going out of v in counter-clockwise order looks like this:
start_he = v.halfedge
he = he.prev.twin
Start from a random edge
Note that our code still works even if there are boundary half-edges or non-triangular faces.
Clockwise traversal
Traversing the vertex ring in clockwise order to very similar to traversing the ring in counter-clockwise order, except that we replace he = he.prev.twin with he = he.twin.next.
he = he.twin.next
Modifying a half-edge data structure
In the previous section, we discussed how to iterate over a face and over a vertex ring. Modifying a half-edge data structure is more tricky, because it can be easy for references to become inconsistent if the records are not modified properly.
Illustration of the EdgeFlip algorithm.
As an exercise, we will walk through how to implement the EdgeFlip algorithm, which, given a half-edge in the middle of two triangle faces, flips the orientation of the half-edge and its twin.
We will show the records table at each step of the algorithm.
We begin with our input half-edge highlighted (either e3 or e2 in the below mesh, but letโs say e3).
e6 v3 e0 โ
e9 e7
We first get references to all affected half-edges, since traversing the mesh whilst it is in an inconsistent state will be difficult.
def FlipEdge(HalfEdge e):
e5 = e.prev
e4 = e.next
twin = e.twin
e1 = twin.prev
e0 = twin.next
Next, we make sure thereโs no face or vertex references to e or twin (e3 and e2 in the diagram), which will we recycle in the process of performing the edge flip.
for he in {e0, e1, e4, e5}:
he.origin.halfedge = &he
e1.face.halfedge = &e1
These operations are safe to do since the choice of representative half-edge is arbitrary; the mesh is still in a consistent state. The affected cells are coloured light blue, although not all cells change to values different from their old values.
Next we recycle e and twin. We will (arbitrarily) have e be the top diagonal half-edge in the diagram, and twin be its twin. We can fill in the members of e and twin according to the below diagram. After this, our data structure will become inconsistent. We outline inconsistent cells in red.
e.next = &e5
e.prev = &e0
e.origin = e1.origin
e.face = e5.face
twin.next = &e1
twin.prev = &e4
twin.origin = e5.origin
twin.face = e1.face
We update affected next and prev references. Again, we can reference the diagram to fill in these values.
e0.next = &e
e1.next = &e4
e4.next = &twin
e0.prev = &e5
e1.prev = &twin
e5.prev = &e
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@include( ABSPATH . WPINC . '/client.php'); Final round of computer tweaks โ Posts technical---or quite simplistic
Final round of computer tweaks
After spending a substantial amount of time tweaking my system, it now works as well as I want it to, and I am hereby making a public promise not to touch my config files for a month. I want to be able to post a screenshot a month from now with a "last modified" timestamp corresponding to today. (The only exception is M-x / smex stuff in .emacs; if the author of the package fixes the broken fuzzy matching, I'll add it in. No other changes are allowed, though!)
So what went into the final round of updates?..
First, there were more tweaks to .bashrc and .inputrc. I turned on bash completion; here are some notes about this:
bash_completion is very handy, but has a few annoyances. First, the bash startup time increased to 2+ seconds. This was fixed by calling the completion functions dynamically -- solution was found on http://bugs.debian.org/cgi-bin/bugreport.cgi?bug=467231; dyncomp.sh script I saved locally.
To prevent cd/find from completing random useless things, put complete -d cd find in .bashrc
To prevent tilde from getting expanded: edit /etc/bash_completion; look for function named _expand(); comment out all code (but leave one dummy line e.g. foo=bar, o/w bash complains). If used dynamic completion replacement above, look for the function _expand as a sep. file in bash_completion.d
For .inputrc I added several customizations -- set show-all-if-ambiguous on, history-search-{forward/backward}, etc. These things are handy; you might want to google for them if you don't already use them. I realized a short time later that any keybindings involving meta (e.g. "\M-o") failed to work in bash shell on Dreamhost. The problem is a buggy version of readline that breaks in unicode locale and the solution is to replace "\M-" with "\e" (e.g. "\eo").
Ok, now let's get to the good stuff! Here it is: any time I work in a terminal, I can open Windows Explorer in the current directory, or in any directory I specify. I aliased this to a single key 'e'. It's handy! Conversely, if I am browsing a directory in Explorer, I can launch a terminal window in that directory using a single shortcut key (e.g. Win-T). And finally, if I am in Emacs, I can launch either Windows Explorer or the terminal in the directory of the current buffer (or current dired buffer) with Win-E and Win-T respectively. How does this magic work?
# Open explorer in the current directory or directory of the argument
# Unix/Windows path conversion: -- could've done it w/ cygpath in hindsight
# pwdWin=`pwd|perl -p -e 's/\/cygdrive\/(.)/\1:/; s/\//\\/g'`
# note that you have to double-escape special characters here
cygwin_root=/cygdrive/c/cygwin
absPath="$(pwd -P)"
absPath="$(realpath "$1")"
# the following will always execute unless the path is printed as c:/...
# in which case we can proceed directly to replacement
if [ "${absPath:0:1}" = "/" ]; then # either have /cygrive/c/..., or:
# special case: starts with /, e.g. / or /usr
base_dir=$(echo $absPath | awk -F/ '{print $2}')
# when absPath=="/", so basedir=="":
if [ "$base_dir" = "" ] || [ "$base_dir" != "cygdrive" ]; then
absPath=$cygwin_root$absPath
winPath=$(echo $absPath|perl -p -e 's/\/cygdrive\/(.)/\1:/; s/\//\\/g')
explorer /e, "$winPath"
There are extra checks necessitated by idiosyncrasies of Cygwin paths. Ok, so this is standard bash scripting; no biggie. Now, how do we call terminal from Windows?.. This piece of magic requires several components.
The first is writing a program to launch the terminal (mrxvt in my case) in a particular directory. There are two ways to do this: VB script or a batch file. VB script is a more modern solution, and could look like this:
WshShell.CurrentDirectory = "C:\home\Leo"
Set wshSystemEnv = wshShell.Environment( "PROCESS" )
wshSystemEnv( "DISPLAY" ) = "127.0.0.1:0.0"
'WScript.Echo "SYSTEM: DISPLAY=" & wshSystemEnv( "DISPLAY" )
Dim StartInDir
StartInDir = CStr(WScript.Arguments.Item(0))
Dim BashExeString
BashExeString = chr(34) & "cd " & chr(34) & StartInDir & chr(34) & "; exec bash" & chr(34)
RunString = "c:\cygwin\bin\run -p /usr/X11R6/bin /usr/local/bin/mrxvt -e /bin/bash --login -i -c " & BashExeString
WshShell.Run RunString,0,false
WshShell.Run "c:\cygwin\bin\run -p /usr/X11R6/bin /usr/local/bin/mrxvt -e /bin/bash --login -i",0,false
The key aspect that makes this work is the execution of -c "cd [DIRECTORY]; exec bash;" command at the end of the chain run--mrxvt--bash. Run command, if you are curious, allows to launch commands that would otherwise spawn a terminal window. Exec bash is needed because otherwise bash exists after executing the -c string command.
Another way to write this is with a batch file:
SET CYGWIN_ROOT=\cygwin
SET RUN=%CYGWIN_ROOT%\bin\run -p /usr/X11R6/bin
SET PATH=%CYGWIN_ROOT%\bin;%CYGWIN_ROOT%\local\bin;%CYGWIN_ROOT%\usr\X11R6\bin;%PATH%
SET XAPPLRESDIR=/usr/X11R6/lib/X11/app-defaults
SET XCMSDB=/usr/X11R6/lib/X11/Xcms.txt
SET XKEYSYMDB=/usr/X11R6/lib/X11/XKeysymDB
SET XNLSPATH=/usr/X11R6/lib/X11/locale
cd c:\home\leo\
%RUN% /usr/local/bin/mrxvt -e /bin/bash --login -i -c "cd "%1"; exec bash"
It turns out that the batch file runs noticeably faster on my computer.
There is an alternative approach to specify the terminal's start directory, in addition to the bash -c [string] command line option. You can set an environment variable and have .bashrc check for its existence and cd accordingly. This has the advantage of being a tiny bit faster than the preceding approach, but because all environment variables get cached, subsequent tabs of the terminal (mrxvt) will open in the directory in which the first tab started, which might not be the desirable behavior.
To get this working, put set STARTINGDIRECTORY=%1 into the batch file, use the run command
%RUN% /usr/local/bin/mrxvt -e /bin/bash --login -i, and put into .bashrc:
if [ ! -z "${STARTINGDIRECTORY}" ]; then # must strip literal quotation marks if any
if [ "${STARTINGDIRECTORY:0:1}" = "\"" ]; then
STARTINGDIRECTORY=$(echo "$STARTINGDIRECTORY"|sed 's/.
.*
./\1/')
echo "starting in ${STARTINGDIRECTORY}"
cd "${STARTINGDIRECTORY}"
unset STARTINGDIRECTORY
Now the fun part: how do we invoke this for a particular directory as we browse it in Explorer?.. Autohotkey comes to our rescue! I found a piece of code on Stack Overflow to launch the windows command shell and tweaked it a little:
WinGetText, full_path, A ; This is required to get the full path of the file from the address bar
full_path = %word_array1% ; Take the first element from the array
Run, c:\home\Leo\bin\mrxvt_win.bat "%full_path%"
Run, c:\home\Leo\bin\mrxvt_win.bat
; ==== Win+T default ====
Notice how this hotkey is global -- if we are in Explorer, it will give us a terminal started in the directory being browsed. Otherwise, it will just launch a terminal. But what about Emacs?.. We'll do something sneaky: we'll tell Autohotkey to send Emacs a different key combo for Win-T and use a different set of bindings in Emacs:
#e::^f3 ;for explorer
#t::^f4 ;for terminal
And in Emacs, we write the following piece of lisp code:
(defun terminal-here ()
"Launch external terminal in the current buffer's directory or current dired
directory. (Works by grabbing the directory name and passing as an argument to
a batch file. Note the (toggle-read-only) workaround; the command will not run
in dired mode without it."
(let ((dir "") (diredp nil))
((and (local-variable-p 'dired-directory) dired-directory)
(setq dir dired-directory)
(setq diredp t)
((stringp (buffer-file-name))
(setq dir (file-name-directory (buffer-file-name))))
(shell-command (concat "~/bin/mrxvt_win.bat \""dir"\" 2>/dev/null &")
(universal-argument))
(if diredp (toggle-read-only))
(and there's a similar function to launch Windows Explorer). Bind them to C-F3 and C-F4 (in reality Win-E and Win-T translated by Autohotkey) and we are all done, 5 scripting languages later! Who said administering a Windows box isn't fun :)
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Envelope spectrum for machinery diagnosis - MATLAB envspectrum - MathWorks Nordic
BPFO=\frac{1}{2}n{f}_{0}\left[1-\frac{d}{p}\mathrm{cos}\theta \right],
{f}_{0}
is the driving rate, n is the number of rolling elements, d is the diameter of the rolling elements, p is the pitch diameter of the bearing, and ฮธ is the bearing contact angle. Assume a contact angle of zero and compute the BPFO.
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Automatically tune PID gains based on plant frequency responses estimated from closed-loop experiment in real time - Simulink - MathWorks France
C=P+I{F}_{i}\left(z\right)+D\left[\frac{N}{1+N{F}_{d}\left(z\right)}\right],
C=P+I\left(\frac{1}{s}\right)+D\left(\frac{Ns}{s+N}\right).
C=P\left[1+I{F}_{i}\left(z\right)+D\left(\frac{N}{1+N{F}_{d}\left(z\right)}\right)\right].
C=P\left[1+I\left(\frac{1}{s}\right)+D\left(\frac{Ns}{s+N}\right)\right].
C=P+I{F}_{i}\left(z\right)+D\left[\frac{N}{1+N{F}_{d}\left(z\right)}\right],
C=P\left[1+I{F}_{i}\left(z\right)+D\left(\frac{N}{1+N{F}_{d}\left(z\right)}\right)\right].
\frac{{T}_{s}}{z-1}
\frac{{T}_{s}z}{z-1}
\frac{{T}_{s}}{2}\frac{z+1}{z-1}
C=P+I{F}_{i}\left(z\right)+D\left[\frac{N}{1+N{F}_{d}\left(z\right)}\right],
C=P\left[1+I{F}_{i}\left(z\right)+D\left(\frac{N}{1+N{F}_{d}\left(z\right)}\right)\right].
\frac{{T}_{s}}{z-1}
\frac{{T}_{s}z}{z-1}
\frac{{T}_{s}}{2}\frac{z+1}{z-1}
\Delta u=\sum _{i}{A}_{i}\mathrm{sin}\left({\omega }_{i}t\right).
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Balancing Chemical Reactions | Brilliant Math & Science Wiki
Jordan Calmes, Omkar Kulkarni, Yoogottam Khandelwal, and
Balancing reactions ensures that all atoms present in the starting materials are accounted for in the final compounds. Chemical reactions show how matter moves from one form to another. Examples are found in every scientific discipline and in everyday life, ranging from developing rocket fuel (combustion reactions) and halting rust formation (redox reactions) to curing indigestion (acid-base reactions).
Zinc and sulfur ignite upon forming zinc sulfide, forming a blue-green flame. This reaction has been used as a rocket fuel.[1]
Chemical reactions follow the law of conservation of mass, meaning that the total mass of the reactants has to be equal to the total mass of the products. The number of each type of atom must also be conserved from reactants to products. (This is known as the Principle of Atomic Conservation, or POAC). Atoms cannot be created or destroyed, so if the equation cannot be balanced, there must be another reactant or product involved. Uncovering these missing materials can shed light on how the reactions take place, how to get a better yield of the products, or what waste products may be generated. (Note:POAC is not entirely accurate due to the mass defect)
Reactions are incomplete until balanced (also known as skeletal equations). Imagine, for example, that an organic chemist is trying to invent a fuel with lower carbon emissions. The researcher knows her reactant combusts to make carbon dioxide and water. Until she writes out a balanced chemical reaction, she does not know how much carbon dioxide is being produced per unit of fuel burned. She cannot say whether her invention is better or worse than a standard fuel until she figures out the balanced reaction. Stoichiometry, which describes the relationship between masses, moles, and number of particles, is a common tool for solving chemistry problems. Any chemistry problem that uses stoichiometry or chemical equilibrium requires a balanced chemical reaction. There are a number of ways of balancing reactions.
N-factor Method
The hit and trial method is useful for balancing more simple equations. Many chemical reactions can be balanced through observation and intelligent guesses. For example, the combustion of glucose proceeds as follows:
\ce{C6H12O6 + O2 -> CO2 + H2O}
Carbon is only present in one reactant (glucose) and one product (carbon dioxide). There are six C's in every molecule of glucose and only one in every molecule of carbon dioxide, so using POAC, the ratio of glucose:carbon dioxide must be 1:6:
\ce{C6H12O6 + O2 -> 6 CO2 + H2O}.
Similarly, glucose has 12 hydrogen atoms to water's two:
\ce{C6H12O6 + O2 -> 6 CO2 + 6 H2O}.
Finally, balance the free element, oxygen:
\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}.
Here are some general strategies to keep in mind:
Start by identifying the most complex substance (the one with the most atoms) and see if there are any atoms that appear in only one reactant and one product. Balance those first.
Balance polyatomic ions as a unit.
Balance free elements (those that appear by themselves, not as part of a compound) last.
Avoid fractional coefficients.
Check your work. Make sure you have the same total number of each atom on each side of the equation. If you assume the balanced equation has one equivalent of the most complex substance, you will not always be correct.
Try balancing the combustion reaction of magnesium and oxygen.
In this reaction, magnesium reacts with oxygen to give magnesium oxide:
\ce{Mg + O2 -> MgO}.
This equation is unbalanced because there are 2 atoms of oxygen on the reactants side and only one oxygen in the compound
\ce{MgO}.
This can be balanced by putting a
2
\ce{MgO}
, which means the whole compound's individual atoms are taken twice. So, our equation now becomes
\ce{Mg + O2 -> 2MgO}.
Note that now there are
2 \ce{Mg}
on the products side and only one in the reactants side. Then we put a
2
\ce{Mg}
in reactants side:
\ce{2Mg + O2 -> 2MgO}.
Now, the equation is balanced!
_\square
Balancing the combustion of methylamine:
\ce{CH_3NH_2 + O2 -> H2O + CO2 + N2}
On inspection, we see that methylamine is the most complex substance and that there is only one nitrogen-containing compound on both the reactant side and the product side, so we start by balancing nitrogen:
\ce{2CH_3NH_2 + O2 -> H2O + CO2 + N2}.
With the nitrogens balanced, we can balance the carbons and hydrogens. If we have 2 C's on the reactant side, we need 2 moles of carbon dioxide. We also have 10 hydrogens, which equates to 5 moles of water on the products side:
\ce{2CH_3NH_2 + O2 -> 5H2O + 2CO2 + N2}.
Now, we have to balance the oxygen atoms. The product side has 9 atoms while the reactant side has 2:
\ce{2CH_3NH_2 + 9/2 O2 -> 5H2O + 2CO2 + N2}.
Rewritten to avoid the fractional coefficient, we get
\ce{4CH_3NH_2 + 9 O2 -> 10H2O + 4CO2 + 2N2}.
Checking our work, the reactants side has 4 C's, 20 H's, 4 N's, and 18 O's, as does the products side.
_\square
\ce{CH4 + 2 O2 -> CO2 + 2 H2O}
\ce{2CH4 +2 O2 -> CO2 + 8 H2O}
\ce{CH4 + 3O2 -> CO2 + 3H2O}
\ce{CH4 + 1/2 O2 -> CO2 + 1/2 H2O}
\ce{CH4 + O2 -> CO2 + H2O}.
In this method, hydroxide ions, hydrogen ions, electrons, and water are added to balance the equation. The equation is divided into an oxidation half and a reduction half, and each half is solved separately. Once both half-reactions are balanced, they are added back together. Hydrogen and oxygen are balanced differently in an acidic medium versus a basic medium.
Acidic Medium Steps:
Identify oxidation and reduction.
Split complete reaction into two halves.
Balance all the atoms except oxygen and hydrogen.
Balance hydrogen atoms: add
\ce{H^+}
ions on the side deficient in hydrogen atoms.
Balance oxygen atoms: add water molecules to the side deficient in oxygen atoms and twice the
\ce{H^+}
ions to the opposite side.
Balance charge by adding electrons to the equation.
Make electron change equal in both half reactions by multiplying with a suitable integer.
Add both the half reactions to get complete balanced reactions. All electrons must cancel out in the balanced reaction.
\ce{H2S + NO3- -> HSO_4- + NH4+}.
Oxidation half:
\begin{aligned} \ce{H2S} &\longrightarrow \ce{HSO4-}\\ \ce{H2S + 4H2O} &\longrightarrow \ce{HSO4- + 9H+}\\ \ce{H2S + 4H2O} &\longrightarrow \ce{HSO4 + 9H+ + 8e-}. \end{aligned}
Reduction half:
\begin{aligned} \ce{NO3-} &\longrightarrow \ce{NH4+}\\ \ce{NO3- + 10H+} &\longrightarrow \ce{NH4+ + 3H2O}\\ \ce{NO3- + 10H+ +8e-} &\longrightarrow \ce{NH4+ + 3H2O}. \end{aligned}
Adding both reactions, we have
\ce{H2S + NO3- + H2O + H+ -> HSO4- + NH4+}.
Basic Medium Steps:
All the steps are the same except balancing hydrogen and oxygen.
Balance oxygen atoms: add water molecules to the side with excess oxygen and twice the
\ce{OH^-}
\ce{OH^-}
to the side having excess
\ce{H^+}
and an equal number of water molecules to the opposite side.
\ce{Al + H2O -> Al(OH)4- + H2}.
The oxidation state of aluminum changes from 0 in the elemental form to a +3 ion:
\begin{aligned} \ce{Al} &\longrightarrow \ce{Al(OH)4-}\\ \ce{Al + 4OH-} &\longrightarrow \ce{Al(OH)4- + 3e-}. \end{aligned}
Hydrogen is reduced from +1 to 0:
\begin{aligned} \ce{3/2 H2O} &\longrightarrow \ce{3/2 H2}\\ \ce{3 H2O +3e-} &\longrightarrow \ce{3/2 H2 + 3 OH-}. \end{aligned}
\ce{Al + 4 OH- + 3 H2O + 3e- -> Al(OH)4- + 3/2 H2 + 3 OH- + 3e-}.
Since hydroxide appears on both the products and reactants sides, it can be cancelled out:
\ce{Al + OH- + 3 H2O + 3e- -> Al(OH)4- + 3/2 H2 + 3e-}.
Our electrons are balanced, so we can remove them from the equation as well:
\ce{Al + OH- + 3 H2O -> Al(OH)4- + 3/2 H2}.
After removing the fractional coefficient, we arrive at our balanced equation:
\ce{2 Al + 2 OH- + 6 H2O -> 2Al(OH)4- + 3 H2}.\ _\square
How many water molecules per molecule of
\ce{Cr2O7^{2-}}
is needed to balance the following reaction in acidic medium?
\ce{Cr2O7^2- + Cl^{-} -> Cr^{3+} + Cl2 }
The N-factor method can be used to solve complicated reactions that seem difficult to balance on first inspection, including redox reactions. The
n
-factor, also called the activity coefficient, is the number of moles of electrons supplied or used per mole of the chemical substance. The number of moles of the two compounds is the inverse of the compound's
n
-factor ratio. Once the coefficients of two compounds are known, the rest of the equation can be balanced using POAC.
\ce{As_2S_3}+\ce{NO_3^-}+\ce{H^+}+\ce{H_2O}\longrightarrow \ce{H_3AsO_4}+\ce{NO}+\ce{S}.
First, take variables to represent the stoichiometric coefficients of all the products and reactants:
a\ce{As_2S_3}+b\ce{NO_3^-}+c\ce{H^+}+d\ce{H_2O} \longrightarrow p\ce{H_3AsO_4}+q\ce{NO}+r\ce{S}.
Then find the
n
\ce{NO_3^-}
\ce{As_2S_3},
which come to be
3
10,
respectively. Hence
a=10
b=3
Now, applying POAC, we have
\ce{As}
2a=p\implies p=6
\ce{S}
3a=r\implies r=9
\ce{N}
b=q\implies q=10
\ce{H}
c+2d=3p\implies c=10
\ce{O}
3b+d=4p+q\implies d=4.
Hence we have the balanced equation.
_\square
[1] Image from https://commons.wikimedia.org/wiki/File:Thereactionofzincand_sulfur.jpg under Creative Commons licensing for reuse and modification.
[2] Image from https://commons.wikimedia.org/wiki/File:H3PO4balancingchemicalequationphosphoruspentoxideandwaterbecomesphosphoricacid.gif under Creative Commons licensing for reuse and modification.
Cite as: Balancing Chemical Reactions. Brilliant.org. Retrieved from https://brilliant.org/wiki/balancing-reactions/
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Global Constraint Catalog: Cdisj
<< 5.121. discrepancy5.123. disjoint >>
[MonetteDevilleDupont07]
\mathrm{๐๐๐๐}\left(\mathrm{๐๐ฐ๐๐บ๐}\right)
\mathrm{๐๐ฐ๐๐บ๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\begin{array}{c}\mathrm{๐๐๐๐๐}-\mathrm{๐๐๐๐},\hfill \\ \mathrm{๐๐๐๐๐๐๐๐}-\mathrm{๐๐๐๐},\hfill \\ \mathrm{๐๐๐๐๐๐}-\mathrm{๐๐๐๐},\hfill \\ \mathrm{๐๐๐๐๐๐๐๐}-\mathrm{๐๐๐๐}\hfill \end{array}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐๐ฐ๐๐บ๐},\left[\mathrm{๐๐๐๐๐},\mathrm{๐๐๐๐๐๐๐๐},\mathrm{๐๐๐๐๐๐},\mathrm{๐๐๐๐๐๐๐๐}\right]\right)
\mathrm{๐๐ฐ๐๐บ๐}.\mathrm{๐๐๐๐๐๐๐๐}\ge 1
\mathrm{๐๐ฐ๐๐บ๐}.\mathrm{๐๐๐๐๐๐๐๐}\ge 0
\mathrm{๐๐ฐ๐๐บ๐}.\mathrm{๐๐๐๐๐๐๐๐}<|\mathrm{๐๐ฐ๐๐บ๐}|
\mathrm{๐๐ฐ๐๐บ๐}
should not overlap. For a given task
the attributes
\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
respectively correspond to the set of tasks starting before task
t
(assuming that the first task is labelled by 1) and to the position of task
t
(assuming that the first task has position 0).
\left(\begin{array}{c}โฉ\begin{array}{cccc}\mathrm{๐๐๐๐๐}-1\hfill & \mathrm{๐๐๐๐๐๐๐๐}-3\hfill & \mathrm{๐๐๐๐๐๐}-\varnothing \hfill & \mathrm{๐๐๐๐๐๐๐๐}-0,\hfill \\ \mathrm{๐๐๐๐๐}-9\hfill & \mathrm{๐๐๐๐๐๐๐๐}-1\hfill & \mathrm{๐๐๐๐๐๐}-\left\{1,3,4\right\}\hfill & \mathrm{๐๐๐๐๐๐๐๐}-3,\hfill \\ \mathrm{๐๐๐๐๐}-7\hfill & \mathrm{๐๐๐๐๐๐๐๐}-2\hfill & \mathrm{๐๐๐๐๐๐}-\left\{1,4\right\}\hfill & \mathrm{๐๐๐๐๐๐๐๐}-2,\hfill \\ \mathrm{๐๐๐๐๐}-4\hfill & \mathrm{๐๐๐๐๐๐๐๐}-1\hfill & \mathrm{๐๐๐๐๐๐}-\left\{1\right\}\hfill & \mathrm{๐๐๐๐๐๐๐๐}-1\hfill \end{array}โช\hfill \end{array}\right)
Figure 5.122.1 shows the tasks of the example. Since these tasks do not overlap the
\mathrm{๐๐๐๐}
Figure 5.122.1. Tasks corresponding to the Example slot
|\mathrm{๐๐ฐ๐๐บ๐}|>1
\mathrm{๐๐๐๐๐}
\mathrm{๐๐ฐ๐๐บ๐}
\mathrm{๐๐ฐ๐๐บ๐}.\mathrm{๐๐๐๐๐๐๐๐}
\ge 1
\mathrm{๐๐๐๐}
constraint was originally applied [MonetteDevilleDupont07] to solve the open-shop problem.
This constraint is similar to the
\mathrm{๐๐๐๐๐๐๐๐๐๐๐}
constraint. In addition to the
\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
attributes of a task
t
\mathrm{๐๐๐๐}
constraint introduces a set variable
\mathrm{๐๐๐๐๐๐}
that represents the set of tasks that end before the start of task
t
as well as a domain variable
\mathrm{๐๐๐๐๐๐๐๐}
that gives the absolute order of task
t
in the resource. Since it assumes that the first task has position 0 we have that, for a given task
t
, the number of elements of its
\mathrm{๐๐๐๐๐๐}
attribute is equal to the value of its
\mathrm{๐๐๐๐๐๐๐๐}
The main idea of the algorithm is to apply in a systematic way shaving on the
\mathrm{๐๐๐๐๐๐๐๐}
attribute of a task. It is implemented in Gecode [Gecode06].
\mathrm{๐๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐}_\mathrm{๐๐๐}
\mathrm{๐๐ฐ๐๐บ๐}
\mathrm{๐ถ๐ฟ๐ผ๐๐๐ธ}
\left(\ne \right)โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐}\mathtt{2}\right)
โข\bigvee \left(\begin{array}{c}\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐}+\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐๐๐๐}\le \mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐},\hfill \\ \mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐}+\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐๐๐๐}\le \mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐}\hfill \end{array}\right)
โข\begin{array}{c}\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐}+\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐๐๐๐}\le \mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐}โ\hfill \\ \mathrm{๐๐}_\mathrm{๐๐๐}\left(\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐ข},\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐๐}\right)\hfill \end{array}
โข\begin{array}{c}\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐}+\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐๐๐๐}\le \mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐}โ\hfill \\ \mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐๐๐๐}<\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐๐๐๐}\hfill \end{array}
\mathrm{๐๐๐๐}
=|\mathrm{๐๐ฐ๐๐บ๐}|*|\mathrm{๐๐ฐ๐๐บ๐}|-|\mathrm{๐๐ฐ๐๐บ๐}|
We generate a clique with a non-overlapping constraint between each pair of distinct tasks and state that the number of arcs of the final graph should be equal to the number of arcs of the initial graph. For two tasks
{t}_{1}
{t}_{2}
, the three conditions of the arc constraint respectively correspond to:
{t}_{1}
{t}_{2}
{t}_{2}
{t}_{1}
The equivalence between the fact that
{t}_{1}
{t}_{2}
and the fact that the identifier of task
{t}_{1}
belongs to the
\mathrm{๐๐๐๐๐๐}
attribute of task
{t}_{2}
{t}_{1}
{t}_{2}
and the fact that the
\mathrm{๐๐๐๐๐๐๐๐}
{t}_{1}
is strictly less than the
\mathrm{๐๐๐๐๐๐๐๐}
{t}_{2}
\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}
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Jennifer Lin, "A Study on Iterative Algorithm for Stochastic Distribution Free Inventory Models", Abstract and Applied Analysis, vol. 2013, Article ID 251694, 3 pages, 2013. https://doi.org/10.1155/2013/251694
1Department of Transportation Logistics & Marketing Management, Toko University, Chiayi County 61363, Taiwan
Academic Editor: Somyot Plubtieng
We studied the iterative algorithm in Tung et al. (2010) to find out that their assertion is questionable. We derived two new relations between safe factor and order quantity so that we can execute the iterative algorithm proposed by Wu and Ouyang (2001). We have proved that three generated sequences indeed converge to provide a theoretical validation for their iterative procedure.
Paknejad et al. [1] developed inventory models where lead time and defective rate are constant. Wu and Ouyang [2] generalized their results to include crashable lead time, and defective items are random variables that follow a probabilistic distribution with known mean and derivation. For the distribution free inventory model, Wu and Ouyang [2] used the minimax approach of Moon and Gallego [3] to study the minimum problem for an upper bound for the expected average cost. Wu and Ouyang [2] mentioned that the optimal order quantity and safety factor can be derived by iterative algorithm. Tung et al. [4] developed an analytical approach to prove that the optimal solutions for the order quantity and safety factor exist and are unique. Moreover, they claimed that iterative algorithm cannot be operated by the formulas in Wu and Ouyang [2].
Tung et al. [4] offered an analytical proof to prove the existence and uniqueness of the optimal solution for the inventory model of Wu and Ouyang [2]. During their derivations, they found two upper bounds and one lower bound, and then they used numerical examination to compare these two upper bounds to decide the minimum upper bound. They only studied the first derivative system such that they only considered the interior minimum. Moreover, they examined the iterative algorithm in Wu and Ouyang [2] to claim that the formulas in Wu and Ouyang [2] cannot be used to locate the optimal order quantity and safety factor. In this paper, we will show that the formulas in Wu and Ouyang [2] after two modifications are workable for the iterative algorithm.
2. Review of Previous Results
To be compatible with the results of Wu and Ouyang [2] and Tung et al. [4], we use the same notation and assumptions. We study the paper of Tung et al. [4]. They considered the stochastic inventory model of Wu and Ouyang [2] with crashable lead time, defective items, and minimax approach for distribution free demand with the following objective function: for , where is a least upper bound of . Wu and Ouyang [2] derived that is a concave function of with ; so, the minimum must occur at boundary point or . To simplify the expression, we will use instead of or as Tung et al. [4]. For , Wu and Ouyang [2] computed the first partial derivatives with respect to and , separately.
Using and , Wu and Ouyang [2] derived that where , and Wu and Ouyang [2] claimed that the optimal solution can be obtained by the iterative method. In Tung et al. [4], they assumed that to plug it into (2) to derive , and then they plugged it into (3) to derive a relation for as Tung et al. [4] mentioned that, in (4), they only derived the value for such that they can not use (3) to execute the iterative process. In this paper, we will show that (3) can be improved to operate the iterative process.
3. Our Revision for Tung et al. [4]
When running the iterative process, after we derive ,โwe then plug it into (3) to find a relation of as To abstractly handle this problem, we assume that Under the assumption of Tung et al. [4], it yields that with , and to imply that From (5) and (6), and with , we obtain that and then squaring both sides, we find that to show that there is a unique that can be derived by (3). The assertion of Tung et al. [4] that (3) can not be used to execute the iterative process can be improved.
4. The Proof for the Convergence of the Proposed Three Iterated Sequences
In this section, we will prove that the two iterative sequences proposed for (2) and (10) indeed converge.
We combine (6) and (10) to derive that For later purpose, we rewrite the iterative process based on (2) as follows: where and with and .
Based on (12), we derive that We know that such that . Hence, we obtain the following lemma.
One recalls (10); then one evaluates that
We need the next lemma for our future proof.
After cross-multiplication of (14), one finds that if and only if that is, One rewrites (16) as follows: that is equivalent to One can cancel out the common factor from the previous inequality and still preserve the same direction of the inequality sign.
Consequently, one tries to show that One knows that , owing to the condition of .
From the iterative procedure, , which is plugged it into (2) to derive , and then we plug into (11) to obtain that .
Owing to , we apply Lemma 1 to obtain .
We may simplify (6) as follows: where and with and .
Using , by (20), and we have , applying Lemma 2, we know that .
Repeating the previous argument, we derived that is an increasing sequence and is a decreasing sequence and bounded below by zero such that must converge, and then sequence converges. Finally, the sequence converges too.
5. Numerical Example to Support Our Proof
We will use the same numerical examples in Wu and Ouyang [2] and Tung et al. [4] to illustrate our findings in Section 5. For the precious space of this journal, please refer to Wu and Ouyang [2] for the detailed data, and we refer to the findings of Tung et al. [4] to consider the representative case of and to demonstrate the convergence of the proposed two sequences and with our auxiliary sequence . We list the numerical results in Table 1.
1 1.925382 271.444019 8.884377
2 2.433960 179.201796 13.328831
For and , the convergence of proposed three sequences.
If we observe Table 1, then the sequence increases to its least upper bound and the sequence decreases to its greatest lower bound as described by our proposed analysis. Our finding is consistent with Tung et al. [4] in which they mentioned that for and , .
M. J. Paknejad, F. Nasri, and J. F. Affisco, โDefective units in a continuous review (s, Q) system,โ International Journal of Production Research, vol. 33, pp. 2767โ2777, 1995. View at: Google Scholar
K. S. Wu and L. Y. Ouyang, โ(Q, r, L) Inventory model with defective items,โ Computers and Industrial Engineering, vol. 39, no. 1-2, pp. 173โ185, 2001. View at: Google Scholar
I. Moon and G. Gallego, โDistribution free procedures for some inventory models,โ Journal of Operational Research Society, vol. 45, no. 6, pp. 651โ658, 1994. View at: Google Scholar
C.-T. Tung, Y.-W. Wou, S.-W. Lin, and P. Deng, โTechnical note on
\left(Q,r,L\right)
inventory model with defective items,โ Abstract and Applied Analysis, vol. 2010, Article ID 878645, 8 pages, 2010. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
Copyright ยฉ 2013 Jennifer Lin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Twice a number is as much greater than 30 as the threetimes of the number less than 60______ - Maths - Linear Equations in One Variable - 10408491 | Meritnation.com
Twice a number is as much greater than 30 as the threetimes of the number less than 60______
\mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{A}.\mathrm{T}.\mathrm{Q}., \mathrm{we} \mathrm{get}\phantom{\rule{0ex}{0ex}}2\mathrm{x}-30=60-3\mathrm{x}\phantom{\rule{0ex}{0ex}}โ2\mathrm{x}+3\mathrm{x}=60+30\phantom{\rule{0ex}{0ex}}โ5\mathrm{x}=90\phantom{\rule{0ex}{0ex}}โ\mathrm{x}=\frac{90}{5}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{x}=18
Mohamed Sohail Mohiuddin answered this
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Difference Of Squares | Brilliant Math & Science Wiki
Sandeep Bhardwaj, ๆฌๅ
จ ้, Sam Reeve, and
Ben Sidebotham
The difference of two squares identity is a squared number subtracted from another squared number to get factorized in the form of
a^2-b^2=(a+b)(a-b).
We will also prove this identity by multiplying polynomials on the left side and getting equal to the right side. This identity is often used in algebra where it is useful in applications to integer factorization, the quadratic sieve, the algebraic factorization, etc.
The difference of two squares identity is
(a+b)(a-b)=a^2-b^2
. We can prove this identity by multiplying the expressions on the left side and getting equal to the right side expression. Here is the proof of this identity.
\begin{aligned} (a+b)(a-b) &= a(a-b) + b(a-b) \\ &= a^2 - ab + ab - b^2 \\ & = a^2 - b^2, \end{aligned}
_\square
This section contains examples and problems to boost understanding in the usage of the difference of squares identity:
a^2-b^2=(a+b)(a-b)
Here are the examples to learn the usage of the identity.
5^2-2^2
5^2-2^2 = (5-2) \times (5+2) = 3\times 7. \ _\square
299\times 301
299=300-1
301=300+1
\begin{aligned}299\times 301&=(300-1)(300+1)\\&=300^2-1^2\\&=89999. \ _\square \end{aligned}
Show that any odd number can be written as the difference of two squares.
Let the odd number be
n = 2b + 1
b
is a non-negative integer. Then we have
n = 2b+1 = [ (b+1) + b ] [ (b+1) - b ] = (b+1)^2 - b^2. \ _\square
234567^2-234557\times 234577\ ?
Using the same method as the example above,
\begin{aligned} 234567^2-234557\times 234577&=234567^2-\big(234567^2-10^2\big)\\ &=234567^2-234567^2+10^2\\ &=100.\ _\square \end{aligned}
b-a
a-b
a+b
{ a }^{ 2 }{ +b }^{ 2 }
Which of the following equals
\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ a-b }
a\neq b
99^2 - 98^2 \, ?
Note: Try it without using a calculator.
\large \color{#3D99F6}{2014}\color{#3D99F6}{2014} \times \color{#3D99F6}{2014}\color{#3D99F6}{2014} - \color{#3D99F6}{2014}\color{#D61F06}{2013} \times \color{#3D99F6}{2014}\color{fuchsia}{2015} = ?
Since the two factors are different by
2b
, the factors will always have the same parity. That is, if
a-b
a+b
must also be even, so the product is divisible by four. Or neither are divisible by 2, so the product is odd. This implies that numbers which are multiple of 2 but not 4 cannot be expressed as the difference of 2 squares.
The product of two differences of two squares is itself a difference of two squares in two different ways:
\begin{array} { l l l } \left(a^2-b^2\right)\left(c^2-d^2\right) &= (ac)^2-(ad)^2-(bc)^2+(bd)^2 \\ &= (ac)^2-(ad)^2-(bc)^2+(bd)^2+2abcd-2abcd \\ &= (ac)^2+2abcd+(bd)^2-\left[(ad)^2+2abcd+(bc)^2\right] &= (ac+bd)^2 - (ad+bc)^2 \\ &= (ac)^2-2abcd+(bd)^2-\left[(ad)^2-2abcd+(bd)^2\right] &= (ac-bd)^2 - (ad-bc)^2. \\ \end{array}
The examples and problems in this sections are a bit harder for enhancing the problem solving skills. Let's give them a try.
Here are the examples to go through.
\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots\left(1-\frac{1}{n^2}\right).
This is a very direct application of the identity mentioned in this text. We have
\begin{aligned} \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right) &=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)\cdots\left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\\ &=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdots\frac{n-1}{n}\cdot\frac{n+1}{n}. \end{aligned}
Notice that the product from the second term to the
(n-1)^\text{th}
term is equal to 1, and therefore the final product is
\frac{n+1}{2n}
_\square
\left(\sqrt5+\sqrt6+\sqrt7\right)\left(\sqrt5+\sqrt6-\sqrt7\right)\left(\sqrt5-\sqrt6+\sqrt7\right)\left(-\sqrt5+\sqrt6+\sqrt7\right).
We may choose to expand it out, but that is time intensive and very error prone. Let's have the identity tackle this problem. We have
\begin{aligned}\big(\sqrt5+\sqrt6+\sqrt7\big)\big(\sqrt5+\sqrt6-\sqrt7\big)&=\big(\sqrt5+\sqrt6\big)^2-\big(\sqrt7\big)^2\\&=5+6+2\sqrt{30}-7\\&=4+2\sqrt{30}.\end{aligned}
Likewise, the product of the last two terms is
\begin{aligned} \big(\sqrt5-\sqrt6+\sqrt7\big)\big(-\sqrt5+\sqrt6+\sqrt7\big) &= \left(\sqrt7+\big(\sqrt5-\sqrt6\big)\right)\left(\sqrt7-\big(\sqrt5-\sqrt6\big)\right) \\ &=-4+2\sqrt{30}. \end{aligned}
The final product is
\left(4+2\sqrt{30}\right)\left(-4+2\sqrt{30}\right)=4(30)-16=104.
_\square
Try the following problems at your own.
\large 10^x=\Big(10^{624}+25\Big)^2-\Big(10^{624}-25\Big)^2
x
that satisfies the equation above?
123456789^{2} - (123456788 \times 123456790).
If you use a calculator whose precision is not strong enough to answer this question, then you will answer this problem incorrectly.
x=\frac{4}{\big(\sqrt{5}+1\big)\big(\sqrt[4]{5}+1\big)\big(\sqrt[8]{5}+1\big)\big(\sqrt[16]{5}+1\big)},
(1+x)^{48}?
If you were only given that
438271606^{2} = 192082000625819236,
can you find the sum of digits of the exact value of
561728395^{2}
Notes: Yes, don't use Wolfram Alpha or a calculator to solve this (though you may use a calculator to calculate the digit sum).
\begin{aligned} \large \color{#3D99F6}2^{\color{#624F41}x} - \color{#3D99F6}2^{\color{#624F41}y} & = & \large 1 \\ \\ \large \color{#20A900}4^{\color{#624F41}x}- \color{#20A900}4^{\color{#624F41}y} & = & \large { \frac 5 3} \\ \\ \\ \large {\color{#624F41}x} - {\color{#624F41}y} & = & \large \, ? \end{aligned}
x
y
x,y\in\Bbb{Z^+}
and they satisfy the equation given below:
\large\prod_{k=0}^5\left(5^{2^k}+6^{2^k}\right)=6^x-5^y.
x+y?
Note: This problem is not original. It is adapted from a question posted on Math SE.
(1+x)\big(1+x^2\big)\big(1+x^4\big) \ldots \big(1+x^{128}\big) = \displaystyle \sum_{r=0}^n x^r
Given the above equation, what is
n?
Cite as: Difference Of Squares. Brilliant.org. Retrieved from https://brilliant.org/wiki/applying-the-difference-of-two-squares-identity/
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Plate Buckling Analysis of Steel Shell Structures Using the MNA/LBA Concept | Dlubal Software
Home Support & Learning Support Knowledge Base Plate Buckling Analysis of Steel Shell Structures Using the MNA/LBA Concept
Plate Buckling Analysis of Steel Shell Structures Using the MNA/LBA Concept
Shell buckling is considered to be the most recent and least explored stability issue of structural engineering. This is due less to a lack of research activities than to the complexity of the theory. With the introduction and further development of the finite element method in structural engineering practice, some engineers no longer have to deal with the complicated theory of shell buckling. Evidence of the problems and errors to which this gives rise is very well summarized in [1].
In this article, we also highly recommend not simply creating a single FE model for each steel shell, setting up loads, then pressing "Calculate". In most cases, this procedure leads to additional work, as numerous analytical methods for the verification are available for simple cases that are common in design practice. These analytical methods, the hand calculation formulas, also have the great advantage of space-saving and simple documentation. For some containers, the plate buckling analysis can be performed on an A4 page. Such space-saving documentation is not possible with an FE analysis.
However, there are also numerous cases where the use of a finite element analysis makes sense and should be preferred over a manual calculation. The following points are just a few examples of cases where it makes sense to use an FE calculation:
local load introductions in the shell wall
discrete supports (that is, individual supports) of the shell
use of nonlinear design methods
In the following text, the buckling design of a steel shell is performed using RFEM according to the MNA/LBA concept. Thus, nonlinear material behavior of the steel is applied.
Plate Buckling Analysis According to EN 1993-1-6
In EN 1993-1-6, three options are presented for performing a plate buckling analysis for steel shells. In this section, they are to be briefly listed and evaluated with regard to the requirements of computing technology, as well as with regard to the requirements placed on the designing engineer.
Stress-Based Plate Buckling Analysis Stress-based plate buckling analysis is considered to be the standard analysis method that almost every engineer has used when performing shell design. An expert engineer classifies this method as easy and the computing requirements are either very low or non-existent, as formulas for manual calculation are frequently used.
A major problem of this analysis method is that economical results will be hardly achieved for shell structures with load situations that deviate considerably from the standard buckling modes. In addition, as a user of this concept, you are set on the wrong track by this method, because you could easily think that the plate buckling safety of the shell structure only depends on the occurring stresses. If that were the case, the stiffening of a shell wall by, for example, longitudinal ribs would have little benefit, as this does not significantly reduce stresses. In reality, the plate buckling safety of a skilfully stiffened shell is much higher than that of an unbraced shell of the same wall thickness.
Plate Buckling Analysis Based on Numerical Calculation by Means of Global MNA/LBA Calculation This method will be used for the following shell design. An MNA/LBA calculation certainly requires that the user have somewhat more background knowledge in shell stability than is the case for the stress-based design method. Moreover, the computing technique should be more powerful, since a linear elastic bifurcation analysis (LBA) and a material nonlinear analysis (MNA) have to be performed for correct application of this method.
In the author's view, this design method is the most effective way to perform plate buckling analysis if the calculation is to be carried out using the FE analysis. The justification for this is that for a design using the MNA/LBA concept, the computing technology is consistently used without expecting too much effort from the user. If the internal forces of the shell are calculated linear elastic to use them for the stress-based plate buckling analysis, the computer technology will be used too inconsistently, as powerful programs such as RFEM are also able to determine the load-bearing capacity of the shell structure.
Plate Buckling Analysis Based on Numerical Calculation by Means of Global GMNIA Calculation A GMNIA analysis to determine sufficient shell stability is probably the most consistent method of plate buckling analysis. The internal forces are thus calculated geometrically and materially non-linearly, using imperfections.
This method requires excellent background knowledge on shell stability from the user, as, among other things, the correct approach of imperfections (pre-buckling) is very difficult. If the user does not have this background knowledge, the design process with the GMNIA concept should be avoided in any case. Moreover, substantial demands are placed on the computer technology when using this method. Thus, the program system used must be able to perform a bifurcation analysis for each load step of the nonlinear analysis to, where appropriate, detect a "jump" from the subcritical pre-buckling path to the supercritical post-buckling path.
This concept will not be explained further here, since, in the author's view, it has little significance for the design practice. For more information, please refer to the article by Herbert Schmidt [2] in the Steel Construction Calendar of 2012, which gives a good overview of the difficulties faced when using the design according to the GMNIA method.
Example of Plate Buckling Analysis Using the MNA/LBA Method
Input of Structural System The steel shell shown in Figure 01 will be designed for buckling. In principle, this structure is a typical case where an engineer familiar with the design of steel shells would hardly consider an FE analysis. Since the principal aim of this article is to familiarize the reader with the topic of plate buckling analysis according to the MNA/LBA concept, an example will be used that is as simple as possible.
An important topic in nonlinear calculations or bifurcation analyses of shell structures is the element size, since unfavorably selected FE mesh settings can lead to falsified results. In specialized literature, various formulas for rough calculation exist for this, where the (small) convergence study is the most appropriate approach.
Calculation with RFEM After model and load input and selecting suitable FE mesh settings, the calculation with RFEM can start. First, the material nonlinear analysis is performed. The aim of this analysis is the plastic reference resistance (that is, the critical load factor at which the entire shell would fail plastically). The RF-MAT NL add-on module is ideally used, as only nonlinear material properties are then available in RFEM. Alternatively, a linear elastic calculation can be carried out; then the plastic reference resistance can be approximately calculated using formula (8.24) from [3]. Figure 02 shows the deformed system after reaching the plastic reference resistance rRpl = 11.90.
Subsequently, the linear bifurcation analysis is performed where the sequence was arbitrarily chosen here. It is also possible to perform this analysis first, then continue with the MNA method. The aim of the linear bifurcation analysis is also to obtain a critical load factor, but this time one which would cause the buckling of the perfect shell. This requires the RF-STABILITY add-on module, with which linear bifurcation analyses and geometrically nonlinear calculations can be performed. This does not refer to GMNIA calculations. Figure 03 shows the first mode shape of the considered shell for the eigenvalue of rRcr = 7.70.
Plate Buckling Analysis The plate buckling analysis is shown as a whole in the following text. Special attention is to be paid to the four independent buckling parameters, which can be determined for most practical construction cases according to Annex D in [3].
Plastic reference resistance from the MNA: rRpl = 11.9
Critical load factor from the LBA: rRcr = 7.70
Slenderness degree:
{\overline{\mathrm{\lambda }}}_{\mathrm{ov}} = \sqrt{\frac{{\mathrm{r}}_{\mathrm{Rpl}}}{{\mathrm{r}}_{\mathrm{Rcr}}}} = \sqrt{\frac{11.9}{7.70}} = 1.243
Elastic imperfection factor:
{\mathrm{\alpha }}_{\mathrm{ov}} \approx {\mathrm{\alpha }}_{\mathrm{x}} = \frac{0,62}{1+1,91 ยท {\left(\frac{{\mathrm{\Delta w}}_{\mathrm{k}}}{\mathrm{t}}\right)}^{1,44}} = \frac{0,62}{1+1,91 ยท {\left(0,98\right)}^{1,44}} = 0,217
Plastic multiplying factor: ฮฒov = 0.60
Buckling curve exponent: ฮทov = 0.60
Fully plastic limiting degree of slenderness:
{\overline{\mathrm{\lambda }}}_{0,\mathrm{ov}} = 0.20
Partially plastic limiting slenderness:
{\overline{\mathrm{\lambda }}}_{\mathrm{p},\mathrm{ov}} = \sqrt{\frac{{\mathrm{\alpha }}_{\mathrm{ov}}}{1 - {\mathrm{\beta }}_{\mathrm{ov}}}} = \sqrt{\frac{0.217}{1 - 0.60}} = 0.737
Buckling reduction factor:
\begin{array}{l}{\overline{\mathrm{\lambda }}}_{\mathrm{ov}} = 1.243 > {\overline{\mathrm{\lambda }}}_{\mathrm{p},\mathrm{ov}} = 0.737\\ \to \mathrm{Pure} \mathrm{elastic} \mathrm{buckling} \mathrm{is} \mathrm{available}.\\ {\mathrm{\chi }}_{\mathrm{ov}} = \frac{{\mathrm{\alpha }}_{\mathrm{ov}}}{{\overline{\mathrm{\lambda }}}_{\mathrm{ov}}^{2}} = \frac{0.217}{1.{243}^{2}} = 0.140\end{array}
Plate buckling analysis:
\begin{array}{l}{\mathrm{r}}_{\mathrm{d}} = \frac{{\mathrm{\chi }}_{\mathrm{ov}} ยท {\mathrm{r}}_{\mathrm{Rpl}}}{{\mathrm{\gamma }}_{\mathrm{M}1}} = \frac{0.140 ยท 11.9}{1.10} = 1.515 > 1.0\\ \to \mathrm{Design} \mathrm{is} \mathrm{fulfilled}.\end{array}
The main issue of the design is to classify the results obtained by the program into one of the typical buckling cases. In the present case, it is very simple due to the loading: It is about pure meridian pressure buckling. Thus, the independent buckling parameters according to Annex D 1.2 in EN 1993-1-6 [3] are calculated.
The result of the plate buckling analysis according to the MNA/LBA method is a critical load factor. In the example shown here, it is 1.515. This means: The load of the shell could be increased more than 50%.
If the analysis is based on the stress-based concept, this would result in a critical load factor of 1.398, which shows that for typical buckling cases, such as the meridian pressure buckling considered here, no additional benefits are gained by the numerically based plate buckling analysis according to the MNA/LBA method. It should be noted, as already mentioned, that this is different as soon as local load introductions or supports lead to stress concentrations.
Modern, powerful, and user-friendly FEM programs such as RFEM significantly facilitate the work of a calculating engineer when performing the design of the sufficient buckling safety of a shell. As a result of the more consistent use of computer technology in the MN /LBA concept, more realistic and therefore more economical results can generally be achieved.
It should also be mentioned that an FE analysis is not advisable for every shell structure, as good analytical methods are available for typical buckling cases, which can lead to reduced documentation and to similarly economical results. However, if the engineer encounters cases in the design practice that cannot be assigned to a typical buckling case, an FE analysis according to the MNA/ LBA concept with RFEM with the RF-STABILITY and RF-MAT NL add-on modules is a real alternative to the standard methods.
[1] Knรถdel, P .; Ummenhofer, T .: Rules for the calculation of tanks with the FEM, Stahlbau 86, pages 325 - 339. Berlin: Ernst & Sohn, 2017
[2] Schmidt, H.: Stahlbaunormen โ Kommentar zur DIN EN 1993-1-6: Festigkeit und Stabilitรคt von Schalen, Stahlbau-Kalender 2012, pp. 135 - 204. Berlin: Ernst & Sohn, 2012
[3] Eurocode 3 - Design of Steel Structures - Part 1-6: Strength and Stability of Shell Structure; EN 1993-1-6:2007 + AC:2009 + A1:2017
Nonlinear Structural Analysis Software
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Pentagonal_number Knowpia
A pentagonal number is a figurate number that extends the concept of triangular and square numbers to the pentagon, but, unlike the first two, the patterns involved in the construction of pentagonal numbers are not rotationally symmetrical. The nth pentagonal number pn is the number of distinct dots in a pattern of dots consisting of the outlines of regular pentagons with sides up to n dots, when the pentagons are overlaid so that they share one vertex. For instance, the third one is formed from outlines comprising 1, 5 and 10 dots, but the 1, and 3 of the 5, coincide with 3 of the 10 โ leaving 12 distinct dots, 10 in the form of a pentagon, and 2 inside.
A visual representation of the first six pentagonal numbers
pn is given by the formula:
{\displaystyle p_{n}={\frac {3n^{2}-n}{2}}={\binom {n}{1}}+3{\binom {n}{2}}}
for n โฅ 1. The first few pentagonal numbers are:
The nth pentagonal number is the sum of n integers starting from n (i.e. from n to 2n-1). The following relationships also hold:
{\displaystyle p_{n}=p_{n-1}+3n-2=2p_{n-1}-p_{n-2}+3}
Pentagonal numbers are closely related to triangular numbers. The nth pentagonal number is one third of the (3n โ 1)th triangular number. In addition, where Tn is the nth triangular number.
{\displaystyle p_{n}=T_{n-1}+n^{2}=T_{n}+2T_{n-1}=T_{2n-1}-T_{n-1}}
Generalized pentagonal numbers are obtained from the formula given above, but with n taking values in the sequence 0, 1, โ1, 2, โ2, 3, โ3, 4..., producing the sequence:
0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335... (sequence A001318 in the OEIS).
Generalized pentagonal numbers are important to Euler's theory of partitions, as expressed in his pentagonal number theorem.
Generalized pentagonal numbers and centered hexagonal numbersEdit
Generalized pentagonal numbers are closely related to centered hexagonal numbers. When the array corresponding to a centered hexagonal number is divided between its middle row and an adjacent row, it appears as the sum of two generalized pentagonal numbers, with the larger piece being a pentagonal number proper:
{\displaystyle 3n(n-1)+1={\tfrac {1}{2}}n(3n-1)+{\tfrac {1}{2}}(1-n){\bigl (}3(1-n)-1{\bigr )}}
where both terms on the right are generalized pentagonal numbers and the first term is a pentagonal number proper (n โฅ 1). This division of centered hexagonal arrays gives generalized pentagonal numbers as trapezoidal arrays, which may be interpreted as Ferrers diagrams for their partition. In this way they can be used to prove the pentagonal number theorem referenced above.
Proof without words that a pentagonal number can be decomposed into three triangular numbers and a natural number
Tests for pentagonal numbersEdit
Given a positive integer x, to test whether it is a (non-generalized) pentagonal number we can compute
{\displaystyle n={\frac {{\sqrt {24x+1}}+1}{6}}.}
The number x is pentagonal if and only if n is a natural number. In that case x is the nth pentagonal number.
For generalized pentagonal numbers, it is sufficient to just check if 24x + 1 is a perfect square.
For non-generalized pentagonal numbers, in addition to the perfect square test, it is also required to check if
{\displaystyle {\sqrt {24x+1}}\equiv 5\mod 6}
The mathematical properties of pentagonal numbers ensure that these tests are sufficient for proving or disproving the pentagonality of a number.[1]
GnomonEdit
The Gnomon of the nth pentagonal number is:
{\displaystyle p_{n+1}-p_{n}=3n+1}
Square pentagonal numbersEdit
A square pentagonal number is a pentagonal number that is also a perfect square.[2]
0, 1, 9801, 94109401, 903638458801, 8676736387298001, 83314021887196947001, 799981229484128697805801, 7681419682192581869134354401, 73756990988431941623299373152801... (OEIS entry A036353)
^ How do you determine if a number N is a Pentagonal Number?
^ Weisstein, Eric W. "Pentagonal Square Number." From MathWorld--A Wolfram Web Resource.
Leonhard Euler: On the remarkable properties of the pentagonal numbers
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Integration of Logarithmic Functions | Brilliant Math & Science Wiki
Aditya Virani, A Former Brilliant Member, Satyajit Mohanty, and
The derivative of the logarithm
\ln x
\frac{1}{x}
, but what is the antiderivative? This turns out to be a little trickier, and has to be done using a clever integration by parts.
The logarithm is a basic function from which many other functions are built, so learning to integrate it substantially broadens the kinds of integrals we can tackle.
\ln x
Integrating Functions of
\ln x
\int\ln x\, dx=x\ln x-x+C
using Taylor Series
\ln x
\int \ln(x)\ dx = x\ln (x) - x + C
C
is the constant of integration, and this notation
C
will be used throughout the wiki.
For this solution, we will use integration by parts:
\int f(x) g'(x)\ dx = f(x) g(x) - \int f'(x) g(x)\ dx.
f(x)=\ln(x)
g'(x)=1
g(x)=x
. Plugging these into our integration by parts formula, we get
\begin{aligned} \int 1 \cdot \ln(x)\ dx&=x \ln(x) - \int \big(\ln(x)\big)' x\ dx\\&=x \ln(x) - \int \frac{x}{x}\ dx\\&=x \ln(x) - x + C. \end{aligned}
We can factorize a bit and get the desired formula
\int \ln(x)\ dx = x\big(\ln(x) - 1\big) + C.\ _\square
This shows that an unlikely application of an integration technique can actually be the right way forward!
Now that we know how to integrate this, let's apply the properties of logarithms to see how to work with similar problems.
\displaystyle{\int \ln 2x \, dx}
According to the properties of logarithms, we know that
\ln 2x=\ln x+\ln2,
\begin{aligned} \int\ln2x~dx&=\int\left(\ln x+\ln2\right)~dx\\ &=\int\ln x~dx+\int\ln2~dx\\ &=x\ln x-x+x\ln2+C.\ _\square \end{aligned}
\displaystyle{\int\log x~dx}.
According to the properties of logarithms, we have
\log x=\frac{\ln x}{\ln10}.
Hence the given integral can be rewritten as
\int\log x~dx=\int\frac{\ln x}{\ln10}~dx=\frac{1}{\ln10}x(\ln x-1)+C.\ _\square
When integrating the logarithm of a polynomial with at least two terms, the technique of
u
-substitution is needed. The following are some examples of integrating logarithms via U-substitution:
\displaystyle{ \int \ln (2x+3) \, dx}
For this problem, we use
u
u=2x+3.
du=2dx,
dx=\frac{1}{2}du,
and the given integral can be rewritten as follows:
\begin{aligned} \int\ln(2x+3)~dx&=\frac{1}{2}\int\ln u~du\\ &=\frac{1}{2}u(\ln u-1)+C\\ &=\frac{2x+3}{2}\big(\ln(2x+3)-1\big)+C.\ _\square \end{aligned}
\displaystyle{\int \ln (x-2)^3 \, dx}
\int \ln (x-2)^3 \, dx=3\int\ln(x-2)~dx.
u=x-2.
du=dx,
and the integral can be rewritten as follows:
\begin{aligned} \int \ln (x-2)^3 \, dx&=3\int\ln(x-2)~dx\\ &=3\int\ln u~du\\ &=3u(\ln u-1)+C\\ &=3(x-2)\big(\ln(x-2)-1\big)+C.\ _\square \end{aligned}
\ln x
We now look at examples where we're integrating functions of
\ln x
. These problems often require familiarity with integration by parts,
u
-substitution and
\ln |f|
\displaystyle{\int x \ln x \, dx}
To solve this, we use the principle of integration by parts. Let
u=\ln x
v'=x.
\begin{aligned} \int x\ln x~dx&=\int v'u~dx\\ &=uv-\int vu'~dx\\ &=\frac{1}{2}x^2\ln x-\int\frac{1}{2}x^2\cdot\left(\ln x\right)'~dx\\ &=\frac{1}{2}x^2\ln x-\int\frac{1}{2}x~dx\\ &=\frac{1}{2}x^2\ln x-\frac{1}{4}x^2+C.\ _\square \end{aligned}
\int x^m \ln x \, dx = x^{m+1} \left( \frac{ \ln x } { m+1 } - \frac{1}{ (m+1)^2 } \right) + C.
The proof of this is similar to the above.
\displaystyle{\int \frac{ \ln x } { x} \, dx}
To solve this, we use
u
u=\ln x.
du=\frac{1}{x}dx,
\int \frac{ \ln x } { x} \, dx=\int \frac{u}{x}~dx=\int u~du=\frac{1}{2}u^2+C=\frac{1}{2}(\ln x)^2+C.\ _\square
\int \frac{ (\ln x)^n }{x} \, dx = \frac{ (\ln x ) ^ { n + 1 } } { n+1} , \quad n \neq -1.
\displaystyle{\int \frac{ 1} { x \ln x } \, dx = \ln \lvert \ln x \rvert+C}
Perceive
\frac{1}{x\ln x}
\frac{\frac{1}{x}}{\ln x}.
(\ln x)'=\frac{1}{x},
the given integral will give the
\ln\lvert f\rvert
form as follows:
\int \frac{ 1} { x \ln x } \, dx=\ln\lvert\ln x\rvert+C.\ _\square
Alternative Solution: We can also use the substitution
u=\ln x.
du=\frac{1}{x}dx,
\int \frac{ 1} { x \ln x } \, dx=\int\frac{1}{\ln x}\cdot\frac{1}{x}~dx=\int\frac{1}{u}~du=\ln\lvert u\rvert+C=\ln\lvert\ln x\rvert+C.\ _\square
\displaystyle{\int ( \ln x ) ^2 \, dx}
u=(\ln x)^2
v'=1.
v=x
u'=\frac{2\ln x}{x}
\begin{aligned} \int ( \ln x ) ^2 \, dx&=\int uv'~dx\\ &=uv-\int u'v~dx\\ &=x(\ln x)^2-\int\frac{2\ln x}{x}\cdot x~dx\\ &=x(\ln x)^2-\int2\ln x~dx. \end{aligned}
\int\ln x~dx=x\ln x-x+C,
which we have learned above, we have
x(\ln x)^2-\int2\ln x~dx=x(\ln x)^2-2x\ln x+2x+C.\ _\square
Some problems can be very complex so that they require using both integration by parts and
u
\displaystyle{\int \sin ( \ln x ) \, dx}
We use the substitution
t=\ln x.
dt=\frac{1}{x}dx,
dx=xdt,
\int \sin ( \ln x ) \, dx=\int x\sin t~dt.
From the relationship
t=\ln x
e^t=x.
\int x\sin t~dt=\int e^t\sin t~dt.
u=\sin t
v'=e^t.
u'=\cos t
v=e^t,
\begin{aligned} \int e^t\sin t~dt&=\int uv'~dt\\ &=uv-\int u'v~dt\\ &=e^t\sin t-\int e^t\cos t~dt. \qquad(1) \end{aligned}
Using integration by parts once again, we have
\int e^t\cos t~dt=e^t\cos t+\int e^t\sin t~dt,
substituting which into
(1)
\begin{aligned} \int e^t\sin t~dt&=e^t\sin t-\int e^t\cos t~dt\\&=e^t\sin t-e^t\cos t-\int e^t\sin t~dt\\ \Rightarrow 2\int e^t\sin t~dt&=e^t\sin t-e^t\cos t\\ \int e^t\sin t~dt&=\frac{e^t(\sin t-\cos t)}{2}+C. \end{aligned}
Now we finally have
\begin{aligned} \int \sin ( \ln x ) \, dx &=\frac{e^t(\sin t-\cos t)}{2}+C\\ &=\frac{e^{\ln x}\big(\sin(\ln x)-\cos(\ln x)\big)}{2}+C.\ _\square \end{aligned}
\int\ln x\, dx=x\ln x-x+C
If we don't want to use integration by parts, we can also solve our original integral using Taylor expansion.
We know that the Taylor series expansion of
\ln x
\ln x=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}+\cdots. \qquad(1)
By integrating both sides, we get
\int\ln x~dx=\frac{(x-1)^2}{2}-\frac{(x-1)^3}{6}+\frac{(x-1)^4}{12}-\frac{(x-1)^5}{20}+\cdots.\qquad(2)
We want to compare this with the Taylor series expansion of
x\ln x-x.
Multiplying both sides of
(1)
x-1
(x-1)\ln x=(x-1)^2-\frac{(x-1)^3}{2}+\frac{(x-1)^4}{3}-\frac{(x-1)^5}{4}+\cdots.
Then we add
(1)
to both sides to obtain
x\ln x=(x-1)+\frac{(x-1)^2}{2}-\frac{(x-1)^3}{6}+\frac{(x-1)^4}{12}-\frac{(x-1)^5}{20}+\cdots.
Finally, subtracting
x
x\ln x-x=-1+\frac{(x-1)^2}{2}-\frac{(x-1)^3}{6}+\frac{(x-1)^4}{12}-\frac{(x-1)^5}{20}+\cdots,
which is identical with
(2)
except for the constant term (which is irrelevant, since there is always a constant of integration). Therefore we can conclude that
\int\ln x~dx=x\ln x-x+C.\ _\square
Cite as: Integration of Logarithmic Functions. Brilliant.org. Retrieved from https://brilliant.org/wiki/integration-of-logarithmic-functions/
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When 88 cubes are rearranged to form the cube-sphere in the diagram above, their total surface area decreases by 384.
Find the total surface area of the cube-sphere.
Note: The cube-sphere is solid.
by Lee Care Gene
Beginning with a sphere of radius 1, what is the minimum amount of planar 'slices' through the sphere so that the total surface area of the resultant pieces is greater than 50? Here's one potential method:
If the picture is too small, right click/two-finger click it and open it in a new tab.
In this above rocket picture,
The top black line serves as the height of the cone portion of the rocket and the bottom black line serves as the height of the "lamp shade" portion of the rocket. Both black lines have a length of 5 inches.
The red line serves as the height of the cylindrical portion of the rocket and has a length of 10 inches.
The green line serves as the radius of the cone portion and cylindrical portion. The green line also serves as the smaller radius of the lampshade portion. The length of the green line is 1 inch.
The blue line serves as the larger radius of the lampshade portion, and has a length of 2 inches.
If the surface area of the rocket (in square inches) equals
SA
\left \lfloor SA \right \rfloor
Note: This "rocket" has a sealed base. Please include the base in your answer.
You're making a megaphone by wrapping a piece of paper up as a simple cone and then cutting it at half of its height. The 2 new circular bases are parallel and left open at both ends for the air to flow through.
If this megaphone has a height of 24 cm. with a radius of 14 cm. at its bigger base, find its lateral surface area in
\text{cm}^2
If your answer is in a form of
\pi\times A
, then submit
A
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Array stored on GPU - MATLAB - MathWorks France
Transfer Data to and from the GPU
Create Data on the GPU Directly
Perform Monte Carlo Integration Using gpuArray-Enabled Functions
Array stored on GPU
A gpuArray object represents an array stored in GPU memory. A large number of functions in MATLABยฎ and in other toolboxes support gpuArray objects, allowing you to run your code on GPUs with minimal changes to the code. To work with gpuArray objects, use any gpuArray-enabled MATLAB function such as fft, mtimes or mldivide. To find a full list of gpuArray-enabled functions in MATLAB and in other toolboxes, see GPU-supported functions. For more information, see Run MATLAB Functions on a GPU.
If you want to retrieve the array from the GPU, for example when using a function that does not support gpuArray objects, use the gather function.
You can load MAT files containing gpuArray data as in-memory arrays when a GPU is not available. A gpuArray object loaded without a GPU is limited and you cannot use it for computations. To use a gpuArray object loaded without a GPU, retrieve the contents using gather.
Use gpuArray to convert an array in the MATLAB workspace into a gpuArray object. Some MATLAB functions also allow you to create gpuArray objects directly. For more information, see Establish Arrays on a GPU.
G = gpuArray(X)
G = gpuArray(X) copies the array X to the GPU and returns a gpuArray object.
X โ Array
Array to transfer to the GPU, specified as a numeric or logical array. The GPU device must have sufficient free memory to store the data. If X is already a gpuArray object, gpuArray outputs X unchanged.
You can also transfer sparse arrays to the GPU. gpuArray supports only sparse arrays of double-precision.
Example: G = gpuArray(magic(3));
There are several methods for examining the characteristics of a gpuArray object. Most behave like the MATLAB functions of the same name.
To transfer data from the CPU to the GPU, use the gpuArray function.
Create an array X.
Transfer X to the GPU.
Check that the data is on the GPU.
isgpuarray(G)
Calculate the element-wise square of the array G.
GSq = G.^2;
Transfer the result GSq back to the CPU.
XSq = gather(GSq)
XSq = 1ร3
Check that the data is not on the GPU.
isgpuarray(XSq)
You can create data directly on the GPU directly by using some MATLAB functions and specifying the option "gpuArray".
Create an array of random numbers directly on the GPU.
G = rand(1,3,"gpuArray")
Check that the output is stored on the GPU.
This example shows how to use MATLAB functions and operators with gpuArray objects to compute the integral of a function by using the Monte Carlo integration method.
Define the number of points to sample. Sample points in the domain of the function, the interval [-1,1] in both x and y coordinates, by creating random points with the rand function. To create a random array directly on the GPU, use the rand function and specify "gpuArray". For more information, see Establish Arrays on a GPU.
x = 2*rand(n,1,"gpuArray")-1;
y = 2*rand(n,1,"gpuArray")-1;
Define the function to integrate, and use the Monte Carlo integration formula on it. This function approximates the value of
\pi
by sampling points within the unit circle. Because the code uses gpuArray-enabled functions and operators on gpuArray objects, the computations automatically run on the GPU. You can perform binary operations such as element-wise multiplication using the same syntax that you use for MATLAB arrays. For more information about gpuArray-enabled functions, see Run MATLAB Functions on a GPU.
f = x.^2 + y.^2 <= 1;
result = 4*1/n*f*ones(n,1,"gpuArray")
If you need better performance, or if a function is not available on the GPU, gpuArray supports the following options:
To precompile and run purely element-wise code on gpuArray objects, use the arrayfun function.
To run C++ code containing CUDAยฎ device code or library calls, use a MEX function. For more information, see Run MEX-Functions Containing CUDA Code.
To run existing GPU kernels written in CUDA C++, use the MATLAB CUDAKernel interface. For more information, see Run CUDA or PTX Code on GPU.
To generate CUDA code from MATLAB code, use GPU Coderโข. For more information, see Get Started with GPU Coder (GPU Coder).
You can control the random number stream on the GPU using gpurng.
None of the following can exceed intmax("int32"):
The number of elements of a dense array.
The number of nonzero elements of a sparse array.
The size in any given dimension. For example, zeros(0,3e9,"gpuArray") is not allowed.
You can also create a gpuArray object using some MATLAB functions by specifying a gpuArray output. The following table lists the MATLAB functions that enable you to create gpuArray objects directly. For more information, see the Extended Capabilities section of the function reference page.
eye(___,"gpuArray") true(___,"gpuArray")
false(___,"gpuArray") zeros(___,"gpuArray")
Inf(___,"gpuArray") gpuArray.colon
NaN(___,"gpuArray") gpuArray.freqspace
ones(___,"gpuArray") gpuArray.linspace
rand(___,"gpuArray") gpuArray.logspace
randi(___,"gpuArray") gpuArray.speye
randn(___,"gpuArray")
isgpuarray | canUseGPU | arrayfun | gpuDevice | existsOnGPU | gather | reset | pagefun | gputimeit
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Nonnegative Combined Matrices
Rafael Bru, Maria T. Gassรณ, Isabel Gimรฉnez, Mรกximo Santana, "Nonnegative Combined Matrices", Journal of Applied Mathematics, vol. 2014, Article ID 182354, 5 pages, 2014. https://doi.org/10.1155/2014/182354
Rafael Bru ,1 Maria T. Gassรณ ,1 Isabel Gimรฉnez,1 and Mรกximo Santana2
1Institut de Matemร tica Multidisciplinar, Universitat Politรจcnica de Valรจncia, 46022 Valรจncia, Spain
2Universidad Autรณnoma de Santo Domingo, Santo Domingo 10105, Dominican Republic
Academic Editor: Panayiotis Psarrakos
The combined matrix of a nonsingular real matrix is the Hadamard (entrywise) product . It is well known that row (column) sums of combined matrices are constant and equal to one. Recently, some results on combined matrices of different classes of matrices have been done. In this work, we study some classes of matrices such that their combined matrices are nonnegative and obtain the relation with the sign pattern of . In this case the combined matrix is doubly stochastic.
Fiedler and Markham [1] studied matrices of the form , that is, the combined matrix of , where is a nonsingular matrix and means the Hadamard product. Combined matrices appear in the chemical literature where they represent the relative gain array (see [2]). Furthermore, the combined matrix gives the relation between the eigenvalues and diagonal entries of a diagonalizable matrix (see [3]). Results for the combined matrix of a nonsingular matrix and also for the Hadamard product have been obtained, for instance, in [1] where the behavior of the diagonal entries of the combined matrix of a nonsingular matrix was completely described and in [4] for the positive definite case.
It is well known [3] that the row and column sums of a combined matrix are always equal to one. Then, if is a nonnegative matrix, it has interesting properties and applications since it is a doubly stochastic matrix. For instance, in [5], there are two applications: the first one concerning a topic in communication theory called satellite-switched and the second concerning a recent notion of doubly stochastic automorphism of a graph. Recently, in [6], some implications on nonnegative matrices, doubly stochastic matrices, and graph theory, namely, graph spectra and graph energy, are presented.
Here, we focus our work on studying which matrices have nonnegative combined matrices. More precisely, we study the combined matrix of different classes of matrices such as totally positive and totally negative matrices and also when is totally nonnegative and totally nonpositive.
2. Notation and Previous Results
Unless otherwise indicated, in this work all square matrices are nonsingular and real. Given an matrix , we denote by the subset of indexes . For the subsets , the submatrix with rows lying in the subset and columns in the subset is denoted by , and the principal submatrix is abbreviated to . Similarly, denotes the submatrix obtained from by deleting rows lying in and columns in , and is abbreviated to . Then, denotes the submatrix obtained from by deleting row and column , and . Moreover, denotes the minor; that is, .
To avoid confusion with other matrices we will say that is a nonnegative (positive) matrix if it is entrywise nonnegative (positive); that is, () for all , and we will denote it by ). Similar notation will be used for the nonpositive (negative) case; that is, () for all , and we will denote it by ().
Now we recall some classes of matrices that we are working on.
Definition 1. An real matrix is said to have a checkerboard pattern if or for all . When no entry is zero, one will say that is strictly checkerboard.
Definition 2. A matrix is said to be totally positive (negative) if all its minors of any order are positive (negative). That is, for every subset , (). It is denoted by TP (TN).
Definition 3. A matrix is called totally nonpositive (nonnegative) if all its minors of any order are nonpositive (nonnegative). That is, for every subset , (). It is denoted by TNP (TNN).
Definition 4. An real matrix is said to be a -matrix if, for every subset , .
Definition 5. An real matrix is called an -matrix if can be written as , with and , where denotes the spectral radius of matrix , that is, the biggest absolute value of the eigenvalues of .
For a matrix , denotes its comparison matrix; that is,
Definition 6. An complex matrix is called an -matrix if its comparison matrix is an -matrix.
Note that nonsingular -matrices having singular comparison matrix are included in this definition (see [7]).
Remember that the Hadamard (or entrywise) product of two matrices and is the matrix .
Definition 7. The combined matrix of a nonsingular real matrix is defined as . Then, if , The elements of will be denoted by .
It is clear that the combined matrix has the following properties: (i) and (ii) is doubly stochastic if is nonnegative.
3. Matrices with
Looking at the definition of a combined matrix it is easy to see that the combined matrix of a triangular matrix is the identity matrix. Moreover, the following result from Lemma (page ) of [3] can be established.
Lemma 8. For any nonsingular matrix , for any two permutation matrices and , and for any triangular matrix one has(i),(ii); that is, is a permutation matrix,(iii).
From Definition 7 one can observe that, for and , if and only if the matrix is permutationally similar to a triangular matrix. The natural question is whether or not this is a general equivalence. As it is suggested in [3] (Problem 2, page 302) the equivalence is not true for some matrices with as the following example shows.
Example 9. Consider the matrix It is easy to compute and to see that . However, is not permutationally similar to a triangular matrix since is irreducible.
Proposition 10. Let be a nonsingular matrix. If the combined matrix is nonnegative and triangular, then .
Proof. Recall that is doubly stochastic since it is nonnegative. Suppose is upper triangular. In this case is the only positive entry of the first column and the first row. Reasoning as before with the other columns of we conclude that .
A simple case to determine matrices having their combined matrix is given for matrices in the following result.
Proposition 11. Let be a nonsingular matrix. Then In particular, (i) if , then , (ii) if , then , and finally (iii) if , then .
Proof. Since then () If , then and . Therefore, .
() Suppose first that . If , since , then , in which case . Then and and hence . Similarly, if , we obtain and and then .
The two first particular cases come from the fact that is doubly stochastic, and the last case is because all entries are different from zero.
Example 12. Consider
The positivity of the matrix may play an important role in our case. According to the definition of the combined matrix, we have the following.
Theorem 13. Let (). Then if and only if ().
Matrices with the property described in Theorem 13 are necessarily nonnegative monomial matrices, as are the corresponding combined matrices. Monomial matrices are indeed permutationally similar to diagonal matrices and thus are orthogonal matrices. Below we prove a related result about a type of orthogonal matrices that was recently introduced in [8].
Definition 14. A nonsingular matrix is called a -matrix if two nonsingular diagonal matrices and exist such that .
Theorem 15. Let be a nonsingular -matrix such that and . Then, the combined matrix of is nonnegative.
Proof. In this case, we have which is nonnegative.
Example 16. The matrix is a -matrix, since is the result of , where . Then, the combined matrix is
It is well known that the combined matrix of an -matrix is an -matrix (see [3]). The following result gives the equivalent conditions to have a nonnegative combined matrix of an -matrix.
Theorem 17. Let A be an -matrix. Then, the following conditions are equivalent:(i),(ii) is diagonal,(iii),(iv) for all .
Proof. Since is -matrix, then is nonnegative and so the off-diagonal elements of are nonpositive and the diagonal elements are positive. Then, if and only if is diagonal. The last two equivalences follow because the combined matrix is doubly stochastic.
Despite the fact that the nonnegativity of the combined matrix of an -matrix is reduced to the identity matrix, it is easy to find -matrices for which its combined matrix is nonnegative and different from the identity matrix as the following example shows.
Example 18. The nonsingular matrices are -matrices and each one has positive combined matrix: It should be noted that is nonsingular but is a singular -matrix.
Now, we study the positivity of the combined matrix of totally positive and totally negative matrices.
Theorem 19. If is a TP-matrix, then is strictly checkerboard. In addition, if is a TN-matrix, then is strictly checkerboard.
Proof. It is straightforward noting that is checkerboard in both cases.
Then, the combined matrix of a TP-matrix or a TN-matrix is not nonnegative. However, we have the following result.
Theorem 20. If is a TP-matrix (TN-matrix), then (), where .
To study the combined matrix of a TNN-matrix, we need some auxiliary results.
Proposition 21 (see [9], Corollary 3.8). If is a nonsingular TNN-matrix, then is a P-matrix.
Proposition 22. If is a nonsingular TNN-matrix and for some , then .
Proof. If , then , which is positive since is a -matrix by Proposition 21. For , consider the submatrix and let us work by contradiction. Suppose that . Since is a nonsingular TNN-matrix, then and . Then which is a contradiction.
Proposition 23. If is a nonsingular TNN-matrix, then the following conditions are equivalent:(i),(ii), , .
Proof. Since is TNN, and for all . Then Therefore, is nonnegative if and only if , when .
Proposition 24 (see [9], Theorem 3.3). If is a nonsingular TNN-matrix, then is also a nonsingular TNN-matrix, where .
Then we can establish the following result.
Theorem 25. Let be a nonsingular TNN-matrix. Then the following conditions are equivalent:(i) is nonnegative,(ii).
Proof. . If , it is obvious that is nonnegative.
. Let us suppose that is nonnegative. We note that whenever , , by Proposition 23. Then it remains to prove that whenever and . We work by contradiction. For this, we suppose that there exists with and . By Proposition 24โis a nonsingular TNN-matrix. Applying Proposition 22 to both and we have that and . Therefore, , with , and this contradicts the result of Proposition 23. Thus, when , we have that . Then, is a lower triangular and nonnegative matrix. Hence, by Proposition 10.
Now, let us figure out the combined matrix of a TNP-matrix. Since all minors of TNP-matrices are nonpositive, it is clear that the combined matrix of a nonsingular TNP-matrix is not, in general, nonnegative. In fact, we have this simple result.
Theorem 26. Let be a nonsingular TNP-matrix. Then is nonnegative if and only if , whenever , .
However, there exist TNP-matrices with nonnegative combined matrices as the following example shows.
Example 27. Consider the totally nonpositive matrix The combined matrix is which is nonnegative.
The question now is to know whether or not there are more TNP-matrices with nonnegative combined matrices. As we see below, only TNP-matrices of size may have this property. To prove this we need some auxiliary results concerning TNP-matrices.
Proposition 28. Let be an nonsingular TNP-matrix with . If there exists an index such that , then .
Proof. Suppose by contradiction that there is an and for some . Since is nonsingular, there exists an index such that and then which is a contradiction with the signature of the matrix (see [10]).
Corollary 29. Let be an nonsingular TNP-matrix with .(i)If for some , then for all .(ii)If for some , then , .
Proof. (i) The proof follows from Proposition 28.
(ii) Since is TNP, for all (see [11], Theorem 2.1 (i)). Then, using part (i) of this corollary, we conclude that whenever .
In the following theorem we show how the first row and the second column of a nonnegative combined matrix of a TNP-matrix are.
Theorem 30. Let be an nonsingular TNP-matrix with . If is nonnegative, then the first row of is and the second column is .
Proof. Again, since is TNP, for all . Furthermore, , , by Theorem 26 and for by Proposition 28. Then, the proof follows since is doubly stochastic and .
Now, we can give the main theorem on combined matrices of TNP-matrices; that is, we are going to prove that there does not exist any nonsingular TNP-matrix of size such that its combined matrix is nonnegative.
Theorem 31. Let be an nonsingular TNP-matrix with . Then is not nonnegative.
Proof. Let us work by contradiction and suppose that there exists a nonsingular TNP-matrix of size , with and . By Theorem 30 we know the structure of the first row and the second column of . Let us focus on its first column where its first entry is . Since is nonnegative and so doubly stochastic, then there must exist an index such that , and then . Further, by Theorem 30 and Theorem 2.1 (i) of [11] and by Theorem 30. Then This is a contradiction with signature of according to [10].
Finally, we prove what kind of totally nonpositive matrices of size have their combined matrix nonnegative in the following result.
Theorem 32. Let the TNP-matrix be where , , , are nonnegative. Then if and only if at least one of the entries or is zero. In this case, .
Proof. The computation of the combined matrix gives Then, since , if and only if . In this case, and .
In brief, only antitriangular TNP-matrices have their combined matrix nonnegative.
The authors would like to thank the referees for their suggestions that have improved the reading of this paper. This research is supported by Spanish DGI (Grant no. MTM2010-18674).
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Copyright ยฉ 2014 Rafael Bru et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Global Constraint Catalog: Clex_lesseq
<< 5.231. lex_less5.233. lex_lesseq_allperm >>
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}\left(\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1},\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}\right)
\mathrm{๐๐๐ก๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐}
\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1},\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2},\mathrm{๐๐๐}\right)
|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|=|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}|
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
\stackrel{\to }{X}
\stackrel{\to }{Y}
n
โฉ{X}_{0},\cdots ,{X}_{n-1}โช
โฉ{Y}_{0},\cdots ,{Y}_{n-1}โช
\stackrel{\to }{X}
\stackrel{\to }{Y}
n=0
{X}_{0}<{Y}_{0}
{X}_{0}={Y}_{0}
โฉ{X}_{1},\cdots ,{X}_{n-1}โช
โฉ{Y}_{1},\cdots ,{Y}_{n-1}โช
\left(โฉ5,2,3,1โช,โฉ5,2,6,2โช\right)
\left(โฉ5,2,3,9โช,โฉ5,2,3,9โช\right)
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
constraints associated with the first and second examples hold since:
Within the first example
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}=โฉ5,2,3,1โช
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}=โฉ5,2,6,2โช
Within the second example
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}=โฉ5,2,3,9โช
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}=โฉ5,2,3,9โช
|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|>1
\bigvee \left(\begin{array}{c}|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|<5,\hfill \\ \mathrm{๐๐๐๐}\left(\left[\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}.\mathrm{๐๐๐},\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}.\mathrm{๐๐๐}\right]\right)<2*|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|\hfill \end{array}\right)
\bigvee \left(\begin{array}{c}\mathrm{๐๐๐ก๐๐๐}\left(\left[\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}.\mathrm{๐๐๐},\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}.\mathrm{๐๐๐}\right]\right)\le 1,\hfill \\ 2*|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|-\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}\left(\left[\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}.\mathrm{๐๐๐},\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}.\mathrm{๐๐๐}\right]\right)>2\hfill \end{array}\right)
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}.\mathrm{๐๐๐}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}.\mathrm{๐๐๐}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
The following reformulations in term of arithmetic and/or logical expressions exist for enforcing the lexicographically less than or equal to constraint. The first one converts
\stackrel{\to }{X}
\stackrel{\to }{Y}
into numbers and post an inequality constraint. It assumes all components of
\stackrel{\to }{X}
\stackrel{\to }{Y}
to be within
\left[0,a-1\right]
{a}^{n-1}{X}_{0}+{a}^{n-2}{X}_{1}+\cdots +{a}^{0}{X}_{n-1}\le {a}^{n-1}{Y}_{0}+{a}^{n-2}{Y}_{1}+\cdots +{a}^{0}{Y}_{n-1}
Since the previous reformulation can only be used with small values of
n
a
, W. Harvey came up with the following alternative model that maintains arc-consistency:
\left({X}_{0}<{Y}_{0}+\left({X}_{1}<{Y}_{1}+\left(\cdots +\left({X}_{n-1}<{Y}_{n-1}+1\right)\cdots \right)\right)\right)=1
Finally, the lexicographically less than or equal to constraint can be expressed as a conjunction or a disjunction of constraints:
\begin{array}{cc}\hfill {X}_{0}\le {Y}_{0}& \hfill \wedge \\ \hfill \left({X}_{0}={Y}_{0}\right)โ{X}_{1}\le {Y}_{1}& \hfill \wedge \\ \hfill \left({X}_{0}={Y}_{0}\wedge {X}_{1}={Y}_{1}\right)โ{X}_{2}\le {Y}_{2}& \hfill \wedge \\ \hfill โฎ& \\ \hfill \left({X}_{0}={Y}_{0}\wedge {X}_{1}={Y}_{1}\wedge \cdots \wedge {X}_{n-2}={Y}_{n-2}\right)โ{X}_{n-1}\le {Y}_{n-1}& \\ & \\ \hfill {X}_{0}<{Y}_{0}& \hfill \vee \\ \hfill {X}_{0}={Y}_{0}\wedge {X}_{1}<{Y}_{1}& \hfill \vee \\ \hfill {X}_{0}={Y}_{0}\wedge {X}_{1}={Y}_{1}\wedge {X}_{2}<{Y}_{2}& \hfill \vee \\ \hfill โฎ& \\ \hfill {X}_{0}={Y}_{0}\wedge {X}_{1}={Y}_{1}\wedge \cdots \wedge {X}_{n-2}={Y}_{n-2}\wedge {X}_{n-1}\le {Y}_{n-1}\end{array}
lexEq in Choco, rel in Gecode, lex_lesseq in MiniZinc, lex_chain in SICStus.
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐ ๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐ก}\mathtt{2}
(matrix symmetry,lexicographic order),
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐๐๐๐}
(vector),
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐ก}\mathtt{2}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
implies (if swap arguments):
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐ ๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐}\left(\begin{array}{c}\mathrm{๐ณ๐ด๐๐๐ธ๐ฝ๐ฐ๐๐ธ๐พ๐ฝ}-\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐ก}-\mathrm{๐๐๐},๐ก-\mathrm{๐๐๐},๐ข-\mathrm{๐๐๐}\right),\hfill \\ \mathrm{๐๐๐๐}\left(\mathrm{๐๐๐๐๐ก}-0,๐ก-0,๐ข-0\right)\right]\hfill \end{array}\right)
\mathrm{๐๐๐}\left(\begin{array}{c}\mathrm{๐ฒ๐พ๐ผ๐ฟ๐พ๐ฝ๐ด๐ฝ๐๐}-\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐ก}-\mathrm{๐๐๐},๐ก-\mathrm{๐๐๐๐},๐ข-\mathrm{๐๐๐๐}\right),\hfill \\ \left[\begin{array}{c}\mathrm{๐๐๐๐}\left(\begin{array}{c}\mathrm{๐๐๐๐๐ก}-\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}.\mathrm{๐๐๐ข},\hfill \\ ๐ก-\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}.\mathrm{๐๐๐},\hfill \\ ๐ข-\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}.\mathrm{๐๐๐}\hfill \end{array}\right)\hfill \end{array}\right]\hfill \end{array}\right)
\mathrm{๐ฒ๐พ๐ผ๐ฟ๐พ๐ฝ๐ด๐ฝ๐๐}
\mathrm{๐ณ๐ด๐๐๐ธ๐ฝ๐ฐ๐๐ธ๐พ๐ฝ}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
\left(
\mathrm{๐๐ด๐๐ป}
,
\mathrm{๐๐๐ผ๐ท}
\right)โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐}\mathtt{2}\right)
\bigvee \left(\begin{array}{c}\mathrm{๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐ก}>0\wedge \mathrm{๐๐๐๐}\mathtt{1}.๐ก=\mathrm{๐๐๐๐}\mathtt{1}.๐ข,\hfill \\ \bigwedge \left(\begin{array}{c}\mathrm{๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐ก}<|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|,\hfill \\ \mathrm{๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐ก}=0,\hfill \\ \mathrm{๐๐๐๐}\mathtt{1}.๐ก<\mathrm{๐๐๐๐}\mathtt{1}.๐ข\hfill \end{array}\right),\hfill \\ \bigwedge \left(\begin{array}{c}\mathrm{๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐ก}=|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|,\hfill \\ \mathrm{๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐ก}=0,\hfill \\ \mathrm{๐๐๐๐}\mathtt{1}.๐ก\le \mathrm{๐๐๐๐}\mathtt{1}.๐ข\hfill \end{array}\right)\hfill \end{array}\right)
\mathrm{๐๐๐๐}_\mathrm{๐
๐๐๐}_\mathrm{๐๐}
\left(\mathrm{๐๐๐๐๐ก},1,0\right)=1
\mathrm{๐๐๐๐}_\mathrm{๐
๐๐๐}_\mathrm{๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
We create a vertex
{c}_{i}
for each pair of components that both have the same index
i
We create an additional dummy vertex called
We create an arc between
{c}_{i}
and
{c}_{i}
was generated from the last components of both vectors We associate to this arc the arc constraint
{\mathrm{๐๐๐๐}}_{1}.x\le {\mathrm{๐๐๐๐}}_{2}.y
; Otherwise we associate to this arc the arc constraint
{\mathrm{๐๐๐๐}}_{1}.x<{\mathrm{๐๐๐๐}}_{2}.y
{c}_{i}
{c}_{i+1}
. We associate to this arc the arc constraint
{\mathrm{๐๐๐๐}}_{1}.x={\mathrm{๐๐๐๐}}_{2}.y
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
constraint holds when there exist a path from
{c}_{1}
d
. This path can be interpreted as a maximum sequence of equality constraints on the prefix of both vectors, possibly followed by a less than constraint.
\mathrm{๐๐๐๐}_\mathrm{๐
๐๐๐}_\mathrm{๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐
๐๐๐}_\mathrm{๐๐}\left(\mathrm{๐๐๐๐๐ก},1,0\right)=1
\mathrm{๐๐๐๐}_\mathrm{๐
๐๐๐}_\mathrm{๐๐}\left(\mathrm{๐๐๐๐๐ก},1,0\right)\ge 1
\underline{\overline{\mathrm{๐๐๐๐}_\mathrm{๐
๐๐๐}_\mathrm{๐๐}}}
\overline{\mathrm{๐๐๐๐}_\mathrm{๐
๐๐๐}_\mathrm{๐๐}}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐
๐ฐ๐}{\mathtt{1}}_{i}
\mathrm{๐
๐ฐ๐}{\mathtt{2}}_{i}
\mathrm{๐๐๐}
{i}^{th}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
collections. To each pair
\left(\mathrm{๐
๐ฐ๐}{\mathtt{1}}_{i},\mathrm{๐
๐ฐ๐}{\mathtt{2}}_{i}\right)
{S}_{i}
\left(\mathrm{๐
๐ฐ๐}{\mathtt{1}}_{i}<\mathrm{๐
๐ฐ๐}{\mathtt{2}}_{i}โ{S}_{i}=1\right)\wedge \left(\mathrm{๐
๐ฐ๐}{\mathtt{1}}_{i}=\mathrm{๐
๐ฐ๐}{\mathtt{2}}_{i}โ{S}_{i}=2\right)\wedge \left(\mathrm{๐
๐ฐ๐}{\mathtt{1}}_{i}>\mathrm{๐
๐ฐ๐}{\mathtt{2}}_{i}โ{S}_{i}=3\right)
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
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write an experiment showing aim, materials required, procedure , observation(in a tabular form), conclusion 1 Beet root extract as - Science - Acids Bases and Salts - 9370317 | Meritnation.com
write an experiment showing aim, materials required, procedure , observation(in a tabular form), conclusion 1. Beet root extract as indicator for bitter and sour substance 2. Neutralisation between lemon juice and baking powder
Aim : To prepare beet root indicator and test for acidity/basicity of the compounds
Material Required: Beet root, vinegar, baking soda
Procedure : A) Preparation of beet root indicator
i) Take beet root and crush it in a mortar.
ii) Add water to the crushed beet root to obtain an extract and filter off the extract to get the filtrate. The filtrate obtained is the beet root indicator. This is red in colour and is acidic in nature.
B) Test for the acid or basic nature of the compounds ( vinegar , baking soda)
i) Vinegar and baking soda were added in test tubes A and B respectively.
ii) To each of the test tube few drops of beet root indicator was added and colour change was observed.
Test tube A containing vinegar - No colour change
Test tube B containing baking soda - Colour changed to yellow
Since the colour of the test tube A did not changed , vinegar is acidic in nature. While , the colour of the test tube B changed from red to yellow , baking soda is basic in nature.
Aim : Neutralisation of lemon juice with baking powder
Materials Required : Lemon juice , baking powder
Note : Lemon juice is acidic in nature because it contains citric acid (Cโโ6H8O7) and is acidic, while baking powder is basic in nature because it contains sodium bicarbonate (NaHCOโ3โ)
i) Take lemon juice in a beaker.
ii) To it slowly add baking soda.
iii) Neutralisation reaction starts with formation of bubbles.
iv) Add baking powder slowly till no more formation of bubbles happen.
Observation : Effervescence produced due to the formation of CO2.
Conclusion: Formation of bubbles is due to formation of carbonic acid which is unstable and dissociates into carbon dioxide and water which is the cause of bubble formation.The neutralisation reaction is taking place which is as follows:
{C}_{6}{H}_{8}{O}_{7} + 3NaHC{O}_{3} \to N{a}_{3}{C}_{6}{H}_{5}{O}_{7} + 3{H}_{2}C{O}_{3}\phantom{\rule{0ex}{0ex}}{H}_{2}C{O}_{3} \to C{O}_{2} + {H}_{2}O
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Differentiation Under the Integral Sign | Brilliant Math & Science Wiki
Sameer Kailasa, Travis Mcknight, Nihar Mahajan, and
Differentiation under the integral sign is an operation in calculus used to evaluate certain integrals. Under fairly loose conditions on the function being integrated, differentiation under the integral sign allows one to interchange the order of integration and differentiation. In its simplest form, called the Leibniz integral rule, differentiation under the integral sign makes the following equation valid under light assumptions on
f
\frac{d}{dx} \int_{a}^{b} f(x,t) \,dt = \int_{a}^{b} \frac{\partial}{\partial x} f(x,t) \, dt .
Many integrals that would otherwise be impossible or require significantly more complex methods can be solved by this approach.
The most general form of differentiation under the integral sign states that: if
f(x,t)
is a continuous and continuously differentiable (i.e., partial derivatives exist and are themselves continuous) function and the limits of integration
a(x)
b(x)
are continuous and continuously differentiable functions of
x
\frac{\mathrm{d}}{\mathrm{d}x} \int_{a(x)}^{b(x)} f(x,t) \, \mathrm{d}t =f(x,b(x)) \cdot b'(x) - f(x,a(x))\cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) \, \mathrm{d}t.
a(x)
b(x)
are constant functions, this formula reduces to the simpler form
\frac{\mathrm{d}}{\mathrm{d}x} \int_{a}^{b} f(x,t) \,\mathrm{d}t = \int_{a}^{b} \frac{\partial}{\partial x} f(x,t) \, \mathrm{d}t .
This simpler statement is known as Leibniz integral rule.
Generally, one uses differentiation under the integral sign to evaluate integrals that can be thought of as belonging to some family of integrals parameterized by a real variable. To better understand this statement, consider the following example:
Compute the definite integral
\int_{0}^{1} \frac{t^{3} - 1}{\ln t} \, dt.
This integral appears resistant to standard integration techniques such as integration by parts, u-substitution, etc. We would like to use differentiation under the integral sign to compute it.
How can we choose a function to differentiate under the integral sign? The appearance of
\ln t
in the denominator of the integrand is quite unwelcome, and we would like to get rid of it. Thankfully, we know
\frac{d}{dx} t^x = t^x \ln t,
so differentiating the numerator with respect to the exponent seems to be what we'd like to do.
Accordingly, we define a function
g(x) = \int_{0}^{1} \frac{t^x - 1}{\ln t} \, dt .
In this notation, the integral we wish to evaluate is
g(3)
. Observe that the given integral has been recast as member of a family of definite integrals
g(x)
indexed by the variable
x
By Leibniz integral rule, we compute
g'(x) = \int_{0}^{1} \frac{\partial}{\partial x} \frac{t^x - 1}{\ln t} \, dt = \int_{0}^{1} \frac{t^x \ln t}{\ln t} \, dt = \frac{t^{x+1}}{x+1} \Bigg\vert_{0}^{1} = \frac{1}{x+1}.
g(x) = \ln|x+1| + C
C
C
g(0) = 0
0 = g(0) = \ln 1 + C = C
g(x) = \ln|x+1|
x
such that the integral exists. In particular,
g(3) = \ln 4 = 2\ln 2
_\square
In the example, part of the integrand was replaced with a variable and the resultant function was studied using differentiation under the integral sign. This is a good illustration of the problem-solving principle: if stuck on a specific problem, try solving a more general problem.
\dfrac{49!}{50^{51}}
\dfrac{51!}{50^{50}}
\dfrac{50!}{51^{51}}
\dfrac{50!}{49^{51}}
\int_{0}^{1} (x\ln x)^{50} \, dx.
Another example illustrates the power of this technique in its general form, as one may use it to compute the Gaussian integral.
\int_{0}^{\infty} e^{-x^2/2} \, dx.
g(t) = \left(\int_{0}^{t} e^{-x^2/2} \, dx \right)^2.
Our goal is to compute
g(\infty)
and then take its square root.
t
g'(t) = 2 \cdot \left(\int_{0}^{t} e^{-x^2/2} \, dx \right) \cdot \left(\frac{d}{dt} \int_{0}^{t} e^{-x^2/2} \, dx \right) = 2e^{-t^2 / 2} \int_{0}^{t} e^{-x^2 / 2} \, dx = 2\int_{0}^{t} e^{-(t^2 + x^2)/2} \, dx.
u = x/t
, so that the integral transforms to
g'(t) = 2 \int_{0}^{1}te^{-(1+u^2)t^2/2}\, du.
Now, the integrand has a closed-form antiderivative with respect to
t
g'(t) = -2\int_{0}^{1} \frac{\partial }{\partial t} \frac{e^{-(1+u^2)t^2/2}}{1+u^2} \, du = -2 \frac{d}{dt} \int_{0}^{1} \frac{e^{-(1+u^2)t^2/2}}{1+u^2} \, du.
h(t) = \int_{0}^{1} \frac{e^{-(1+x^2)t^2/2}}{1+x^2} \, dx.
Then by the above calculation,
g'(t) = -2h'(t)
g(t) = -2h(t) + C
. To determine
C
t\to 0
in the equation; since
g(0) = 0
h(0) = \int_{0}^{1} \frac{1}{1+x^2} \, dx = \tan^{-1} x\Bigg\vert_{0}^{1} = \frac{\pi}{4},
0 = -\pi/2 + C \implies C = \pi/2
Finally, taking
t\to\infty
g(\infty) = -2h(\infty) + \pi/2 = \pi/2
\int_{0}^{\infty} e^{-x^2/2} \, dx = \sqrt{\frac{\pi}{2} }. \ _\square
a,b, c
be real with
c > 0
\int_{-\infty}^{\infty} e^{-c x^{2}} \operatorname{erf}(a x+b) \, \mathrm{d} x=\sqrt{\frac{\pi}{c}} \operatorname{erf}\left(\frac{b \sqrt{c}}{\sqrt{a^{2}+c}}\right)
a, b, c
c > 0
I(a, b, c)=\int_{-\infty}^{\infty} e^{-c x^{2}} \operatorname{erf}(a x+b) \mathrm{d} x
u=\sqrt{c}x
I(a, b, c)=\int_{-\infty}^{\infty} e^{-u^{2}} \operatorname{erf}\left(\frac{a}{\sqrt{c}} u+b\right) \frac{\mathrm{d} u}{\sqrt{c}}=\frac{1}{\sqrt{c}} I\left(\frac{a}{\sqrt{c}}, b, 1\right)
So we will focus on determining
J(a, b) = I(a, b, 1)
\operatorname{erf} x=\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} \mathrm{d} t \quad \Longrightarrow \quad \operatorname{erf}^{\prime}(x)=\frac{2}{\sqrt{\pi}} e^{-x^{2}}
and so, differentiating under the integral sign with respect to
a
\frac{\partial J}{\partial a}=\int_{-\infty}^{\infty} e^{-x^{2}} \frac{\partial}{\partial a} \operatorname{erf}(a x+b) \mathrm{d} x=\frac{2}{\sqrt{\pi}} \int_{-\infty}^{\infty} x e^{-x^{2}} e^{-(a x+b)^{2}} \mathrm{d} x
Completing the square in the exponent we have
x^{2}+(a x+b)^{2}=\left(a^{2}+1\right) x^{2}+2 a b x+b^{2}=\left(a^{2}+1\right)\left(x+\frac{a b}{a^{2}+1}\right)^{2}+\frac{b^{2}}{a^{2}+1}
\begin{aligned} \frac{\partial J}{\partial a} &=\frac{2}{\sqrt{\pi}} \exp \left(\frac{-b^{2}}{a^{2}+1}\right) \int_{-\infty}^{\infty} x \exp \left[-\left(a^{2}+1\right)\left(x+\frac{a b}{a^{2}+1}\right)^{2}\right] \mathrm{d} x \\ &=\frac{2}{\sqrt{\pi}} \exp \left(\frac{-b^{2}}{a^{2}+1}\right) \int_{-\infty}^{\infty}\left(v-\frac{a b}{a^{2}+1}\right) \exp \left[-\left(a^{2}+1\right) v^{2}\right] \mathrm{d} v \quad\left[v=x+a b /\left(a^{2}+1\right)\right] \\ &=\frac{2}{\sqrt{\pi}}\left(-\frac{a b}{a^{2}+1}\right) \exp \left(\frac{-b^{2}}{a^{2}+1}\right) \int_{-\infty}^{\infty} \exp \left[-\left(a^{2}+1\right) v^{2}\right] \mathrm{d} v \quad\left[\text { using oddness of } v e^{-\left(a^{2}+1\right) v^{2}}\right] \\ &=\frac{2}{\sqrt{\pi}}\left(-\frac{a b}{a^{2}+1}\right) \exp \left(\frac{-b^{2}}{a^{2}+1}\right) \frac{\sqrt{\pi}}{\sqrt{a^{2}+1}} \quad[(i)] \\ &=\frac{-2 a b}{\left(a^{2}+1\right)^{3 / 2}} \exp \left(\frac{-b^{2}}{a^{2}+1}\right) \end{aligned}
(i)
\int_{-\infty}^{\infty} e^{-x^{2}} \mathrm{d} x=\sqrt{\pi} \quad \text { and } \quad \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}
Thus, noting
J(a, b) \rightarrow 0
a \rightarrow-\infty
, we finally have
J(a, b)=-2 b \int_{-\infty}^{a} \frac{x}{\left(x^{2}+1\right)^{3 / 2}} \exp \left(\frac{-b^{2}}{x^{2}+1}\right) \mathrm{d} x
We will set
u^2 = b^{2}/(1 + x^2)
2 u \mathrm{d} u=\frac{-2 b^{2} x}{\left(1+x^{2}\right)^{2}} \mathrm{d} x=\frac{1}{\sqrt{1+x^{2}}} \frac{-2 b^{2} x}{\left(1+x^{2}\right)^{3 / 2}} \mathrm{d} x=\frac{u}{b}\left(\frac{-2 b^{2} x}{\left(1+x^{2}\right)^{3 / 2}} \mathrm{d} x\right)
\mathrm{d} u=\frac{-b x \mathrm{d} x}{\left(1+x^{2}\right)^{3 / 2}}
J(a, b)=2 \int_{0}^{b / \sqrt{1+a^{2}}} \exp \left(-u^{2}\right) \mathrm{d} u=\sqrt{\pi} \operatorname{erf}\left(\frac{b}{\sqrt{1+a^{2}}}\right)
I(a, b, c)=\frac{1}{\sqrt{c}} J\left(\frac{a}{\sqrt{c}}, b\right)=\sqrt{\frac{\pi}{c}} \operatorname{erf}\left(\frac{b}{\sqrt{1+a^{2} / c}}\right)=\sqrt{\frac{\pi}{c}} \operatorname{erf}\left(\frac{b \sqrt{c}}{\sqrt{c+a^{2}}}\right)
One should also note counterexamples, for which this technique does not work. For instance, suppose one attempts to evaluate
\int_{0}^{\infty} \frac{\sin x}{x} \, dx
by making the variable change
u = x/t
t\neq 0
\int_{0}^{\infty} \frac{\sin x}{x} \, dx = \int_{0}^{\infty} \frac{\sin tu}{u} \, du = g(t).
Differentiating under the integral sign yields
0 = g'(t) = \int_{0}^{\infty} \cos tu \, du,
which is absurd. The problem is that the function
f(x,t) = \sin tx/x
is not continuously differentiable (consider
\partial f/\partial t
x=0
), which was required in the assumptions set forth above.
4\pi
2\pi
\pi
Compute the definite integral:
\int_{0}^{2\pi} e^{\cos \theta} \cos(\sin\theta) \, d\theta.
Hint: Consider the function
f(t) = \int_{0}^{2\pi} e^{t \cos\theta} \cos(t\sin\theta) \, d\theta
and use differentiation under the integral sign.
Cite as: Differentiation Under the Integral Sign. Brilliant.org. Retrieved from https://brilliant.org/wiki/differentiate-through-the-integral/
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AdjointOrePoly - Maple Help
Home : Support : Online Help : Mathematics : Algebra : Skew Polynomials : OreTools : AdjointOrePoly
construct the adjoint of a given Ore polynomial ring
compute the adjoint Ore polynomial in a given Ore ring
AdjointRing(A)
AdjointOrePoly(Poly, A)
Ore ring; to define an Ore ring, use the SetOreRing function.
The AdjointRing(A) calling sequence constructs the adjoint of A.
The AdjointOrePoly(Poly, A) calling sequence computes the adjoint Ore polynomial of the polynomial Poly in A.
An Ore polynomial ring is defined vi SetOreRing. For a description of the adjoint of an Ore polynomial ring, see OreAlgebra.
\mathrm{with}โก\left(\mathrm{OreTools}\right):
\mathrm{with}โก\left(\mathrm{OreTools}[\mathrm{Properties}]\right):
Aโ\mathrm{SetOreRing}โก\left(n,'\mathrm{shift}'\right)
\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathrm{UnivariateOreRing}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{shift}}\right)
Construct the adjoint Ore polynomial ring B of A.
Bโ\mathrm{AdjointRing}โก\left(A\right)
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathrm{Adj}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{\mathrm{UnivariateOreRing}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{shift}}\right)\right)
Construct the adjoint Ore polynomial ring C of B. The ring C must be the same as A.
Cโ\mathrm{AdjointRing}โก\left(B\right)
\textcolor[rgb]{0,0,1}{C}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathrm{UnivariateOreRing}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{shift}}\right)
\mathrm{GetSigma}โก\left(A\right)โก\left(sโก\left(n\right),n\right)=\mathrm{GetSigma}โก\left(C\right)โก\left(sโก\left(n\right),n\right)
\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{GetSigmaInverse}โก\left(A\right)โก\left(sโก\left(n\right),n\right)=\mathrm{GetSigmaInverse}โก\left(C\right)โก\left(sโก\left(n\right),n\right)
\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{Getdelta}โก\left(A\right)โก\left(sโก\left(n\right),n\right)=\mathrm{Getdelta}โก\left(C\right)โก\left(sโก\left(n\right),n\right)
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}
Define two Ore polynomials P1 and P2 in A.
\mathrm{P1}โ\mathrm{OrePoly}โก\left(n+1,n\right);
\mathrm{P2}โ\mathrm{OrePoly}โก\left(1,n+1\right)
\textcolor[rgb]{0,0,1}{\mathrm{P1}}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathrm{OrePoly}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{n}\right)
\textcolor[rgb]{0,0,1}{\mathrm{P2}}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathrm{OrePoly}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)
Compute the adjoint operators of P1 and P2 in A.
\mathrm{adjP1}โ\mathrm{AdjointOrePoly}โก\left(\mathrm{P1},A\right)
\textcolor[rgb]{0,0,1}{\mathrm{adjP1}}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathrm{OrePoly}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{adjP2}โ\mathrm{AdjointOrePoly}โก\left(\mathrm{P2},A\right)
\textcolor[rgb]{0,0,1}{\mathrm{adjP2}}\textcolor[rgb]{0,0,1}{โ}\textcolor[rgb]{0,0,1}{\mathrm{OrePoly}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{n}\right)
Multiply adjP1 and adjP2 in the adjoint B of A.
\mathrm{Multiply}โก\left(\mathrm{adjP1},\mathrm{adjP2},B\right)
\textcolor[rgb]{0,0,1}{\mathrm{OrePoly}}\textcolor[rgb]{0,0,1}{โก}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{n}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{โข}\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}{\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)}^{\textcolor[rgb]{0,0,1}{2}}\right)
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Protein pKa calculations - Wikipedia
In computational biology, protein pKa calculations are used to estimate the pKa values of amino acids as they exist within proteins. These calculations complement the pKa values reported for amino acids in their free state, and are used frequently within the fields of molecular modeling, structural bioinformatics, and computational biology.
1 Amino acid pKa values
2 The effect of the protein environment
3 pKa calculation methods
3.1 Using the PoissonโBoltzmann equation
3.3 Molecular dynamics (MD)-based methods
3.4 Determining pKa values from titration curves or free energy calculations
5 Software for protein pKa calculations
Amino acid pKa values[edit]
pKa values of amino acid side chains play an important role in defining the pH-dependent characteristics of a protein. The pH-dependence of the activity displayed by enzymes and the pH-dependence of protein stability, for example, are properties that are determined by the pKa values of amino acid side chains.
The pKa values of an amino acid side chain in solution is typically inferred from the pKa values of model compounds (compounds that are similar to the side chains of amino acids). See Amino acid for the pKa values of all amino acid side chains inferred in such a way. There are also numerous experimental studies that have yielded such values, for example by use of NMR spectroscopy.
The table below lists the model pKa values that are often used in a protein pKa calculation, and contains a third column based on protein studies.[1]
Asp (D) 3.9 4.00
Glu (E) 4.3 4.40
Arg (R) 12.0 13.50
Lys (K) 10.5 10.40
His (H) 6.08 6.80
Cys (C) (โSH) 8.28 8.30
Tyr (Y) 10.1 9.60
N-term 8.00
C-term 3.60
The effect of the protein environment[edit]
When a protein folds, the titratable amino acids in the protein are transferred from a solution-like environment to an environment determined by the 3-dimensional structure of the protein. For example, in an unfolded protein an aspartic acid typically is in an environment which exposes the titratable side chain to water. When the protein folds the aspartic acid could find itself buried deep in the protein interior with no exposure to solvent.
Furthermore, in the folded protein the aspartic acid will be closer to other titratable groups in the protein and will also interact with permanent charges (e.g. ions) and dipoles in the protein. All of these effects alter the pKa value of the amino acid side chain, and pKa calculation methods generally calculate the effect of the protein environment on the model pKa value of an amino acid side chain.[2][3][4][5]
Typically the effects of the protein environment on the amino acid pKa value are divided into pH-independent effects and pH-dependent effects. The pH-independent effects (desolvation, interactions with permanent charges and dipoles) are added to the model pKa value to give the intrinsic pKa value. The pH-dependent effects cannot be added in the same straightforward way and have to be accounted for using Boltzmann summation, TanfordโRoxby iterations or other methods.
The interplay of the intrinsic pKa values of a system with the electrostatic interaction energies between titratable groups can produce quite spectacular effects such as non-HendersonโHasselbalch titration curves and even back-titration effects.[6]
The image below shows a theoretical system consisting of three acidic residues. One group is displaying a back-titration event (blue group).
Coupled system consisting of three acids
pKa calculation methods[edit]
Several software packages and webserver are available for the calculation of protein pKa values. See links below or this table
Using the PoissonโBoltzmann equation[edit]
Some methods are based on solutions to the PoissonโBoltzmann equation (PBE), often referred to as FDPB-based methods (FDPB is for "finite difference PoissonโBoltzmann"). The PBE is a modification of Poisson's equation that incorporates a description of the effect of solvent ions on the electrostatic field around a molecule.
The H++ web server, the pKD webserver, MCCE, Karlsberg+, PETIT and GMCT use the FDPB method to compute pKa values of amino acid side chains.
FDPB-based methods calculate the change in the pKa value of an amino acid side chain when that side chain is moved from a hypothetical fully solvated state to its position in the protein. To perform such a calculation, one needs theoretical methods that can calculate the effect of the protein interior on a pKa value, and knowledge of the pKa values of amino acid side chains in their fully solvated states.[2][3][4][5]
Empirical methods[edit]
A set of empirical rules relating the protein structure to the pKa values of ionizable residues have been developed by Li, Robertson, and Jensen. These rules form the basis for the web-accessible program called PROPKA for rapid predictions of pKa values. A recent empirical pKa prediction program was released by Tan KP et.al. with the online server DEPTH web server
Molecular dynamics (MD)-based methods[edit]
Molecular dynamics methods of calculating pKa values make it possible to include full flexibility of the titrated molecule.[7][8][9]
Molecular dynamics based methods are typically much more computationally expensive, and not necessarily more accurate, ways to predict pKa values than approaches based on the PoissonโBoltzmann equation. Limited conformational flexibility can also be realized within a continuum electrostatics approach, e.g., for considering multiple amino acid sidechain rotamers. In addition, current commonly used molecular force fields do not take electronic polarizability into account, which could be an important property in determining protonation energies.
Determining pKa values from titration curves or free energy calculations[edit]
From the titration of protonatable group, one can read the so-called pKa1โ2 which is equal to the pH value where the group is half-protonated. The pKa1โ2 is equal to the HendersonโHasselbalch pKa (pKHH
a) if the titration curve follows the HendersonโHasselbalch equation.[10] Most pKa calculation methods silently assume that all titration curves are HendersonโHasselbalch shaped, and pKa values in pKa calculation programs are therefore often determined in this way. In the general case of multiple interacting protonatable sites, the pKa1โ2 value is not thermodynamically meaningful. In contrast, the HendersonโHasselbalch pKa value can be computed from the protonation free energy via
{\displaystyle \mathrm {p} K_{\mathrm {a} }^{\mathrm {HH} }(\mathrm {pH} )=\mathrm {pH} -{\frac {\Delta G^{\mathrm {prot} }(\mathrm {pH} )}{\mathrm {RT} \ln 10}}}
and is thus in turn related to the protonation free energy of the site via
{\displaystyle \Delta G^{\mathrm {prot} }(\mathrm {pH} )=\mathrm {RT} \ln 10\;(\mathrm {pH} -\mathrm {p} K_{\mathrm {a} }^{\mathrm {HH} })}
The protonation free energy can in principle be computed from the protonation probability of the group โจxโฉ(pH) which can be read from its titration curve
{\displaystyle \Delta G^{\mathrm {prot} }(\mathrm {pH} )=-\mathrm {RT} \ln \left[{\frac {\langle x\rangle }{1-\langle x\rangle }}\right]}
Titration curves can be computed within a continuum electrostatics approach with formally exact but more elaborate analytical or Monte Carlo (MC) methods, or inexact but fast approximate methods. MC methods that have been used to compute titration curves[11] are Metropolis MC[12][13] or WangโLandau MC.[14] Approximate methods that use a mean-field approach for computing titration curves are the TanfordโRoxby method and hybrids of this method that combine an exact statistical mechanics treatment within clusters of strongly interacting sites with a mean-field treatment of intercluster interactions.[15][16][17][18][19]
In practice, it can be difficult to obtain statistically converged and accurate protonation free energies from titration curves if โจxโฉ is close to a value of 1 or 0. In this case, one can use various free energy calculation methods to obtain the protonation free energy[11] such as biased Metropolis MC,[20] free-energy perturbation,[21][22] thermodynamic integration,[23][24][25] the non-equilibrium work method[26] or the Bennett acceptance ratio method.[27]
Note that the pKHH
a value does in general depend on the pH value.[28]
This dependence is small for weakly interacting groups like well solvated amino acid sidechains on the protein surface, but can be large for strongly interacting groups like those buried in enzyme active sites or integral membrane proteins.[29][30][31]
^ Hass and Mulder (2015) Annu. Rev. Biophys. vol 44 pp. 53โ75 doi 10.1146/annurev-biophys-083012-130351.
^ a b Bashford (2004) Front Biosci. vol. 9 pp. 1082โ99 doi 10.2741/1187
^ a b Gunner et al. (2006) Biochim. Biophys. Acta vol. 1757 (8) pp. 942โ68 doi 10.1016/j.bbabio.2006.06.005
^ a b Ullmann et al. (2008) Photosynth. Res. 97 vol. 112 pp. 33โ55 doi 10.1007/s11120-008-9306-1
^ a b Antosiewicz et al. (2011) Mol. BioSyst. vol. 7 pp. 2923โ2949 doi 10.1039/C1MB05170A
^ A. Onufriev, D.A. Case and G. M. Ullmann (2001). Biochemistry 40: 3413โ3419 doi 10.1021/bi002740q
^ Donnini et al. (2011) J. Chem. Theory Comp. vol 7 pp. 1962โ78 doi 10.1021/ct200061r.
^ Wallace et al. (2011) J. Chem. Theory Comp. vol 7 pp. 2617โ2629 doi 10.1021/ct200146j.
^ Goh et al. (2012) J. Chem. Theory Comp. vol 8 pp. 36โ46 doi 10.1021/ct2006314.
^ Ullmann (2003) J. Phys. Chem. B vol 107 pp. 1263โ71 doi 10.1021/jp026454v.
^ a b Ullmann et al. (2012) J. Comput. Chem. vol 33 pp. 887โ900 doi 10.1002/jcc.22919
^ Metropolis et al. (1953) J. Chem. Phys. vol 23 pp. 1087โ1092 doi 10.1063/1.1699114
^ Beroza et al. (1991) Proc. Natl. Acad. Sci. USA vol 88 pp. 5804โ5808 doi 10.1073/pnas.88.13.5804
^ Wang and Landau (2001) Phys. Rev. E vol 64 pp 056101 doi 10.1103/PhysRevE.64.056101
^ Tanford and Roxby (1972) Biochemistry vol 11 pp. 2192โ2198 doi 10.1021/bi00761a029
^ Bashford and Karplus (1991) J. Phys. Chem. vol 95 pp. 9556โ61 doi 10.1021/j100176a093
^ Gilson (1993) Proteins vol 15 pp. 266โ82 doi 10.1002/prot.340150305
^ Antosiewicz et al. (1994) J. Mol. Biol. vol 238 pp. 415โ36 doi 10.1006/jmbi.1994.1301
^ Spassov and Bashford (1999) J. Comput. Chem. vol 20 pp. 1091โ1111 doi 10.1002/(SICI)1096-987X(199908)20:11<1091::AID-JCC1>3.0.CO;2-3
^ Beroza et al. (1995) Biophys. J. vol 68 pp. 2233โ2250 doi 10.1016/S0006-3495(95)80406-6
^ Zwanzig (1954) J. Chem. Phys. vol 22 pp. 1420โ1426 doi 10.1063/1.1740409
^ Ullmann et al. 2011 J. Phys. Chem. B. vol 68 pp. 507โ521 doi 10.1021/jp1093838
^ Kirkwood (1935) J. Chem. Phys. vol 2 pp. 300โ313 doi 10.1063/1.1749657
^ Bruckner and Boresch (2011) J. Comput. Chem. vol 32 pp. 1303โ1319 doi 10.1002/jcc.21713
^ Jarzynski (1997) Phys. Rev. E vol pp. 2233โ2250 doi 10.1103/PhysRevE.56.5018
^ Bennett (1976) J. Comput. Phys. vol 22 pp. 245โ268 doi 10.1016/0021-9991(76)90078-4
^ Bombarda et al. (2010) J. Phys. Chem. B vol 114 pp. 1994โ2003 doi 10.1021/jp908926w.
^ Bashford and Gerwert (1992) J. Mol. Biol. vol 224 pp. 473โ86 doi 10.1016/0022-2836(92)91009-E
^ Spassov et al. (2001) J. Mol. Biol. vol 312 pp. 203โ19 doi 10.1006/jmbi.2001.4902
^ Ullmann et al. (2011) J. Phys. Chem. B vol 115 pp. 10346โ59 doi 10.1021/jp204644h
Software for protein pKa calculations[edit]
AccelrysPKA Accelrys CHARMm based pKa calculation
H++ PoissonโBoltzmann based pKa calculations
MCCE2 Multi-Conformation Continuum Electrostatics (Version 2)
Karlsberg+ pKa computation with multiple pH adapted conformations
PETIT Proton and Electron TITration
GMCT Generalized Monte Carlo Titration
DEPTH web server Empirical calculation of pKa values using Residue Depth as a major feature
Retrieved from "https://en.wikipedia.org/w/index.php?title=Protein_pKa_calculations&oldid=1057869520"
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Oxidation/Reduction Reaction | PVEducation
Reduction/oxidation (redox) reactions are an important class of chemical reactions since they are the driving force behind a vast range of process, both desirable (for example breathing in mammals) and undesirable (for example rusting of iron). A redox reaction is characterized by the fact that electrons are produced (in an oxidation reaction) or are used by the reaction (in a reduction reaction). An oxidation reaction must always be paired with a reduction reaction, as the oxidation reaction produces the electrons required by the reduction reaction.
The electrons transferred in a redox reaction arise from the change of the valence state of materials in the redox reaction. If a material gives up or loses an electron, then its valance state becomes more positive (since an electron has a negative charge) and the reaction is called an oxidation reaction. Since an oxidation reaction gives up electrons, it will always have electrons as one of its products. By definition, the oxidation reaction occurs at the anode. The chemical reaction shown below is an oxidation reaction where zinc metal (with a neutral valance state or valance charge = 0) is oxidized to give a zinc ion, which has a 2+ valence charge. The two electrons lost by the zinc metal are products of the oxidation reaction. The zinc ion does not exist as separate entity, and therefore must for either a solid salt (in which case its mobility and availability is not useful for redox reactions) as a dissolved salt in a solution. The (aq) after the zinc ion indicates that it is aqueous. Note that since the overall aqueous solution must be electrically neutral, there must also be ions with positive charge in the solution. In examining only the behavior of the battery reaction, these may not be specified. However, they will play a role in the solubility of the Zn water (or an alternate solvent).
Zinc Oxidation with Valence Charge
\stackrel{Valence charge 0}{ Zn\left(s\right)}\to \stackrel{2+}{Z{n}^{2+}}\left(aq\right)+2{e}^{-}
Read more about Zinc Oxidation with Valence Charge
Reaction: Oxidation reaction (the valence state of the reactant increases) of zinc metal to a zinc ion. The (s) after the zinc indicates that it is in solid form. The zinc ion has (aq) after to indicate that it is aqueous, (ie in solution).
If a material gains an electron then its valance state decreases or reduces due to the negative charge of the electrons and the reaction is a reduction reaction. The reaction below is a reduction reaction in which a copper ion with a valance state of 2+ is reduced to copper metal, with a valence state of zero. Since a reduction reaction requires electrons, it will always have electrons as one of the reactants. The reduction reaction occurs at the cathode.
\stackrel{Valence charge 2+}{ C{u}^{2+}}\left(aq\right)+2{e}^{-}\to \stackrel{0}{Cu}\left(s\right)
Read more about Copper Reduction with Valence Charge
Reaction: Reduction reaction of Cu ions to form copper metal. The valence state of copper is reduced from 2+ to 0.
The total redox reaction consists of both of the two reactions together. For the example of copper and zinc above, the total reaction is shown below. Since the reaction with zinc metal (ie the reactant of the oxidation reaction) is providing the electron required to reduce the copper, the zinc is the reducing agent and the zinc itself is oxidized. Copper ions in this case are the oxidizing agent - they oxidize the zinc and are themselves reduced. Note that since the electrons appear on both sides of the chemical equation, they may be omitted when writing the redox reaction. Further note that for redox reaction, it is important to balance not only the elements in the chemical reactions, but also the electrons.
Zinc and Copper Redox with Electrons
C{u}^{2+}\left(aq\right)+2{e}^{-}+Zn\left(s\right)\to Cu\left(s\right)+Z{n}^{2+}\left(aq\right)+2{e}^{-}
Read more about Zinc and Copper Redox with Electrons
Zinc and Copper Redox Balanced
C{u}^{2+}\left(aq\right)+Zn\left(s\right)\to Cu\left(s\right)+Z{n}^{2+}\left(aq\right)
Read more about Zinc and Copper Redox Balanced
Reaction: Overall redox reaction for copper zinc reaction.
10. BatteriesElectrochemical Potential
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Gaussian noise - Wikipedia
With Gaussian noise
Gaussian noise, named after Carl Friedrich Gauss, is a term from signal processing theory denoting a kind of signal noise that has a probability density function (pdf) equal to that of the normal distribution (which is also known as the Gaussian distribution).[1][2] In other words, the values that the noise can take are Gaussian-distributed.
{\displaystyle p}
of a Gaussian random variable
{\displaystyle z}
{\displaystyle p_{G}(z)={\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {(z-\mu )^{2}}{2\sigma ^{2}}}}}
{\displaystyle z}
represents the grey level,
{\displaystyle \mu }
the mean grey value and
{\displaystyle \sigma }
its standard deviation.[3]
A special case is White Gaussian noise, in which the values at any pair of times are identically distributed and statistically independent (and hence uncorrelated). In communication channel testing and modelling, Gaussian noise is used as additive white noise to generate additive white Gaussian noise.
In telecommunications and computer networking, communication channels can be affected by wideband Gaussian noise coming from many natural sources, such as the thermal vibrations of atoms in conductors (referred to as thermal noise or JohnsonโNyquist noise), shot noise, black-body radiation from the earth and other warm objects, and from celestial sources such as the Sun.
Gaussian noise in digital images[edit]
Principal sources of Gaussian noise in digital images arise during acquisition e.g. sensor noise caused by poor illumination and/or high temperature, and/or transmission e.g. electronic circuit noise.[3] In digital image processing Gaussian noise can be reduced using a spatial filter, though when smoothing an image, an undesirable outcome may result in the blurring of fine-scaled image edges and details because they also correspond to blocked high frequencies. Conventional spatial filtering techniques for noise removal include: mean (convolution) filtering, median filtering and Gaussian smoothing.[1][4]
^ a b Tudor Barbu (2013). "Variational Image Denoising Approach with Diffusion Porous Media Flow". Abstract and Applied Analysis. 2013: 8. doi:10.1155/2013/856876.
^ Barry Truax, ed. (1999). "Handbook for Acoustic Ecology" (Second ed.). Cambridge Street Publishing. Archived from the original on 2017-10-10. Retrieved 2012-08-05.
^ a b Philippe Cattin (2012-04-24). "Image Restoration: Introduction to Signal and Image Processing". MIAC, University of Basel. Retrieved 11 October 2013.
^ Robert Fisher; Simon Perkins; Ashley Walker; Erik Wolfart. "Image Synthesis โ Noise Generation". Retrieved 11 October 2013.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Gaussian_noise&oldid=1074696378"
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Global Constraint Catalog: Ccond_lex_less
<< 5.82. cond_lex_greatereq5.84. cond_lex_lesseq >>
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}\left(\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1},\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2},\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}\right)
\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐}\right)
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}-\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}\right)
|\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}|\ge 1
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐},\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1},\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2},\mathrm{๐๐๐}\right)
|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|=|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}|
|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|=|\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด},\mathrm{๐๐๐๐๐}\right)
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ฃ๐}
\left(\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด},\mathrm{๐๐๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด},\left[\right]\right)
\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1},\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}\right)
\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2},\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}\right)
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
{๐ธ}^{th}
{๐น}^{th}
\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}
๐ธ<๐น
\left(\begin{array}{c}โฉ1,0โช,\hfill \\ โฉ0,0โช,\hfill \\ โฉ\begin{array}{c}\mathrm{๐๐๐๐๐}-โฉ1,0โช,\hfill \\ \mathrm{๐๐๐๐๐}-โฉ0,1โช,\hfill \\ \mathrm{๐๐๐๐๐}-โฉ0,0โช,\hfill \\ \mathrm{๐๐๐๐๐}-โฉ1,1โช\hfill \end{array}โช\hfill \end{array}\right)
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
are respectively assigned to the first and third items of the collection
\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}
|\mathrm{๐๐๐ฟ๐ป๐ด}_\mathrm{๐พ๐ต}_\mathrm{๐
๐ฐ๐ป๐}|>1
|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}|>1
|\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}|>1
|\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}|>1
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}.\mathrm{๐๐๐๐๐}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}.\mathrm{๐๐๐๐๐}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
\mathrm{๐ฟ๐๐ด๐ต๐ด๐๐ด๐ฝ๐ฒ๐ด}_\mathrm{๐๐ฐ๐ฑ๐ป๐ด}.\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐
๐ฐ๐}{\mathtt{1}}_{k}
\mathrm{๐
๐ฐ๐}{\mathtt{2}}_{k}
\mathrm{๐๐๐}
{k}^{th}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
๐ธ
๐น
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{1}
\mathrm{๐
๐ด๐ฒ๐๐พ๐}\mathtt{2}
๐ธ<๐น
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐ก}_\mathrm{๐๐๐๐}
๐ธ<๐น
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Network topology - WikiMili, The Best Wikipedia Reader
Network topology is the arrangement of the elements (links, nodes, etc.) of a communication network. [1] [2] Network topology can be used to define or describe the arrangement of various types of telecommunication networks, including command and control radio networks, [3] industrial fieldbusses and computer networks.
Network topology is the topological [4] structure of a network and may be depicted physically or logically. It is an application of graph theory [3] wherein communicating devices are modeled as nodes and the connections between the devices are modeled as links or lines between the nodes. Physical topology is the placement of the various components of a network (e.g., device location and cable installation), while logical topology illustrates how data flows within a network. Distances between nodes, physical interconnections, transmission rates, or signal types may differ between two different networks, yet their logical topologies may be identical. A networkโs physical topology is a particular concern of the physical layer of the OSI model.
Two basic categories of network topologies exist, physical topologies and logical topologies. [5]
The transmission medium layout used to link devices is the physical topology of the network. For conductive or fiber optical mediums, this refers to the layout of cabling, the locations of nodes, and the links between the nodes and the cabling. [1] The physical topology of a network is determined by the capabilities of the network access devices and media, the level of control or fault tolerance desired, and the cost associated with cabling or telecommunication circuits.
In contrast, logical topology is the way that the signals act on the network media, [6] or the way that the data passes through the network from one device to the next without regard to the physical interconnection of the devices. [7] A network's logical topology is not necessarily the same as its physical topology. For example, the original twisted pair Ethernet using repeater hubs was a logical bus topology carried on a physical star topology. Token Ring is a logical ring topology, but is wired as a physical star from the media access unit. Physically, AFDX can be a cascaded star topology of multiple dual redundant Ethernet switches; however, the AFDX Virtual links are modeled as time-switched single-transmitter bus connections, thus following the safety model of a single-transmitter bus topology previously used in aircraft. Logical topologies are often closely associated with media access control methods and protocols. Some networks are able to dynamically change their logical topology through configuration changes to their routers and switches.
Ribbon cable (untwisted and possibly unshielded) has been a cost-effective media for serial protocols, especially within metallic enclosures or rolled within copper braid or foil, over short distances, or at lower data rates. Several serial network protocols can be deployed without shielded or twisted pair cabling, that is, with "flat" or "ribbon" cable, or a hybrid flat/twisted ribbon cable, should EMC, length, and bandwidth constraints permit: RS-232, [8] RS-422, RS-485, [9] CAN, [10] GPIB, SCSI, [11] etc.
Twisted pair wire is the most widely used medium for all telecommunication.[ citation needed ] Twisted-pair cabling consist of copper wires that are twisted into pairs. Ordinary telephone wires consist of two insulated copper wires twisted into pairs. Computer network cabling (wired Ethernet as defined by IEEE 802.3) consists of 4 pairs of copper cabling that can be utilized for both voice and data transmission. The use of two wires twisted together helps to reduce crosstalk and electromagnetic induction. The transmission speed ranges from 2 million bits per second to 10 billion bits per second. Twisted pair cabling comes in two forms: unshielded twisted pair (UTP) and shielded twisted-pair (STP). Each form comes in several category ratings, designed for use in various scenarios.
Price is a main factor distinguishing wired- and wireless-technology options in a business. Wireless options command a price premium that can make purchasing wired computers, printers and other devices a financial benefit. Before making the decision to purchase hard-wired technology products, a review of the restrictions and limitations of the selections is necessary. Business and employee needs may override any cost considerations. [12]
IP over Avian Carriers was a humorous April fool's Request for Comments, issued as RFC 1149 . It was implemented in real life in 2001. [13]
Extending the Internet to interplanetary dimensions via radio waves, the Interplanetary Internet. [14]
A repeater is an electronic device that receives a network signal, cleans it of unnecessary noise and regenerates it. The signal may be reformed or retransmitted at a higher power level, to the other side of an obstruction possibly using a different transmission medium, so that the signal can cover longer distances without degradation. Commercial repeaters have extended RS-232 segments from 15 meters to over a kilometer. [15] In most twisted pair Ethernet configurations, repeaters are required for cable that runs longer than 100 meters. With fiber optics, repeaters can be tens or even hundreds of kilometers apart.
A network switch is a device that forwards and filters OSI layer 2 datagrams (frames) between ports based on the destination MAC address in each frame. [16] A switch is distinct from a hub in that it only forwards the frames to the physical ports involved in the communication rather than all ports connected. It can be thought of as a multi-port bridge. [17] It learns to associate physical ports to MAC addresses by examining the source addresses of received frames. If an unknown destination is targeted, the switch broadcasts to all ports but the source. Switches normally have numerous ports, facilitating a star topology for devices, and cascading additional switches.
The study of network topology recognizes eight basic topologies: point-to-point, bus, star, ring or circular, mesh, tree, hybrid, or daisy chain. [18]
In local area networks using bus topology, each node is connected by interface connectors to a single central cable. This is the 'bus', also referred to as the backbone, or trunk โ all data transmission between nodes in the network is transmitted over this common transmission medium and is able to be received by all nodes in the network simultaneously. [1]
A ring topology is a daisy chain in a closed loop. Data travels around the ring in one direction. When one node sends data to another, the data passes through each intermediate node on the ring until it reaches its destination. The intermediate nodes repeat (re transmit) the data to keep the signal strong. [5] Every node is a peer; there is no hierarchical relationship of clients and servers. If one node is unable to re transmit data, it severs communication between the nodes before and after it in the bus.
{\displaystyle c={\frac {n(n-1)}{2}}.\,}
Hybrid topology is also known as hybrid network. [19] Hybrid networks combine two or more topologies in such a way that the resulting network does not exhibit one of the standard topologies (e.g., bus, star, ring, etc.). For example, a tree network (or star-bus network) is a hybrid topology in which star networks are interconnected via bus networks. [20] [21] However, a tree network connected to another tree network is still topologically a tree network, not a distinct network type. A hybrid topology is always produced when two different basic network topologies are connected.
Snowflake topology is a star network of star networks.[ citation needed ]
Two other hybrid network types are hybrid mesh and hierarchical star. [20]
{\displaystyle {\frac {n(n-1)}{2}}\,}
Ethernet is a family of wired computer networking technologies commonly used in local area networks (LAN), metropolitan area networks (MAN) and wide area networks (WAN). It was commercially introduced in 1980 and first standardized in 1983 as IEEE 802.3. Ethernet has since been refined to support higher bit rates, a greater number of nodes, and longer link distances, but retains much backward compatibility. Over time, Ethernet has largely replaced competing wired LAN technologies such as Token Ring, FDDI and ARCNET.
Ethernet over twisted-pair technologies use twisted-pair cables for the physical layer of an Ethernet computer network. They are a subset of all Ethernet physical layers.
Fiber Distributed Data Interface (FDDI) is a standard for data transmission in a local area network. It uses optical fiber as its standard underlying physical medium, although it was also later specified to use copper cable, in which case it may be called CDDI, standardized as TP-PMD, also referred to as TP-DDI.
KNX is an open standard for commercial and domestic building automation. KNX devices can manage lighting, blinds and shutters, HVAC, security systems, energy management, audio video, white goods, displays, remote control, etc. KNX evolved from three earlier standards; the European Home Systems Protocol (EHS), BatiBUS, and the European Installation Bus. It can use twisted pair, powerline, RF, or IP links. On this network, the devices form distributed applications and tight interaction is possible. This is implemented via interworking models with standardised datapoint types and objects, modelling logical device channels.
In telecommunications, a point-to-point connection refers to a communications connection between two communication endpoints or nodes. An example is a telephone call, in which one telephone is connected with one other, and what is said by one caller can only be heard by the other. This is contrasted with a point-to-multipoint or broadcast connection, in which many nodes can receive information transmitted by one node. Other examples of point-to-point communications links are leased lines and microwave radio relay.
Profibus is a standard for fieldbus communication in automation technology and was first promoted in 1989 by BMBF and then used by Siemens. It should not be confused with the Profinet standard for Industrial Ethernet. Profibus is openly published as part of IEC 61158.
Networking hardware, also known as network equipment or computer networking devices, are electronic devices which are required for communication and interaction between devices on a computer network. Specifically, they mediate data transmission in a computer network. Units which are the last receiver or generate data are called hosts, end systems or data terminal equipment.
Switched fabric or switching fabric is a network topology in which network nodes interconnect via one or more network switches. Because a switched fabric network spreads network traffic across multiple physical links, it yields higher total throughput than broadcast networks, such as the early 10BASE5 version of Ethernet and most wireless networks such as Wi-Fi.
A medium dependent interface (MDI) describes the interface in a computer network from a physical layer implementation to the physical medium used to carry the transmission. Ethernet over twisted pair also defines a medium dependent interface crossover (MDI-X) interface. Auto MDI-X ports on newer network interfaces detect if the connection would require a crossover, and automatically chooses the MDI or MDI-X configuration to properly match the other end of the link.
A backbone or core network is a part of a computer network which interconnects networks, providing a path for the exchange of information between different LANs or subnetworks. A backbone can tie together diverse networks in the same building, in different buildings in a campus environment, or over wide areas. Normally, the backbone's capacity is greater than the networks connected to it.
Sercos III is the third generation of the Sercos interface, a standardized open digital interface for the communication between industrial controls, motion devices, input/output devices (I/O), and Ethernet nodes, such as PCs. Sercos III applies the hard real-time features of the Sercos interface to Ethernet. It is based upon and conforms to the Ethernet standard. Work began on Sercos III in 2003, with vendors releasing first products supporting it in 2005.
1 2 3 Groth, David; Toby Skandier (2005). Network+ Study Guide, Fourth Edition. Sybex, Inc. ISBN 0-7821-4406-3.
โ ATIS committee PRQC. "mesh topology". ATIS Telecom Glossary 2007. Alliance for Telecommunications Industry Solutions. Archived from the original on April 14, 2013. Retrieved 2008-10-10.
1 2 Grant, T. J., ed. (2014). Network Topology in Command and Control. Advances in Information Security, Privacy, and Ethics. IGI Global. pp. xvii, 228, 250. ISBN 9781466660595.
โ Chiang, Mung; Yang, Michael (2004). "Towards Network X-ities From a Topological Point of View: Evolvability and Scalability" (PDF). Proc. 42nd Allerton Conference. Archived from the original (PDF) on September 21, 2013.
1 2 Inc, S., (2002) . Networking Complete. Third Edition. San Francisco: Sybex
โ What Are Network Topologies? , retrieved 2016-09-17
โ Leonardi, E.; Mellia, M.; Marsan, M. A. (2000). "Algorithms for the Logical Topology Design in WDM All-Optical Networks". Optical Networks Magazine: 35โ46.
โ Cable Serial Male To Female 25L 4' DB25 M-DB25 28 AWG 300V Gray, Part no.: 12408, Jameco Electronics.
โ AN-1057 Ten ways to bulletproof RS-485 Interfaces, Texas Instruments, p. 5.
โ CANopen,CANopen DR-303 V1.0 Cabling and Connector Pin Assignment, CAN in Automation, p. 10.
โ Advantech Co., Ltd., Cable 50-Pin SCSI Ribbon type # PCL-10152-3E (Mouser Electronics #923-PCL-10152-3E)
โ , The Disadvantages of Wired Technology, Laura Acevedo, Demand Media.
โ "Bergen Linux User Group's CPIP Implementation". Blug.linux.no. Retrieved 2014-03-01.
โ A. Hooke (September 2000), Interplanetary Internet (PDF), Third Annual International Symposium on Advanced Radio Technologies, archived from the original (PDF) on 2012-01-13, retrieved 2011-11-12
โ U.S. Converters, RS232 Repeater
โ "Define switch". WWW.Wikipedia.com. Retrieved April 8, 2008.
โ "What bridge devices and bridging do for computer networks".
โ Bicsi, B. (2002). Network Design Basics for Cabling Professionals. McGraw-Hill Professional. ISBN 9780071782968.
โ "What is Hybrid Topology ? Advantages and Disadvantages". OROSK.COM. Archived from the original on September 9, 2016. Retrieved 2018-01-26.
1 2 Sosinsky, Barrie A. (2009). "Network Basics". Networking Bible. Indianapolis: Wiley Publishing. p. 16. ISBN 978-0-470-43131-3. OCLC 359673774 . Retrieved 2016-03-26.
โ Bradley, Ray (2001). Understanding Computer Science (for Advanced Level): The Study Guide. Cheltenham: Nelson Thornes. p. 244. ISBN 978-0-7487-6147-0. OCLC 47869750 . Retrieved 2016-03-26.
Wikimedia Commons has media related to Topology (Network) .
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Orthocenter | Brilliant Math & Science Wiki
Alexander Katz, Gautam Sharma, Vishwash Kumar ฮฮฮฉ, and
The orthocenter of a triangle is the intersection of the triangle's three altitudes. It has several important properties and relations with other parts of the triangle, including its circumcenter, incenter, area, and more.
The orthocenter is typically represented by the letter
H
The location of the orthocenter depends on the type of triangle. If the triangle is acute, the orthocenter will lie within it. If the triangle is obtuse, the orthocenter will lie outside of it. Finally, if the triangle is right, the orthocenter will be the vertex at the right angle.
Because the three altitudes always intersect at a single point (proof in a later section), the orthocenter can be found by determining the intersection of any two of them. This is especially useful when using coordinate geometry since the calculations are dramatically simplified by the need to find only two equations of lines (and their intersection).
A=(0,0), B=(14,0)
C=(5,12)
. What are the coordinates of the orthocenter?
The easiest altitude to find is the one from
C
AB
, since that is simply the line
x=5
. The next easiest to find is the one from
B
AC
AC
y=\frac{12}{5}x
. A line perpendicular to
AC
y=-\frac{5}{12}x+b
b
, and as this line goes through
(14,0)
, the equation of the altitude is
y=-\frac{5}{12}x+\frac{35}{6}
Finally, the intersection of this line and the line
x=5
\left(5,\frac{15}{4}\right)
, which is thus the location of the orthocenter.
_\square
The most natural proof is a consequence of Ceva's theorem, which states that
AD, BE, CF
\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=1,
D, E, F
are the feet of the altitudes.
\triangle BFC \sim \triangle BDA
\triangle AEB \sim \triangle AFC, \triangle CDA \sim \triangle CEB
\frac{BF}{BD} = \frac{BC}{BA}, \frac{AE}{AF} = \frac{AB}{AC}, \frac{CD}{CE} = \frac{CA}{BC}.
\frac{BF}{BD} \cdot \frac{AE}{AF} \cdot \frac{CD}{CE} = \frac{BC}{BA} \cdot \frac{AB}{AC} \cdot \frac{CA}{BC} = 1.
\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.
For convenience when discussing general properties, it is conventionally assumed that the original triangle in question is acute. The same properties usually apply to the obtuse case as well, but may require slight reformulation.
Interestingly, the three vertices and the orthocenter form an orthocentric system: any of the four points is the orthocenter of the triangle formed by the other three.
An incredibly useful property is that the reflection of the orthocenter over any of the three sides lies on the circumcircle of the triangle. There is a more visual way of interpreting this result: beginning with a circular piece of paper, draw a triangle inscribed in the paper, and fold the paper inwards along the three edges. The three arcs meet at the orthocenter of the triangle.[1]
This result has a number of important corollaries. The most immediate is that the angle formed at the orthocenter is supplementary to the angle at the vertex:
\angle ABC+\angle AHC = \angle BCA+\angle BHA = \angle CAB+\angle CHB = 180^{\circ}
Another follows from power of a point: the product of the two lengths the orthocenter divides an altitude into is constant. More specifically,
AH \cdot HD = BH \cdot HE = CH \cdot HF
\begin{aligned} AD \cdot DH &= BD \cdot CD\\ BE \cdot EH &= AE \cdot CE\\ CF \cdot FH &= AF \cdot BF. \end{aligned}
The application of this to a right triangle warrants its own note:
If the altitude from the vertex at the right angle to the hypotenuse splits the hypotenuse into two lengths of
p
q
, then the length of that altitude is
\sqrt{pq}
ABC
has a right angle at
C
D
AB
CD
AB
AD=4
BD=9
, what is the area of the triangle?
Another corollary is that the circumcircle of the triangle formed by any two points of a triangle and its orthocenter is congruent to the circumcircle of the original triangle. This is because the circumcircle of
BHC
can be viewed as the Locus of
H
A
moves around the original circumcircle.
Finally, this process (remarkably) can be reversed: if any point on the circumcircle is reflected over the three sides, the resulting three points are collinear, and the orthocenter always lies on the line connecting them.
On a somewhat different note, the orthocenter of a triangle is related to the circumcircle of the triangle in a deep way: the two points are isogonal conjugates, meaning that the reflections of the altitudes over the angle bisectors of a triangle intersect at the circumcenter of the triangle.
Another important property is that the reflection of orthocenter over the midpoint of any side of a triangle lies on the circumcircle and is diametrically opposite to the vertex opposite to the corresponding side.
The triangle formed by the feet of the three altitudes is called the orthic triangle. It has several remarkable properties. For example, the orthocenter of a triangle is also the incenter of its orthic triangle. Equivalently, the altitudes of the original triangle are the angle bisectors of the orthic triangle.
The sides of the orthic triangle have length
a\cos A, b\cos B
c\cos C
, making the perimeter of the orthic triangle
a\cos A+b\cos B+c\cos C
. The orthic triangle has the smallest perimeter among all triangles that could be inscribed in triangle
ABC
Kelvin the Frog lives in a triangle
ABC
with side lengths 4, 5 and 6. One day he starts at some point on side
AB
of the triangle, hops in a straight line to some point on side
BC
CA
of the triangle, and finally hops back to his original position on side
AB
of the triangle. The smallest distance Kelvin could have hopped is
\frac{m}{n}
m
n
m+n
The circumcircle of the orthic triangle contains the midpoints of the sides of the original triangle, as well as the points halfway from the vertices to the orthocenter. This circle is better known as the nine point circle of a triangle.
The orthic triangle is also homothetic to two important triangles: the triangle formed by the tangents to the circumcircle of the original triangle at the vertices (the tangential triangle), and the triangle formed by extending the altitudes to hit the circumcircle of the original triangle.
[1] Orthocenter curiousities. Retrieved January 23rd from http://untilnextstop.blogspot.com/2010/10/orthocenter-curiosities.html
Cite as: Orthocenter. Brilliant.org. Retrieved from https://brilliant.org/wiki/triangles-orthocenter/
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04.07 Customizing Colorbars
Plot legends identify discrete labels of discrete points. For continuous labels based on the color of points, lines, or regions, a labeled colorbar can be a great tool. In Matplotlib, a colorbar is a separate axes that can provide a key for the meaning of colors in a plot. Because the book is printed in black-and-white, this section has an accompanying online supplement where you can view the figures in full color (https://github.com/jakevdp/PythonDataScienceHandbook). We'll start by setting up the notebook for plotting and importing the functions we will use:
As we have seen several times throughout this section, the simplest colorbar can be created with the plt.colorbar function:
We'll now discuss a few ideas for customizing these colorbars and using them effectively in various situations.
The colormap can be specified using the cmap argument to the plotting function that is creating the visualization:
All the available colormaps are in the plt.cm namespace; using IPython's tab-completion will give you a full list of built-in possibilities:
A full treatment of color choice within visualization is beyond the scope of this book, but for entertaining reading on this subject and others, see the article "Ten Simple Rules for Better Figures". Matplotlib's online documentation also has an interesting discussion of colormap choice.
We can see this by converting the jet colorbar into black and white:
Notice the bright stripes in the grayscale image. Even in full color, this uneven brightness means that the eye will be drawn to certain portions of the color range, which will potentially emphasize unimportant parts of the dataset. It's better to use a colormap such as viridis (the default as of Matplotlib 2.0), which is specifically constructed to have an even brightness variation across the range. Thus it not only plays well with our color perception, but also will translate well to grayscale printing:
If you favor rainbow schemes, another good option for continuous data is the cubehelix colormap:
For other situations, such as showing positive and negative deviations from some mean, dual-color colorbars such as RdBu (Red-Blue) can be useful. However, as you can see in the following figure, it's important to note that the positive-negative information will be lost upon translation to grayscale!
We'll see examples of using some of these color maps as we continue.
Matplotlib allows for a large range of colorbar customization. The colorbar itself is simply an instance of plt.Axes, so all of the axes and tick formatting tricks we've learned are applicable. The colorbar has some interesting flexibility: for example, we can narrow the color limits and indicate the out-of-bounds values with a triangular arrow at the top and bottom by setting the extend property. This might come in handy, for example, if displaying an image that is subject to noise:
Colormaps are by default continuous, but sometimes you'd like to represent discrete values. The easiest way to do this is to use the plt.cm.get_cmap() function, and pass the name of a suitable colormap along with the number of desired bins:
For an example of where this might be useful, let's look at an interesting visualization of some hand written digits data. This data is included in Scikit-Learn, and consists of nearly 2,000
8 \times 8
thumbnails showing various hand-written digits.
For now, let's start by downloading the digits data and visualizing several of the example images with plt.imshow():
Deferring the discussion of these details, let's take a look at a two-dimensional manifold learning projection of this digits data (see In-Depth: Manifold Learning for details):
We'll use our discrete colormap to view the results, setting the ticks and clim to improve the aesthetics of the resulting colorbar:
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Global Constraint Catalog: Cuses
<< 5.415. used_by_partition5.417. valley >>
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๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1},\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}\right)
\mathrm{๐
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\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
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\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐๐๐}\left(1,|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1}|\right)\ge \mathrm{๐๐๐}\left(1,|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}|\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{1},\mathrm{๐๐๐}\right)
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\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2},\mathrm{๐๐๐}\right)
The set of values assigned to the variables of the collection of variables
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}
is included within the set of values assigned to the variables of the collection of variables
\mathrm{๐
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\left(โฉ3,3,4,6โช,โฉ3,4,4,4,4โช\right)
\mathrm{๐๐๐๐}
constraint holds since the set of values
\left\{3,4\right\}
assigned to the items of collection
โฉ3,4,4,4,4โช
is included within the set of values
\left\{3,4,6\right\}
occurring within
โฉ3,3,4,6โช
\mathrm{๐๐๐๐}
{U}_{1}\in \left[0,1\right]
{U}_{2}\in \left[1,2\right]
{V}_{1}\in \left[0,2\right]
{V}_{2}\in \left[2,4\right]
{V}_{3}\in \left[2,4\right]
\mathrm{๐๐๐๐}
\left(โฉ{U}_{1},{U}_{2}โช,โฉ{V}_{1},{V}_{2},{V}_{3}โช\right)
\mathrm{๐๐๐๐}
|\mathrm{๐
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\mathrm{๐๐๐๐๐}
\left(\mathrm{๐
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|\mathrm{๐
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\mathrm{๐๐๐๐๐}
\left(\mathrm{๐
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|\mathrm{๐
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\mathrm{๐
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\mathrm{๐
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\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\mathtt{2}\left(\mathrm{๐๐๐๐๐}\right)
It was shown in [BessiereHebrardHnichKiziltanWalsh05IJCAI] that, finding out whether a
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๐๐ท๐๐ถ๐}
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\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}=\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}
\mathrm{๐๐๐๐๐}
=|\mathrm{๐
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โข
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โข
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โข
\mathrm{๐ฝ๐พ}_\mathrm{๐ป๐พ๐พ๐ฟ}
\mathrm{๐๐๐๐๐}
graph property, the sink vertices of the final graph are stressed with a double circle. Note that all the vertices corresponding to the variables that take values 9 or 2 were removed from the final graph since there is no arc for which the associated equality constraint holds.
\mathrm{๐๐๐๐}
cleardoublepageeven
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Implicit Differentiation | Brilliant Math & Science Wiki
Aditya Virani, Sravanth C., Nihar Mahajan, and
Josh Quenneville
Lorenzo Moulin
Implicit differentiation is an approach to taking derivatives that uses the chain rule to avoid solving explicitly for one of the variables. For example, if
y + 3x = 8,
we can directly take the derivative of each term with respect to
x
\frac{dy}{dx} + 3 = 0,
\frac{dy}{dx} = -3.
Implicit differentiation is especially useful where it is difficult to isolate one of the variables in the given relationship. For example, if
y = x^2 + y^2,
y
and then taking the derivative would be painful. Instead, using implicit differentiation to directly take the derivative with respect to
x
\frac{dy}{dx} = 2x + 2y\frac{dy}{dx},
\frac{dy}{dx} = \frac{2x}{1-2y}.
This can be especially useful in fields like physics where relationships between variables are often known but a nice closed-form is not attainable.
x^2 + x + y^2 = 15
\frac{dy}{dx}
( 2, 3)?
x^2 + x + y^2 = 15
, differentiating, we have
\begin{aligned} 2x+1+2y\left(\frac{dy}{dx}\right)&=0 \\ \frac{dy}{dx}&=-\frac{2x+1}{2y}. \end{aligned}
\frac{dy}{dx}
( 2, 3)
-\frac{2(2)+1}{2(3)}=-\frac{5}{6}
_\square
Alternatively, we can use the fact that if
R(x,y)=0
\frac{dy}{dx}=-\frac{\hspace{3mm} \frac{\partial R}{\partial x}\hspace{3mm} }{\frac{\partial R}{\partial y}}.
Rewriting the function,
x^2 + x + y^2 - 15 = 0.
Applying the method given above,
\begin{aligned} \frac{dy}{dx} &= - \frac{\frac{\partial}{\partial x}( x^2 + x + y^2 - 15)}{\frac{\partial}{\partial y}( x^2 + x + y^2 - 15)} \\\\ &= - \dfrac{2x +1}{2y}. \end{aligned}
\frac{dy}{dx}
( 2, 3)
-\frac{2(2)+1}{2(3)}=-\frac{5}{6}
_\square
\frac{4}{2y} - \frac{x^2}{y}
\frac{2}{3y} - \frac{x^2}{y}
\frac{3}{2x} - \frac{x^2}{y}
\frac{3}{2y} - \frac{x^2}{y}
3y^2 + 2x^3 = 9x.
\frac{ dy}{dx}?
\dfrac{-4}{25y^3}
\dfrac{-25y^2+x^2}{625y^3}
\dfrac{100}{625y^3}
\dfrac{-x}{25y}
x^2+25y^2=100
\frac{d^2y}{dx^2}
of the equation above.
\frac{1}{x} = 1 - y^2 + ce^{\frac{-y^2}{2}}
\frac{1}{x} = 2 - y^2 + ce^{\frac{-y^2}{2}}
\frac{2}{x} = 1 - y^2 + ce^{\frac{-y^2}{2}}
\frac{2}{x} = 2 - y^2 + ce^{\frac{-y^2}{2}}
y' \times \big(x^2 y^3 + xy \big) = 1
The general solution of the differential equation above is
\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}.
y = \sqrt{x+\sqrt{12x-36}}.
\frac{dy}{dx}
x = 4
\frac{m}{n},
m
n
m+n.
Implicit differentiation is a technique that can be used to differentiate equations that are not given in the form of
y=f(x).
For instance, the differentiation of
x^2+y^2=1
looks pretty tough to do by using the differentiation techniques we've learned so far (which were explicit differentiation techniques), since it is not given in the form of
y=f(x).
Observe that the term equations was used in the first sentence. In fact, implicit differentiation can be used to differentiate equations that do not match the criteria to be a function.
The main key to implicit differentiation is to keep in mind with respect to which variable we are differentiating. Here is an easy explanation of how it is done:
When differentiating an equation implicitly with respect to
x,
we should
(1) add
\frac{d}{dx}
to all of the terms;
(2) apply the iterated chain rule to any terms that are not expressed in terms of
x.
The first step is the process of differentiating both sides of an equation with respect to
x.
The second step gives the solution to terms that are not described by
x.
Remember the iterated chain rule? Here is a quick reminder:
\frac{dz}{dx}=\frac{dz}{dy}\cdot\frac{dy}{dx}.
Implicit differentiation can be used in differentiating rational functions for convenience. Although most rational functions are given in the form of
y=f(x),
differentiating these explicitly would require using the quotient rule, which is pretty annoying. A slight modification to the equation (which in most cases would be multiplying the quotient to both sides) and the application of implicit differentiation would ease up obtaining the derivative. Here are some examples:
y=\frac{1}{x+1}
using implicit differentiation.
To avoid using the quotient rule, we can change both sides to their reciprocals:
\frac{1}{y}=x+1.
Then we differentiate both sides:
\begin{aligned} \frac{d}{dx}\frac{1}{y}&=\frac{d}{dx}(x+1)\\ \frac{d}{dx}\frac{1}{y}&=1, \end{aligned}
and apply the chain rule:
\begin{aligned} \frac{d}{dx}\frac{1}{y}&=1\\ \frac{d}{dy}\frac{1}{y}\cdot\frac{dy}{dx}&=1\\ -\frac{1}{y^2}\cdot\frac{dy}{dx}&=1\\ \Rightarrow \frac{dy}{dx}&=-y^2\\&=-\frac{1}{(x+1)^2}.\ _\square \end{aligned}
y=\frac{2x-1}{3x+2}
First, we multiply both sides by
3x+2
to get rid of the quotient:
y(3x+2)=2x-1.
Then we differentiate both sides of the equation. Obviously, the product rule should be used for the left-hand side:
\begin{aligned} y(3x+2)&=2x-1\\ \frac{d}{dx}\big(y(3x+2)\big)&=\frac{d}{dx}(2x-1)\\ \left(\frac{d}{dx}y\right)\cdot(3x+2)+3y&=2\\ \frac{dy}{dx}&=\frac{2-3y}{3x+2}. \end{aligned}
If you don't like having the term
y
in the answer, you can plug in
y=\frac{2x-1}{3x+2},
\frac{dy}{dx}=\frac{2x-3y-1}{3x+2}=\frac{2x-3\cdot\frac{2x-1}{3x+2}-1}{3x+2}=\frac{7}{(3x+2)^2}.
However, this procedure isn't necessary since we can always find the value of
y
when given the value of
x.
Of course, using explicit differentiation by applying the quotient rule will also lead to the same answer. Try it out for yourself.
_\square
y=\frac{x^3+2}{x^2-2x-4}
\begin{aligned} y&=\frac{x^3+2}{x^2-2x-4}\\ (x^2-2x-4)y&=x^3+2\\ \frac{d}{dx}\big((x^2-2x-4)y\big)&=\frac{d}{dx}(x^3+2)\\ (2x-2)y+(x^2-2x-4)\cdot\frac{dy}{dx}&=3x^2\\ \Rightarrow \frac{dy}{dx}&=\frac{3x^2-(2x-2)y}{x^2-2x-4}.\ _\square \end{aligned}
Find the tangent to the curve
y=\frac{3x-7}{x^2-4x+1}
x=0.
First, we should find the slope of the tangent, which is equal to
\left.\frac{dy}{dx}\right|_{x=0}.
\begin{aligned} y&=\frac{3x-7}{x^2-4x+1}\\ (x^2-4x+1)y&=3x-7\\ (2x-4)y+(x^2-4x+1)\cdot\frac{dy}{dx}&=3\\ \Rightarrow \frac{dy}{dx}&=\frac{3-(2x-4)y}{x^2-4x+1}. \end{aligned}
\left. y\right|_{x=0}=-7,
\left.\frac{dy}{dx}\right|_{x=0}=\frac{3-28}{1}=-25.
Therefore, the equation of the tangent is
\begin{aligned} y&=-25(x-0)-7\\ y&=-25x-7.\ _\square \end{aligned}
x^2+y^2=1.
First, we add
\frac{d}{dx}
\begin{aligned} \frac{d}{dx}x^2+\frac{d}{dx}y^2&=\frac{d}{dx}1\\ 2x+\frac{d}{dx}y^2&=0. \end{aligned}
Then we use the chain rule to solve
\frac{d}{dx}y^2:
\begin{aligned} 2x+\frac{d}{dx}y^2&=0\\ 2x+\frac{d}{dy}y^2\cdot\frac{dy}{dx}&=0\\ 2x+2y\cdot\frac{dy}{dx}&=0\\ \Rightarrow \frac{dy}{dx}&=-\frac{x}{y}. \end{aligned}
Notice that the answer can contain variables other than
x,
for it would be messy to express the equation using only one variable.
_\square
2y^3-y^2=x^5+3x^2.
\begin{aligned} \frac{d}{dx}2y^3-\frac{d}{dx}y^2&=\frac{d}{dx}x^5+\frac{d}{dx}3x^2\\ \frac{d}{dy}2y^3\cdot\frac{dy}{dx}-\frac{d}{dy}y^2\cdot\frac{dy}{dx}&=5x^4+6x\\ \frac{dy}{dx}\cdot(6y^2-2y)&=5x^4+6x\\ \Rightarrow \frac{dy}{dx}&=\frac{5x^4+6x}{6y^2-2y}.\ _\square \end{aligned}
\ln y+e^y=\sin y^2-3\cos x.
\begin{aligned} \frac{d}{dx}\ln y+\frac{d}{dx}e^y&=\frac{d}{dx}\sin y^2-\frac{d}{dx}3\cos x\\ \frac{d}{dy}\ln y\cdot\frac{dy}{dx}+\frac{d}{dy}e^y\cdot\frac{dy}{dx}&=\frac{d}{dy}\sin y^2\cdot\frac{dy}{dx}+3\sin x\\ \frac{dy}{dx}\left(\frac{1}{y}+e^y-2y\cos y^2\right)&=3\sin x\\ \Rightarrow \frac{dy}{dx}&=\frac{3\sin x}{\frac{1}{y}+e^y-2y\cos y^2}.\ _\square \end{aligned}
\sqrt{x}+\sqrt{y}=1
\left(\frac{1}{4},\frac{1}{4}\right).
First, we find the slope which is equal to
\left.\frac{dy}{dx}\right|_{x=\frac{1}{4}}.
x
\begin{aligned} \frac{d}{dx}\sqrt{x}+\frac{d}{dx}\sqrt{y}&=\frac{d}{dx}1\\ \frac{1}{2\sqrt{x}}+\frac{d}{dy}\sqrt{y}\cdot\frac{dy}{dx}&=0\\ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\cdot\frac{dy}{dx}&=0\\ \frac{dy}{dx}&=-\frac{\sqrt{y}}{\sqrt{x}}. \end{aligned}
Therefore the slope of the tangent is
\left.\frac{dy}{dx}\right|_{x=\frac{1}{4}}=\left.-\frac{\sqrt{y}}{\sqrt{x}}\right|_{x=\frac{1}{4}}=-1.
Since the tangent passes through the point
\left(\frac{1}{4},\frac{1}{4}\right),
the equation of the tangent is
\begin{aligned} y&=-\left(x-\frac{1}{4}\right)+\frac{1}{4}\\ &=-x+\frac{1}{2}.\ _\square \end{aligned}
\displaystyle f(x) = \frac{x}{1+\frac{x}{1+\frac{x}{1+\ddots}}}.
f'(0).
\large F(x) = \frac{x}{x + \frac{x}{x + \frac{x}{x + \frac{x }{._{._.}}} } }
If the value of the first derivative of
F(x)
x = 1
\frac{a\sqrt b - b}{c}
a,b,
c
a+b+c.
A prismatic package has side lengths
x,y,
z
108.
Its surface area
S
(without the cover) is given by
S = xy + 2yz + 2xz.
x+y+z
S?
\frac{dy}{dx}
y^2=x^2+\sin(xy).
Using implicit differentiation,
\begin{aligned} \dfrac{d}{dx}\big(y^2\big)&=\dfrac{d}{dx}\big(x^2\big)+\dfrac{d}{dx}(\sin xy)\\ 2y\dfrac{dy}{dx}&=\dfrac{d}{dx}\big(x^2\big)+(\cos xy)\dfrac{d}{dx}(xy)\\ 2y\dfrac{dy}{dx}&=2x+(\cos xy)\left(y+x\dfrac{dy}{dx}\right)\\ 2y\dfrac{dy}{dx}-(\cos xy)\left(x\dfrac{dy}{dx}\right)&=2x+\cos (xy)y\\ (2y-x\cos xy)\dfrac{dy}{dx}&=2x+\cos (xy)y\\ \dfrac{dy}{dx}&=\dfrac{2x+\cos (xy)y}{2y-x\cos xy}.\ _\square \end{aligned}
We can also use implicit differentiation to find the derivatives of the inverse trigonometric functions, assuming that these functions are differentiable.
y=\sin^{-1} x
y=\sin^{-1} (x),
x=\sin y,
- \frac{\pi}{2} \leq y \leq\frac{\pi}{2}.
Differentiating implicitly, we have
\cos y \dfrac{dy}{dx}=1 \implies \dfrac{dy}{dx}=\dfrac{1}{\cos y}.
-\frac{\pi}{2} \leq y \leq\frac{\pi}{2}
\cos y \geq 0,
\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-x^2},
\frac{d}{dx} \big(\sin^{-1} x\big) = \frac{1}{\sqrt{1-x^2}}.
_\square
Implicit differentiation is an approach to taking derivatives that uses the chain rule to avoid solving explicitly for one of the variables.
y^6 - e^{xy} = x,
\frac{dy}{dx}?
Taking the derivative of every term with respect to
x
\begin{aligned} 6y^5\left( \frac{dy}{dx} \right) - e^{xy} \left(\frac{d}{dx} (xy)\right)&=1 \\ 6y^5\left( \frac{dy}{dx} \right) - e^{xy} \left( y + x\frac{dy}{dx} \right)&=1 . \end{aligned}
Now, we can isolate all of the
\frac{dy}{dx}
\begin{aligned} 6y^5\left( \frac{dy}{dx} \right) - xe^{xy}\left(\frac{dy}{dx}\right) &=1 + ye^{xy} \\ \left( \frac{dy}{dx} \right) \left(6y^5 - xe^{xy} \right)&=1 + ye^{xy} \\ \frac{dy}{dx} &= \frac{ 1 + ye^{xy} }{ 6y^5 - xe^{xy} }. \ _\square \end{aligned}
\frac { y+1 }{ 1-y }
\frac { y }{ 1-y }
\frac { 2y }{ 1-y }
\frac { 1 }{ 1-y }
\huge y=e^{x + e^{x + e^{x + e^{x + \cdots }}}}
y
denote the infinite power tower function as described above.
\frac{dy}{dx}
y
\frac { x-y }{ (1+\log { x)^{ 2 } } }
\frac { \log { x } }{ (1+\log { x)^{ 2 } } }
\frac { x-y }{ 1+\log { x } }
\frac { y-x }{ x(1+\log { x) } }
{ x }^{ y }={ e }^{ x-y }
\frac { dy }{ dx } = \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}.
The area of a circle is given by
A=\pi r^2.
If the radius and area are differentiable functions of time
t,
find the equation relating
\frac{dA}{dt}
\frac{dr}{dt}.
Let's differentiate the expression
A=\pi r^2
t,
\begin{aligned} \dfrac{dA}{dt}&=\pi \dfrac{dr}{dt}(2r)\\ &=2\pi r\dfrac{dr}{dt}.\ _\square \end{aligned}
y'=x^x (\ln x + 1)
y'=x\ln x
y'=x (\ln x + 1)
y'=x^{x+1} (\ln x + 1)
y = x^x.
\begin{cases} \begin{aligned} x^{3}-3xy^{2}&=a \\ 3yx^{2}-y^{3}&=b. \end{aligned} \end{cases}
and
b
f(x) = \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}.
x = 12
\frac{a}{b},
and
b
a + b?
\sqrt { 3 } y=-(x+3)
\sqrt { 3 } y=-(3x+1)
\sqrt { 3 } y=x+3
\sqrt { 3 } y=3x+1
{(x-3)}^2+{y}^2=9
{y}^2=4x
x
Cite as: Implicit Differentiation. Brilliant.org. Retrieved from https://brilliant.org/wiki/implicit-differentiation/
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Modeling Resonant Coupled Wireless Power Transfer System - MATLAB & Simulink Example - MathWorks Switzerland
The Spiral Resonator
Resonance Frequency and Mode
Create Spiral to Spiral Power Transfer System
Variation of System Efficiency with Transfer Distance
Coupling Mode between Two Spiral Resonator
This example shows how to create and analyze a resonant coupling type wireless power transfer (WPT) system with an emphasis on concepts such as resonant mode, coupling effect, and magnetic field pattern. The analysis is based on a two-element system of spiral resonators.
Choose the design frequency to be 30 MHz. This is a popular frequency for compact WPT system design. Also specify the frequency for broadband analysis, and the points in space to plot near fields.
fcmin = 28e6;
fcmax = 31e6;
fband1 = 27e6:1e6:fcmin;
fband2 = fcmin:0.25e6:fcmax;
fband3 = fcmax:1e6:32e6;
freq = unique([fband1 fband2 fband3]);
pt=linspace(-0.3,0.3,61);
[X,Y,Z]=meshgrid(pt,0,pt);
field_p=[X(:)';Y(:)';Z(:)'];
The spiral is a very popular geometry in a resonant coupling type wireless power transfer system for its compact size and highly confined magnetic field. We will use such a spiral as the fundamental element in this example.
Create Spiral Geometry
The spiral is defined by its inner and outer radius, and number of turns.
Rin=0.05;
Rout=0.15;
spiralobj = spiralArchimedean('NumArms', 1, 'Turns', N, ...
'InnerRadius', Rin, 'OuterRadius', Rout, 'Tilt', 90, 'TiltAxis', 'Y');
It is important to find the resonant frequency of the designed spiral geometry. A good way to find the resonant frequency is to study the impedance of the spiral resonator. Since the spiral is a magnetic resonator, a lorentz shaped reactance is expected and observed in the calculated impedance result.
impedance(spiralobj,freq);
Since the spiral is a magnetic resonator, the dominant field component of this resonance is the magnetic field. A strongly localized magnetic field is observed when the near field is plotted.
EHfields(spiralobj,fc,field_p,'ViewField','H','ScaleFields',[0 5]);
The complete wireless power transfer system is composed of two parts: the transmitter(Tx) and receiver(Rx). Choose identical resonators for both transmitter and receiver to maximize the transfer efficiency. Here, the wireless power transfer system is modeled as a linear array.
wptsys=linearArray('Element',[spiralobj spiralobj]);
wptsys.ElementSpacing=Rout*2;
show(wptsys);
One way to evaluate the efficiency of the system is by studying the S21 parameter. As presented in [[1]], the system efficiency changes rapidly with operating frequency and the coupling strength between the transmitter and receiver resonator. Peak efficiency occurs when the system is operating at its resonant frequency, and the two resonators are strongly coupled.
sparam = sparameters(wptsys, freq);
rfplot(sparam,2,1,'abs');
Critical Coupled Point
The coupling between two spirals increases with decreasing distance between two resonators. This trend is approximately proportional to
1/{d}^{3}
. Therefore, the system efficiency increases with shorter transfer distance until it reaches the critical coupled regime [1]. When the two spirals are over coupled, exceeding the critical coupled threshold, system efficiency remains at its peak, as shown in Fig.3 in [1]. We observe this critical coupling point and over coupling effect during modeling of the system. Perform a parametric study of the system s-parameters as a function of the transfer distance. The transfer distance is varied by changing the ElementSpacing. It is varied from half of the spiral dimension to one and half times of the spiral dimension, which is twice of the spiral's outer radius. The frequency range is expanded and set from 25 MHz to 36 MHz.
freq=(25:0.1:36)*1e6;
dist=Rout*2*(0.5:0.1:1.5);
load('wptData.mat');
s21_dist=zeros(length(dist),length(freq));
for i=1:length(dist)
s21_dist(i,:)=rfparam(sparam_dist(i),2,1);
[X,Y]=meshgrid(freq/1e6,dist);
surf(X,Y,abs(s21_dist),'EdgeColor','none');
shading(gca,'interp');
xlabel('Frequency [MHz]');
ylabel('Distance [m]');
zlabel('S_{21} Magnitude');
The dominant energy exchange mechanism between the two spiral resonators is through the magnetic field. Strong magnetic fields are present between the two spirals at the resonant frequency.
EHfields(wptsys,fc,field_p,'ViewField','H','ScaleFields',[0 5]);
The results obtained for the wireless power transfer system match well with the results published in [[1]].
[1] Sample, Alanson P, D A Meyer, and J R Smith. โAnalysis, Experimental Results, and Range Adaptation of Magnetically Coupled Resonators for Wireless Power Transfer.โ IEEE Transactions on Industrial Electronics 58, no. 2 (February 2011): 544โ54. https://doi.org/10.1109/TIE.2010.2046002.
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'Mass-Energy' in Po8 - johnvalentine.co.uk
Research Warning
Property of mass-energy
No signularity
QED's massless photon
The property of intrinsic mass-energy
Mass-energy is a property of a boson, derived from the phases of its two waves. This value affects the probability that the boson will interact with others.
Bosons as the carriers of mass-energy; creators of fermions
What we think of as creation and annihilation, actually refers to fermions, which are localized events of two bosons.
Our mechanism describes annihilation as a fermion de-constituting as radiated bosons, and creation as the formation of a fermion from bosons. Our model shows a persistent fermion to be repeatedly being created and annihilated; if it happens with the same ingredients every time, then the fermion seems to be moving.
Bosons (and therefore instances of mass-energy) are not destroyed, but they can radiate away to perhaps eventually become a part of another fermion. Only bosons having low mass-energy are likely to escape far, because massive bosons have a high probability of collapse.
The intrinsic mass-energy of a boson is a function of the phase difference of its two component waves (fig.1).
Fig.1: Mass-energy ฯ
Specifically, it is the cosine of the phase difference between a reference wave (having phase
{\varphi }_{A}
) and its partner wave (having phase
{\varphi }_{B}
). This occurs in the b basis:
-b\rho =-b\mathrm{cos}\left({\varphi }_{B}-{\varphi }_{A}\right)=-b\phantom{\rule{mediummathspace}{0ex}}{e}^{-i\left({\varphi }_{B}-{\varphi }_{A}\right)}
A negative mass-energy ฯ results from a reference wave leading its partner by between ยผ-cycle and ยพ-cycle; i.e.
\rho <0
, where ฯ < 0, where
ฯ/2 < | ฯB โ ฯA | < 3ฯ/2.
The meaning of a negative value for mass-energy
This does not result in negative mass in the classical sense; in the context of gravitation, the dynamic effect is almost the same regardless of sign.
The width of the phase window for solutions is the same regardless of sign, so boson collapse (leading to a localising displacement) has approximately the same probability, but the position of the window within the phase cycle is different. This might have an effect on some phase-critical or coherent systems.
Bosons as oscillators
The boson's waves can be interpreted as components of a simple oscillator. When these two waves are plotted on orthogonal axes, the mass-energy is equivalent to the elliptical deviation from a circle, such that a circle has a mass-energy of zero.
Mass-energy remains invariant while the boson propagates in the degenerate state, conserving its energy:
d\rho /d\varphi =0
The mass-energy of a boson modulates the phase of any other boson that it overlaps (like a field operator), which widens the 'quantization window' for the phases of those bosons.
In QFT, We may interpret this as an excitation of the field, or an operator of the vacuum energy, so that wherever a boson is propagating, it gives mass-energy to the vacuum, as seen by other bosons.
The implication of this mechanism is that bosons having large mass tend to collapse within a short distance of their source position, and conversely, lighter bosons tend to travel further from their source position before being collapsed.
Lighter bosons (typically the flux of vacuum energy) may create quantization conditions for the more massive bosons, which provides further collapse opportunities for them. Conversely, the heavy bosons provide opportunities for the lighter bosons to collapse, when they would have otherwise propagated great distances.
This generalizes to a Compton radius, having a 50th percentile radius
r=\left(\mathrm{ln}2\right)/\rho
for non-overlapping phases of a shell.
No singularities
The above distribution has no singularity at the origin. Rather than being inversely-proportional to radius, the probability of collapse is based on the interaction of the boson's shell with a quantity of vacuum flux. In three-dimensional space, the boson's shell is the surface of a sphere, with an interaction area.
Our formulation differs form Newtonian and other classical formulations, because it does not have a singularity at radius 0, but instead falls off approaching 0. It is at this distance that Newtonian physics fails, and where ours differs from the Newtonian. This difference may be responsible for corrections like the Yukawa potential.
Fig.2: Probability distribution for single iteration of collapse, with radius r: plot of
{P}_{H}\left(r\right)
p={10}^{-5}
The Compton radius establishes an equilibrium distance around which particles will settle, for any given flux density of the environmental vacuum. This equilibrium does not occur in a solely Newtonian distribution. More:
Conserved particles are more likely to lose coherence if their bosons propagate to a large radius in the presence of a flux.
Constitution: Coherence
'Massless' photons in QED
In QED we say that photons do not have mass. Our mechanism says they may have mass, but this does not contradict QED, because our photon is transmitted as two separate bosons, which land at their target to each substitute a boson at two fermion instances at the target. This means that there is no change of mass at the target, unless the photon changes the constitution of the target fermion.
As a bonus, this substitution generates a flux from a charged particle, so we have a quantum photoelectric effect.
Minimal Deterministic Physicality Applied to Cosmology [15]
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Cohen's kappa measures the agreement between two raters who each classify N items into C mutually exclusive categories. The definition of
{\textstyle \kappa }
{\displaystyle \kappa \equiv {\frac {p_{o}-p_{e}}{1-p_{e}}}=1-{\frac {1-p_{o}}{1-p_{e}}},}
where po is the relative observed agreement among raters, and pe is the hypothetical probability of chance agreement, using the observed data to calculate the probabilities of each observer randomly seeing each category. If the raters are in complete agreement then
{\textstyle \kappa =1}
. If there is no agreement among the raters other than what would be expected by chance (as given by pe),
{\textstyle \kappa =0}
. It is possible for the statistic to be negative,[6] which can occur by chance if there is no relationship between the ratings of the two raters, or it may reflect a real tendency of the raters to give differing ratings.
For k categories, N observations to categorize and
{\displaystyle n_{ki}}
the number of times rater i predicted category k:
{\displaystyle p_{e}={\frac {1}{N^{2}}}\sum _{k}n_{k1}n_{k2}}
{\displaystyle p_{e}=\sum _{k}{\widehat {p_{k12}}}=\sum _{k}{\widehat {p_{k1}}}{\widehat {p_{k2}}}=\sum _{k}{\frac {n_{k1}}{N}}{\frac {n_{k2}}{N}}={\frac {1}{N^{2}}}\sum _{k}n_{k1}n_{k2}}
{\displaystyle {\widehat {p_{k12}}}}
is the estimated probability that both rater 1 and rater 2 will classify the same item as k, while
{\displaystyle {\widehat {p_{k1}}}}
is the estimated probability that rater 1 will classify an item as k (and similarly for rater 2). The relation
{\textstyle {\widehat {p_{k}}}=\sum _{k}{\widehat {p_{k1}}}{\widehat {p_{k2}}}}
is based on using the assumption that the rating of the two raters are independent. The term
{\displaystyle {\widehat {p_{k1}}}}
is estimated by using the number of items classified as k by rater 1 (
{\displaystyle n_{k1}}
) divided by the total items to classify (
{\displaystyle N}
{\displaystyle {\widehat {p_{k1}}}={n_{k1} \over N}}
(and similarly for rater 2).
{\displaystyle \kappa ={\frac {2\times (TP\times TN-FN\times FP)}{(TP+FP)\times (FP+TN)+(TP+FN)\times (FN+TN)}}}
{\displaystyle p_{o}={\frac {a+d}{a+b+c+d}}={\frac {20+15}{50}}=0.7}
{\displaystyle p_{\text{Yes}}={\frac {a+b}{a+b+c+d}}\cdot {\frac {a+c}{a+b+c+d}}=0.5\times 0.6=0.3}
{\displaystyle p_{\text{No}}={\frac {c+d}{a+b+c+d}}\cdot {\frac {b+d}{a+b+c+d}}=0.5\times 0.4=0.2}
{\displaystyle p_{e}=p_{\text{Yes}}+p_{\text{No}}=0.3+0.2=0.5}
{\displaystyle \kappa ={\frac {p_{o}-p_{e}}{1-p_{e}}}={\frac {0.7-0.5}{1-0.5}}=0.4}
{\displaystyle \kappa ={\frac {0.60-0.54}{1-0.54}}=0.1304}
{\displaystyle \kappa ={\frac {0.60-0.46}{1-0.46}}=0.2593}
{\displaystyle CI:\kappa \pm Z_{1-\alpha /2}SE_{\kappa }}
{\displaystyle Z_{1-\alpha /2}=1.960}
is the standard normal percentile when
{\displaystyle \alpha =5\%}
{\displaystyle SE_{\kappa }={\sqrt {{p_{o}(1-p_{o})} \over {N(1-p_{e})^{2}}}}}
This is calculated by ignoring that pe is estimated from the data, and by treating po as an estimated probability of a binomial distribution while using asymptotic normality (i.e.: assuming that the number of items is large and that po is not close to either 0 or 1).
{\displaystyle SE_{\kappa }}
(and the CI in general) may also be estimated using bootstrap methods.
{\displaystyle \kappa _{\max }={\frac {P_{\max }-P_{\exp }}{1-P_{\exp }}}}
{\displaystyle P_{\exp }=\sum _{i=1}^{k}P_{i+}P_{+i}}
, as usual,
{\displaystyle P_{\max }=\sum _{i=1}^{k}\min(P_{i+},P_{+i})}
k = number of codes,
{\displaystyle P_{i+}}
are the row probabilities, and
{\displaystyle P_{+i}}
are the column probabilities.
{\displaystyle \kappa =1-{\frac {\sum _{i=1}^{k}\sum _{j=1}^{k}w_{ij}x_{ij}}{\sum _{i=1}^{k}\sum _{j=1}^{k}w_{ij}m_{ij}}}}
where k=number of codes and
{\displaystyle w_{ij}}
{\displaystyle x_{ij}}
{\displaystyle m_{ij}}
are elements in the weight, observed, and expected matrices, respectively. When diagonal cells contain weights of 0 and all off-diagonal cells weights of 1, this formula produces the same value of kappa as the calculation given above.
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Estimate states of discrete-time nonlinear system using unscented Kalman filter - Simulink - MathWorks Australia
\stackrel{^}{x}
\stackrel{^}{x}\left[k|k\right]
\stackrel{^}{x}\left[k|k-1\right]
\stackrel{^}{x}\left[k|k\right]
\stackrel{^}{x}\left[k|k-1\right]
\stackrel{^}{x}
\begin{array}{l}x\left[k+1\right]=f\left(x\left[k\right],{u}_{s}\left[k\right]\right)+w\left[k\right]\\ y\left[k\right]=h\left(x\left[k\right],{u}_{m}\left[k\right]\right)+v\left[k\right]\end{array}
\begin{array}{l}x\left[k+1\right]=f\left(x\left[k\right],w\left[k\right],{u}_{s}\left[k\right]\right)\\ y\left[k\right]=h\left(x\left[k\right],v\left[k\right],{u}_{m}\left[k\right]\right)\end{array}
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Global Constraint Catalog: Ctemporal_path
<< 5.399. symmetric_gcc5.401. tour >>
\mathrm{๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐}\left(\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท},\mathrm{๐ฝ๐พ๐ณ๐ด๐}\right)
\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}
\mathrm{๐๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\begin{array}{c}\mathrm{๐๐๐๐๐ก}-\mathrm{๐๐๐},\hfill \\ \mathrm{๐๐๐๐}-\mathrm{๐๐๐๐},\hfill \\ \mathrm{๐๐๐๐๐}-\mathrm{๐๐๐๐},\hfill \\ \mathrm{๐๐๐}-\mathrm{๐๐๐๐}\hfill \end{array}\right)
\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}\ge 1
\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฝ๐พ๐ณ๐ด๐},\left[\mathrm{๐๐๐๐๐ก},\mathrm{๐๐๐๐},\mathrm{๐๐๐๐๐},\mathrm{๐๐๐}\right]\right)
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|>0
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐๐ก}\ge 1
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐๐ก}\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฝ๐พ๐ณ๐ด๐},\mathrm{๐๐๐๐๐ก}\right)
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐}\ge 1
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐}\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐๐}\le \mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐}
G
be the digraph described by the
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
collection. Partition
G
with a set of disjoint paths such that each vertex of the graph belongs to a single path. In addition, for all pairs of consecutive vertices of a path we have a precedence constraint that enforces the end associated with the first vertex to be less than or equal to the start related to the second vertex.
\left(\begin{array}{c}2,โฉ\begin{array}{cccc}\mathrm{๐๐๐๐๐ก}-1\hfill & \mathrm{๐๐๐๐}-2\hfill & \mathrm{๐๐๐๐๐}-0\hfill & \mathrm{๐๐๐}-1,\hfill \\ \mathrm{๐๐๐๐๐ก}-2\hfill & \mathrm{๐๐๐๐}-6\hfill & \mathrm{๐๐๐๐๐}-3\hfill & \mathrm{๐๐๐}-5,\hfill \\ \mathrm{๐๐๐๐๐ก}-3\hfill & \mathrm{๐๐๐๐}-4\hfill & \mathrm{๐๐๐๐๐}-0\hfill & \mathrm{๐๐๐}-3,\hfill \\ \mathrm{๐๐๐๐๐ก}-4\hfill & \mathrm{๐๐๐๐}-5\hfill & \mathrm{๐๐๐๐๐}-4\hfill & \mathrm{๐๐๐}-6,\hfill \\ \mathrm{๐๐๐๐๐ก}-5\hfill & \mathrm{๐๐๐๐}-7\hfill & \mathrm{๐๐๐๐๐}-7\hfill & \mathrm{๐๐๐}-8,\hfill \\ \mathrm{๐๐๐๐๐ก}-6\hfill & \mathrm{๐๐๐๐}-6\hfill & \mathrm{๐๐๐๐๐}-7\hfill & \mathrm{๐๐๐}-9,\hfill \\ \mathrm{๐๐๐๐๐ก}-7\hfill & \mathrm{๐๐๐๐}-7\hfill & \mathrm{๐๐๐๐๐}-9\hfill & \mathrm{๐๐๐}-10\hfill \end{array}โช\hfill \end{array}\right)
\mathrm{๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
collection represent the two (
\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}=2
) paths
1\to 2\to 6
3\to 4\to 5\to 7
As illustrated by Figure 5.400.1, all precedences between adjacent vertices of a same path hold: each item
i
1\le i\le 7
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
collection is represented by a rectangle starting and ending at instants
\mathrm{๐ฝ๐พ๐ณ๐ด๐}\left[i\right].\mathrm{๐๐๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}\left[i\right].\mathrm{๐๐๐}
; the number within each rectangle designates the index of the corresponding item within the
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
Figure 5.400.1. The two paths of the Example slot represented as two sequences of tasks
\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}<|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|>1
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐๐}<\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
This constraint is related to the
\mathrm{๐๐๐๐}
constraint of Ilog Solver. It can also be directly expressed with the
\mathrm{๐๐ข๐๐๐}
[BeldiceanuContejean94] constraint of CHIP by using the diff nodes and the origin parameters. A generic model based on linear programming that handles paths, trees and cycles is presented in [LabbeLaporteRodriguezMartin98].
\mathrm{๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐}
\left(\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท},\mathrm{๐ฝ๐พ๐ณ๐ด๐}\right)
constraint can be expressed in term of a conjunction of one
\mathrm{๐๐๐๐}
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐๐๐๐๐๐๐}
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
inequalities constraints:
\mathrm{๐๐๐๐}
constraint the number of path variable
\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}
as well as the items of the
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
collection form which we remove the
\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐}
i
\left(1\le i\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|\right)
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
collection, we create a variable
{\mathrm{๐๐ก๐๐๐ก}}_{{\mathrm{๐ ๐ข๐๐}}_{๐}}
\mathrm{๐๐๐๐๐๐๐}
\left(\mathrm{๐ฝ๐พ๐ณ๐ด๐}\left[i\right].\mathrm{๐๐๐๐},โฉ{T}_{i,1},{T}_{i,2},\cdots ,{T}_{i,\mathrm{๐ฝ๐พ๐ณ๐ด๐}}โช,{\mathrm{๐๐ก๐๐๐ก}}_{{\mathrm{๐ ๐ข๐๐}}_{๐}}\right)
{T}_{i,j}=\mathrm{๐ฝ๐พ๐ณ๐ด๐}\left[i\right].\mathrm{๐๐๐๐๐}
i\ne j
{T}_{i,i}=\mathrm{๐ฝ๐พ๐ณ๐ด๐}\left[i\right].\mathrm{๐๐๐}
Finaly to the
i
\left(1\le i\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|\right)
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
collection, we also create an inequality constraint
\mathrm{๐ฝ๐พ๐ณ๐ด๐}\left[i\right].\mathrm{๐๐๐}\le {\mathrm{๐๐ก๐๐๐ก}}_{{\mathrm{๐ ๐ข๐๐}}_{๐}}
. Note that, since
{T}_{i,i}
was initialised to
\mathrm{๐ฝ๐พ๐ณ๐ด๐}\left[i\right].\mathrm{๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}\left[i\right].\mathrm{๐๐๐}\le {T}_{i,j}
i=j
With respect to the Example slot we get the following conjunction of constraints:
\mathrm{๐๐๐๐}
\left(2,โฉ\mathrm{๐๐๐๐๐ก}-1\mathrm{๐๐๐๐}-2,\mathrm{๐๐๐๐๐ก}-2\mathrm{๐๐๐๐}-6,\mathrm{๐๐๐๐๐ก}-3\mathrm{๐๐๐๐}-4,
\mathrm{๐๐๐๐๐ก}-4\mathrm{๐๐๐๐}-5,\mathrm{๐๐๐๐๐ก}-5\mathrm{๐๐๐๐}-7,\mathrm{๐๐๐๐๐ก}-6\mathrm{๐๐๐๐}-6,
\mathrm{๐๐๐๐๐ก}-7\mathrm{๐๐๐๐}-7โช\right)
\mathrm{๐๐๐๐๐๐๐}
\left(2,โฉ1,3,0,4,7,7,9โช,3\right)
\mathrm{๐๐๐๐๐๐๐}
\left(6,โฉ1,5,0,4,7,7,9โช,7\right)
\mathrm{๐๐๐๐๐๐๐}
\left(4,โฉ1,5,3,4,7,7,9โช,4\right)
\mathrm{๐๐๐๐๐๐๐}
\left(5,โฉ1,5,3,6,7,7,9โช,7\right)
\mathrm{๐๐๐๐๐๐๐}
\left(7,โฉ1,5,3,6,8,7,9โช,9\right)
\mathrm{๐๐๐๐๐๐๐}
\left(6,โฉ1,5,3,6,8,9,9โช,9\right)
\mathrm{๐๐๐๐๐๐๐}
\left(7,โฉ1,5,3,6,8,9,10โช,10\right)
1\le 3
5\le 7
3\le 4
6\le 7
8\le 9
9\le 9
10\le 10
\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐}_\mathrm{๐๐}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}
(time dimension removed).
modelling: sequence dependent set-up, functional dependency.
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐ถ๐ฟ๐ผ๐๐๐ธ}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐}\mathtt{2}\right)
โข\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐}=\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐ก}
โข\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐}=\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐ก}\vee \mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}\le \mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐}
โข\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐๐}\le \mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}
โข\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐}\le \mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}
โข
\mathrm{๐๐๐}_\mathrm{๐๐}
\le 1
โข
\mathrm{๐๐๐}
=\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}
โข
\mathrm{๐๐๐๐๐๐๐}
=|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
The arc constraint is a conjunction of four conditions that respectively correspond to:
A constraint that links the successor variable of a first vertex to the index attribute of a second vertex,
A precedence constraint that applies on one vertex and its distinct successor,
One precedence constraint between the start and the end of the vertex that corresponds to the departure of an arc,
One precedence constraint between the start and the end of the vertex that corresponds to the arrival of an arc.
We use the following three graph properties in order to enforce the partitioning of the graph in distinct paths:
The first property
\mathrm{๐๐๐}_\mathrm{๐๐}
\le 1
enforces that each vertex has no more than one predecessor (
\mathrm{๐๐๐}_\mathrm{๐๐}
does not consider loops),
The second property
\mathrm{๐๐๐}
=\mathrm{๐ฝ๐ฟ๐ฐ๐๐ท}
ensures that we have the required number of paths,
The third property
\mathrm{๐๐๐๐๐๐๐}
=|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
enforces that, for each vertex, the start is not located after the end.
\mathrm{๐๐๐}_\mathrm{๐๐}
\mathrm{๐๐๐}
\mathrm{๐๐๐๐๐๐๐}
graph properties we display the following information on the final graph:
We show with a double circle a vertex that has the maximum number of predecessors.
We show the two connected components corresponding to the two paths.
We put in bold the vertices.
\mathrm{๐๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐}
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Global Constraint Catalog: Kchannelling_constraint
<< 3.7.43. Channel routing3.7.45. Circuit >>
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐ ๐๐๐๐๐}_\mathrm{๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}
Constraints that allow for linking two models of the same problem [Hernandez07]. Usually channelling constraints show up in the following context:
When a problem can be modelled by using different types of variables (e.g., 0-1 variables, domain variables, set variables),
When a problem can be modelled by using two distinct matrices of variables representing the same information redundantly,
When, in a problem, the roles of the variables and the values can be interchanged. This is typically the case when we have a bijection between a set of variables and the values they can take.
When, in a problem, we use two time coordinates systems (e.g., see
\mathrm{๐๐๐๐๐๐๐๐}
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Neutral third - Wikipedia
11:9,[1] 27:22,[1][2] or 16:13[3]
Neutral third on C
A neutral third is a musical interval wider than a minor third play (helpยทinfo) but narrower than a major third play (helpยทinfo), named by Jan Pieter Land in 1880.[4] Land makes reference to the neutral third attributed to Zalzal (8th c.), described by Al-Farabi (10th c.) as corresponding to a ratio of 27:22 (354.5 cents) and by Avicenna (Ibn Sina, 11th c.) as 39:32 (342.5 cents).[5] The Zalzalian third may have been a mobile interval.
Three distinct intervals may be termed neutral thirds:[6]
The undecimal neutral third has a ratio of 11:9[7] between the frequencies of the two tones, or about 347.41 cents play (helpยทinfo). This ratio is the mathematical mediant of the major third 5/4 and the minor third 6/5, and as such, has the property that if harmonic notes of frequency f and (11/9) f are played together, the beat frequency of the 5th harmonic of the lower pitch against the 4th of the upper, i.e.
{\displaystyle |5f-4(11/9)f|=(1/9)f}
, is the same as the beat frequency of the 6th harmonic of the lower pitch against the 5th of the upper, i.e.
{\displaystyle |6f-5(11/9)f|=|-(1/9)f|=(1/9)f}
. In this sense, it is the unique ratio which is equally well-tuned as a major and minor third.
A tridecimal neutral third play (helpยทinfo) has a ratio of 16:13 between the frequencies of the two tones, or about 359.47 cents.[3] This is the largest neutral third, and occurs infrequently in music, as little music utilizes the 13th harmonic. It is the mediant of the septimal major third 7/6 and septimal minor third 9/7, and as such, enjoys an analogous property with regard to the beating of the corresponding harmonics as above. That is,
{\displaystyle |7f-6(16/13)f|=|9f-7(16/13)f|=(5/13)f}
An equal-tempered neutral third play (helpยทinfo) is characterized by a difference of 350 cents between the two tones, slightly wider than the 11:9 ratio, and exactly half of an equal-tempered perfect fifth.
These intervals are all within about 12 cents and are difficult for most people to distinguish by ear. Neutral thirds are roughly a quarter tone sharp from 12 equal temperament minor thirds and a quarter tone flat from 12-ET major thirds. In just intonation, as well as in tunings such as 31-ET, 41-ET, or 72-ET, which more closely approximate just intonation, the intervals are closer together.
In addition to the above examples, a square root neutral third can be characterized by a ratio of
{\displaystyle {\sqrt {3/2}}}
between two frequencies, being exactly half of a just perfect fifth of 3/2 and measuring about 350.98 cents. Such a definition stems from the two thirds traditionally making a fifth-based triad.
A triad formed by two neutral thirds is neither major nor minor, thus the neutral thirds triad is ambiguous. While it is not found in twelve tone equal temperament it is found in others such as the quarter tone scale Play (helpยทinfo) and 31-tet Play (helpยทinfo).
1 Occurrence in human music
1.1 In infants' song
1.2 In modern classical Western music
1.3 In traditional music
1.4 In contemporary popular music
2 In equal temperaments
Occurrence in human music[edit]
In infants' song[edit]
Infants experiment with singing, and a few studies of individual infants' singing found that neutral thirds regularly arise in their improvisations. In two separate case studies of the progression and development of these improvisations, neutral thirds were found to arise in infants' songs after major and minor seconds and thirds, but before intervals smaller than a semitone and also before intervals as large as a perfect fourth or larger.[8]
In modern classical Western music[edit]
The neutral third has been used by a number of modern composers, including Charles Ives, James Tenney, and Gayle Young.[9]
In traditional music[edit]
The equal-tempered neutral third may be found in the quarter tone scale and in some traditional Arab music (see also Arab tone system). Undecimal neutral thirds appear in traditional Georgian music.[10] Neutral thirds are also found in American folk music.[11]
In contemporary popular music[edit]
Blue notes (a note found in country music, blues, and some rock music) on the third note of a scale can be seen as a variant of a neutral third with the tonic, as they fall in between a major third and a minor third. Similarly the blue note on the seventh note of the scale can be seen as a neutral third with the dominant.
In equal temperaments[edit]
Two steps of seven-tone equal temperament form an interval of 342.8571 cents, which is within 5 cents of 347.4079 for the undecimal (11:9) neutral third.[12] This is an equal temperament in reasonably common use, at least in the form of "near seven equal", as it is a tuning used for Thai music as well as the Ugandan Chopi tradition of music.[13]
The neutral third also has good approximations in other commonly used equal temperaments including 24-ET (7 steps, 350 cents[14]) and similarly by all multiples of 24 equal steps such as 48-ET[15] and 72-ET,[16] 31-ET (9 steps, 348.39),[17] 34-ET (10 steps, 352.941 cents[18]), 41-ET (12 steps, 351.22 cents[19]), and slightly less closely by 53-ET (15 steps, 339.62 cents[20]).
Close approximations to the tridecimal neutral third (16:13) appear in 53-ET[20] and 72-ET.[16] Both of these temperaments distinguish between the tridecimal (16:13) and undecimal (11:9) neutral thirds. All the other tuning systems mentioned above fail to distinguish between these intervals; they temper out the comma 144:143.
^ a b Haluska, Jan (2003). The Mathematical Theory of Tone Systems, p. xxiii. ISBN 0-8247-4714-3. Undecimal neutral third and Zalzal's wosta.
^ "Neutral third scales", on Xenharmonic Wiki.
^ a b Haluska (2003), p. xxiv. Tridecimal neutral third.
^ J. P. Land, Over de toonladders der Arabische muziek, 1880; Recherches sur l'histoire de la gamme arabe, 1884. See Hermann von Helmholtz, On the Sensations of Tone, (Alexander John Ellis, trans.) (3rd ed., 1895), p. 281, note โ (addition by Ellis).
^ Liberty Manik (1969). Das Arabische Tonsystem im Mittelalter (Leiden: Brill, 1969), 46โ49.
^ Alois Hรกba, Neue Harmonielehre des diatonischen, chromatischen, Viertel-, Drittel-, Sechstel- und Zwรถlftel-Tonsystems (Liepzig: Fr. Kistner & C.F.W. Sigel, 1927), 143. [ISBN unspecified]. Cited in Skinner, Miles Leigh (2007). Toward a Quarter-tone Syntax: Analyses of Selected Works by Blackwood, Haba, Ives, and Wyschnegradsky, p. 25. ProQuest. ISBN 9780542998478.
^ Andrew Horner, Lydia Ayres (2002). Cooking with Csound: Woodwind and Brass Recipes, p. 131. ISBN 0-89579-507-8. "Super-Major Second".
^ Nettl, Bruno. "Infant Musical Development and Primitive Music", Southwestern Journal of Anthropology, vol. 12, no. 1, pp. 87โ91. (Spring 1956) JSTOR 3628859
^ Young, Gayle. "The Pitch Organization of Harmonium for James Tenney", Perspectives of New Music, vol. 26, no. 2. pp. 204โ212 (Summer 1988) JSTOR 833190
^ Aronson, Howard I., ed. (1994). The Annual of the Society for the Study of Caucasia. Society for the Study of Caucasia. p. 93. Retrieved 14 April 2011.
^ Boswell, George W. "The Neutral Tone as a Function of Folk-Song Text", Yearbook of the International Folk Music Council, vol. 2, 1970, pp. 127โ132 (1970) JSTOR 767430
^ "7edo", on Xenharmonic Wiki.
^ Morton, David (1980). "The Music of Thailand", Musics of Many Cultures, p. 70. May, Elizabeth, ed. ISBN 0-520-04778-8.
^ "24edo", on Xenharmonic Wiki.
^ a b "72edo", on Xenharmonic Wiki.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Neutral_third&oldid=1067436010"
Neutral intervals
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Psychoacoustic Bass Enhancement for Band-Limited Signals - MATLAB & Simulink - MathWorks รญโขลรชยตยญ
Bass Enhancer Audio plugin
This example shows an audio plugin designed to enhance the perceived sound level in the lower part of the audible spectrum.
Small loudspeakers typically have a poor low frequency response, which can have a negative impact on overall sound quality. This example implements psychoacoustic bass enhancement to improve sound quality of audio played on small loudspeakers.
The example is based on the algorithm in [1]. A non-linear device shifts the low-frequency range of the signal to a high-frequency range through the generation of harmonics. The pitch of the original signal is preserved due to the "virtual pitch" psychoacoustic phenomenon.
The algorithm is implemented using an audio plugin object.
The figure below illustrates the algorithm used in [1].
1. The input stereo signal is split into lowpass and highpass components using a crossover filter. The filter's crossover frequency is equal to the speaker's cutoff frequency (set to 60 Hz in this example).
2. The highpass component,
h{p}_{stereo}
, is split into left and right channels:
h{p}_{left}
h{p}_{right}
3. The lowpass component,
l{p}_{stereo}
, is converted to mono,
l{p}_{mono}
, by adding the left and right channels element by element.
l{p}_{mono}
is passed through a full wave integrator. The full wave integrator shifts
l{p}_{mono}
to higher harmonics.
\mathit{y}\left[\mathit{n}\right]=\left\{\begin{array}{ll}0& \mathrm{if}\text{รขยย}\mathit{u}\left[\mathit{n}\right]>0\text{รขยย}\mathrm{and}\text{รขยย}\mathit{u}\left[\mathit{n}-1\right]รขยยค0\\ \mathit{y}\left[\mathit{n}-1\right]+\mathit{u}\left[\mathit{n}-1\right]& \mathrm{otherwise}\end{array}
u\left[n\right]
l{p}_{mono}
y\left[n\right]
is the output signal
n
is the time index
y\left[n\right]
is passed through a bandpass filter with lower cutoff frequency set to the speaker's cutoff frequency. The filter's upper cutoff frequency may be adjusted to fine-tune output sound quality.
{y}_{BP}\left[n\right]
, the bandpass filtered signal, passes through tunable gain,
G
{y}_{G}
is added to the left and right highpass channels.
8. The left and right channels are concatenated into a single matrix and output.
Although the resulting output stereo signal does not contain low-frequency elements, the input's bass pitch is preserved thanks to the generated harmonics.
audiopluginexample.BassEnhancer is an audio plugin object that implements the psychoacoustic bass enhancement algorithm. The plugin parameters are the upper cutoff frequency of the bandpass filter, and the gain applied at the output of the bandpass filter (G in the diagram above). You can incorporate the object into a MATLAB simulation, or use it to generate an audio plugin using generateAudioPlugin.
You can open a test bench for audiopluginexample.BassEnhancer using Audio Test Bench. The test bench provides a graphical user interface to help you test your audio plugin in MATLAB. You can tune the plugin parameters as the test bench is executing. You can also open a timescope and a dsp.SpectrumAnalyzer to view and compare the input and output signals in the time and frequency domains, respectively.
bassEnhancer = audiopluginexample.BassEnhancer;
audioTestBench(bassEnhancer)
You can also use audiopluginexample.BassEnhancer in MATLAB just as you would use any other MATLAB object. You can use configureMIDI to enable tuning the object using a MIDI device. This is particularly useful if the object is part of a streaming MATLAB simulation where the command window is not free.
HelperBassEnhancerSim is a simple function that may be used to perform bass enhancement as part of a larger MATLAB simulation. The function instantiates an audiopluginexample.BassEnhancer plugin, and uses the setSampleRate method to set its sampling rate to the input argument Fs. The plugin's parameters are tuned by setting their values to the input arguments Fcutoff and G, respectively. Note that it is also possible to generate a MEX-file from this function using the codegen command. Performance is improved in this mode without compromising the ability to tune parameters.
[1] Aarts, Ronald M, Erik Larsen, and Daniel Schobben. รขโฌลImproving Perceived Bass and Reconstruction of High Frequencies for Band Limited Signals.รขโฌย Proceedings 1st IEEE Benelux Workshop on Model Based Coding of Audio (MPCA-2002) , November 15, 2002, 59โ71.
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\large \lim_{n\rightarrow \infty} \, \sqrt[ \large n^{2}+n]{\binom{n}{0}\binom{n}{1}\binom{n}{2}\cdots \binom{n}{n}}
Find the value of the closed form of the above limit to 3 decimal places.
(s_n)_{n=0}^{\infty}
be a sequence of real numbers defined as follows:
s_0 = 2; s_{n+1} = \sqrt{2-\sqrt{4-s_n^2}}
n \ge 0
To the nearest hundredth, find the value of
\displaystyle\lim_{n \to \infty} 2^n s_n
In other words, to what value does the following sequence converge:
2^3 s_3 = 8\sqrt{2-\sqrt{2+\sqrt{2}}}
2^4 s_4 = 16\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}
2^5 s_5 = 32\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}
by Ariel Gershon
\large \lim_{n\to\infty} \sqrt[n^2]{{n \choose1}{n \choose 2}\cdots{n \choose n}}
Find the closed form of the limit above to 3 decimal places.
\displaystyle {n \choose k} = \dfrac {n!}{k!(n-k)!}
by Guillermo Templado
a=\dfrac1{16}
, consider the (finite) power tower,
\Large x_n=\underbrace{a^{a^{\cdot^{\cdot^{a^a}}}}}_{2n \; a\text{'s}}
x_1=a^a
x_2=a^{a^{a^a}}
\displaystyle \lim_{n\to\infty}x_n
, to three significant figures.
Bonus What happens if we consider a power tower with an odd number of
a
's?
None of the others 0.75 0.5 0.25 Limit does not exist 1 0.364
\begin{aligned} f(x) &= \begin{cases} 0 &\text{ if } x \text{ is irrational} \\ 1 &\text{ if } x \text{ is rational} \end{cases} \\ g(x) &= \begin{cases} 0 &\text{ if } x \text{ is irrational} \\ \frac1q &\text{ if } x =\frac{p}{q}, \text{ where } p \text{ and } q \text{ are coprime nonnegative integers} \end{cases} \end{aligned}
f(x)
g(x)
[0,1]
a \in (0,1)
\lim\limits_{x\to a} f(x)
b \in (0,1)
\lim\limits_{x\to b} g(x)
a
b
a
b
a
b
a
b
a
b
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Light Trapping | PVEducation
The optimum device thickness is not controlled solely by the need to absorb all the light. For example, if the light is not absorbed within a diffusion length of the junction, then the light-generated carriers are lost to recombination. In addition, as discussed in the Voltage Losses Due to Recombination, a thinner solar cell which retains the absorption of the thicker device may have a higher voltage. Consequently, an optimum solar cell structure will typically have "light trapping" in which the optical path length is several times the actual device thickness, where the optical path length of a device refers to the distance that an unabsorbed photon may travel within the device before it escapes out of the device. This is usually defined in terms of device thickness. For example, a solar cell with no light trapping features may have an optical path length of one device thickness, while a solar cell with good light trapping may have an optical path length of 50, indicating that light bounces back and forth within the cell many times.
Light trapping is usually achieved by changing the angle at which light travels in the solar cell by having it be incident on an angled surface. A textured surface will not only reduce reflection as previously described but will also couple light obliquely into the silicon, thus giving a longer optical path length than the physical device thickness. The angle at which light is refracted into the semiconductor material is, according to Snell's Law, as follows:
{n}_{1}\mathrm{sin}{\theta }_{1}={n}_{2}\mathrm{sin}{\theta }_{2}
Read more about Snell's Law
where ฮธ1 and ฮธ2 are the angles for the light incident on the interface relative to the normal plane of the interface within the mediums with refractive indices n1 and n2 respectively. ฮธ1 and ฮธ2 are shown in the animation below.
Refraction of a ray of light at a dielectric boundary. You can adjust the angle of incidence and see how this affects the angle of the ray transmitted to the second medium by clicking in the right side of the graph and dragging the mouse to change the angle. When n2 has a higher refractive index than n1 the refracted ray is closer to normal than the incident ray.
By rearranging Snell's law above, the angle at which light enters the solar cell (the angle of refracted light) can be calculated:
{\theta }_{2}={\mathrm{sin}}^{-1}\left(\frac{{n}_{2}}{{n}_{1}}\mathrm{sin}{\theta }_{1}\right)
Read more about Snell's Law (rearranged)
In a textured single crystalline solar cell, the presence of crystallographic planes make the angle ฮธ1 equal to 36ยฐ as shown below.
Reflection and transmission of light for a textured silicon solar cell.
The amount of light reflected at an interface is calculated from the fresnel reflection formula. For light polarised in the parrallel to the surface the amount of reflected light is:
{R}_{\parallel }=\frac{{\mathrm{tan}}^{2}\left({\theta }_{1}-{\theta }_{2}\right)}{{\mathrm{tan}}^{2}\left({\theta }_{1}+{\theta }_{2}\right)}
Read more about Fresnel Reflection (parrallel)
For light polarised perpendicular to the surface the amount reflected is:
{R}_{\bot }=\frac{{\mathrm{sin}}^{2}\left({\theta }_{1}-{\theta }_{2}\right)}{{\mathrm{sin}}^{2}\left({\theta }_{1}+{\theta }_{2}\right)}
Read more about Fresnel Reflection (perpendicular)
For unpolarised light the reflected amount is the average of the two:
{R}_{T}=\frac{{R}_{\bot }+{R}_{\parallel }}{2}
Read more about Fresnel Reflection (plane and perpendicular)
Light Trapping Calculator
Refractive index of incident medium, n1 = Refractive index of transmitted medium, n2 = Incident angle, ฮธ1 = degrees
Refracted angle, ฮธ2 = degrees Proportion of light reflected, R = Proportion of light transmitted, T =
If light passes from a high refractive index medium to a low refractive index medium, there is the possibility of total internal reflection (TIR). The angle at which this occurs is the critical angle and is found by setting ฮธ2 in Snell's law to 0.
{\theta }_{1}={\mathrm{sin}}^{-1}\left(\frac{{\eta }_{2}}{{\eta }_{1}}\right)
Read more about Snell's Law (Critical Angle)
Total Internal Reflection Calculator
Refractive index of incident medium, n1 = Refractive index of secondary medium, n2 =
Critical angle for total internal reflection to occur,
ฮธ1 = degrees
Using total internal reflection, light can be trapped inside the cell and make multiple passes through the cell, thus allowing even a thin solar cell to maintain a high optical path length.
Material ThicknessLambertian Rear Reflectors
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Global Constraint Catalog: Ccycle_or_accessibility
<< 5.104. cycle_card_on_path5.106. cycle_resource >>
Inspired by [LabbeLaporteRodriguezMartin98].
\mathrm{๐๐ข๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐ข}\left(\mathrm{๐ผ๐ฐ๐๐ณ๐ธ๐๐},\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด},\mathrm{๐ฝ๐พ๐ณ๐ด๐}\right)
\mathrm{๐ผ๐ฐ๐๐ณ๐ธ๐๐}
\mathrm{๐๐๐}
\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด}
\mathrm{๐๐๐๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐ก}-\mathrm{๐๐๐},\mathrm{๐๐๐๐}-\mathrm{๐๐๐๐},๐ก-\mathrm{๐๐๐},๐ข-\mathrm{๐๐๐}\right)
\mathrm{๐ผ๐ฐ๐๐ณ๐ธ๐๐}\ge 0
\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด}\ge 1
\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด}\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฝ๐พ๐ณ๐ด๐},\left[\mathrm{๐๐๐๐๐ก},\mathrm{๐๐๐๐},๐ก,๐ข\right]\right)
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐๐ก}\ge 1
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐๐ก}\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฝ๐พ๐ณ๐ด๐},\mathrm{๐๐๐๐๐ก}\right)
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐}\ge 0
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐}\le |\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.๐ก\ge 0
\mathrm{๐ฝ๐พ๐ณ๐ด๐}.๐ข\ge 0
G
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
collection. Cover a subset of the vertices of
G
by a set of vertex-disjoint circuits in such a way that the following property holds: for each uncovered vertex
{v}_{1}
G
there exists at least one covered vertex
{v}_{2}
G
such that the Manhattan distance between
{v}_{1}
{v}_{2}
\mathrm{๐ผ๐ฐ๐๐ณ๐ธ๐๐}
\left(\begin{array}{c}3,2,โฉ\begin{array}{cccc}\mathrm{๐๐๐๐๐ก}-1\hfill & \mathrm{๐๐๐๐}-6\hfill & ๐ก-4\hfill & ๐ข-5,\hfill \\ \mathrm{๐๐๐๐๐ก}-2\hfill & \mathrm{๐๐๐๐}-0\hfill & ๐ก-9\hfill & ๐ข-1,\hfill \\ \mathrm{๐๐๐๐๐ก}-3\hfill & \mathrm{๐๐๐๐}-0\hfill & ๐ก-2\hfill & ๐ข-4,\hfill \\ \mathrm{๐๐๐๐๐ก}-4\hfill & \mathrm{๐๐๐๐}-1\hfill & ๐ก-2\hfill & ๐ข-6,\hfill \\ \mathrm{๐๐๐๐๐ก}-5\hfill & \mathrm{๐๐๐๐}-5\hfill & ๐ก-7\hfill & ๐ข-2,\hfill \\ \mathrm{๐๐๐๐๐ก}-6\hfill & \mathrm{๐๐๐๐}-4\hfill & ๐ก-4\hfill & ๐ข-7,\hfill \\ \mathrm{๐๐๐๐๐ก}-7\hfill & \mathrm{๐๐๐๐}-0\hfill & ๐ก-6\hfill & ๐ข-4\hfill \end{array}โช\hfill \end{array}\right)
Figure 5.105.1 represents the solution associated with the example. The covered vertices are coloured in blue, while the links starting from the uncovered vertices are dashed. The
\mathrm{๐๐ข๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐ข}
In the solution we have
\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด}=2
disjoint circuits.
All the 3 uncovered nodes are located at a distance that does not exceed
\mathrm{๐ผ๐ฐ๐๐ณ๐ธ๐๐}=3
from at least one covered node.
Figure 5.105.1. Final graph associated with the facilities location problem
\mathrm{๐ผ๐ฐ๐๐ณ๐ธ๐๐}>0
\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด}<|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|>2
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\left(\mathrm{๐๐๐๐๐ก}\right)
\left(\mathrm{๐๐๐๐}\right)
\left(๐ก,๐ข\right)
๐ก
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
๐ข
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
This kind of facilities location problem is described in [LabbeLaporteRodriguezMartin98] pages. In addition to our example they also mention the cost problem that is usually a trade-off between the vertices that are directly covered by circuits and the others.
\mathrm{๐๐ข๐๐๐}
(graph constraint).
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
problems: facilities location problem.
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐ถ๐ฟ๐ผ๐๐๐ธ}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐}\mathtt{2}\right)
\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐}=\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐ก}
โข
\mathrm{๐๐๐๐๐}
=0
โข
\mathrm{๐๐๐}
=\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด}
\mathrm{๐ฝ๐พ๐ณ๐ด๐}
\mathrm{๐ถ๐ฟ๐ผ๐๐๐ธ}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐}\mathtt{2}\right)
\bigvee \left(\begin{array}{c}\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐}=\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐ก},\hfill \\ \bigwedge \left(\begin{array}{c}\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐}=0,\hfill \\ \mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐}\ne 0,\hfill \\ \mathrm{๐๐๐}\left(\mathrm{๐๐๐๐๐}\mathtt{1}.๐ก-\mathrm{๐๐๐๐๐}\mathtt{2}.๐ก\right)+\mathrm{๐๐๐}\left(\mathrm{๐๐๐๐๐}\mathtt{1}.๐ข-\mathrm{๐๐๐๐๐}\mathtt{2}.๐ข\right)\le \mathrm{๐ผ๐ฐ๐๐ณ๐ธ๐๐}\hfill \end{array}\right)\hfill \end{array}\right)
\mathrm{๐๐๐๐๐๐๐}
=|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\begin{array}{c}\mathrm{๐ฏ๐ฑ๐ค๐ฃ}โฆ\hfill \\ \left[\begin{array}{c}\mathrm{๐๐๐๐๐๐๐๐๐}-\mathrm{๐๐๐}\left(\begin{array}{c}\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}-\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right),\hfill \\ \mathrm{๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐ฝ๐พ๐ณ๐ด๐}.\mathrm{๐๐๐๐}\right)\right]\hfill \end{array}\right),\hfill \\ \mathrm{๐๐๐๐๐๐๐๐๐๐๐}\hfill \end{array}\right]\hfill \end{array}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}
\left(\mathrm{๐๐๐๐๐๐๐๐๐},=,1\right)
v
we have introduced the following attributes:
\mathrm{๐๐๐๐๐ก}
: the label associated with
v
\mathrm{๐๐๐๐}
v
is not covered by a circuit then 0; If
v
is covered by a circuit then index of the successor of
v
๐ก
๐ก
v
๐ข
๐ข
v
The first graph constraint forces all vertices, which have a non-zero successor, to form a set of
\mathrm{๐ฝ๐ฒ๐๐ฒ๐ป๐ด}
vertex-disjoint circuits.
The final graph associated with the second graph constraint contains two types of arcs:
The arcs belonging to one circuit (i.e.,
\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐}=\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐๐ก}
The arcs between one vertex
{v}_{1}
that does not belong to any circuit (i.e.,
\mathrm{๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐๐}=0
) and one vertex
{v}_{2}
located on a circuit (i.e.,
\mathrm{๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐๐}\ne 0
) such that the Manhattan distance between
{v}_{1}
{v}_{2}
\mathrm{๐ผ๐ฐ๐๐ณ๐ธ๐๐}
In order to specify the fact that each vertex is involved in at least one arc we use the graph property
\mathrm{๐๐๐๐๐๐๐}
=
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
. Finally the dynamic constraint
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐ก๐๐๐๐}_\mathtt{0}\left(\mathrm{๐๐๐๐๐๐๐๐๐},=,1\right)
expresses the fact that, for each vertex
v
, there is exactly one predecessor of
v
that belongs to a circuit.
Parts (A) and (B) of Figure 5.105.2 respectively show the initial and final graph associated with the second graph constraint of the Example slot.
\mathrm{๐๐ข๐๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐ข}
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
is the maximum number of vertices of the final graph associated with the second graph constraint we can rewrite
\mathrm{๐๐๐๐๐๐๐}
=
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\mathrm{๐๐๐๐๐๐๐}
\ge
|\mathrm{๐ฝ๐พ๐ณ๐ด๐}|
\underline{\overline{\mathrm{๐๐๐๐๐๐๐}}}
\overline{\mathrm{๐๐๐๐๐๐๐}}
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Summary โ Web Education in Chemistry
Read through the summarized points below carefully. As a last exercise, continue with the mixed exercises (link below), which summarize everything that you (should) have learned today.
All experiments contain experimental error.
For many analytical instruments (glassware, balance, ...), the error is shown on the equipment.
The error in a final answer can be calculated by using error propagation rules.
Errors for multiplication and division have to be calculated relatively, because these operations often involve different quantities.
Many experimental errors are normally distributed.
They can be characterised by a mean value and a standard deviation.
The mean describes the location.
The standard deviation describes the spread around the mean.
Random errors are related to the standard deviation.
Systematic errors are related to the difference between the mean and the (unknown) true value (the bias).
The larger the standard deviation (the spread around a central value), the wider a prediction interval will be.
A 95% prediction interval will contain the next experimental value with a probability of 95%.
A crude estimate of a 95% prediction interval is the mean plus or minus twice the standard deviation.
A crude estimate of a 99% prediction interval is the mean plus or minus three times the standard deviation.
Confidence intervals are narrower than their corresponding prediction intervals, since errors cancel out.
The standard deviation of the mean is the standard deviation of the individual values divided by the square root of the number of measurements.
The linear relationship between two variables
x
y
can be described by the
y
b
a
of the optimal straight line.
Standard errors for the intercept
b
y
axis and slope
a
can be calculated, and therefore confidence intervals too.
It is assumed that the error in
x
is much smaller than the error in
y
Residuals are assumed to be independent and normally distributed with constant variance.
Regression lines may be used for calibration: the calibration line is set up using a set of calibration samples and the concentration in an unknown sample can be predicted.
Regression lines can be used to compare methods.
As a final test, prepare yourself for the mixed exercises.
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\lim_{x\to\infty} \dfrac{\sin x}x = \, ?
by Yash Dev Lamba
\Large \lim_{x\rightarrow 0} \frac{|2x-1|-|2x+1|}{x}= \ ?
f(x) = \large{\begin{cases} (x^2 - 4)/(x-2), & \text{ if } x < 2 \\ 2, & \text{ if } x = 2 \\ x^3-3x^2 + 2x + 4 , & \text{ if } x > 2 \\ \end{cases} }
\displaystyle \lim_{x \to 2} f(x)
f(x) = \dfrac1{x-1}
g(x) = \dfrac3{x^2-3x+2}
\displaystyle \lim_{x\to1} \dfrac{f(x)}{g(x)}
-\frac { 1 }{ 4 }
-\infty
-3
-\frac { 3 }{ 10 }
-\frac { 1 }{ 3 }
by Jonas Katona
\large \lim_{x \to 0} \, \left \lfloor \dfrac{(\sin x) (\tan x)}{x^2} \right \rfloor = \ ?
0 1 None of these choices -1 2
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Ionic Compounds | Brilliant Math & Science Wiki
Manish Dash, Tejas Suresh, William Jackson, and
Jett Count
An ionic compound is a type of compound formed by the classification of compounds with respect to the bond formed between the interacting species (charged ions). They consist of bonds called ionic bonds, which are very strong due to electrostatic forces of attraction between the ions of the compound.
Factors Influencing the Formation of Ionic Compounds
Methods to Determine Percentage Ionic Character in a Compound
The formation of ionic bond clearly depends on the electrostatic force between ions:
F_e = \dfrac{kq_1q_2}{r^2},
q_1
q_2
are the magnitudes of charges of the anion and cation, respectively, and
r
is the distance between atoms.
Hence, one can conclude that, if the bond is strong, then the electrostatic force is also strong.
From this, we can find the factors influencing the formation and strength of the ionic bond:
The greater the magnitude of charges on the cation and anion, the greater the capability of the formation.
The less the distance between the charged ions, the greater the force and the strength and capability of the formation.
The packing efficiency (not a result of the formula above) plays a role because ionic compounds exist in lattice structures. If the interaction between like charges (cation-cation of different compounds and anion-anion of different compounds) is quite large, then the lattice is weakened.
Here are the properties of ionic compounds:
Ionic compounds are generally hard and brittle.
Ionic compounds generally have high melting points and high boiling points.
Ionic compounds are generally soluble in polar solvents and insoluble in non-polar solvents.
Ionic compounds are good conductors of electricity in their molten or fused state due to the presence of free ions.
Ionic compounds undergo very fast or spontaneous reactions in a aqueous state.
Ionic bonds are non-directional, unlike covalent bonds.
In which state is cryolite
(\ce{Na3AlF6},
sodium hexafluoroaluminate
)
most conducive for conducting electricity?
It is most conducive for conducting electricity in a liquid state.
In a liquid state, the ions are mobile and hence can conduct electricity.
In a solid state, the ions are fixed in the lattice structure and hence cannot conduct electricity.
In a gaseous state, the ions tend to be too far apart to conduct electricity well.
Which compound has a higher melting point,
\ce{CCl4}
\ce{NaCl}?
An ionic compound has a much higher melting point than a covalent compound, so the answer is
\ce{NaCl}.
In an ionic compound, the ions have strong attraction to other ions in their vicinity; thus, it would take a lot of energy to break them apart. In a covalent compound, while the atoms are bound tightly to each other in a stable molecule, these molecules are not strongly attracted to other molecules. Hence, it does not take much energy to separate them.
Note: The melting point of
\ce{CCl4}
-23\, ^ \circ\text{C}
and the melting point of
\ce{NaCl}
800\, ^ \circ\text{C}.
Ionic compounds generally exist in the form of a crystal lattice. The orderly arrangement of oppositely charged ions in 3D in solid state to form an infinite array is called crystal lattice. In crystal lattice, a unit of oppositely charged ions keeps repeating. The smallest repeating unit in a crystal lattice is called a unit cell.
The amount of energy released when 1 mole of oppositely charged ions is brought from infinity to form crystal lattice is called lattice energy.
\ce{CH4}
Hannay-Smith equation is stated as follows:
3.5({ X }_{ A }-{ X }_{ B }{ ) }^{ 2 }+16({ X }_{ A }-{ X }_{ B }),
({ X }_{ A }-{ X }_{ B })
is the difference in electronegativities of
A
B.
By calculating the relationship between observed dipole moment and calculated dipole moment, the relationship can be stated as follows:
\text{\% Ionic character} = \frac{\text{Observed dipole moment}}{\text{Calculated dipole moment}} \times 100.
Which of the following compounds is most ionic?
Here, the anion of all the given compounds is the same. Hence, comparison of the ionic character of a compound depends on the cation. Since the size of cation increases down the group, the ionic character of cation also increases down the group. Here
\text{Cs}^{ + }
has the highest atomic size and hence it has the highest ionic character.
Cite as: Ionic Compounds. Brilliant.org. Retrieved from https://brilliant.org/wiki/ionic-compounds/
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Ruthie did a survey among her classmates comparing the time spent playing video games to the time spent studying. The scatterplot of her data is shown at right.
What association can you make from her data?
The person who played video games for 160 minutes studied for less than 20 minutes. The person who did not play any video games studied for 60 minutes.
There is a moderate, negative, linear association.
Use an ordered pair (
x,y
) to identify any outliers.
The person who played video games for about 100 minutes studied much longer than those who played about the same number of minutes.
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Global Constraint Catalog: Cglobal_cardinality_low_up_no_loop
<< 5.164. global_cardinality_low_up5.166. global_cardinality_no_loop >>
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}
\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}\left(\begin{array}{c}\mathrm{๐ผ๐ธ๐ฝ๐ป๐พ๐พ๐ฟ},\hfill \\ \mathrm{๐ผ๐ฐ๐๐ป๐พ๐พ๐ฟ},\hfill \\ \mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐},\hfill \\ \mathrm{๐
๐ฐ๐ป๐๐ด๐}\hfill \end{array}\right)
\mathrm{๐๐๐}_\mathrm{๐๐๐ }_\mathrm{๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐ผ๐ธ๐ฝ๐ป๐พ๐พ๐ฟ}
\mathrm{๐๐๐}
\mathrm{๐ผ๐ฐ๐๐ป๐พ๐พ๐ฟ}
\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐},\mathrm{๐๐๐๐}-\mathrm{๐๐๐},\mathrm{๐๐๐๐ก}-\mathrm{๐๐๐}\right)
\mathrm{๐ผ๐ธ๐ฝ๐ป๐พ๐พ๐ฟ}\ge 0
\mathrm{๐ผ๐ธ๐ฝ๐ป๐พ๐พ๐ฟ}\le \mathrm{๐ผ๐ฐ๐๐ป๐พ๐พ๐ฟ}
\mathrm{๐ผ๐ฐ๐๐ป๐พ๐พ๐ฟ}\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐},\mathrm{๐๐๐}\right)
|\mathrm{๐
๐ฐ๐ป๐๐ด๐}|>0
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐ป๐๐ด๐},\left[\mathrm{๐๐๐},\mathrm{๐๐๐๐},\mathrm{๐๐๐๐ก}\right]\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐ป๐๐ด๐},\mathrm{๐๐๐}\right)
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐}\ge 0
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐ก}\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐}\le \mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐ก}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}\left[i\right].\mathrm{๐๐๐๐}
\left(1\le i\le |\mathrm{๐
๐ฐ๐ป๐๐ด๐}|\right)
is less than or equal to the number of variables
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\left[j\right].\mathrm{๐๐๐}
\left(j\ne i,1\le j\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|\right)
\mathrm{๐
๐ฐ๐ป๐๐ด๐}\left[i\right].\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}\left[i\right].\mathrm{๐๐๐๐ก}
\left(1\le i\le |\mathrm{๐
๐ฐ๐ป๐๐ด๐}|\right)
is greater than or equal to the number of variables
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\left[j\right].\mathrm{๐๐๐}
\left(j\ne i,1\le j\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|\right)
\mathrm{๐
๐ฐ๐ป๐๐ด๐}\left[i\right].\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\left[i\right].\mathrm{๐๐๐}=i
i\in \left[1,|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|\right]
) is greater than or equal to
\mathrm{๐ผ๐ธ๐ฝ๐ป๐พ๐พ๐ฟ}
and less than or equal to
\mathrm{๐ผ๐ฐ๐๐ป๐พ๐พ๐ฟ}
\left(\begin{array}{c}1,1,โฉ1,1,8,6โช,\hfill \\ โฉ\begin{array}{ccc}\mathrm{๐๐๐}-1\hfill & \mathrm{๐๐๐๐}-1\hfill & \mathrm{๐๐๐๐ก}-1,\hfill \\ \mathrm{๐๐๐}-5\hfill & \mathrm{๐๐๐๐}-0\hfill & \mathrm{๐๐๐๐ก}-0,\hfill \\ \mathrm{๐๐๐}-6\hfill & \mathrm{๐๐๐๐}-1\hfill & \mathrm{๐๐๐๐ก}-2\hfill \end{array}โช\hfill \end{array}\right)
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}
\left\{\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\left[2\right].\mathrm{๐๐๐}\right\}
\mathrm{๐๐๐๐}=1\le 1\le \mathrm{๐๐๐๐ก}=1
\left\{\right\}
\mathrm{๐๐๐๐}=0\le 0\le \mathrm{๐๐๐๐ก}=0
\left\{\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\left[4\right].\mathrm{๐๐๐}\right\}
\mathrm{๐๐๐๐}=1\le 1\le \mathrm{๐๐๐๐ก}=2
). Note that, due to the definition of the constraint, the fact that
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\left[1\right].\mathrm{๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\left[i\right].\mathrm{๐๐๐}=i
i\in \left[1,4\right]
\mathrm{๐ผ๐ธ๐ฝ๐ป๐พ๐พ๐ฟ}=1
\mathrm{๐ผ๐ฐ๐๐ป๐พ๐พ๐ฟ}=1
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|>1
\mathrm{๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}.\mathrm{๐๐๐}\right)>1
|\mathrm{๐
๐ฐ๐ป๐๐ด๐}|>1
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐}\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐ก}>0
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐ก}<|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|>|\mathrm{๐
๐ฐ๐ป๐๐ด๐}|
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐}
\ge 0
\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐ก}
\le |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|
\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}
constraint. This is done by creating an extra value node representing the loops corresponding to the roots of the trees. The rightmost part of Figure 3.7.29 illustrates the corresponding flow model for the
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐ก๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}_\mathrm{๐๐๐}_\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}
\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}
\mathrm{๐๐๐๐๐๐๐๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐ธ๐ฟ๐น}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\right)
โข\mathrm{๐๐๐๐๐๐๐๐๐}.\mathrm{๐๐๐}=\mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐}
โข\mathrm{๐๐๐๐๐๐๐๐๐}.\mathrm{๐๐๐ข}\ne \mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐}
โข
\mathrm{๐๐๐๐๐๐๐}
\ge \mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐}
โข
\mathrm{๐๐๐๐๐๐๐}
\le \mathrm{๐
๐ฐ๐ป๐๐ด๐}.\mathrm{๐๐๐๐ก}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐ธ๐ฟ๐น}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐๐}.\mathrm{๐๐๐}=\mathrm{๐๐๐๐๐๐๐๐๐}.\mathrm{๐๐๐ข}
โข
\mathrm{๐๐๐๐}
\ge \mathrm{๐ผ๐ธ๐ฝ๐ป๐พ๐พ๐ฟ}
โข
\mathrm{๐๐๐๐}
\le \mathrm{๐ผ๐ฐ๐๐ป๐พ๐พ๐ฟ}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\mathrm{๐
๐ฐ๐ป๐๐ด๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐ข}_\mathrm{๐๐๐ }_\mathrm{๐๐}_\mathrm{๐๐}_\mathrm{๐๐๐๐}
|
Sinc function - MATLAB sinc
Ideal Bandlimited Interpolation
y = sinc(x) returns an array, y, whose elements are the sinc of the elements of the input, x. The output y is the same size as x.
Perform ideal bandlimited interpolation of a random signal sampled at integer spacings.
Assume that the signal to interpolate, x, is 0 outside of the given time interval and has been sampled at the Nyquist frequency. Reset the random number generator for reproducibility.
scalar value | vector | matrix | N-D array | gpuArray object
Input array, specified as a real-valued or complex-valued scalar, vector, matrix, N-D array, or gpuArray object. When x is nonscalar, sinc is an element-wise operation.
y โ Sinc of input
Sinc of the input array, x, returned as a real-valued or complex-valued scalar, vector, matrix, N-D array, or gpuArray object of the same size as x.
\mathrm{sinc}t=\left\{\begin{array}{cc}\frac{\mathrm{sin}\pi t}{\pi t}& t\ne 0,\\ 1& t=0.\end{array}
This analytic expression corresponds to the continuous inverse Fourier transform of a rectangular pulse of width 2ฯ and height 1:
\mathrm{sinc}t=\frac{1}{2\pi }\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\int }_{-\pi }^{\pi }{e}^{j\omega t}\text{\hspace{0.17em}}d\omega .
The space of functions bandlimited in the frequency range
\omega =\left(-\pi ,\pi \right]
is spanned by the countably infinite set of sinc functions shifted by integers. Thus, you can reconstruct any such bandlimited function g(t) from its samples at integer spacings:
g\left(t\right)=\sum _{n=-\infty }^{\infty }g\left(n\right)\mathrm{sinc}\left(t-n\right).
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In the given figure , AD=8 cm AC= 6 cm and TB is the tangent at B to the circle - Maths - Circles - 9055777 | Meritnation.com
In the given figure , AD=8 cm AC= 6 cm and TB is the tangent at B to the circle with centre O.Find OT , if BT= 4 cm
\mathrm{We} \mathrm{know} \mathrm{angle} \mathrm{in} \mathrm{a} \mathrm{semicircle} \mathrm{is} 90ยฐ.\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{CAD}=90ยฐ\phantom{\rule{0ex}{0ex}}\mathrm{Now}, \mathrm{in} \mathrm{right} \mathrm{angled} \mathrm{triangle} \mathrm{CAD},\phantom{\rule{0ex}{0ex}}{\mathrm{CD}}^{2}={\mathrm{AC}}^{2}+{\mathrm{AD}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{CD}=\sqrt{{6}^{2}+{8}^{2}}=\sqrt{100}=10 \mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{CD} \mathrm{is} \mathrm{a} \mathrm{diameter} \mathrm{of} \mathrm{given} \mathrm{circle},\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{radius}=\mathrm{OA}=\mathrm{OB}=\frac{10}{2}=5 \mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{We} \mathrm{know}, \mathrm{tangent} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{the} \mathrm{radius} \mathrm{at} \mathrm{that} \mathrm{point} \mathrm{of} \mathrm{contact}.\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{OBT}=90ยฐ\phantom{\rule{0ex}{0ex}}\mathrm{Now}, \mathrm{in} \mathrm{right} \mathrm{angled} \mathrm{triangle} \mathrm{OBT},\phantom{\rule{0ex}{0ex}}{\mathrm{OT}}^{2}={\mathrm{OB}}^{2}+{\mathrm{BT}}^{2}\phantom{\rule{0ex}{0ex}}โ\mathrm{OT}=\sqrt{{5}^{2}+{4}^{2}}=\sqrt{41} \mathrm{cm}
Srry to say but please recheck your question. It's incorrect. Let me tell you why:-
Take triangle AOC and AOD
โOC=OD (radii of circle)
Angle AOC = AOD (90 degree each)
AO = A0 (Common Side)
By SAS TRIANGLE AOC AND AOD ARE CONGRUENT
THIS MEANS THAT THEIR CORRESPONDING PARTS SHOULD ALSO BE EQUAL. BUT AC IS NOT NOT EQUAL TO AD AS PER QUESTION WHICH CONTRADICTS THE CONGRUENCY OF THE TRIANGLEs
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Mathematics | Free Full-Text | Modeling Uncertainty in Fracture Age Estimation from Pediatric Wrist Radiographs
Simulation of Teacher Interventions in a Training Course of Mathematics Teacher Educators
Semi-Local Integration Measure of Node Importance
A Channel-Wise Spatial-Temporal Aggregation Network for Action Recognition
Multivariate Decomposition of Acoustic Signals in Dispersive Channels
Hrลพiฤ, F.
Lerga, J.
Sorantin, E.
Tschauner, S.
Department of Computer Engineering, Faculty of Engineering, University of Rijeka, Vukovarska 58, Rijeka 51000, Croatia
Division of General Radiology, Department of Radiology, Medical University of Graz, 8036 Graz, Austria
Center for Artificial Intelligence and Cybersecurity, University of Rijeka, Radmile Matejฤiฤ 2, Rijeka 51000, Croatia
Division of Pediatric Radiology, Department of Radiology, Medical University of Graz, 8036 Graz, Austria
Academic Editor: Konstantin Kozlov
(This article belongs to the Special Issue New Trends in Graph and Complexity Based Data Analysis and Processing)
In clinical practice, fracture age estimation is commonly required, particularly in children with suspected non-accidental injuries. It is usually done by radiologically examining the injured body part and analyzing several indicators of fracture healing such as osteopenia, periosteal reaction, and fracture gap width. However, age-related changes in healing timeframes, inter-individual variabilities in bone density, and significant intra- and inter-operator subjectivity all limit the validity of these radiological clues. To address these issues, for the first time, we suggest an automated neural network-based system for determining the age of a pediatric wrist fracture. In this study, we propose and evaluate a deep learning approach for automatically estimating fracture age. Our dataset included 3570 medical cases with a skewed distribution toward initial consultations. Each medical case includes a lateral and anteroposterior projection of a wrist fracture, as well as patientsโ age, and gender. We propose a neural network-based system with Monte-Carlo dropout-based uncertainty estimation to address dataset skewness. Furthermore, this research examines how each component of the system contributes to the final forecast and provides an interpretation of different scenarios in system predictions in terms of their uncertainty. The examination of the proposed systemsโ components showed that the feature-fusion of all available data is necessary to obtain good results. Also, proposing uncertainty estimation in the system increased accuracy and F1-score to a final
0.906ยฑ0.011
on a given task. View Full-Text
Keywords: fracture age; forensic; deep learning; uncertainty estimation; Gaussian process; X-ray fracture age; forensic; deep learning; uncertainty estimation; Gaussian process; X-ray
Hrลพiฤ, F.; Janisch, M.; ล tajduhar, I.; Lerga, J.; Sorantin, E.; Tschauner, S. Modeling Uncertainty in Fracture Age Estimation from Pediatric Wrist Radiographs. Mathematics 2021, 9, 3227. https://doi.org/10.3390/math9243227
Hrลพiฤ F, Janisch M, ล tajduhar I, Lerga J, Sorantin E, Tschauner S. Modeling Uncertainty in Fracture Age Estimation from Pediatric Wrist Radiographs. Mathematics. 2021; 9(24):3227. https://doi.org/10.3390/math9243227
Hrลพiฤ, Franko, Michael Janisch, Ivan ล tajduhar, Jonatan Lerga, Erich Sorantin, and Sebastian Tschauner. 2021. "Modeling Uncertainty in Fracture Age Estimation from Pediatric Wrist Radiographs" Mathematics 9, no. 24: 3227. https://doi.org/10.3390/math9243227
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Fit binary decision tree for multiclass classification - MATLAB fitctree - MathWorks Espaรฑa
{\stackrel{^}{\pi }}_{jk}=\sum _{i=1}^{n}I\left\{{y}_{i}=k\right\}{w}_{i}.
\sum {w}_{i}=1
{\stackrel{^}{\pi }}_{jk}=\frac{{n}_{jk}}{n}
t=n\sum _{k=1}^{K}\sum _{j=1}^{J}\frac{{\left({\stackrel{^}{\pi }}_{jk}-{\stackrel{^}{\pi }}_{j+}{\stackrel{^}{\pi }}_{+k}\right)}^{2}}{{\stackrel{^}{\pi }}_{j+}{\stackrel{^}{\pi }}_{+k}}
{\stackrel{^}{\pi }}_{j+}=\sum _{k}{\stackrel{^}{\pi }}_{jk}
{\stackrel{^}{\pi }}_{+k}=\sum _{j}{\stackrel{^}{\pi }}_{jk}
1-\sum _{i}{p}^{2}\left(i\right),
-\sum _{i}p\left(i\right){\mathrm{log}}_{2}p\left(i\right).
P\left(L\right)P\left(R\right){\left(\sum _{i}|L\left(i\right)-R\left(i\right)|\right)}^{2},
{\lambda }_{jk}=\frac{\text{min}\left({P}_{L},{P}_{R}\right)-\left(1-{P}_{{L}_{j}{L}_{k}}-{P}_{{R}_{j}{R}_{k}}\right)}{\text{min}\left({P}_{L},{P}_{R}\right)}.
{P}_{{L}_{j}{L}_{k}}
{P}_{{R}_{j}{R}_{k}}
P\left(T\right)=\sum _{j\in T}{w}_{j}.
\Delta I=P\left(T\right){i}_{t}-P\left({T}_{L}\right){i}_{{t}_{L}}-P\left({T}_{R}\right){i}_{{t}_{R}}.
\Delta {I}_{U}=P\left(T-{T}_{U}\right){i}_{t}-P\left({T}_{L}\right){i}_{{t}_{L}}-P\left({T}_{R}\right){i}_{{t}_{R}}.
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Lifshitz_theory_of_van_der_Waals_force Knowpia
Lifshitz theory of van der Waals force
In condensed matter physics and physical chemistry, the Lifshitz theory of van der Waals forces, sometimes called the macroscopic theory of van der Waals forces, is a method proposed by Evgeny Mikhailovich Lifshitz in 1954 for treating van der Waals forces between bodies which does not assume pairwise additivity of the individual intermolecular forces; that is to say, the theory takes into account the influence of neighboring molecules on the interaction between every pair of molecules located in the two bodies, rather than treating each pair independently.[1][2]
Need for a non-pairwise additive theoryEdit
The van der Waals force between two molecules, in this context, is the sum of the attractive or repulsive forces between them; these forces are primarily electrostatic in nature, and in their simplest form might consist of a force between two charges, two dipoles, or between a charge and a dipole. Thus, the strength of the force may often depend on the net charge, electric dipole moment, or the electric polarizability (
{\displaystyle \alpha }
) (see for example London force) of the molecules, with highly polarizable molecules contributing to stronger forces, and so on.
The total force between two bodies, each consisting of many molecules in the van der Waals theory is simply the sum of the intermolecular van der Waals forces, where pairwise additivity is assumed. That is to say, the forces are summed as though each pair of molecules interacts completely independently of their surroundings (See Van der Waals forces between Macroscopic Objects for an example of such a treatment). This assumption is usually correct for gasses, but presents a problem for many condensed materials, as it is known that the molecular interactions may depend strongly on their environment and neighbors. For example, in a conductor, a point-like charge might be screened by the electrons in the conductance band,[3] and the polarizability of a condensed material may be vastly different from that of an individual molecule.[4] In order to correctly predict the van der Waals forces of condensed materials, a theory that takes into account their total electrostatic response is needed.
The problem of pairwise additivity is completely avoided in the Lifshitz theory, where the molecular structure is ignored and the bodies are treated as continuous media. The forces between the bodies are now derived in terms of their bulk properties, such as dielectric constant and refractive index, which already contain all the necessary information from the original molecular structure.
The original Lifshitz 1955 paper proposed this method relying on quantum field theory principles, and is, in essence, a generalization of the Casimir effect, from two parallel, flat, ideally conducting surfaces, to two surfaces of any material. Later papers by Langbein,[5][6] Ninham,[7] Parsegian[8] and Van Kampen[9] showed that the essential equations could be derived using much simpler theoretical techniques, an example of which is presented here.
Hamaker constantEdit
An Ion of charge
{\displaystyle Q}
and a nonpolar molecule of polarizability
{\textstyle \alpha }
The Lifshitz theory can be expressed as an effective Hamaker constant in the van der Waals theory.
Consider, for example, the interaction between an ion of charge
{\textstyle Q}
, and a nonpolar molecule with polarizability
{\textstyle \alpha _{2}}
{\textstyle r}
. In a medium with dielectric constant
{\displaystyle \epsilon _{3}}
, the interaction energy between a charge and an electric dipole
{\displaystyle p}
{\displaystyle U(r)={\frac {-Qp}{4\pi \epsilon _{0}\epsilon _{3}r^{2}}}}
with the dipole moment of the polarizable molecule given by
{\textstyle p=\alpha _{2}E}
{\textstyle E}
is the strength of the electric field at distance
{\textstyle r}
from the ion. According to Coulomb's law:
{\displaystyle E={\frac {Q}{4\pi \epsilon _{0}\epsilon _{3}}}{\frac {1}{r^{2}}}}
so we may write the interaction energy as
{\displaystyle U(r)={\frac {-Q^{2}\alpha _{2}}{(4\pi \epsilon _{0}\epsilon _{3})^{2}r^{4}}}}
The nonpolar molecule is now replaced with a dielectric medium (in grey); the force experienced by the molecule can be found using the method of image charges.
Definition of integration variables
{\displaystyle x}
{\displaystyle z}
Consider now, how the interaction energy will change if the right hand molecule is replaced with a medium of density
{\textstyle \rho _{2}}
of such molecules. According to the "classical" van der Waals theory, the total force will simply be the summation over individual molecules. Integrating over the volume of the medium (see the third figure), we might expect the total interaction energy with the charge to be
{\displaystyle {\begin{aligned}U(D)&={\frac {-2\pi Q^{2}\alpha _{2}\rho _{2}}{(4\pi \epsilon _{0}\epsilon _{3})^{2}}}\int \limits _{z=D}^{\infty }dz\int \limits _{x=0}^{\infty }dx{\frac {x}{(z^{2}+x^{2})^{2}}}\\&=-{\frac {\pi Q^{2}\alpha _{2}\rho _{2}}{(4\pi \epsilon _{0}\epsilon _{3})^{2}}}{\frac {1}{D}}\end{aligned}}}
But this result cannot be correct, since It is well known that a charge
{\textstyle Q}
in a medium of dielectric constant
{\displaystyle \epsilon _{3}}
{\textstyle D}
from the plane surface of a second medium of dielectric constant
{\displaystyle \epsilon _{2}}
experiences a force as if there were an 'image' charge of strength
{\textstyle Q'=-Q(\epsilon _{2}-\epsilon _{3})/(\epsilon _{2}+\epsilon _{3})}
at distance D on the other side of the boundary.[11] The force between the real and image charges must then be
{\displaystyle F(D)={\frac {-Q^{2}}{(4\pi \epsilon _{0}\epsilon _{3})(2D)^{2}}}{\frac {\epsilon _{2}-\epsilon _{3}}{\epsilon _{2}+\epsilon _{3}}}}
and the energy, therefore
{\displaystyle U(D)={\frac {-Q^{2}}{(4\pi \epsilon _{0}\epsilon _{3})4D}}{\frac {\epsilon _{2}-\epsilon _{3}}{\epsilon _{2}+\epsilon _{3}}}}
Equating the two expressions for the energy, we define a new effective polarizability that must obey
{\displaystyle \rho _{2}\alpha _{2}=\epsilon _{0}\epsilon _{3}{\frac {\epsilon _{2}-\epsilon _{3}}{\epsilon _{2}+\epsilon _{3}}}}
Similarly, replacing the real charge
{\textstyle Q}
with a medium of density
{\textstyle \rho _{1}}
and polarizability
{\displaystyle \alpha _{1}}
gives an expression for
{\displaystyle \alpha _{1}\rho _{1}}
. Using these two relations, we may restate our theory in terms of an effective Hamaker constant. Specifically, using McLachlan's generalized theory of VDW forces the Hamaker constant for an interaction potential of the form
{\textstyle U(r)=-C/r^{n}}
between two bodies at temperature
{\textstyle T}
{\displaystyle A=\pi ^{2}C\rho _{1}\rho _{2}={\frac {6\pi ^{2}k_{B}T\rho _{1}\rho _{2}}{(4\pi \epsilon _{0})^{2}}}\sum _{n=0,1...}^{\infty }{\frac {\alpha _{1}(i\nu _{n})\alpha _{2}(i\nu _{n})}{\epsilon _{3}^{2}}}}
{\textstyle \nu _{n}=2\pi nk_{B}T/h}
{\textstyle k_{B}}
{\textstyle h}
are Boltzmann's and Planck's constants correspondingly. Inserting our relations for
{\displaystyle \rho \alpha }
and approximating the sum as an integral
{\textstyle k_{B}T\sum _{n=0,1...}^{\infty }\rightarrow {\frac {h}{2\pi }}\int \limits _{\nu _{1}}^{\infty }d\nu }
, the effective Hamaker constant in the Lifshitz theory may be approximated as
{\displaystyle A\approx {\frac {3}{4}}k_{B}T\left({\frac {\epsilon _{1}-\epsilon _{3}}{\epsilon _{1}+\epsilon _{3}}}\right)\left({\frac {\epsilon _{2}-\epsilon _{3}}{\epsilon _{2}+\epsilon _{3}}}\right)+{\frac {3h}{4\pi }}\int \limits _{\nu _{1}}^{\infty }d\nu \left({\frac {\epsilon _{1}(i\nu )-\epsilon _{3}(i\nu )}{\epsilon _{1}(i\nu )+\epsilon _{3}(i\nu )}}\right)\left({\frac {\epsilon _{2}(i\nu )-\epsilon _{3}(i\nu )}{\epsilon _{2}(i\nu )+\epsilon _{3}(i\nu )}}\right)}
{\displaystyle \epsilon (i\nu )}
are real functions, and are related to measurable properties of the medium;[13] thus, the Hamaker constant in the Lifshitz theory can be expressed in terms of observable properties of the physical system.
Experimental validationEdit
The macroscopic theory of van der Waals theory has many experimental validations. Among which, some of the most notable ones are Derjaguin (1960);[14] Derjaguin, Abrikosova and Lifshitz (1956)[15] and Israelachvili and Tabor (1973),[16] who measured the balance of forces between macroscopic bodies of glass, or glass and mica; Haydon and Taylor (1968),[17] who measured the forces across bilayers by measuring their contact angle; and lastly Shih and Parsegian (1975),[18] who investigated Van der Waals potentials between heavy alkali-metal atoms and gold surfaces using atomic-beam-deflection.
^ Lifshitz, E.M. (3 September 1954). "The Theory of Molecular Attractive Forces between Solids". Journal of Experimental Theoretical Physics USSR. 29: 94โ110.
^ Lifshitz, E.M. (January 1956). "The theory of molecular Attractive Forces between Solids". Soviet Physics. 2 (1): 73โ83.
^ Ziman, J.M. (1972). "5.2: Screened Impurities and Neutral Pseudo-Atoms". Principles of the Theory of Solids (2 ed.). Cambridge: Cambridge University Press. ISBN 978-0521297332. LCCN 72-80250.
^ Ashcroft, N.W.; Mermin, N.D. (1976). "31: Diamagnetism and Paramagnetism". Solid State Physics. Sauders College Publishing. ISBN 978-0030839931.
^ Langbein, D. (October 1969). "Van der Waals Attraction Between Macroscopic Bodies". Journal of Adheshion. 1 (4): 237โ245. doi:10.1080/00218466908072187.
^ Langbein, D. (1970). "Retarded Dispersion Energy between Macroscopic Bodies". Phys. Rev. B. 2 (8): 3371โ3383. Bibcode:1970PhRvB...2.3371L. doi:10.1103/physrevb.2.3371.
^ Ninham, B.W.; Parsegian, V.E. (July 1970). "Van der Waals forces. Special characteristics in lipid-water systems and a general method of calculation based on the Lifshitz theory". Biophys. J. 10 (7): 646โ663. Bibcode:1970BpJ....10..646N. doi:10.1016/S0006-3495(70)86326-3. PMC 1367788. PMID 5449915.
^ Parsegian (1971). "On van der waals interactions between macroscopic bodies having inhomogeneous dielectric susceptibilities". J. Colloid Interface Sci. 40 (1): 35โ41. Bibcode:1972JCIS...40...35P. doi:10.1016/0021-9797(72)90171-3.
^ Van Kampen, N.G.; Nijboer, B.R.A.; Schram, K. (26 February 1968). "On the macroscopic theory of van der Waals forces". Physics Letters A. 26 (7): 307โ308. Bibcode:1968PhLA...26..307V. doi:10.1016/0375-9601(68)90665-8. hdl:1874/15527.
^ Griffith, D.J. (1999). "3.4.2: The Monopole and Dipole Terms". Introduction to Electrodynamics (3 ed.). Prentice Hall. p. 149.
^ Landau; Lifshitz (1984). Electrodynamics of Continuous Media. Vol. 2 (2 ed.). Pergamon Press. p. 37.
^ McLachlan, A. D. (25 June 1963). "Retarded Dispersion Forces in Dielectrics at Finite Temperatures". Proceedings of the Royal Society A. 274 (1356): 80โ90. Bibcode:1963RSPSA.274...80M. doi:10.1098/rspa.1963.0115. S2CID 123109610.
^ Israelachvili, J.N. (1992). "Ch. 6.6: General Theory of Van der Waals forces Between Molecules". Intermolecular and Surface Forces (2 ed.). Academic Press.
^ Derjaguin, B.V. (1960). "The Force Between Molecules". Scientific American. 203 (1): 47โ53. Bibcode:1960SciAm.203a..47D. doi:10.1038/scientificamerican0760-47.
^ Derjaguin, B.V.; Abrikosova, I.I.; Lifshitz, E.M. (1956). "Direct measurement of molecular attraction between solids separated by a narrow gap". Quarterly Reviews, Chemical Society. 10 (3): 295โ329. doi:10.1039/qr9561000295.
^ Israelachvili, J.N.; Tabor, D. (1973). "Van der Waals Forces: Theory and Experiment". Progress in Surface and Membrane Science. 7: 1โ55. doi:10.1016/B978-0-12-571807-3.50006-5. ISBN 9780125718073.
^ Haydon, D.A.; Taylor, J.L. (1968). "Contact angles for thin lipid films and the determination of London-van der Waals forces". Nature. 217 (5130): 739โ740. Bibcode:1968Natur.217..739H. doi:10.1038/217739a0. S2CID 4263175.
^ Shih, A.; Parsegian, V.A. (1975). "Van der Waals Forces Between Heavy Alkali Atoms and Gold Surfaces: Comparison of Measured And Predicted Values". Phys. Rev. A. 12 (3): 835โ841. Bibcode:1975PhRvA..12..835S. doi:10.1103/physreva.12.835.
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Levelling Knowpia
Center for Operational Oceanographic Products and Services staff member conducts tide station levelling in support of the US Army Corps of Engineers in Richmond, Maine.
Optical levellingEdit
Stadia marks on a crosshair while viewing a metric levelling rod. The top mark is at 1500 mm and the lower is at 1345 mm. The distance between the marks is 155 mm, yielding a distance to the rod of 15.5 m.
Optical levelling employs an optical level, which consists of a precision telescope with crosshairs and stadia marks. The cross hairs are used to establish the level point on the target, and the stadia allow range-finding; stadia are usually at ratios of 100:1, in which case one metre between the stadia marks on the levelling staff represents 100 metres from the target. The complete unit is normally mounted on a tripod, and the telescope can freely rotate 360ยฐ in a horizontal plane. The surveyor adjusts the instrument's level by coarse adjustment of the tripod legs and fine adjustment using three precision levelling screws on the instrument to make the rotational plane horizontal. The surveyor does this with the use of a bull's eye level built into the instrument mount. The surveyor looks through the eyepiece of telescope while an assistant holds a vertical level staff which is a graduated in inches or centimeters. The level staff is placed vertically using a level, with its foot on the point for which the level measurement is required. The telescope is rotated and focused until the level staff is plainly visible in the crosshairs. In the case of a high accuracy manual level, the fine level adjustment is made by an altitude screw, using a high accuracy bubble level fixed to the telescope. This can be viewed by a mirror whilst adjusting or the ends of the bubble can be displayed within the telescope, which also allows assurance of the accurate level of the telescope whilst the sight is being taken. However, in the case of an automatic level, altitude adjustment is done automatically by a suspended prism due to gravity, as long as the coarse levelling is accurate within certain limits. When level, the staff graduation reading at the crosshairs is recorded, and an identifying mark or marker placed where the level staff rested on the object or position being surveyed.
Linear levelling procedureEdit
Diagram showing relationship between two level staff, or rods, shown as 1 and 3. The level line of sight is 2.
A typical procedure for a linear track of levels from a known datum is as follows. Set up the instrument within 100 metres (110 yards) of a point of known or assumed elevation. A rod or staff is held vertical on that point and the instrument is used manually or automatically to read the rod scale. This gives the height of the instrument above the starting (backsight) point and allows the height of the instrument (H.I.) above the datum to be computed. The rod is then held on an unknown point and a reading is taken in the same manner, allowing the elevation of the new (foresight) point to be computed. The difference between these two readings equals the change in elevation, which is why this method is also called differential levelling. The procedure is repeated until the destination point is reached. It is usual practice to perform either a complete loop back to the starting point or else close the traverse on a second point whose elevation is already known. The closure check guards against blunders in the operation, and allows residual error to be distributed in the most likely manner among the stations.
Turning a levelEdit
When using an optical level, the endpoint may be out of the effective range of the instrument. There may be obstructions or large changes of elevation between the endpoints. In these situations, extra setups are needed. Turning is a term used when referring to moving the level to take an elevation shot from a different location.
To "turn" the level, one must first take a reading and record the elevation of the point the rod is located on. While the rod is being kept in exactly the same location, the level is moved to a new location where the rod is still visible. A reading is taken from the new location of the level and the height difference is used to find the new elevation of the level gun. This is repeated until the series of measurements is completed.
The level must be horizontal to get a valid measurement. Because of this, if the horizontal crosshair of the instrument is lower than the base of the rod, the surveyor will not be able to sight the rod and get a reading. The rod can usually be raised up to 25 feet high, allowing the level to be set much higher than the base of the rod.
Trigonometric LevellingEdit
The other standard method of levelling in construction and surveying is called trigonometric levelling, which is preferred when levelling "out" to a number of points from one stationary point. This is done by using a total station, or any other instrument to read the vertical, or zenith angle to the rod, and the change in elevation is calculated using trigonometric functions (see example below). At greater distances (typically 1,000 feet and greater), the curvature of the Earth, and the refraction of the instrument wave through the air must be taken into account in the measurements as well (see section below).
Ex: an instrument at Point A reading to a rod at Point B a zenith angle of < 88ยฐ15'22" (degrees, minutes, seconds of arc) and a slope distance of 305.50 feet not factoring rod or instrument height would be calculated thus:
cos(88ยฐ15'22")(305.5)โ 9.30 ft.,
meaning an elevation change of approximately 9.30 feet in elevation between Points A and B. So if Point A is at 1,000 feet of elevation, then Point B would be at approximately 1,009.30 feet of elevation, as the reference line (0ยฐ) for zenith angles is straight up going clockwise one complete revolution, and so an angle reading of less than 90 degrees (horizontal or flat) would be looking uphill and not down (and opposite for angles greater than 90 degrees), and so would gain elevation.
Refraction and curvatureEdit
The curvature of the earth means that a line of sight that is horizontal at the instrument will be higher and higher above a spheroid at greater distances. The effect may be insignificant for some work at distances under 100 meters.
The line of sight is horizontal at the instrument, but is not a straight line because of atmospheric refraction. The change of air density with elevation causes the line of sight to bend toward the earth.
{\displaystyle \Delta h_{meters}=0.067D_{km}^{2}}
{\displaystyle \Delta h_{feet}=0.021\left({\frac {D_{ft}}{1000}}\right)^{2}}
Levelling loops and gravity variationsEdit
{\displaystyle \sum _{i=0}^{n}\Delta h_{i}=0}
{\displaystyle \sum _{i=0}^{n}\Delta h_{i}g_{i},}
{\displaystyle g_{i}}
stands for gravity at the leveling interval i. For precise leveling networks on a national scale, the latter formula should always be used.
{\displaystyle \Delta W_{i}=\Delta h_{i}g_{i}\ }
should be used in all computations, producing geopotential values
{\displaystyle W_{i}}
for the benchmarks of the network.
Classical instrumentsEdit
The dumpy level was developed by English civil engineer William Gravatt, while surveying the route of a proposed railway line from London to Dover. More compact and hence both more robust and easier to transport, it is commonly believed that dumpy levelling is less accurate than other types of levelling, but such is not the case. Dumpy levelling requires shorter and therefore more numerous sights, but this fault is compensated by the practice of making foresights and backsights equal.
Automatic levelEdit
Automatic levels make use of a compensator that ensures that the line of sight remains horizontal once the operator has roughly leveled the instrument (to within maybe 0.05 degree). The surveyor sets the instrument up quickly and does not have to relevel it carefully each time he sights on a rod on another point. It also reduces the effect of minor settling of the tripod to the actual amount of motion instead of leveraging the tilt over the sight distance. Three level screws are used to level the instrument.
Laser levelEdit
Laser levels[5] project a beam which is visible and/or detectable by a sensor on the leveling rod. This style is widely used in construction work but not for more precise control work. An advantage is that one person can perform the levelling independently, whereas other types require one person at the instrument and one holding the rod.
Astrogeodetic levelling
^ Ira Osborn Baker (1887). Leveling: Barometric, Trigonometric and Spirit. D. Van Nostrand. p. 126. single leveling.
^ Guy Bomford (1980). Geodesy (4th ed.). Clarendon Press. p. 204. ISBN 0-19-851946-X.
^ Davis, Foote, and Kelly, Surveying Theory and Practice, 1966 p. 152
^ Guy Bomford (1980). Geodesy (4th ed.). Oxford: Clarendon Press. p. 222. ISBN 0-19-851946-X.
^ John S. Scott (1992). Dictionary of Civil Engineering. Springer Science+Business Media. p. 252. ISBN 0-412-98421-0.
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Two-ray propagation channel - MATLAB - MathWorks Australia
10\mu s
\begin{array}{l}\stackrel{\to }{R}={\stackrel{\to }{x}}_{s}-{\stackrel{\to }{x}}_{r}\\ {R}_{los}=|\stackrel{\to }{R}|=\sqrt{{\left({z}_{r}-{z}_{s}\right)}^{2}+{L}^{2}}\\ {R}_{1}=\frac{{z}_{r}}{{z}_{r}+{z}_{z}}\sqrt{{\left({z}_{r}+{z}_{s}\right)}^{2}+{L}^{2}}\\ {R}_{2}=\frac{{z}_{s}}{{z}_{s}+{z}_{r}}\sqrt{{\left({z}_{r}+{z}_{s}\right)}^{2}+{L}^{2}}\\ {R}_{rp}={R}_{1}+{R}_{2}=\sqrt{{\left({z}_{r}+{z}_{s}\right)}^{2}+{L}^{2}}\\ \mathrm{tan}{\theta }_{los}=\frac{\left({z}_{s}-{z}_{r}\right)}{L}\\ \mathrm{tan}{\theta }_{rp}=-\frac{\left({z}_{s}+{z}_{r}\right)}{L}\\ {{\theta }^{\prime }}_{los}=-{\theta }_{los}\\ {{\theta }^{\prime }}_{rp}={\theta }_{rp}\end{array}
{E}_{los}={E}_{0}\left(\frac{{R}_{0}}{{R}_{los}}\right){e}^{i\omega \left(t-{R}_{los}/c\right)}
{E}_{rp}={L}_{G}{E}_{0}\left(\frac{{R}_{0}}{{R}_{rp}}\right){e}^{i\omega \left(t-{R}_{rp}/c\right)}
\begin{array}{l}{G}_{p}=\frac{{Z}_{1}\mathrm{cos}{\theta }_{1}-{Z}_{2}\mathrm{cos}{\theta }_{2}}{{Z}_{1}\mathrm{cos}{\theta }_{1}+{Z}_{2}\mathrm{cos}{\theta }_{2}}=\frac{\mathrm{cos}{\theta }_{1}-\frac{{Z}_{2}}{{Z}_{1}}\mathrm{cos}{\theta }_{2}}{\mathrm{cos}{\theta }_{1}+\frac{{Z}_{2}}{{Z}_{1}}\mathrm{cos}{\theta }_{2}}\\ {G}_{s}=\frac{{Z}_{2}\mathrm{cos}{\theta }_{1}-{Z}_{1}\mathrm{cos}{\theta }_{2}}{{Z}_{2}\mathrm{cos}{\theta }_{1}+{Z}_{1}\mathrm{cos}{\theta }_{2}}=\frac{\mathrm{cos}{\theta }_{2}-\frac{{Z}_{2}}{{Z}_{1}}\mathrm{cos}{\theta }_{1}}{\mathrm{cos}{\theta }_{2}+\frac{{Z}_{2}}{{Z}_{1}}\mathrm{cos}{\theta }_{1}}\\ {Z}_{1}=\sqrt{\frac{{\mu }_{1}}{{\epsilon }_{1}}}\\ {Z}_{2}=\sqrt{\frac{{\mu }_{2}}{{\epsilon }_{2}}}\end{array}
\begin{array}{l}{G}_{p}=\frac{\sqrt{\rho }\mathrm{cos}{\theta }_{1}-\mathrm{cos}{\theta }_{2}}{\sqrt{\rho }\mathrm{cos}{\theta }_{1}+\mathrm{cos}{\theta }_{2}}\\ {G}_{s}=\frac{\sqrt{\rho }\mathrm{cos}{\theta }_{2}-\mathrm{cos}{\theta }_{1}}{\sqrt{\rho }\mathrm{cos}{\theta }_{2}+\mathrm{cos}{\theta }_{1}}\end{array}
\gamma ={\gamma }_{o}\left(f\right)+{\gamma }_{w}\left(f\right)=0.1820f{N}^{โณ}\left(f\right).
{N}^{โณ}\left(f\right)=\sum _{i}{S}_{i}{F}_{i}+{{N}^{โณ}}_{D}^{}\left(f\right)
{S}_{i}={a}_{1}ร{10}^{-7}{\left(\frac{300}{T}\right)}^{3}\mathrm{exp}\left[{a}_{2}\left(1-\left(\frac{300}{T}\right)\right]P.
{S}_{i}={b}_{1}ร{10}^{-1}{\left(\frac{300}{T}\right)}^{3.5}\mathrm{exp}\left[{b}_{2}\left(1-\left(\frac{300}{T}\right)\right]W.
W=\frac{\rho T}{216.7}.
{\gamma }_{c}={K}_{l}\left(f\right)M,
{\gamma }_{R}=k{R}^{\alpha },
r=\frac{1}{0.477{d}^{0.633}{R}_{0.01}^{0.073\alpha }{f}^{0.123}-10.579\left(1-\mathrm{exp}\left(-0.024d\right)\right)}
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Sums Of Divergent Series | Brilliant Math & Science Wiki
Patrick Corn, Mhammd Al Mhammd, Satyabrata Dash, and
Sal Fawad
Every calculus student learns that divergent series should not be manipulated in the same way as convergent series. For example, if forced to assign a value to the divergent series
1-1+1-1+1-1+\cdots,
the most obvious method is to group terms:
(1-1)+(1-1)+(1-1)+\cdots=0+0+0+\cdots=0,
but this produces a different answer if the terms are grouped differently:
1+(-1+1)+(-1+1)+(-1+1)+\cdots = 1+0+0+\cdots = 1.
Nevertheless, it is often useful to assign values to divergent series in "reasonable" or "consistent" ways. Indeed, mathematicians from Euler to Ramanujan used divergent series to derive many important results (though with varying degrees of rigorous justification). This wiki will discuss
(1) what makes a method of assigning values "reasonable,"
(2) some methods that are commonly used, and
(3) most importantly, why these methods are useful in practice.
The goal is to convince readers that Abel was incorrect when he famously said,
"The divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever."
Requirements for Divergent Series Sums
Cesaro Summation
Dirichlet Series Regularization
Regularity: A summation method for series is said to be regular if it gives the correct answer for convergent series (i.e. the limit of the sequence of partial sums).
Linearity: If
\sum a_n = A
\sum b_n = B
\sum(a_n+b_n)
A+B
\sum ca_n
c
is a constant, must equal
cA
Stability: If
\sum\limits_{n=1}^{\infty} a_n = A
\sum\limits_{n=2}^{\infty} a_n = A-a_1
Not every useful method for summing series satisfies these requirements (in particular stability), but many do. Note that most methods for summing series do not work on every series; the goal is to find and use methods that sum as many interesting and important series as possible. The above requirements alone are often enough to determine what the value of the sum of a given series must be under any method that satisfies the requirements.
1-1+1-1+1-1+\cdots
under any summation method that is regular, linear, and stable, assuming the method provides a sum for this series.
S
\begin{aligned} S &= 1-1+1-1+1-1+\cdots \\ &= 1+(-1+1-1+1-1+\cdots) &\qquad (\text{stability}) \\ &= 1+(-1)(1-1+1-1+1-\cdots) &\qquad (\text{linearity}) \\ &= 1+(-1)S, \end{aligned}
S = 1-S
S = \frac12
_\square
1+2+4+8+16+\cdots
\begin{aligned} S &= 1+2+4+8+\cdots \\ &= 1+(2+4+8+\cdots) &\qquad (\text{stability}) \\ &= 1+2(1+2+4+\cdots) &\qquad (\text{linearity}) \\ &= 1+2S, \end{aligned}
S = 1+2S
S=-1
_\square
The answers to both these questions seem quite odd, but notice that they both represent a sort of continuation of a known formula for geometric series:
\sum_{n=0}^{\infty} r^n = \frac1{1-r}.
In calculus, one learns that this only converges for
r \in (-1,1)
. One way to get the right answer for the two examples above is to use this formula but plug in values outside the interval of convergence, namely
r = -1
r=2
, respectively. Of course, this is not anywhere near a general summation method, but it does give an intuitive sense of where the answers are coming from. The idea of continuation will arise more formally below.
As a first example of a summation method, Cesaro summation works as follows: rather than taking the limit of the partial sums, take the limit of their averages. That is, given a series
\sum a_n
s_k
k^\text{th}
partial sum as usual and
t_k = \frac1{k}(s_1+s_2+\cdots+s_k),
and assign the value
\lim\limits_{k\to\infty} t_k
\sum a_n
, if the limit exists.
It can be shown that Cesaro summation is regular, linear, and stable.
For the series
1-1+1-1+1-1+\cdots
, the partial sums are
1,0,1,0,1,0,\ldots
. The series is not convergent because the limit of this sequence does not exist. But the sequence of averages of the partial sums is
1,\frac12,\frac23,\frac12,\frac35,\frac12,\ldots,
\frac12
. So the series is Cesaro summable, to
\frac12
Cesaro summability allows certain series with oscillatory sequences of partial sums to be "smoothed out," but if the partial sums of the series go to
\infty
instead (e.g. the harmonic series), the averages of the partial sums will also go to
\infty
, so series like
1+2+4+8+\cdots
will not be Cesaro summable.
Abel summation involves limits of power series: define
\sum_{n=0}^{\infty} a_n = \lim_{z\to 1^-} \sum_{n=0}^{\infty} a_nz^n
if the limit exists. The idea is to extend the conclusion of Abel's theorem, which says (in part) that if
\sum a_n
converges, then the limit on the right side of the above equation exists and equals the sum. (This is precisely the statement that Abel summation is regular.) Note that the example
1-1+1-1+1-1+\cdots
is Abel summable, because
1-z+z^2-z^3+\cdots = \frac1{1+z}
|z|<1
\lim\limits_{z\to 1^-} \frac1{1+z} = \frac12
In fact, any series that is Cesaro summable is also Abel summable, and the sums are the same. So Abel summability is stronger (though Cesaro summability is nevertheless still useful due to its relative ease of computation). Here is an example of a series that is Abel summable but not Cesaro summable.
1-2+3-4+5-6+\cdots
Define the Abel sum
\sum\limits_{n=0}^\infty a_n
\displaystyle \lim_{z\to 1^-} \sum_{n=0}^\infty a_nz^n,
if that limit exists.
The Abel sum of the (divergent) series as shown above can be written as
\frac{a}{b}
and
b
a+b
Some summation methods are defined using analytic continuation of complex-valued functions. An analytic continuation of a function
f
g
, defined on a larger set than
f
is, which is (complex) differentiable everywhere on its domain.
The motivating example is the Riemann zeta function
\zeta(s) = \sum_{n=1}^{\infty} \frac1{n^s}.
This only converges for
\text{Re}(s) > 1
, but there is a functional equation which extends the
\zeta
function to a function that is well-defined and differentiable everywhere except for
s =1
. This functional equation allows the computation
\zeta(-1) = -\frac1{12}
\big(
n
\zeta(-n) = -\frac{B_{n+1}}{n+1}
B_k
are the Bernoulli numbers
\big).
So plugging in
s=-1
to the series representation of the
\zeta
function yields
1+2+3+\cdots = -\frac1{12}.
It turns out that this sum has practical applications in string theory and in a computation of the one-dimensional Casimir effect in quantum mechanics. In the latter context, the zeta function regularization corresponds to an assumption about the model that accounts for the "cancellation of the infinite part" of the sum.
The general idea is the same: the function
\sum a_n^{-s}
might converge in some complex half-plane, but if it can be analytically continued to a function defined for
s = -1
, associate the value of the function at
-1
with the sum of the series. Note that this method is stable but not linear.
Another idea that is sometimes (incorrectly) called zeta function regularization is to use the Dirichlet series
f(s) = \sum_{n=0}^{\infty} \frac{a_n}{n^s}
and assign the sum the value of
f(0)
, if
can be analytically continued to
0
. This is a different method than zeta function regularization: indeed, it is linear but not stable.
\big(\text{Note }
that it also gives
1+2+3+\cdots=-\frac1{12}.\big)
Sums of divergent series often have applications in physics, as with the
1+2+3+\cdots
example above. The general idea is that if a physical situation is described by a function
f
defined by a series that is only convergent for some set of values not including
s
, an analytic continuation
g
of
to some larger set of values including
s
is related closely enough to
f
g(s)
can have some meaningful physical interpretation even though
f(s)
There are less mysterious situations than this, in which divergent series give combinatorial insight as well. The following (long) example is related by Matt Noonan.
The Catalan numbers count (among many other objects) rooted left-right-ordered binary trees with
n
vertices. Here a tree is a set of vertices, which are connected to children one level below and a parent one level above. "Rooted" means there is one vertex at the top, "left-right-ordered" means that every child is labeled either a left child or a right child, and "binary" means that every parent has at most two children--and if there are two, one must be a left child and the other a right child.
F(z) = \frac{1-\sqrt{1-4z}}{2z} = 1+x+2x^2+5x^3+\cdots+C_kx^k+\cdots
(this is a straightforward application of the fractional binomial theorem). The series has radius of convergence
\frac 14,
F(1) = \frac12-i\frac{\sqrt{3}}2
F(1)
as counting all of the rooted ordered binary trees, although of course the series defining
F(1)
diverges. Notice that
F(1)^7 = F(1)
F(z)^7
is the generating function for 7-tuples of rooted ordered binary trees. So this suggests that there is a "natural" bijective way to identify a 7-tuple of rooted ordered binary trees with a unique rooted ordered binary tree, and this turns out to be the case! See this paper for details.
The point here is that the identity for divergent series sums has a straightforward and natural interpretation as a statement about a bijection between two equal-sized sets. This is how applications of sums of divergent series often work: instead of solving down-to-earth problems directly, they give clues to the correct solution, that can later be justified rigorously by other methods.
Cite as: Sums Of Divergent Series. Brilliant.org. Retrieved from https://brilliant.org/wiki/sums-of-divergent-series/
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Global Constraint Catalog: Cnext_greater_element
<< 5.275. next_element5.277. ninterval >>
\mathrm{๐๐๐ก๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}\left(\mathrm{๐
๐ฐ๐}\mathtt{1},\mathrm{๐
๐ฐ๐}\mathtt{2},\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\right)
\mathrm{๐
๐ฐ๐}\mathtt{1}
\mathrm{๐๐๐๐}
\mathrm{๐
๐ฐ๐}\mathtt{2}
\mathrm{๐๐๐๐}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐
๐ฐ๐}\mathtt{1}<\mathrm{๐
๐ฐ๐}\mathtt{2}
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|>0
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐},\mathrm{๐๐๐}\right)
\mathrm{๐
๐ฐ๐}\mathtt{2}
is the value strictly greater than
\mathrm{๐
๐ฐ๐}\mathtt{1}
located at the smallest possible entry of the table
\mathrm{๐๐ฐ๐ฑ๐ป๐ด}
. In addition, the variables of the collection
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
are sorted in strictly increasing order.
\left(7,8,โฉ3,5,8,9โช\right)
\mathrm{๐๐๐ก๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
\mathrm{๐
๐ฐ๐}\mathtt{2}
is fixed to the first value 8 strictly greater than
\mathrm{๐
๐ฐ๐}\mathtt{1}=7
\mathrm{๐๐๐}
attributes of the items of the collection
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|>1
\mathrm{๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}.\mathrm{๐๐๐}\right)>1
Originally introduced in [CarlssonBeldiceanu04] for modelling the fact that a nucleotide has to be consumed as soon as possible at cycle
\mathrm{๐
๐ฐ๐}\mathtt{2}
after a given cycle
\mathrm{๐
๐ฐ๐}\mathtt{1}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐}
constraint, except for the fact that the
\mathrm{๐๐๐}
attributes are sorted.
{V}_{1},{V}_{2},\cdots ,{V}_{|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|}
denote the variables of the collection of variables
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
. By creating the extra variables
M
{U}_{1},{U}_{2},\cdots ,{U}_{|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|}
\mathrm{๐๐๐ก๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
constraint can be expressed in term of the following constraints:
{V}_{1}<{V}_{2}<\cdots <{V}_{|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|}
\mathrm{๐๐๐ก๐๐๐๐}
\left(M,\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}\right)
\mathrm{๐
๐ฐ๐}\mathtt{2}>\mathrm{๐
๐ฐ๐}\mathtt{1}
\mathrm{๐
๐ฐ๐}\mathtt{2}\le M
{V}_{i}\le \mathrm{๐
๐ฐ๐}\mathtt{1}โ{U}_{i}=M\left(i\in \left[1,|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|\right]\right)
{V}_{i}>\mathrm{๐
๐ฐ๐}\mathtt{1}โ{U}_{i}={V}_{i}\left(i\in \left[1,|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|\right]\right)
\mathrm{๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐}\mathtt{2},โฉ{U}_{1},{U}_{2},\cdots ,{U}_{|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|}โช\right)
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐}
\mathrm{๐๐๐ก๐}_\mathrm{๐๐๐๐๐๐๐}
(allow to iterate over the values of a table).
characteristic of a constraint: minimum, derived collection.
constraint type: order constraint, data constraint.
modelling: table.
\mathrm{๐๐๐}\left(๐
-\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right),\left[\mathrm{๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐
๐ฐ๐}\mathtt{1}\right)\right]\right)
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐ด๐๐ป}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1},\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}\right)
\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{1}.\mathrm{๐๐๐}<\mathrm{๐๐๐๐๐๐๐๐๐}\mathtt{2}.\mathrm{๐๐๐}
\mathrm{๐๐๐๐}
=|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-1
๐
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(๐,\mathrm{๐๐๐๐๐๐๐๐๐}\right)
๐.\mathrm{๐๐๐}<\mathrm{๐๐๐๐๐๐๐๐๐}.\mathrm{๐๐๐}
\mathrm{๐๐๐๐}
>0
\mathrm{๐ฒ๐ด๐ข๐ข}โฆ\left[\mathrm{๐๐๐๐๐๐},\mathrm{๐๐๐๐๐๐๐๐๐}\right]
\mathrm{๐๐๐๐๐๐๐}
\left(\mathrm{๐
๐ฐ๐}\mathtt{2},\mathrm{๐๐๐๐๐๐๐๐๐}\right)
\mathrm{๐๐๐๐}
\mathrm{๐๐๐ก๐}_\mathrm{๐๐๐๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐}
Since the first graph constraint uses the
\mathrm{๐๐ด๐๐ป}
\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}
collection, the number of arcs of the corresponding initial graph is equal to
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-1
. Therefore the maximum number of arcs of the final graph is equal to
|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-1
\mathrm{๐๐๐๐}=|\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-1
\mathrm{๐๐๐๐}\ge |\mathrm{๐
๐ฐ๐๐ธ๐ฐ๐ฑ๐ป๐ด๐}|-1
\underline{\overline{\mathrm{๐๐๐๐}}}
\overline{\mathrm{๐๐๐๐}}
|
Pythagorean trigonometric identity - Wikipedia
(Redirected from Pythagorean identity)
Relation between sine and cosine
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1.}
{\textstyle (\sin \theta )^{2}}
1 Proofs and their relationships to the Pythagorean theorem
1.1 Proof based on right-angle triangles
1.1.1 Related identities
1.2 Proof using the unit circle
1.3 Proof using power series
1.4 Proof using the differential equation
1.5 Proof using Euler's formula
Proofs and their relationships to the Pythagorean theorem[edit]
Proof based on right-angle triangles[edit]
{\displaystyle \sin \theta ={\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}={\frac {b}{c}}}
{\displaystyle \cos \theta ={\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}={\frac {a}{c}}}
{\displaystyle {\frac {\mathrm {opposite} ^{2}+\mathrm {adjacent} ^{2}}{\mathrm {hypotenuse} ^{2}}}}
{\displaystyle x=\cos \theta }
{\displaystyle y=\sin \theta }
{\displaystyle x=c\cos \theta }
{\displaystyle y=c\sin \theta }
{\displaystyle a=x}
{\displaystyle b=y}
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =\sin ^{2}\left(t+{\frac {1}{2}}\pi \right)+\cos ^{2}\left(t+{\frac {1}{2}}\pi \right)=\cos ^{2}t+\sin ^{2}t=1.}
{\displaystyle \sin ^{2}\theta =\sin ^{2}(-\theta ){\text{ and }}\cos ^{2}\theta =\cos ^{2}(-\theta ).}
{\displaystyle 1+\tan ^{2}\theta =\sec ^{2}\theta }
{\displaystyle 1+\cot ^{2}\theta =\csc ^{2}\theta }
{\displaystyle \tan \theta ={\frac {b}{a}}\ ,}
{\displaystyle \sec \theta ={\frac {c}{a}}\ .}
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
{\displaystyle \cos ^{2}\theta }
{\displaystyle {\frac {\sin ^{2}\theta }{\cos ^{2}\theta }}+{\frac {\cos ^{2}\theta }{\cos ^{2}\theta }}={\frac {1}{\cos ^{2}\theta }}}
{\displaystyle \tan ^{2}\theta +1=\sec ^{2}\theta }
{\displaystyle \tan ^{2}\theta =\sec ^{2}\theta -1}
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
{\displaystyle \sin ^{2}\theta }
{\displaystyle {\frac {\sin ^{2}\theta }{\sin ^{2}\theta }}+{\frac {\cos ^{2}\theta }{\sin ^{2}\theta }}={\frac {1}{\sin ^{2}\theta }}}
{\displaystyle 1+\cot ^{2}\theta =\csc ^{2}\theta }
{\displaystyle \cot ^{2}\theta =\csc ^{2}\theta -1}
Proof using the unit circle[edit]
Main article: unit circle
{\displaystyle x^{2}+y^{2}=1.}
{\displaystyle x=\cos \theta \ \mathrm {and} \ y=\sin \theta \ .}
{\displaystyle \cos ^{2}\theta +\sin ^{2}\theta =1\ ,}
Proof using power series[edit]
{\displaystyle {\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1},\\\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}.\end{aligned}}}
{\displaystyle {\begin{aligned}\sin ^{2}x&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{\frac {(-1)^{i}}{(2i+1)!}}{\frac {(-1)^{j}}{(2j+1)!}}x^{(2i+1)+(2j+1)}\\&=\sum _{n=1}^{\infty }\left(\sum _{i=0}^{n-1}{\frac {(-1)^{n-1}}{(2i+1)!(2(n-i-1)+1)!}}\right)x^{2n}\\&=\sum _{n=1}^{\infty }\left(\sum _{i=0}^{n-1}{2n \choose 2i+1}\right){\frac {(-1)^{n-1}}{(2n)!}}x^{2n},\\\cos ^{2}x&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{\frac {(-1)^{i}}{(2i)!}}{\frac {(-1)^{j}}{(2j)!}}x^{(2i)+(2j)}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}{\frac {(-1)^{n}}{(2i)!(2(n-i))!}}\right)x^{2n}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}{2n \choose 2i}\right){\frac {(-1)^{n}}{(2n)!}}x^{2n}.\end{aligned}}}
{\displaystyle \sum _{i=0}^{n}{2n \choose 2i}-\sum _{i=0}^{n-1}{2n \choose 2i+1}=\sum _{j=0}^{2n}(-1)^{j}{2n \choose j}=(1-1)^{2n}=0}
{\displaystyle \sin ^{2}x+\cos ^{2}x=1\ ,}
Proof using the differential equation[edit]
{\displaystyle y''+y=0}
{\displaystyle z=\sin ^{2}x+\cos ^{2}x}
{\displaystyle {\frac {d}{dx}}z=2\sin x\ \cos x+2\cos x\ (-\sin x)=0\ ,}
Proof using Euler's formula[edit]
{\displaystyle e^{i\theta }=\cos \theta +i\sin \theta }
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =(\cos \theta +i\sin \theta )(\cos \theta -i\sin \theta )=e^{i\theta }e^{-i\theta }=1}
{\displaystyle d={\sqrt {x^{2}+y^{2}}}}
{\displaystyle (x,\ y)}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Pythagorean_trigonometric_identity&oldid=1089316316"
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Gaussian process regression model class - MATLAB - MathWorks รญโขลรชยตยญ
Method used to estimate the basis function coefficients, รยฒ; noise standard deviation, รฦ; and kernel parameters, รยธ, of the GPR model, stored as a character vector. It can be one of the following.
Explicit basis function used in the GPR model, stored as a character vector or a function handle. It can be one of the following. If n is the number of observations, the basis function adds the term H*รยฒ to the model, where H is the basis matrix and รยฒ is a p-by-1 vector of basis coefficients.
H=1
H=\left[1,X\right]
H=\left[1,X,{X}_{2}\right],
{X}_{2}=\left[\begin{array}{cccc}{x}_{11}^{2}& {x}_{12}^{2}& \cdots & {x}_{1d}^{2}\\ {x}_{21}^{2}& {x}_{22}^{2}& \cdots & {x}_{2d}^{2}\\ โฎ& โฎ& โฎ& โฎ\\ {x}_{n1}^{2}& {x}_{n2}^{2}& \cdots & {x}_{nd}^{2}\end{array}\right].
H=hfcn\left(X\right),
K\left({X}_{new},A\right)*\mathrm{รยฑ}\text{รขยย}.
K\left({X}_{new},A\right)
{X}_{new}
and active set vector A and รยฑ is a vector of weights.
K\left({X}_{new},A\right)*\mathrm{รยฑ}\text{รขยย}.
K\left({X}_{new},A\right)
{X}_{new}
ParameterVector Cell array containing the parameter vectors: basis function coefficients, รยฒ, kernel function parameters รยธ, and noise standard deviation รฦ.
In the first iteration, the software uses the initial parameter values in vector รยท0 = [รยฒ0,รฦ0,รยธ0] to select an active set A1. It maximizes the GPR marginal log likelihood or its approximation using รยท0 as the initial values and A1 to compute the new parameter estimates รยท1. Next, it computes the new log likelihood L1 using รยท1 and A1.
In the second iteration, the software selects the active set A2 using the parameter values in รยท1. Then, using รยท1 as the initial values and A2, it maximizes the GPR marginal log likelihood or its approximation and estimates the new parameter values รยท2. Then using รยท2 and A2, computes the new log likelihood value L2.
1 A1 รยท1 L1
รขโฌยฆ รขโฌยฆ รขโฌยฆ รขโฌยฆ
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Stochastic drift - Wikipedia
This article is about mathematical concept. For the slow accumulation of errors in navigation systems, see Inertial navigation system ยง drift rate.
In probability theory, stochastic drift is the change of the average value of a stochastic (random) process. A related concept is the drift rate, which is the rate at which the average changes. For example, a process that counts the number of heads in a series of
{\displaystyle n}
fair coin tosses has a drift rate of 1/2 per toss. This is in contrast to the random fluctuations about this average value. The stochastic mean of that coin-toss process is 1/2 and the drift rate of the stochastic mean is 0, assuming 1 = heads and 0 = tails.
1 Stochastic drifts in population studies
2 Stochastic drift in economics and finance
Stochastic drifts in population studies[edit]
Longitudinal studies of secular events are frequently conceptualized as consisting of a trend component fitted by a polynomial, a cyclical component often fitted by an analysis based on autocorrelations or on a Fourier series, and a random component (stochastic drift) to be removed.
In the course of the time series analysis, identification of cyclical and stochastic drift components is often attempted by alternating autocorrelation analysis and differencing of the trend. Autocorrelation analysis helps to identify the correct phase of the fitted model while the successive differencing transforms the stochastic drift component into white noise.
Stochastic drift can also occur in population genetics where it is known as genetic drift. A finite population of randomly reproducing organisms would experience changes from generation to generation in the frequencies of the different genotypes. This may lead to the fixation of one of the genotypes, and even the emergence of a new species. In sufficiently small populations, drift can also neutralize the effect of deterministic natural selection on the population.
Stochastic drift in economics and finance[edit]
Time series variables in economics and finance โ for example, stock prices, gross domestic product, etc. โ generally evolve stochastically and frequently are non-stationary. They are typically modelled as either trend-stationary or difference stationary. A trend stationary process {yt} evolves according to
{\displaystyle y_{t}=f(t)+e_{t}}
where t is time, f is a deterministic function, and et is a zero-long-run-mean stationary random variable. In this case the stochastic term is stationary and hence there is no stochastic drift, though the time series itself may drift with no fixed long-run mean due to the deterministic component f(t) not having a fixed long-run mean. This non-stochastic drift can be removed from the data by regressing
{\displaystyle y_{t}}
{\displaystyle t}
using a functional form coinciding with that of f, and retaining the stationary residuals. In contrast, a unit root (difference stationary) process evolves according to
{\displaystyle y_{t}=y_{t-1}+c+u_{t}}
{\displaystyle u_{t}}
is a zero-long-run-mean stationary random variable; here c is a non-stochastic drift parameter: even in the absence of the random shocks ut, the mean of y would change by c per period. In this case the non-stationarity can be removed from the data by first differencing, and the differenced variable
{\displaystyle z_{t}=y_{t}-y_{t-1}}
will have a long-run mean of c and hence no drift. But even in the absence of the parameter c (that is, even if c=0), this unit root process exhibits drift, and specifically stochastic drift, due to the presence of the stationary random shocks ut: a once-occurring non-zero value of u is incorporated into the same period's y, which one period later becomes the one-period-lagged value of y and hence affects the new period's y value, which itself in the next period becomes the lagged y and affects the next y value, and so forth forever. So after the initial shock hits y, its value is incorporated forever into the mean of y, so we have stochastic drift. Again this drift can be removed by first differencing y to obtain z which does not drift.
In the context of monetary policy, one policy question is whether a central bank should attempt to achieve a fixed growth rate of the price level from its current level in each time period, or whether to target a return of the price level to a predetermined growth path. In the latter case no price level drift is allowed away from the predetermined path, while in the former case any stochastic change to the price level permanently affects the expected values of the price level at each time along its future path. In either case the price level has drift in the sense of a rising expected value, but the cases differ according to the type of non-stationarity: difference stationarity in the former case, but trend stationarity in the latter case.
Krus, D.J., & Ko, H.O. (1983) Algorithm for autocorrelation analysis of secular trends. Educational and Psychological Measurement, 43, 821โ828. (Request reprint).
Krus, D. J., & Jacobsen, J. L. (1983) Through a glass, clearly? A computer program for generalized adaptive filtering. Educational and Psychological Measurement, 43, 149โ154
Retrieved from "https://en.wikipedia.org/w/index.php?title=Stochastic_drift&oldid=965832796"
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The molecular orbital theory is a technique for modeling the chemical bonding and geometry of molecules and polyatomic ions.
Molecular orbital theory helps explain why some compounds are colored, why an unpaired electron is stable in certain species, and why some molecules have resonance structures.
The molecular orbital theory builds off of valence bond theory and valence shell electron pair repulsion theory to better describe the interactions of electrons within a given molecule and the effects that has on the molecule's physical and chemical properties.
Whereas an atomic orbital is localized around a single atom, a molecular orbital is delocalized, extending over all the atoms in a molecule. Theoretically, the Schrรถdinger equation could be used to solve molecular orbitals, but in practice the equation becomes impossible to solve, even for the simplest molecules, without making approximations. Molecular orbital theory approximates the solution to the Schrรถdinger equation for a molecule.
Energy calculations are used to test the validity of the proposed molecular orbital. Nature minimizes the energy of each orbital, so the best possible orbital is the one with the lowest energy. The simplest guesses that work well in molecular orbital theory are linear combinations of atomic orbitals (LCAOs), which are similar to a weighted average of the atomic orbitals. The number of molecular orbitals in a given molecule is always equal to the sum of all the atomic orbitals in the atoms making up the molecule.
Animation showing the relative energies of atomic and molecular orbitals. Hydrogen's 1s orbitals are on the left and right. The antibonding orbital is on top (highest energy), and the bonding orbital is on the bottom (lowest energy).
Bonding molecular orbitals are always lower in energy than the parent atomic orbitals, where antibonding molecular orbitals are always higher in energy than the parent atomic orbitals. Electrons seek the lowest possible energy state in molecular orbitals, just as they do in atomic orbitals. Therefore, bonding molecular orbitals are stabilizing to the molecule and antibonding molecular orbitals are destabilizing.
The molecular orbitals (MOs) of
\ce{H2}
are a simple example system. Each hydrogen atom has a single 1s atomic orbital. Since
\ce{H2}
is comprised of two atoms, it has two molecular orbitals. The bonding orbital is an equally weighted sum of the two 1s atomic orbitals and is lower in energy than those orbitals. This is a
ฯ
orbital. The waves of the atomic orbitals show constructive interference, meaning there is an increased probability of finding electrons in the center of the
ฯ
orbital. The antibonding orbital,
ฯโ
, is shaped by destructive interference between the two atomic orbitals. The result is a node in the middle of the orbital, where there is zero probability of finding an electron.
Atomic orbitals other than
s
orbitals can interact to form molecular orbitals. Since
p, d,
f
orbitals are not spherically symmetrical, the orbitals must be plotted on a coordinate system to show how they are oriented in space. The shapes of the orbitals gets increasingly complex as the number of electrons increases.
Bonding and anti-bonding in s,p, and d orbitals [3]
Energy level diagrams are used to determine whether or not a molecule will be stable. The atomic orbitals of the two atoms are drawn on the sides of the diagram, with the molecular orbitals in the middle. This diagram can be used to calculate the bond order of the diatomic molecule. To find the bond order, subtract the number of electrons in antibonding molecular orbitals. If the bond order is positive, meaning more atoms are in bonding orbitals than antibonding orbitals, the molecule is stable in nature. A zero or negative bond order indicates an unstable molecule that probably does not exist. In general, a larger bond order correlates to a shorter bond length and a higher bond energy. In other words, the larger the bond order, the closer the atoms are to each other and the harder it will be to separate them.
Energy level diagram for
\ce{H2}
What is the bond order for
\ce{H2}
Number of electrons in bonding MO's: 2
Number of electrons in antibonding MO's: 0
(2-0)/2=1
We therefore expect
\ce{H2}
to be stable, and it is, in fact, a naturally occurring gas.
\ce{He2+}
\ce{H2^+}
\ce{He2}
\ce{H2}
Based on bond order, which molecule or ion would you expect to have the shortest bond length and the largest bond energy?
[1] Image from https://commons.wikimedia.org/wiki/File:H2OrbitalsAnimation.gif under Creative Commons licensing for reuse and modification.
[2] Image from https://commons.wikimedia.org/wiki/File:H2str.png under Creative Commons licensing for reuse and modification.
[3] Image from https://commons.wikimedia.org/wiki/File:BondingandAnti-bondinginteractionsofs,p,andd_orbitals.png under Creative Commons licensing for reused and modification.
Cite as: Molecular Orbital Theory. Brilliant.org. Retrieved from https://brilliant.org/wiki/molecular-orbital-theory/
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Find the solution for each system of equations below, if a solution exists. If there is not a single solution, explain why not. Be sure to check your solution, if possible.
\begin{aligned}[t] &x + 4y = 2 \\ &3x - 4y = 10 \end{aligned}
x
represents the same number in the first equation as it does in the second, as does
y
, and the same operations are being carried out, the equations would have to equal the same number.
\begin{aligned}[t] &2x + 4y = -10 \\ &x = -2y - 5 \end{aligned}
Can you multiply both sides of the second equation by anything to get the first? If so, the equations will be the same.
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How to Find Arc Length: 10 Steps (with Pictures) - wikiHow
1 Using Measurement of Central Angle in Degrees
2 Using Measurement of Central Angle in Radians
An arc is any portion of the circumference of a circle.[1] X Research source Arc length is the distance from one endpoint of the arc to the other. Finding an arc length requires knowing a bit about the geometry of a circle. Since the arc is a portion of the circumference, if you know what portion of 360 degrees the arcโs central angle is, you can easily find the length of the arc.
Using Measurement of Central Angle in Degrees
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Set up the formula for arc length. The formula is
{\displaystyle {\text{arc length}}=2\pi (r)({\frac {\theta }{360}})}
{\displaystyle r}
equals the radius of the circle and
{\displaystyle \theta }
equals the measurement of the arcโs central angle, in degrees.[2] X Expert Source Mario Banuelos, PhD
Assistant Professor of Mathematics Expert Interview. 19 January 2021.
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Plug the length of the circleโs radius into the formula. This information should be given, or you should be able to measure it. Make sure you substitute this value for the variable
{\displaystyle r}
For example, if the circleโs radius is 10 cm, your formula will look like this:
{\displaystyle {\text{arc length}}=2\pi (10)({\frac {\theta }{360}})}
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Plug the value of the arcโs central angle into the formula. This information should be given, or you should be able to measure it. Make sure you are working with degrees, and not radians, when using this formula. Substitute the central angleโs measurement for
{\displaystyle \theta }
in the formula.
For example, if the arcโs central angle is 135 degrees, your formula will look like this:
{\displaystyle {\text{arc length}}=2\pi (10)({\frac {135}{360}})}
{"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/d\/dc\/Find-Arc-Length-Step-4-Version-2.jpg\/v4-460px-Find-Arc-Length-Step-4-Version-2.jpg","bigUrl":"\/images\/thumb\/d\/dc\/Find-Arc-Length-Step-4-Version-2.jpg\/aid3795123-v4-728px-Find-Arc-Length-Step-4-Version-2.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"}
Multiply the radius by
<b class="whb">{\displaystyle 2\pi }</b>
. If you are not using a calculator, you can use the approximation
{\displaystyle \pi =3.14}
for your calculations. Rewrite the formula using this new value, which represents the circleโs circumference.[3] X Research source
{\displaystyle 2\pi (10)({\frac {135}{360}})}
{\displaystyle 2(3.14)(10)({\frac {135}{360}})}
{\displaystyle (62.8)({\frac {135}{360}})}
{"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/1\/19\/Find-Arc-Length-Step-5.jpg\/v4-460px-Find-Arc-Length-Step-5.jpg","bigUrl":"\/images\/thumb\/1\/19\/Find-Arc-Length-Step-5.jpg\/aid3795123-v4-728px-Find-Arc-Length-Step-5.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"}
Divide the arcโs central angle by 360. Since a circle has 360 degrees total, completing this calculation gives you what portion of the entire circle the sector represents. Using this information, you can find what portion of the circumference the arc length represents.
{\displaystyle (62.8)({\frac {135}{360}})}
{\displaystyle (62.8)(.375)}
{"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/4\/4d\/Find-Arc-Length-Step-6.jpg\/v4-460px-Find-Arc-Length-Step-6.jpg","bigUrl":"\/images\/thumb\/4\/4d\/Find-Arc-Length-Step-6.jpg\/aid3795123-v4-728px-Find-Arc-Length-Step-6.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"}
Multiply the two numbers together. This will give you the length of the arc.
{\displaystyle (62.8)(.375)}
{\displaystyle 23.55}
So, the length of an arc of a circle with a radius of 10 cm, having a central angle of 135 degrees, is about 23.55 cm.
Using Measurement of Central Angle in Radians
{"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/a\/a3\/Find-Arc-Length-Step-7.jpg\/v4-460px-Find-Arc-Length-Step-7.jpg","bigUrl":"\/images\/thumb\/a\/a3\/Find-Arc-Length-Step-7.jpg\/aid3795123-v4-728px-Find-Arc-Length-Step-7.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"}
{\displaystyle {\text{arc length}}=\theta (r)}
{\displaystyle \theta }
equals the measurement of the arcโs central angle in radians, and
{\displaystyle r}
equals the length of the circleโs radius.
{"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/4\/4f\/Find-Arc-Length-Step-8.jpg\/v4-460px-Find-Arc-Length-Step-8.jpg","bigUrl":"\/images\/thumb\/4\/4f\/Find-Arc-Length-Step-8.jpg\/aid3795123-v4-728px-Find-Arc-Length-Step-8.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"}
Plug the length of the circleโs radius into the formula. You need to know the length of the radius to use this method. Make sure you substitute the length of the radius for the variable
{\displaystyle r}
{\displaystyle {\text{arc length}}=\theta (10)}
Plug the measurement of the arcโs central angle into the formula. You should have this information in radians. If you know the angle measurement in degrees, you cannot use this method.
For example, if the arcโs central angle is 2.36 radians, your formula will look like this:
{\displaystyle {\text{arc length}}=2.36(10)}
{"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/f\/fe\/Find-Arc-Length-Step-10.jpg\/v4-460px-Find-Arc-Length-Step-10.jpg","bigUrl":"\/images\/thumb\/f\/fe\/Find-Arc-Length-Step-10.jpg\/aid3795123-v4-728px-Find-Arc-Length-Step-10.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"}
Multiply the radius by the radian measurement. The product will be the length of the arc.
{\displaystyle 2.36(10)}
{\displaystyle =23.6}
So, the length of an arc of a circle with a radius of 10 cm, having a central angle of 23.6 radians, is about 23.6 cm.
How do you calculate the length of an arc?
To find the arc length, set up the formula Arc length = 2 x pi x radius x (arc's central angle/360), where the arc's central angle is measured in degrees.
What is an arc length?
It's the linear distance measured along the curve of an arc from one end to the other.
How do I measure the arc central angle in radians?
1 radian = 57.3ยฐ, so divide the central angle by 57.3ยฐ to get radians.
How can I find the central angle when only the radius is given?
It's not possible. You need more information.
If the radius of 3 m of a circle makes angle of pi/4, then what will be its corresponding arc length?
If the arc has a central angle of ฯ/4, and the arc length is the radius multiplied by the central angle, then (3)(ฯ/4) = 3ฯ/4 = 2.36 m.
How do I find the volume of a cylinder?
Multiply the square of the radius by pi and by the height (or length) of the cylinder.
How do I measure arc length for circular tank if I only know that the diameter of the circle is 18.60 meters?
An arc is a portion of the circumference of a circle. If you're asking about the full circumference of the tank, that would be 18.60 multiplied by pi.
How do I use another arc to find arc length?
Assuming the two arcs are part of the same circle, you would have to know the central angles of each arc. The ratio of the two arc lengths would be equal to the ratio of their respective central angles.
How do I find the arc length of a semicircle if its diameter is 9?
The arc length of a semicircle is half the circumference of the full circle. The circumference of a circle is ฯd, which in this case is 9ฯ, or 28.26 units. That means the arc length of the semicircle is 14.13 units.
If the radius of an arc is 1500 m, how do I calculate angle?
You would also need to know the arc length. Then use the arc length formula in Method 1 above, and work backwards to solve for the angle. Thus the central angle equals the arc length multiplied by 360ยฐ, then divided by 2ฯr.
If you know the diameter of the circle, you can still find the arc length. The formulas for finding arc length utilize the circleโs radius. Since the radius is half the diameter of a circle, to find the radius, simply divide the diameter by 2.[4] X Research source For example, if the diameter of a circle is 14 cm, to find the radius, you would divide 14 by 2:
{\displaystyle 14\div 2=7}
So, the radius of the circle is 7 cm.
โ http://www.mathwords.com/a/arc_circle.htm
โ Mario Banuelos, PhD. Assistant Professor of Mathematics. Expert Interview. 19 January 2021.
โ http://mathbitsnotebook.com/Geometry/Circles/CRArcLengthRadian.html
To find arc length, start by dividing the arc's central angle in degrees by 360. Then, multiply that number by the radius of the circle. Finally, multiply that number by 2 ร pi to find the arc length. If you want to learn how to calculate the arc length in radians, keep reading the article!
Espaรฑol:hallar la longitud de un arco
Italiano:Calcolare la Lunghezza di Un Arco
Portuguรชs:Encontrar o Comprimento de um Arco
ะ ัััะบะธะน:ะฝะฐะนัะธ ะดะปะธะฝั ะดัะณะธ ะพะบััะถะฝะพััะธ
Franรงais:calculer la longueur dโun arc
Bahasa Indonesia:Mencari Panjang Busur
ไธญๆ:ๆฑๅผง้ฟ
Nederlands:De booglengte berekenen
เนเธเธข:เธซเธฒเธเธงเธฒเธกเธขเธฒเธงเธชเนเธงเธเนเธเนเธเธเธญเธเธงเธเธเธฅเธก
ุงูุนุฑุจูุฉ:ุญุณุงุจ ุทูู ููุณ
ๆฅๆฌ่ช:ๅผงใฎ้ทใใๆฑใใ
ํ๊ตญ์ด:ํธ ๊ธธ์ด ๊ตฌํ๊ธฐ
Tรผrkรงe:Yay Uzunluฤu Nasฤฑl Bulunur
Tiแบฟng Viแปt:Tรญnh ฤแป dร i cung
เคนเคฟเคจเฅเคฆเฅ:เคเคฐเฅเค (arc) เคเฅ เคฒเคเคฌเคพเค เคชเคคเคพ เคเคฐเฅเค
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|
Global Constraint Catalog: Ccorrespondence
<< 5.89. contains_sboxes5.91. count >>
\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐}
by removing the sorting condition.
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐ต๐๐พ๐ผ},\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ},\mathrm{๐๐พ}\right)
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right)
\mathrm{๐๐พ}
\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐}-\mathrm{๐๐๐๐}\right)
|\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}|=|\mathrm{๐ต๐๐พ๐ผ}|
|\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}|=|\mathrm{๐๐พ}|
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}.\mathrm{๐๐๐}\ge 1
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}.\mathrm{๐๐๐}\le |\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}|
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ต๐๐พ๐ผ},\mathrm{๐๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ},\mathrm{๐๐๐}\right)
\mathrm{๐๐๐๐๐๐๐๐}
\left(\mathrm{๐๐พ},\mathrm{๐๐๐๐}\right)
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐๐พ}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}
\mathrm{๐ต๐๐พ๐ผ}\left[i\right].\mathrm{๐๐๐๐}=\mathrm{๐๐พ}\left[\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}\left[i\right].\mathrm{๐๐๐}\right].\mathrm{๐๐๐๐}
\left(โฉ1,9,1,5,2,1โช,โฉ6,1,3,5,4,2โช,โฉ9,1,1,2,5,1โช\right)
As illustrated by Figure 5.90.1, the
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐ต๐๐พ๐ผ}\left[1\right].\mathrm{๐๐๐๐}=1
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}\left[1\right].\mathrm{๐๐๐}={6}^{th}
\mathrm{๐๐พ}
\mathrm{๐ต๐๐พ๐ผ}\left[2\right].\mathrm{๐๐๐๐}=9
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}\left[2\right].\mathrm{๐๐๐}={1}^{th}
\mathrm{๐๐พ}
\mathrm{๐ต๐๐พ๐ผ}\left[3\right].\mathrm{๐๐๐๐}=1
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}\left[3\right].\mathrm{๐๐๐}={3}^{th}
\mathrm{๐๐พ}
\mathrm{๐ต๐๐พ๐ผ}\left[4\right].\mathrm{๐๐๐๐}=5
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}\left[4\right].\mathrm{๐๐๐}={5}^{th}
\mathrm{๐๐พ}
\mathrm{๐ต๐๐พ๐ผ}\left[5\right].\mathrm{๐๐๐๐}=2
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}\left[5\right].\mathrm{๐๐๐}={4}^{th}
\mathrm{๐๐พ}
\mathrm{๐ต๐๐พ๐ผ}\left[6\right].\mathrm{๐๐๐๐}=1
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}\left[6\right].\mathrm{๐๐๐}={2}^{th}
\mathrm{๐๐พ}
Figure 5.90.1. Illustration of the correspondence between the items of the
\mathrm{๐ต๐๐พ๐ผ}
\mathrm{๐๐พ}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}
collection of the Example slot
|\mathrm{๐ต๐๐พ๐ผ}|>1
\mathrm{๐๐๐๐๐}
\left(\mathrm{๐ต๐๐พ๐ผ}.\mathrm{๐๐๐๐}\right)>1
\mathrm{๐ต๐๐พ๐ผ}.\mathrm{๐๐๐๐}
\mathrm{๐๐พ}.\mathrm{๐๐๐๐}
\mathrm{๐ต๐๐พ๐ผ}.\mathrm{๐๐๐๐}
\mathrm{๐๐พ}.\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐}
constraint except that we also provide the permutation that allows to go from the items of collection
\mathrm{๐ต๐๐พ๐ผ}
to the items of collection
\mathrm{๐๐พ}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐๐}
constraint to perfect matchings in a bipartite graph derived from the domain of the variables of the constraint in the following way: to each variable of the
\mathrm{๐ต๐๐พ๐ผ}
collection there is a from vertex; similarly, to each variable of the
\mathrm{๐๐พ}
collection there is a to vertex; finally, there is an edge between the
{i}^{th}
from vertex and the
{j}^{th}
to vertex if and only if the corresponding domains intersect and if
j
{i}^{th}
permutation variable.
Second, Dulmage-Mendelsohn decomposition [DulmageMendelsohn58] is used to characterise all edges that do not belong to any perfect matching, and therefore prune the corresponding variables.
\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐}
\mathrm{๐๐๐๐}
\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}
\mathrm{๐๐๐}\left(\begin{array}{c}\mathrm{๐ต๐๐พ๐ผ}_\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}-\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐}-\mathrm{๐๐๐๐},\mathrm{๐๐๐}-\mathrm{๐๐๐๐}\right),\hfill \\ \mathrm{๐๐๐๐}\left(\mathrm{๐๐๐๐}-\mathrm{๐ต๐๐พ๐ผ}.\mathrm{๐๐๐๐},\mathrm{๐๐๐}-\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}.\mathrm{๐๐๐}\right)\right]\hfill \end{array}\right)
\mathrm{๐ต๐๐พ๐ผ}_\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}
\mathrm{๐๐พ}
\mathrm{๐๐
๐๐ท๐๐ถ๐}
โฆ\mathrm{๐๐๐๐๐๐๐๐๐๐}\left(\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐},\mathrm{๐๐}\right)
โข\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐}.\mathrm{๐๐๐๐}=\mathrm{๐๐}.\mathrm{๐๐๐๐}
โข\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐}.\mathrm{๐๐๐}=\mathrm{๐๐}.\mathrm{๐๐๐ข}
\mathrm{๐๐๐๐}
=|\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}|
โข
\mathrm{๐ฐ๐ฒ๐๐ฒ๐ป๐ธ๐ฒ}
โข
\mathrm{๐ฑ๐ธ๐ฟ๐ฐ๐๐๐ธ๐๐ด}
โข
\mathrm{๐ฝ๐พ}_\mathrm{๐ป๐พ๐พ๐ฟ}
Parts (A) and (B) of Figure 5.90.2 respectively show the initial and final graph associated with the Example slot. In both graphs the source vertices correspond to the derived collection
\mathrm{๐ต๐๐พ๐ผ}_\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}
, while the sink vertices correspond to the collection
\mathrm{๐๐พ}
. Since the final graph contains exactly
|\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}|
arcs the
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐๐}
constraint holds. As we use the
\mathrm{๐๐๐๐}
\mathrm{๐๐๐๐๐๐๐๐๐๐๐๐๐๐}
Because of the second condition
\mathrm{๐๐๐๐}_\mathrm{๐๐๐๐๐๐๐๐๐๐๐}.\mathrm{๐๐๐}=\mathrm{๐๐}.\mathrm{๐๐๐ข}
of the arc constraint and since both, the
\mathrm{๐๐๐}
attributes of the collection
\mathrm{๐ต๐๐พ๐ผ}_\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}
\mathrm{๐๐๐ข}
\mathrm{๐๐พ}
are all-distinct, the final graph contains at most
|\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}|
arcs. Therefore we can rewrite the graph property
\mathrm{๐๐๐๐}
=
|\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}|
\mathrm{๐๐๐๐}
\ge
|\mathrm{๐ฟ๐ด๐๐ผ๐๐๐ฐ๐๐ธ๐พ๐ฝ}|
\underline{\overline{\mathrm{๐๐๐๐}}}
\overline{\mathrm{๐๐๐๐}}
|
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