url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
epimedyumsatis.com | 1,716,611,532,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00019.warc.gz | 190,250,805 | 21,740 | # On the net Casinos: Arithmetic of Bonuses
On the internet casino gamers are aware that the latter ones offer you several bonuses. “No cost-load” appears to be beautiful, having said that, are they truly helpful these bonuses? Are they rewarding for gamblers? The answer to this issue is dependent upon plenty of situations. Mathematics will help us response this concern.
Let us begin with a standard reward on kiss918 apk deposit: you transfer \$100 and obtain \$a hundred extra, which Will probably be possible to have acquiring staked \$3000. It is actually a typical example of reward on the primary deposit. The sizes of a deposit and bonus can be distinctive, and also the demanded stake costs, but something continues to be unchangeable – the level of the reward is available for withdrawal following the needed wager. Till this instant it can be not possible to withdraw funds, for a rule.
For those who are going to Perform in the web casino for years and alternatively insistently, this bonus can assist you, it can really be considered free of charge revenue. For those who Perform slots with 95% fork out-outs, a bonus will help you to make on typical extra 2000 \$ of stakes (\$one hundred/(one-0,ninety five)=\$2000), following that the quantity of reward might be about. But there could be complications, such as, if you just want to take a look at a casino, with no taking part in for some time, if you like roulette or other games, forbidden by casinos’ rules for successful again bonuses. In many casinos you won’t be permitted to withdraw dollars or will only return a deposit, if a wager is not really designed within the game titles authorized inside the On line casino. If you are keen on roulette or blackjack, and a reward is often gained back only by taking part in slots, make the needed \$3000 of stakes, in the midst of 95% of shell out-outs you may drop on regular \$3000*(one-0,95)=\$one hundred fifty. As the thing is, you not merely drop the reward but also just take out of the pocket \$50, In cases like this it is better to refuse the reward. Anyway, if blackjack and poker are authorized for successful back the reward with a casino’s earnings only about 0,5%, so it may be predicted that immediately after successful again the reward you’ll have \$one hundred-3000*0,005=\$eighty five from the casino’s funds.
“sticky” or “phantom” bonuses: | 527 | 2,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.940978 |
https://www.hindawi.com/archive/2014/680760/ | 1,511,128,116,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2017-47/segments/1510934805809.59/warc/CC-MAIN-20171119210640-20171119230640-00786.warc.gz | 815,567,316 | 73,800 | • Views 298
• Citations 0
• ePub 18
• PDF 186
`International Journal of Partial Differential EquationsVolume 2014 (2014), Article ID 680760, 8 pageshttp://dx.doi.org/10.1155/2014/680760`
Research Article
## Variational Statement and Domain Decomposition Algorithms for Bitsadze-Samarskii Nonlocal Boundary Value Problem for Poisson’s Two-Dimensional Equation
1Ilia Vekua Institute of Applied Mathematics of Ivane Javakhishvili, Tbilisi State University, 2 University Street, 0186 Tbilisi, Georgia
2Georgian Technical University, 77 Kostava Avenue, 0175 Tbilisi, Georgia
3Caucasus University, 10 Politkovskaya Street, 0186 Tbilisi, Georgia
Received 10 February 2014; Accepted 4 June 2014; Published 19 June 2014
#### Abstract
The Bitsadze-Samarskii nonlocal boundary value problem is considered. Variational formulation is done. The domain decomposition and Schwarz-type iterative methods are used. The parallel algorithm as well as sequential ones is investigated.
#### 1. Introduction
In applied sciences different problems with nonlocal boundary conditions arise very often. In some nonlocal problems, unlike classical boundary value problems, instead of boundary conditions, the dependence between the value of an unknown function on the boundary and some of its values inside of the domain is given.
Modern investigation of nonlocal elliptic boundary value problems originates from Bitsadze and Samarskii work [1], in which by means of the method of integral equations the theorems are proved on the existence and uniqueness of a solution for the second order multidimensional elliptic equations in rectangular domains. Some classes of problems for which the proposed method works are given.
Many works are devoted to the investigation of nonlocal problems for elliptic equations (see, e.g., [218] and references therein).
It is known how a great role takes place in the variational formulation of classical and nonlocal boundary value problems in modern mathematics (see, e.g., [1315, 1927]).
It is also well known that in order to find the approximate solutions, it is important to construct useful economical algorithms. For constructing such algorithms, the method of domain decomposition has a great importance (see, e.g., [23, 28, 29]).
In the work [6] the iterative method of proving the existence of a solution of Bitsadze-Samarskii problem for Laplace equation was proposed. This iterative method is based on the idea of Schwarz alternating method [30, pages 249–254]. It should be noted that the usage of Schwarz alternating method not only gives us the existence of a solution, but also allows finding effective algorithms for numerical resolution of such problems. By this approach the nonlocal problem reduces to classical Dirichlet problems on whole domain that yields the possibility to apply the already developed effective methods for numerical resolution of these problems. In [7, 1113, 15] using Schwarz alternating method and domain decomposition algorithms Bitsadze-Samarskii nonlocal problem is studied for Laplace equation. The domain decomposition algorithms are more economical than the method which was proposed in [6].
In the work [6] the reduction of nonlocal problem to the sequence of Dirichlet problems is studied. For investigating author used Schwarz lemma but not domain decomposition. At first domain decomposition for Bitsadze-Samarskii nonlocal boundary value problem was introduced in [7]. In the abstract [11] the convergence of the domain decomposition method for the second order nonlinear elliptic equation is given. In [12] the domain decomposition sequential and parallel algorithms are fixed. In [13] the sequential and parallel iterative algorithms are given. In [13] attention is devoted to the operator decomposition method and to the possibility of the variational formulation of the problem as well. In [15] the convergence of the domain decomposition parallel algorithm is fixed. Note that in the works [7, 1113, 15] results are mainly given without proof.
In the works [9, 10, 16, 21, 22, 3133] different methods are displayed for study of nonlocal problems in the theory of ordinary differential equations and in the theory of equations with partial derivatives.
The present work is devoted to the variational formulation and domain decomposition and Schwarz-type iterative methods for Bitsadze-Samarskii nonlocal boundary value problem for Poisson’s two-dimensional equation. Here we investigate the parallel algorithm as well as sequential ones. The rate of convergence is presented too.
The outline of this paper is as follows. In Section 2 for the Poisson equation in a rectangle we state Bitsadze-Samarskii nonlocal problem. In Section 3 the variational formulation of this problem is discussed. The convergence of the Schwarz-type iterative sequential algorithm is studied in Section 4. The same question for parallel algorithm is considered in Section 5. In Section 6 some conclusions are given.
#### 2. Formulation of Problem
In the plane , let us consider the rectangle , where and are the given positive constants. By we denote the boundary of the rectangle and by the intersection of the line with the set .
Consider the nonlocal Bitsadze-Samarskii boundary value problem [1] where is the Laplace operator, , ; is a given function, ; and is an unknown function.
Uniqueness of the solution of problem (1)–(3) follows from the extremum principle. It is known [1] that if is a continuous function on , then there exists unique regular solution of problem (1)–(3) .
#### 3. Variational Statement of Problem
We use usual and Sobolev spaces and . Let us denote by the vector space of all real functions satisfying the following conditions: is defined almost everywhere on , the boundary value is defined almost everywhere on , and , .
Functions and are assumed as the same element of if almost everywhere on and almost everywhere on .
Let and the operator which extends elements of as follows:
Operator associates to every function of the vector space the following function in such a way that the function is the odd function with respect to the variable almost everywhere for almost all .
Let us define on vector space the scalar product
Introducing the scalar product (5), let us denote vector space by where the norm is defined as follows:
The following statements take place [24].
Theorem 1. The norm defined in by the formula is equivalent to the norm .
Theorem 2. Space is complete with the metric .
Let the domain of definition of the operator be the vector space of the elements defined on for which the following conditions are fulfilled:(1), , , ;(2), .
Theorem 3. The vector space is dense in the space .
Thus, the operator acts from the dense vector space of the Hilbert space to the space .
Theorem 4. Operator is positively defined on the vector space .
To show the symmetry of the operator we use the following two lemmas, whose proofs are not difficult.
Lemma 5. For an arbitrary function of the vector space the following identity is valid:
Lemma 6. For two arbitrary functions and of the vector space we have
The scalar product given by (5) can be represented in the form
In the case of the scalar product (5) we have the positively defined operator , but it is not symmetric.
As is positive definite operator defined on the vector space which is dense in the space , for the problem (1)–(3) we can use the standard way of the variational formulation [27].
Let us introduce the new scalar product on :
Denote by the corresponding norm and by the corresponding metric. By we denote the space obtained after completion of by the metric .
The following statement is true [24].
Theorem 7. The function belongs to the space if and only if the following relations are fulfilled:
Thus, functions of the space satisfy boundary conditions. For every function there exists a unique function in the space , which minimizes the quadratic functional
For any function the following relation is fulfilled
The function from the space which minimizes the functional (14) is called the generalized solution of the equation .
If the function is sufficiently smooth then is a solution in a classical sense of problem (1)–(3).
#### 4. Domain Decomposition and Sequential Algorithm
In this section and next sections, for simplicity, let us consider Laplace equation with nonlocal (3) and again for simplicity homogeneous Dirichlet (2) conditions. So, we study the following problem:
For problem (15)–(17) let us consider the following sequential iterative procedure:
Here we utilize the following notations: where is a fixed point of the interval , , , and is any continuous function on the segment , which satisfies the following conditions: .
The iterative procedure (18) reduces our nonlocal nonclassical problem (15)–(17) to the investigation of the sequence of classical Dirichlet boundary value problems on every step of the iteration.
The following statement takes place.
Theorem 8. The sequential iterative process (18) converges to a solution of problem (15)–(17) uniformly in the domain , and the following estimations are true: where is a constant independent of functions , , , and constant depends on .
Proof. Note that solving problem (18), we get two sequences of harmonic functions, which are defined on the domains and , respectively.
We have the following relations:if and , then , that is ;if and , then , that is .
Let us introduce the notations:
If , then, for the harmonic function , applying the lemma from [30, pages 250–254] to the domain , we have where and depends only on domain .
If , then, for the harmonic function , applying the extremum principle [30, pages 218] to the domain , we have
If for some index , then from the extremum principle we get . So, the inequality (22) is also true.
If for some index , then from the extremum principle we get again . So, the inequality (23) is also true.
From the estimations (22) and (23) obtained above it follows that
This means that the sequences and tend to zero, and we obtain the uniform convergence of the series in the domains and , respectively.
According to Weierstrass theorem [30, pages 232, 233], the functions and are harmonic ones, defined, respectively, in and , and satisfy the condition (16). As for domain , we have the following relations:if and , then ;if and , then .
The latter difference tends to zero uniformly.
Again, according to the extremum principle, we obtain that the functions and coincide with each other in the domain and define a regular harmonic function on , which represents the solution of the problem (15)–(17).
Now, let us estimate the rate of convergence of the iterative process (18). Using the triangle inequality, the extremum principle and the second inequality of (24), we have Thus,
If in this inequality we tend to , then we obtain in the domain
Analogous estimation is true for in the domain .
Consequently, we get
This completes the proof of Theorem 8.
#### 5. Domain Decomposition and Parallel Algorithm
Algorithm (18) for the solution of the problem (15) has a sequential form. Now, let us consider one more approach to the solution of the problem (15)–(17). In this case the search of approximate solutions on domains and will be carried out not by means of a sequential algorithm but in a parallel way.
Consider the following overlapping parallel iterative process: where and are any continuous functions on the segment , which satisfy the following conditions: , .
The following statement takes place.
Theorem 9. The parallel iterative process (30) converges to a solution of the problem (15)–(17) uniformly in the domain , and the following estimations are true: where is a constant independent of functions , , and , and constant depends on , .
Proof. Let us prove this theorem in a similar way as Theorem 8 was proven. We should note that the boundary value problems (30) are simultaneously solved in domains and , respectively. Sequences of harmonic functions , are defined on and , respectively.
The following relations are satisfied:if and , then ; that is, ;if and , then ; that is, .
If we introduce the notation and, for the harmonic function , apply the lemma from [30, pages 250–254] to the domain , we will have the estimation where and depends on domain only.
If , then from the extremum principle we get that and the inequality (33) is clear.
If , then, for the harmonic function , from the extremum principle in the domain we have
If , then, using again the extremum principle, we have and the inequality (34) holds.
From the estimations (33) and (34), we obtain That means that the sequences and tend to zero.
Thus, in this case, the series analogous to the series from (25) are also uniformly convergent. The corresponding harmonic functions and are defined in and , respectively, and satisfy condition (16). As for the common part of these domains, we haveif and , then ;if and , then and this difference tends to zero, when . Thus, the functions and again coincide in the domain and define a regular harmonic function on , which represents the solution of the problem (15)–(17).
Let us estimate the rate of convergence of the constructed sequences.
We should remark that from the second inequality of (35) we have the following: if , then if , then
Using the triangle inequality, we get
Taking into account inequalities (36) and (37) for and , we obtain from (36)
If and , then from (38) we have
If and , then and at last, supposing and , we have
If in the obtained inequalities we tend to the limit, when , we get the following estimations.
If and , then if and , then if and , then and if and , then
Let us estimate and . We have
So, for any , we obtain the following relation in the domain :
Analogous estimation is true for in the domain
Theorems similar to Theorems 8 and 9 are valid for the sequential as well as parallel algorithms for multigrid domain decomposition case too.
#### 6. Conclusion
Because of importance of nonlocal problems many scientific articles are devoted for their investigation. We gave one problem to illustrate the variational formulation and domain decomposition of the problem. Nonlocal problems for the second and the fourth order ordinary and partial differential equations are also studied by authors. Nonlocal boundary value problems with some kind of nonlocal integral conditions are studied as well (see, e.g., [15, 2426] and references therein).
#### Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
#### Acknowledgments
The first author thanks Fulbright Visiting Scholar Program (Grant no. AY 2012-2013, USA) and the Shota Rustaveli National Science Foundation (Grant no. DI/16/4-120/11, Georgia) for the financial support and the Naval Postgraduate School in Monterey, CA, USA, for hosting him during the nine months of his tenure in 2012-2013. The second author thanks the Shota Rustaveli National Scientific Foundation (Grant no. YS/40/5-106/12, Georgia) for the financial support and the Naval Postgraduate School in Monterey, CA, USA, for hosting him during the four months of his tenure in 2013.
#### References
1. A. V. Bitsadze and A. A. Samarskii, “On some simplified generalization of the linear elliptic problems,” Doklady Akademii Nauk SSSR, vol. 185, pp. 739–740, 1969 (Russian).
2. A. Ashyralyev, “On well-posedness of the nonlocal boundary value problems for elliptic equations,” Numerical Functional Analysis and Optimization, vol. 24, no. 1-2, pp. 1–15, 2003.
3. A. Ashyralyev and E. Ozturk, “The numerical solution of the Bitsadze-Samarskii nonlocal boundary value problems with the Dirichlet-Neumann condition,” Abstract and Applied Analysis, vol. 2012, Article ID 730804, 13 pages, 2012.
4. A. Ashyralyev, O. Tetikoglu, and F. Songul, “FDM for elliptic equations with Bitsadze-Samarskii-Dirichlet conditions,” Abstract and Applied Analysis, vol. 2012, Article ID 454831, 22 pages, 2012.
5. G. Berikelashvili, D. Gordeziani, and S. Kharibegashvili, “Finite difference scheme for one mixed problem with integral condition,” in Proceedings of the 2nd WSEAS International Conference on Finite Differences, Finite Elements, Finite Volumes, Boundary Elements, pp. 118–120, 2009.
6. D. G. Gordeziani, “On one method of solving the Bitsadze-Samarskii boundary value problem,” in Proceedings of the Industrial and Applied Mathematics Seminar, pp. 39–41, 1970, 1970 (Russian).
7. T. A. Jangveladze, “On one iterative method of solution of Bitsadze-Samarskii boundary value problem,” Manuscript, Faculty of Applied Mathematics and Cybernetics of Tbilisi State University, Tbilisi, Georgia, 1975 (Georgian).
8. D. G. Gordeziani, On the Methods of Solution for One Class of Non-Local Boundary Value Problems, University Press, Tbilisi, Georgia, 1981 (Russian).
9. A. K. Gushin and V. P. Mikhailov, “On the continuity of the solutions of a class of non-local problems for an elliptic equation,” Sbornik: Mathematics, vol. 186, no. 2, pp. 197–219, 1995 (Russian).
10. V. A. Il’in and E. I. Moiseev, “Two-dimensional nonlocal boundary value problem for Poisson’s operator in the differential and difference forms,” Mathematical Modelling, vol. 2, pp. 139–156, 1990 (Russian).
11. T. A. Jangveladze, “Domain decomposition method for Bitsadze-Samarskii boundary value problem for second order nonlinear elliptic equation,” in Proceedings of the Postgraduate Students' Scientific Conference of Ivane Javakhishvili Tbilisi State University, p. 7, Tbilisi State University, Tbilisi, Georgia, 1977.
12. T. A. Jangveladze and Z. V. Kiguradze, “Domain decomposition for Bitsadze-Samarskii boundary value problem,” Reports of Enlarged Sessions of the Seminar of I. Vekua Institute of Applied Mathematics, vol. 16, pp. 16–19, 2001.
13. T. Jangveladze, Z. Kiguradze, and G. Lobjanidze, “On decomposition methods and variational formulation for Bitsadze-Samarskii nonlocal boundary value problem for two-dimensional second order elliptic equations,” in Proccedings of the 15th WSEAS International Conference on Applied Mathematics (MATH '10), pp. 116–121, Athens, Greece, December 2010.
14. J. Mo and C. Ouyang, “A class of nonlocal boundary value problems of nonlinear elliptic systems in unbounded domains,” Acta Mathematica Scientia, vol. 21, no. 1, pp. 93–97, 2001.
15. Z. V. Kiguradze, “Domain decomposition and parallel algorithm for Bitsadze-Samarskii boundary value problem,” Reports of Enlarged Sessions of the Seminar of I. Vekua Institute of Applied Mathematics, vol. 10, pp. 49–51, 1995.
16. B. P. Paneyakh, “On some nonlocal boundary value problems for linear differential operators,” Matematicheskie Zametki, vol. 35, no. 3, pp. 425–433, 1984 (Russian).
17. M. P. Sapagovas and P. U. Chegis, “On some boundary value problems with nonlocal conditions,” Differential Equations, vol. 23, pp. 1268–1274, 1987 (Russian).
18. A. L. Skubachevskii, “On a spectrum of some nonlocal elliptic boundary value problems,” Sbornik: Mathematics, vol. 117, pp. 548–558, 1982 (Russian).
19. T. A. Jangveladze and G. B. Lobjanidze, “On a variational statement of a nonlocal boundary value problem for a fourth-order ordinary differential equation,” Differentsial'nye Uravneniya, vol. 45, no. 3, pp. 325–333, 2009 (Russian), English translation in Differential Equations vol. 45, no. 3, pp. 335–343, 2009.
20. T. A. Jangveladze and G. B. Lobjanidze, “On a nonlocal boundary value problem for a fourth-order ordinary differential equation,” Differentsial'nye Uravneniya, vol. 47, no. 2, pp. 181–188, 2011 (Russian), English translation in Differential Equations vol. 47, no. 2, pp. 179–186, 2011.
21. G. L. Karakostas and P. Ch. Tsamatos, “Sufficient conditions for the existence of nonnegative solutions of a nonlocal boundary value problem,” Applied Mathematics Letters, vol. 15, no. 4, pp. 401–407, 2002.
22. I. Kiguradze and T. Kiguradze, “Conditions for the well-posedness of nonlocal problems for higher order linear differential equations with singularities,” Georgian Mathematical Journal, vol. 18, no. 4, pp. 735–760, 2011.
23. P. L. Lions, “On the Schwarz alternating method I,” in First International Symposium on Domain Decomposition Methods for Partial Differential Equations, R. Glowinski, G. H. Golub, G. A. Meurant, and J. Periaux, Eds., pp. 2–42, SIAM, Philadelphia, Pa, USA, 1988.
24. G. Lobjanidze, “On the variational formulation of Bitsadze-Samarskii problem for the equation $-\Delta u+\lambda u=f$,” Reports of Enlarged Sessions of the Seminar of I. Vekua Institute of Applied Mathematics, vol. 18, pp. 39–42, 2003.
25. G. Lobjanidze, “On variational formulation of some nonlocal boundary value problems by symmetric continuation operation of a function,” Journal of Applied Mathematics and Mechanics, vol. 12, pp. 15–22, 2006.
26. G. Lobjanidze, “On variational formulation of some nonlocal boundary value problem,” Reports of Enlarged Sessions of the Seminar of I. Vekua Institute of Applied Mathematics, vol. 25, pp. 80–85, 2011.
27. K. Rektorys, Variational Methods in Mathematics, Science and Engineering, Springer, Prague, Czech Republic, 1980.
28. S. Nepomnyaschikh, “Domain decomposition methods,” Radon Series on Computational and Applied Mathematics, vol. 1, pp. 89–160, 2007.
29. A. Quarteroni and A. Valli, Domain Decomposition Methods for Partial Differential Equations, Oxford University Press, Oxford, UK, 1999.
30. R. Courant and D. Hilbert, Methods of Mathematical Physics, vol. 2, Leningrad, Moscow, Russia, 1951 (Russian).
31. Ch. P. Gupta and S. I. Trofimchuk, “A sharper condition for the solvability of a three-point second order boundary value problem,” Journal of Mathematical Analysis and Applications, vol. 205, no. 2, pp. 586–597, 1997.
32. A. Lomtatidze and L. Malaguti, “On a nonlocal boundary value problem for second order nonlinear singular differential equations,” Georgian Mathematical Journal, vol. 7, no. 1, pp. 133–154, 2000.
33. R. Ma and N. Castaneda, “Existence of solutions of nonlinear m-point boundary-value problems,” Journal of Mathematical Analysis and Applications, vol. 256, no. 2, pp. 556–567, 2001. | 5,392 | 22,199 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-47 | latest | en | 0.874719 |
http://web.eecs.utk.edu/~plank/plank/classes/cs302/Notes/DynamicProgramming/Apples.html | 1,511,489,143,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807056.67/warc/CC-MAIN-20171124012912-20171124032912-00726.warc.gz | 315,272,180 | 2,806 | CS302 Lecture Notes - Dynamic ProgrammingExample program #7: Counting Apples
• November 18, 2009
• James S. Plank
• Directory: /home/plank/cs302/Notes/DynamicProgramming
This is Dumitru's Intermediate problem, and it is really no harder than the subsequence problem above. The problem is as follows:
"You are given a table composed of rows x cols cells, each having a certain quantity of apples. You start from the upper-left corner. At each step you can go down or right one cell. Find the maximum number of apples you can collect."
You can view this as a graph problem -- the "table" is a graph, and there are only edges going down and right. Your job is to find the maximum weight path through the graph, where the weight of a path is the sum of all the node weights on the path.
Unlike finding the minimum weight path, finding the maximum weight path is an "NP-Complete" problem -- its solution is exponential (or at least, that's the best we can do with current knowledge). However, if we simply view it as a dynamic program, we can solve it without worrying about its running time complexity. Spotting the recursion here is pretty easy. The maximum weight path to the cell at row r and column c is equal to the number apples in the cell, plus the maximum of the maximum weight path to the cell to the left, and the cell above the given cell. If we then find the maximum weight path to the lower right-hand cell, we will have found the maximum weight path through the graph.
The code is in apples1.cpp
```#include #include #include using namespace std; typedef vector IArray; class Apple { public: int rows; int cols; vector apples; int find_max(int r1, int c1); }; int Apple::find_max(int r, int c) { int a; int r1, r2; a = apples[r][c]; if (r == 0 && c == 0) return a; if (r == 0) return a + find_max(r, c-1); if (c == 0) return a + find_max(r-1, c); r1 = find_max(r, c-1); r2 = find_max(r-1, c); return (r1 > r2) ? a+r1 : a+r2; } main(int argc, char **argv) { int r, c; Apple a; if (argc != 3) { cerr << "usage: apples1 rows cols -- apples on standard input\n"; exit(1); } a.rows = atoi(argv[1]); a.cols = atoi(argv[2]); a.apples.resize(a.rows); for (r = 0; r < a.rows; r++) a.apples[r].resize(a.cols); for (r = 0; r < a.rows; r++) { for (c = 0; c < a.cols; c++) { cin >> a.apples[r][c]; if (cin.fail()) { cerr << "Not enough apples\n"; exit(1); } } } cout << a.find_max(a.rows-1, a.cols-1) << endl; } ```
We can see it working on some small examples:
```UNIX> cat a1.txt
5 10
6 4
UNIX> apples1 2 2 < a1.txt
19
UNIX> cat a2.txt
18 32 88
03 85 29
64 89 88
UNIX> apples1 3 3 < a2.txt
312
UNIX> calc 18+32+85+89+88
312.000000
UNIX>
```
Let's try a bigger example:
```UNIX> cat a3.txt
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
UNIX> apples1 10 20 < a3.txt
390
UNIX>
```
Is that right? The best path will be to take the top row all the way to the right and then drop down. That path will have a weight of (19*20)/2 + 20*10 = 390. Yes, that is right. Let's try it with two times the number of rows:
```UNIX> cat a3.txt a3.txt | apples1 20 20
```
It hangs, so we must memoize. Again, quite straightforward: (In apples2.cpp):
``` typedef vector IArray; int Apple::find_max(int r, int c) { int a; int r1, r2; int retval; if (cache[r][c] != -1) return cache[r][c]; a = apples[r][c]; if (r == 0 && c == 0) { retval = a; } else if (r == 0) { retval = a + find_max(r, c-1); } else if (c == 0) { retval = a + find_max(r-1, c); } else { r1 = find_max(r, c-1); r2 = find_max(r-1, c); if (r1 > r2) { retval = a+r1; } else { retval = a+r2; } } cache[r][c] = retval; return retval; } ```
```UNIX> cat a3.txt a3.txt | apples2 20 20
590
UNIX> cat a3.txt a3.txt a3.txt a3.txt a3.txt a3.txt a3.txt a3.txt a3.txt a3.txt | apples2 100 20
2190
```
Nice. In apples3.cpp, we remove the recursion. We do so by starting at the beginning of the cache and filling in to the higher values:
```int Apple::find_max() { int r1, r2; int retval; int r, c; cache[0][0] = apples[0][0]; for (r = 1; r < rows; r++) cache[r][0] = apples[r][0] + cache[r-1][0]; for (c = 1; c < cols; c++) cache[0][c] = apples[0][c] + cache[0][c-1]; for (r = 1; r < rows; r++) { for (c = 1; c < cols; c++) { r1 = cache[r][c-1]; r2 = cache[r-1][c]; if (r1 > r2) { cache[r][c] = apples[r][c]+r1; } else { cache[r][c] = apples[r][c]+r2; } } } return cache[rows-1][cols-1]; } ```
As with the maximum subsequence problem, we can reduce the cache size to two rows. I won't do it here -- see if you can do it yourself! | 1,837 | 5,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-47 | latest | en | 0.849535 |
https://housemetric.co.uk/house-price-analysis/CA4-0/Cotehill | 1,718,700,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00705.warc.gz | 269,825,331 | 12,303 | # House prices in CA4 0 (Cotehill)
This article reveals price per square metre data and various charts to help you understand current housing market in Cotehill, Scotby (CA4 0).
## Defining 'CA4 0'
This analysis is limited to properties whose postcode starts with "CA4 0", this is also called the postcode sector. It is shown in red on the map above. There are no official names for postcode sectors so I've just labelled it Cotehill.
You can click on the map labels to change to a neighbouring sector, or you can enter a different postcode sector (e.g. CM23 4) below.
FYI, a postcode sector is the full postcode without the last two letters.
## Price per square metre
Knowing the average house price in Cotehill is not much use. However, knowing average price per square metre can be quite useful. Price per sqm allows some comparison between properties of different size. We define price per square metre as the sold price divided by the internal area of a property:
£ per sqm = price ÷ internal area
E.g. Walla Crag, Southwaite Road, Cotehill, Scotby, sold for £225,000 on Mar-2024. Given the internal area of 67 square metres, the price per sqm is £3,358.
England & Wales have been officially metric since 1965. However house price per square foot is prefered by some estate agents and those of sufficiently advanced age ;-) They may want to convert square meters on this page to square feet.
The chart below is called a histogram, it helps you see the distribution of this house price per sqm data. To make this chart we put the sales data into a series of £ per sqm 'buckets' (e.g. £2,800-£3,000, £3,000-£3,200, £3,200-£3,400 etc...) we then count the number of sales with within in each bucket and plot the results. The chart is based on 40 sales in Cotehill (CA4 0) that took place in the last two years.
##### Distribution of £ per sqm for Cotehill, Scotby
Distribution of £ per sqm house prices in Cotehill
You can see the spread of prices above. This is because although internal area is a key factor in determining valuation, it is not the only factor. Many factors other than size affect desirability; these factors could be condition, aspect, garden size, negotiating power of the vendor etc.
The spread of prices will give you a feel of the typical range to expect in Cotehill (CA4 0). Of the 40 transactions, half were sold for between £2,280 and £3,130 per square metre. The median, or 'middle', price per square metre in 'CA4 0' is £2,680. Notably, only 25% of properties that sold recently were valued at more than £3,130 sqm. For anything to be valued more than this means it has to be more desireable than the clear majority of homes.
## Price map for Cotehill, Scotby
Do have a look at the interactive price map I created. I find it useful and I am sure it will help you in exploring Scotby. You can zoom in all the way to individual properties and then all the way back out to see the whole country. The colours show the current estimated property values.
House price heatmap for Cotehill, Scotby
## Comparison with neighbouring postcode sectors
The table below shows how 'CA4 0' compares to neighbouring postcode sectors.
Postcode sector Lower quartile Middle quartile Upper quartile
CA4 8 Scotby £2,090 sqm £2,470 sqm £2,850 sqm
CA4 0 Cotehill £2,280 sqm £2,680 sqm £3,130 sqm
CA4 9 Armathwaite £2,640 sqm £2,920 sqm £3,150 sqm
## Will Scotby house prices drop in 2024?
I cannot tell the future and don't believe anyone who says they can. I can however plot price trends - I have done this in the chart below for CA4 0 (Cotehill) compared with both the wider area CA4 and inflation (CPIH from the Office of National Statistics). The dashed trend lines in the chart show the average over time.
##### Historic price per square metre in Cotehill,Scotby
House price trends for Cotehill
For the most recent sales activity, rather than a summarized average, it is better to see the underlying data. This is shown in the chart below, where blue dots represent individual sales, click on them to see details. If there is an obvious trend you should be able to spot it here amid the noise from outliers.
##### Most recent CA4 0 sales
Recent trends for Cotehill
Data from Land Registry comes in gradually over time. I update it every month but it takes about 5 months for the majority of sales for Scotby to be recorded. Disclaimer: I do not verify and cannot guarantee the accuracy of any data shown. Outliers exist in the data, typically these are where the EPC registry records the internal area incorrectly, sometimes although very rarely the Land Registry price paid data can be wrong. The data provided throughout this website about Scotby and any other area, is not financial advice. Any information provided does not and cannot ever take in to account the particular financial situation, objectives or property needs of either you or anyone reading this information.
## Street level data
Street Avg size Avg £sqm Recent sales
Front Street, Cotehill, CA4 0D 103 sqm £1,863 6
Scalesceugh Villas, Cotehill, CA4 0B 81 sqm £3,790 5
Barrock Park, Cotehill, CA4 0J 77 sqm £2,322 5
Coopers Close, Cotehill, CA4 0J 88 sqm £2,584 5
## Raw data
Our analysis of Cotehill is derived from what is essentially a big table of sold prices from Land Registry with added property size information. Below are three rows from this table to give you an idea. | 1,337 | 5,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-26 | latest | en | 0.925404 |
https://ropnerssnaric.web.app/360.html | 1,656,815,913,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00769.warc.gz | 533,350,860 | 5,695 | # Nvideo lecture on parallel algorithms book pdf
Parallel systems a parallel system is a parallel algorithm plus a specified parallel architecture. The results of our gpu algorithm were obtained on a nvidia gpu gtx480 with. Preface this rep ort con tains the lecture notes used b. Introduction to parallel algorithms and architectures. Part of the lecture notes in computer science book series lncs, volume 11657.
About this book there is a software gap between hardware potential and the performance that can. When a thread encounters a parallel algorithm, it spreads the work. Scalability is the ability of a parallel system to take. Free computer, mathematics, technical books and lecture notes, etc. Throughout our presentation, we use the following terminology. This is a draft of a paper that will appear in acms computing surveys in the 50thaniversary issue, and is a condensed version of a chapter that will appear in the crc handbook on computer science. The number of processors is denoted with pn, also dependent on the input. Pipelines are often used to process data that stream into an application such as video or audio frames, or financial data. One of our primary measures of goodness of a parallel system will be its scalability. Find materials for this course in the pages linked along the left. Design and analysis of parallel algorithms murray cole e mail. The aim of this book is to provide a rigorous yet accessible treatment of parallel algorithms, including theoretical models of parallel computation, parallel algorithm design for homogeneous and heterogeneous platforms, complexity and performance analysis, and fundamental. In computer science, a parallel algorithm, as opposed to a traditional serial algorithm, is an algorithm which can do multiple operations in a given time.
Lecture notes in computer science including subseries lecture. The emphasis is on the application of the pram parallel random access machine model of parallel computation, with all its variants, to algorithm analysis. The resource consumption in parallel algorithms is both processor cycles on each processor and also the communication overhead between the processors. This tutorial provides an introduction to the design and analysis of. What are some good books to learn parallel algorithms. Algorithms in which several operations may be executed simultaneously are referred to as parallel algorithms. Parallel algorithms the parallel algorithms usually divide the problem into more symmetrical or asymmetrical subproblems and pass them to many processors and put the results back together at one end.
Written by an authority in the field, this book provides an introduction to the design and analysis of parallel algorithms. Mathematical abstraction in a simple programming tool for parallel embedded systems. Parallel algorithms and data structures for interactive data. Parallel algorithms pram p processors, each with a ram, local registers global memory of m locations each processor can in one step do a ram op or readwrite to one global memory location synchronous parallel steps various con. Reference book for parallel computing and parallel algorithms. Lecture notes introduction to algorithms electrical. An introduction to the thrust parallel algorithms library. These algorithms are well suited to todays computers, which basically perform operations in a sequential fashion. Parallel algorithms, pa study materials, engineering class handwritten notes, exam notes, previous year questions, pdf free download. Some important concepts date back to that time, with lots of theoretical activity between 1980 and 1990. Unlike sequential algorithms, parallel algorithms cannot be analyzed very well in isolation. Course notes parallel algorithms wism 459, 20192020.
Parallel algorithm video tutorial by ayush agrawal. In the field of computer science, we have mainly two types of algorithms, i. Segmented scan and related primitives also provide the necessary support for the atten. The total time total number of parallel steps is denoted with tn and it is a function of the input size n. Parallel algorithms two closely related models of parallel computation. This paper presents a parallel system for processing streaming video from. Parallel algorithms, pa study materials, engineering class handwritten notes, exam notes, previous year questions, pdf. Arrays trees hypercubes provides an introduction to the expanding field of parallel algorithms and architectures. Also wanted to know that from which reference book or papers are the concepts in the udacity course on parallel computing taught the history of parallel computing goes back far in the past, where the current interest in gpu computing was not yet predictable. These are lecture notes, homework questions, and exam questions from algorithms courses the author taught at the university of illinois. Cs 1762fall, 2011 2 introduction to parallel algorithms 1. In particular, attention must be paid to the division of work among the different processors solving a problem in parallel and to the communication between them. Slides 1922 presenting a ptas for parallel machine scheduling were skipped and are not examinable.
These algorithms are well suited to todays computers, which basically perform operations in a. On the other hand, important algorithms, such as graph algorithms, dynamic programming, and finitestate machine processing involve fine. Analysis of relationship between simdprocessing features used in nvidia gpus and nec sxaurora tsubasa vector. It has been a tradition of computer science to describe serial algorithms in abstract machine models, often the one known as randomaccess machine. A read is counted each time someone views a publication summary such as the title, abstract, and list of authors, clicks on a figure, or views or downloads the fulltext. If have the pdf link to download please share with me.
E cient parallel scan algorithms for gpus shubhabrata sengupta university of california, davis mark harris michael garland nvidia corporation abstract scan and segmented scan algorithms are crucial building blocks for a great many data parallel algorithms. Most popular books for data structures and algorithms for free downloads. Introduction to parallel algorithms covers foundations of parallel computing. These paradigms make it possible to discover and exploit the parallelism inherent in many classical graph problems. Suitable parallel algorithms and systems software are needed to realise the capabilities of parallel computers. Focusing on algorithms for distributedmemory parallel architectures, this book presents a. Oct 06, 2017 parallel algorithms by henri casanova, et al. The aim of this book is to provide a rigorous yet accessible treatment of parallel algorithms, including theoretical models of parallel computation, parallel algorithm design for homogeneous and heterogeneous platforms, complexity and performance analysis, and fundamental notions of.
It features a systematic approach to the latest design techniques, providing analysis and implementation details for each parallel algorithm described in the book. Therefore, the applications of parallel clustering algorithms and the clustering algorithms for parallel. If youre looking for a free download links of algorithms and parallel computing pdf, epub, docx and torrent then this site is not for you. Hello everyone i need notes or a book of parallel algorithm for preparation of exam. Focusing on algorithms for distributedmemory parallel architectures, parallel algorithms presents a rigorous yet accessible treatment of theoretical models of parallel computation, parallel algorithm design for homogeneous and heterogeneous platforms, complexity and performance analysis, and essential notions of scheduling. We abandon attempts to force sequential algorithms into parallel environments for such attempts usually result in transforming a good uniprocessor algorithm into ahopclcssly greecly parallel algorithm. Given the potentially prohibitive cost of manual parallelization using a lowlevel. This course is ab out distributed algorithms distributed algorithms include a wide range of parallel algorithms whic h can b e classied b yav ariet y of attributes in terpro cess comm unication. What are the best video lectures on algorithms to learn. E cient parallel scan algorithms for gpus shubhabrata sengupta university of california, davis mark harris michael garland nvidia corporation abstract scan and segmented scan algorithms are crucial building blocks for a great many dataparallel algorithms. Parallel algorithm models data parallel each task performs similar operations on different data typically statically map tasks to processes task graph use task dependency graph to promote locality or reduce interactions masterslave one or more master processes generating tasks allocate tasks to slave processes. Distributed algorithms lecture notes for f all nancy a lync h boaz p attshamir jan uary. Most of todays algorithms are sequential, that is, they specify a sequence of steps in which each step consists of a single operation.
Nptel provides elearning through online web and video courses various streams. Circuits logic gates andornot connected by wires important measures number of gates depth clock cycles in synchronous circuit pram p processors, each with a ram, local. A few papers were also covered, that i personally feel give some very important and useful techniques that should be in the toolbox of every algorithms researcher. Principles of parallel algorithm design concurrent and mul isbn. Analysis of parallel algorithms is usually carried out under the assumption that an unbounded number of processors is available. This note concentrates on the design of algorithms and the rigorous analysis of their efficiency. In traditional algorithms, we execute them in a single device, from starting to end while in parallel algorithms, algorithms are executed on different devices and then combined to get the final result.
This book focuses on parallel computation involving the most popular network architectures, namely, arrays, trees, hypercubes, and some closely related networks. Download algorithms and parallel computing pdf ebook. Optimization book by papadimitriou and steiglitz, as well as the network flow book by ahuja, magnanti and orlin and the edited book on approximation algorithms by hochbaum. Devising algorithms which allowmany processorsto work collectively to solve the same problems, butfaster biggermore re nedproblems in the same time. Parallel algorithms are highly useful in processing huge volumes of data in quick time. Circuits logic gates andornot connected by wires important measures number of gates depth clock cycles in synchronous circuit pram p processors, each with a ram, local registers global memory of m locations. Principles of parallel algorithm design concurrent and mul on a figure, or views or downloads the fulltext. Free algorithm books for download best for programmers. The presentation of coles parallel merge sort algorithm is taken from the book by gibbons and. Focusing on algorithms for distributedmemory parallel architectures, this book presents. The subject of this chapter is the design and analysis of parallel algorithms. Scalable collation and presentation of callpath profile data with cube.
1036 1106 157 54 525 1118 1404 697 617 809 909 923 491 731 893 984 383 1106 122 1561 519 287 375 844 1394 1311 1066 209 834 1356 49 737 390 322 715 1193 848 1464 | 2,152 | 11,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | longest | en | 0.922551 |
https://www.30daygmatsuccess.com/gmat-skills-in-real-life-topic-1/ | 1,726,260,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00413.warc.gz | 595,382,276 | 6,442 | # GMAT Skills in Real Life: Topic 1
Supposedly more than a test of basic math and grammar skills, the GMAT claims to accurately assess your ability to:
– interpret graphic data
– reason quantitatively
– recognize which information is relevant
– reason and evaluate arguments
– analyze information from a variety of sources
– develop strategies and make decisions
Clearly, these are all skills that are useful in business school and corporate America, but most GMAT test-takers still wonder about the connection between the test and reality. They view it as an unfair and pointless barrier blocking the way to the next phase of their lives. And they certainly don’t take their studies and test results as an indication of their business acumen. Nor should they!
Nonetheless, it can be interesting to think about how GMAT questions might reflect real life concerns. Take percent change for example. A common GMAT trap on both quantitative questions and critical reasoning is to present information in percentages and draw a conclusion based on concrete numbers. Doing so can be misleading.
Say that the price of a stock decreases 50 percent over the last year and then rebounds 80 percent this year. You may think you’ve earned money on your investment. However, if you invested \$1,000 dollars and the price decreased to \$500, 80 percent of \$500 is \$400, so you now have only \$900. You lost money.
Likewise, if your \$1,000 investment increases by 200 percent, you now have \$3,000 but if it then decreases by 50 percent, you drop all the way back down to \$1,500, not to \$2,500 (sell while it’s hot!). If you’ve read the chapter of my book on assigning values, you might understand better now why it can be important to do so. Percent change can be very deceptive.
As a side note, don’t forget that 200 percent of and 200 percent more than are calculated differently. While of signifies multiplication, more than signifies addition. So 1,000 is 200 percent of 500 and 1,500 is 200 percent more than 500. That’s quite a discrepancy if you are negotiating the salary increase you hope to obtain once you finish your MBA studies!
Updated: | 475 | 2,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.93408 |
http://math.stackexchange.com/questions/176093/proof-that-if-s-n-leq-t-n-for-n-geq-n-then-liminf-n-rightarrow-infty | 1,462,487,094,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861623301.66/warc/CC-MAIN-20160428164023-00199-ip-10-239-7-51.ec2.internal.warc.gz | 181,328,656 | 19,012 | # Proof that if $s_n \leq t_n$ for $n \geq N$, then $\liminf_{n \rightarrow \infty} s_n \leq \liminf_{n \rightarrow \infty} t_n$
This is half of Theorem 3.19 from Baby Rudin. Rudin claims the proof is trivial. What I've come up with so far doesn't seem trivial, however, and is probably also wrong (my problem with it is pointed out below). This makes me wonder whether I'm overlooking some useful fact and/or using an unprofitable approach.
Theorem. If $s_n \leq t_n$ for $n \geq N$, where $N$ is fixed, then
$$\liminf_{n \rightarrow \infty} s_n \leq \liminf_{n \rightarrow \infty} t_n.$$
Proof. Suppose that $s_n \leq t_n$ if $n \geq N$, but that
$$\liminf_{n \rightarrow \infty} s_n > \liminf_{n \rightarrow \infty} t_n.$$
Let $E_s$ denote the set of all subsequential limits of $\{s_n\}$, and let $E_t$ denote the set of all subsequential limits of $\{t_n\}$. Then
$$\inf E_s > \inf E_t.$$
This implies that there exists some $x \in E_t$ such that $\inf E_s > x > \inf E_t$, since otherwise $\inf E_t$ would not be the greatest lower bound of $E_t$. Hence some subsequence of $\{t_n\}$, say $\{t_{n_i}\}$, converges to $x < \inf E_s$.
Lemma. (from Rudin) If $x < \liminf_{n \rightarrow \infty} s_n$, then there exists an integer $N$ such that if $n \geq N$, then $s_n > x$.
By the lemma, there exists an integer $N_0$ such that if $n \geq N_0$, then $s_n > x$.
Now, let $\epsilon = \inf_{n \geq N_0} \{s_n - x\}$. This is where I think my proof breaks down. Can't $\epsilon$ be zero? Then, since $\{t_{n_i}\}$ converges to $x$, there exists an integer $N_1$ such that if $n_i \geq N_1$, then $|t_{n_i} - x| < \epsilon$. But this means that, if $n_i \geq \max\{N, N_0, N_1\}$, we have both
$$s_{n_i} > t_{n_i},$$
as well as $s_{n_i} \geq t_{n_i}$, a contradiction.
-
I've been there ; sometimes reading proofs in analysis that are considered "trivial" makes it a little harsh on yourself because you can't see why it is straight-forward when you don't have experience with some of the tools. Actually in this case it's straightforward, if you draw a little picture, you shouldn't see any problem believing the result, and the reason why it is so easy to believe is if you think about the proof given by copper.hat, which is pretty standard ; when limits exists, for inequalities it makes sense to "evaluate limits on both sides" ; the same remains true for liminfs. – Patrick Da Silva Jul 28 '12 at 7:01
The idea is just that "if I'm always smaller than this guy for every $n$, then I'm not gonna be bigger than him no matter how big $n$ grows". Letting $n$ "grow" is essentially taking limits. I wanted to give you an intuitive reasoning behind the proof ; see copper.hat's answer for details. – Patrick Da Silva Jul 28 '12 at 7:02
I interpret 'trivial' or 'obvious' as 'it has been proved'. – copper.hat Jul 28 '12 at 7:18
i don't quite get why $\inf E_s > \inf E_t$, i think it may not true in general. – Mathematics Jul 28 '12 at 8:22
Thanks for the encouragement and intuition, Patrick. – Jefferson Huang Jul 28 '12 at 13:12
You are making it too hard for yourself.
By definition, $\liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k\geq n} a_k$. Also, notice that the sequence $\inf_{k\geq n} a_k$ is non-decreasing.
So, if $s_n \leq t_n$, then clearly $\inf_{k\geq n} s_k \leq t_n$, for any $n$. From this it follows that $\inf_{k\geq n} s_k \leq \inf_{k\geq n} t_k$, for any $n$ (if not, then $\inf_{k\geq n} s_k > t_{k'}$, for some $k' \geq n$, which is an immediate contradiction).
Since both sides are non-decreasing, we have $\inf_{k\geq n} s_k \leq \lim_{n' \to \infty} \inf_{k\geq n'} t_k = \liminf_{n \to \infty} t_n$, and then we have $\lim_{n' \to \infty} \inf_{k\geq n'} s_k = \liminf_{n \to \infty} s_n \leq \liminf_{n \to \infty} t_n$, as desired.
-
Thanks, copper.hat! – Jefferson Huang Jul 28 '12 at 13:08
If $a = \lim \inf_{n \to \infty} a_n$ and $b = \lim_{n \to \infty} \inf_{k \geq n} a_k$ then the fact that $a=b$ is not simply by definition, at least not by the definition given in Rudin, and needs to be proven. By Rudin's definition $a$ is the infimum of the set of all subequential limits whereas $b$ is the convergence point of the sequence $\{t_n\}$ where $t_n = \inf_{k \geq n} a_k$. Intuitively I can see why these two are the same but I am finding a rigorious proof to be nontrivial. – kyp4 Aug 5 '15 at 1:55 | 1,453 | 4,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2016-18 | longest | en | 0.83144 |
https://manifold.markets/SimonGrayson/champions-league-semi-finals-which | 1,716,128,796,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00657.warc.gz | 331,159,898 | 34,647 | Champions League Semi Finals - which team will have the biggest winning margin?
31
101
380
resolved May 8
100%99.5%
Dortmund
0.0%
Bayern Munich
0.3%
0.1%
PSG
There are 4 teams remaining in the 2023/4 Champions League. The next stage is the semi finals.
Each tie will be played over two legs, with the first leg being played on 30th April and 1st May and the second leg being played on 7th/8th May.
Which of the 4 teams will have the biggest winning margin over the two legs?
Market notes:
• This will only count the winning margin in terms of goals without using away goals or any other kid of tie breaker. So a team who scores 10 and concedes 5 will be considered the same as a team who scores 5 and concedes 0.
• The market is only considering the overall margin across two legs rather than the results of the individual matches. So a team who loses 1-0 and then wins 6-0 has the same five goal winning margin as a team who won their games 2-0 and 3-0.
• If both semi-finals see the same winning margin, the market will resolve 50/50 to each of the two winners.
• Extra time and penalty shootouts aren't relevant to this market since a game only goes to extra time and penalties if it is tied after two legs.
Get Ṁ600 play money
#NameTotal profit
1Ṁ118
2Ṁ74
3Ṁ52
4Ṁ34
5Ṁ22
Sort by:
Dortmund won by 2 goals over two legs.
Madrid won by 1 goal over two legs.
This market resolves to Borussia Dortmund!
@SimonGrayson can resolve to Dortmund
bought Ṁ500 Dortmund YES
.
So if both ties go to extra time then the 4 teams each resolve at 25% being equal even if one match is decided by a single penalty whereas the other has a 5 goal difference in extra time?
@ChristopherRandles Yes - according to the rules as I wrote them, if both games go to extra time, no one had a winning margin above zero so it resolves 25/25/25/25.
I wrote the rules like that for the rounds with more than two games and it was meant to be more about saying that winning by one goal in normal time is better than winning by multiple goals in extra time… I should probably rewrite them for future tournaments! | 541 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-22 | latest | en | 0.975964 |
https://www.goconqr.com/mindmap/4268357/forces-and-motion | 1,556,145,545,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578663470.91/warc/CC-MAIN-20190424214335-20190425000335-00154.warc.gz | 709,000,284 | 10,744 | # Forces and motion
Mind Map by , created over 3 years ago
## IGCSE physics forces and motion
563
70
0
Created by Catarina Borges over 3 years ago
IGCSE Physics Heat Transfer
IGCSE Physics Mock Exam
physics mock exam
Physics circuit symbols
French Beginner
Forces and their effects
IGCSE Physics formulas
P2 Flash cards - motion
Forces and Motion Practice Test
Forces and motion
1 Speed
1.1 Formula speed=distance/time
1.2 Unit m/s
1.3 Velocity
1.3.1 Speed of something and its direction of travel
1.3.2 Velocity can be shown using an arrow ⟶
1.3.3 Velocity is a quantatie that may have direction as well as magnitude and so it is called a vector
2 Accelaration
2.1 average acceleration=change in velocity/time taken
2.2 Unit m/s²
2.3 negative acceleration is called deceleration
2.4 Uniform acceleration means constant acceleration
3 Motion Graphs
3.1 Distance-time graphs
3.1.1 Gradient of the line is numerically equal to the speed
3.1.2 Different objects can show if the object is at a steady speed or stationary
3.1.3 The graph can also show that the object is accelerating or deccelerating
3.2 Speed-time graphs
3.2.1 Gradient of the line is equal to the acceleration
3.2.2 Area under the line is equal to the distance travelled
3.3 Uniform and non-uniform acceleration
3.3.1 Uniform
3.3.2 Non-uniform
4 Free fall
4.1 Unit g
4.1.1 On earth 10m/s²
4.1.2 g decreases as the object is moved away from each
4.2 Acceleration of free fall- Same for all objects falling near the earth's surface, either light or heavy
4.3 When throwing a ball at the air...
4.3.1 Upward velocity of 30m/s is equal to downward velocity of -30m/s
4.3.2 Whatever the ball is travelling up or down it is gaining downward velocity at the rate of 10m/s (g)
5 Forces in balance
5.1 Force
5.1.1 Unit Newton(N)
5.1.2 Small forces can be measured using a spring balance. The greater the force the more the spring is stretched and higher it is the reading on the scale
5.1.3 Examples of different forces
5.1.3.1 Tension
5.1.3.2 Upthrust
5.1.3.3 Weight
5.1.3.4 Friction
5.1.3.5 Air resistance
5.1.4 If many forces in an object are in balance, they cancel each other out
5.1.4.1 With balanced forces on it, an object is either at rest, or moving at a steady velocity
5.1.5 Terminal velocity
5.1.5.1 Example: a skydiver falling from a hovering helicopter
5.1.5.1.1 As her speed increases, the air resistance on her also increases. Eventually, it is enough to balance her weight, and she gains no more speed. She is than at her terminal velocity.
5.2 Newton's first law of motion (if no external force is acting on it the object will)
5.2.1 if stationary, remain stationary
5.2.2 if moving, keep moving at a steady speed in a straight line
6 Force, mass and acceleration
6.1 Inertia
6.1.1 Resistance to a change in velocity or direction
6.1.2 The more mass something has, the more inertia it has
6.2 Resultant force
6.2.1 When two forces are unbalanced. Together, they are equivalent to a single force. This is called the resultant force
6.2.2 Resultant force is equal to zero when forces are balanced, and there is no acceleration
6.3 Resultant force= mass x acceleration
6.3.1 F=ma
6.3.2 Newton's second law of motion
7 Force, weight and gravity
7.1 Gravitattional force
7.1.1 All masses attract each other
7.1.2 The greater the masses, the stronger the force
7.1.3 The closer the masses, the stronger the force
7.1.4 Gravitational field strength (g)
7.1.4.1 Region in which a mass experiences a force due to gravitational attraction
7.1.4.2 weight= mass x g
7.1.4.2.1 w=mg
7.2 Weight
7.2.1 Measured in newtons
7.2.2 Near the earth's surface, an object of mass 1 kg has a weight of 9.8 N
7.2.3 Weight can change from place to place, but mass remains the same | 1,089 | 3,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-18 | latest | en | 0.847693 |
https://math.stackexchange.com/questions/687956/proving-left-lfloor-n-frac-log-b-log-a-right-rfloor-left-lfloor-fra | 1,621,246,896,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992159.64/warc/CC-MAIN-20210517084550-20210517114550-00153.warc.gz | 388,240,494 | 38,829 | # Proving $\left\lfloor n\frac{\log (b)}{\log (a)}\right\rfloor =\left\lfloor \frac{\log \left(b^n+1\right)}{\log (a)}\right\rfloor$
Inspired by this question, I'd like to know how one would go about proving the below more general equation?
$$n \in \mathbb{N},\;a \in \mathbb{N},\;b \in \mathbb{N}$$ $$b^n+1 \notin a^{\mathbb{N}+1}$$
$$\left\lfloor n\frac{\log (b)}{\log (a)}\right\rfloor =\left\lfloor \frac{\log \left(b^n+1\right)}{\log (a)}\right\rfloor$$
Empirically the above appears, prima facie, to be the case. If it's not, apologies, I tried quite a few values of $a$, $b$ and $n$ and came up with the above assumptions.
• Here's an idea: can you prove $\lfloor \log(b^n)\rfloor = \lfloor \log(b^n+1)\rfloor$? If yes, then you can use $\lfloor x/n\rfloor =\lfloor \lfloor x\rfloor /n\rfloor$ given certain conditions I don't remember exactly at the moment. It shouldn't be too difficult to prove it altogether. – Ian Mateus Feb 24 '14 at 1:44
• Hmm... what about $(a,b,n)=(2,3,1)$? ;-) Maybe you wanted to ask for $b^n+1$ not being a power of $a$? – Peter Košinár Feb 24 '14 at 2:25
• @PeterKošinár Oops, thanks for pointing that out! Fixed. – Max Feb 24 '14 at 2:41
Even more is actually true for positive integers $x$ and $a>1$, such that $(x+1)$ is not a power of $a$:
$$\left\lfloor\frac{\log x}{\log a}\right\rfloor = \left\lfloor\frac{\log (x+1)}{\log a}\right\rfloor$$
In order to see why this is so, let $m$ be the largest power of $a$ which doesn't exceed $x$. This means that we have $$a^m \leq x < a^{m+1}$$ and also $$m \leq \log_a x < m + 1$$ Since $\log_a x = \frac{\log x}{\log a}$, we have $$\left\lfloor\frac{\log x}{\log a}\right\rfloor = m$$
The right-hand side floor expression is clearly not smaller than $m$, so we only need to prove it cannot be larger either. Assume (for a proof by contradiction) that this was the case to get $$\frac{\log(x+1)}{\log a} \geq \left\lfloor \frac{\log(x+1)}{\log a}\right\rfloor \geq m+1$$ In the light of inequality $x<a^{m+1}$, the only possibility would be $x+1=a^{m+1}$. But that's precisely what we've forbidden by the not-a-power condition, thus completing our proof.
• Brilliant, thank you very much @Peter! – Max Feb 24 '14 at 2:58 | 775 | 2,214 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-21 | latest | en | 0.820915 |
https://manpag.es/f32/3+stgsy2.f | 1,591,302,428,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347458095.68/warc/CC-MAIN-20200604192256-20200604222256-00599.warc.gz | 421,832,035 | 6,332 | pdf
# stgsy2.f
stgsy2.f
## SYNOPSIS
Functions/Subroutines
subroutine stgsy2 (TRANS, IJOB, M, N, A, LDA, B, LDB, C, LDC, D, LDD, E, LDE, F, LDF, SCALE, RDSUM, RDSCAL, IWORK, PQ, INFO)
STGSY2
solves the generalized Sylvester equation (unblocked algorithm).
## Function/Subroutine Documentation
subroutine stgsy2 (character TRANS, integer IJOB, integer M, integer N, real, dimension( lda, * ) A, integer LDA, real, dimension( ldb, * ) B, integer LDB, real, dimension( ldc, * ) C, integer LDC, real, dimension( ldd, * ) D, integer LDD, real, dimension( lde, * ) E, integer LDE, real, dimension( ldf, * ) F, integer LDF, real SCALE, real RDSUM, real RDSCAL, integer, dimension( * ) IWORK, integer PQ, integer INFO)
STGSY2
solves the generalized Sylvester equation (unblocked algorithm).
Purpose:
STGSY2 solves the generalized Sylvester equation:
A * R - L * B = scale * C (1)
D * R - L * E = scale * F,
using Level 1 and 2 BLAS. where R and L are unknown M-by-N matrices,
(A, D), (B, E) and (C, F) are given matrix pairs of size M-by-M,
N-by-N and M-by-N, respectively, with real entries. (A, D) and (B, E)
must be in generalized Schur canonical form, i.e. A, B are upper
quasi triangular and D, E are upper triangular. The solution (R, L)
overwrites (C, F). 0 <= SCALE <= 1 is an output scaling factor
chosen to avoid overflow.
In matrix notation solving equation (1) corresponds to solve
Z*x = scale*b, where Z is defined as
Z = [ kron(In, A) -kron(B**T, Im) ] (2)
[ kron(In, D) -kron(E**T, Im) ],
Ik is the identity matrix of size k and X**T is the transpose of X.
kron(X, Y) is the Kronecker product between the matrices X and Y.
In the process of solving (1), we solve a number of such systems
where Dim(In), Dim(In) = 1 or 2.
If TRANS = ’T’, solve the transposed system Z**T*y = scale*b for y,
which is equivalent to solve for R and L in
A**T * R + D**T * L = scale * C (3)
R * B**T + L * E**T = scale * -F
This case is used to compute an estimate of Dif[(A, D), (B, E)] =
sigma_min(Z) using reverse communicaton with SLACON.
STGSY2 also (IJOB >= 1) contributes to the computation in STGSYL
of an upper bound on the separation between to matrix pairs. Then
the input (A, D), (B, E) are sub-pencils of the matrix pair in
STGSYL. See STGSYL for details.
Parameters:
TRANS
TRANS is CHARACTER*1
= ’N’, solve the generalized Sylvester equation (1).
= ’T’: solve the ’transposed’ system (3).
IJOB
IJOB is INTEGER
Specifies what kind of functionality to be performed.
= 0: solve (1) only.
= 1: A contribution from this subsystem to a Frobenius
norm-based estimate of the separation between two matrix
pairs is computed. (look ahead strategy is used).
= 2: A contribution from this subsystem to a Frobenius
norm-based estimate of the separation between two matrix
pairs is computed. (SGECON on sub-systems is used.)
Not referenced if TRANS = ’T’.
M
M is INTEGER
On entry, M specifies the order of A and D, and the row
dimension of C, F, R and L.
N
N is INTEGER
On entry, N specifies the order of B and E, and the column
dimension of C, F, R and L.
A
A is REAL array, dimension (LDA, M)
On entry, A contains an upper quasi triangular matrix.
LDA
LDA is INTEGER
The leading dimension of the matrix A. LDA >= max(1, M).
B
B is REAL array, dimension (LDB, N)
On entry, B contains an upper quasi triangular matrix.
LDB
LDB is INTEGER
The leading dimension of the matrix B. LDB >= max(1, N).
C
C is REAL array, dimension (LDC, N)
On entry, C contains the right-hand-side of the first matrix
equation in (1).
On exit, if IJOB = 0, C has been overwritten by the
solution R.
LDC
LDC is INTEGER
The leading dimension of the matrix C. LDC >= max(1, M).
D
D is REAL array, dimension (LDD, M)
On entry, D contains an upper triangular matrix.
LDD
LDD is INTEGER
The leading dimension of the matrix D. LDD >= max(1, M).
E
E is REAL array, dimension (LDE, N)
On entry, E contains an upper triangular matrix.
LDE
LDE is INTEGER
The leading dimension of the matrix E. LDE >= max(1, N).
F
F is REAL array, dimension (LDF, N)
On entry, F contains the right-hand-side of the second matrix
equation in (1).
On exit, if IJOB = 0, F has been overwritten by the
solution L.
LDF
LDF is INTEGER
The leading dimension of the matrix F. LDF >= max(1, M).
SCALE
SCALE is REAL
On exit, 0 <= SCALE <= 1. If 0 < SCALE < 1, the solutions
R and L (C and F on entry) will hold the solutions to a
slightly perturbed system but the input matrices A, B, D and
E have not been changed. If SCALE = 0, R and L will hold the
solutions to the homogeneous system with C = F = 0. Normally,
SCALE = 1.
RDSUM
RDSUM is REAL
On entry, the sum of squares of computed contributions to
the Dif-estimate under computation by STGSYL, where the
scaling factor RDSCAL (see below) has been factored out.
On exit, the corresponding sum of squares updated with the
contributions from the current sub-system.
If TRANS = ’T’ RDSUM is not touched.
NOTE: RDSUM only makes sense when STGSY2 is called by STGSYL.
RDSCAL
RDSCAL is REAL
On entry, scaling factor used to prevent overflow in RDSUM.
On exit, RDSCAL is updated w.r.t. the current contributions
in RDSUM.
If TRANS = ’T’, RDSCAL is not touched.
NOTE: RDSCAL only makes sense when STGSY2 is called by
STGSYL.
IWORK
IWORK is INTEGER array, dimension (M+N+2)
PQ
PQ is INTEGER
On exit, the number of subsystems (of size 2-by-2, 4-by-4 and
8-by-8) solved by this routine.
INFO
INFO is INTEGER
On exit, if INFO is set to
=0: Successful exit
<0: If INFO = -i, the i-th argument had an illegal value.
>0: The matrix pairs (A, D) and (B, E) have common or very
close eigenvalues.
Author:
Univ. of Tennessee
Univ. of California Berkeley
NAG Ltd.
Date:
December 2016
Contributors:
Bo Kagstrom and Peter Poromaa, Department of Computing Science, Umea University, S-901 87 Umea, Sweden.
Definition at line 276 of file stgsy2.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code.
pdf | 1,799 | 5,956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-24 | latest | en | 0.75586 |
https://www.airmilescalculator.com/distance/stl-to-bff/ | 1,627,569,095,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153860.57/warc/CC-MAIN-20210729140649-20210729170649-00210.warc.gz | 629,379,042 | 29,985 | # Distance between St Louis, MO (STL) and Scottsbluff, NE (BFF)
Flight distance from St Louis to Scottsbluff (St. Louis Lambert International Airport – Western Nebraska Regional Airport) is 730 miles / 1175 kilometers / 635 nautical miles. Estimated flight time is 1 hour 52 minutes.
Driving distance from St Louis (STL) to Scottsbluff (BFF) is 827 miles / 1331 kilometers and travel time by car is about 14 hours 21 minutes.
## Map of flight path and driving directions from St Louis to Scottsbluff.
Shortest flight path between St. Louis Lambert International Airport (STL) and Western Nebraska Regional Airport (BFF).
## How far is Scottsbluff from St Louis?
There are several ways to calculate distances between St Louis and Scottsbluff. Here are two common methods:
Vincenty's formula (applied above)
• 730.230 miles
• 1175.192 kilometers
• 634.553 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 728.642 miles
• 1172.635 kilometers
• 633.172 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A St. Louis Lambert International Airport
City: St Louis, MO
Country: United States
IATA Code: STL
ICAO Code: KSTL
Coordinates: 38°44′55″N, 90°22′12″W
B Western Nebraska Regional Airport
City: Scottsbluff, NE
Country: United States
IATA Code: BFF
ICAO Code: KBFF
Coordinates: 41°52′26″N, 103°35′45″W
## Time difference and current local times
The time difference between St Louis and Scottsbluff is 1 hour. Scottsbluff is 1 hour behind St Louis.
CDT
MDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 128 kg (282 pounds).
## Frequent Flyer Miles Calculator
St Louis (STL) → Scottsbluff (BFF).
Distance:
730
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
730
Round trip? | 534 | 2,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.823887 |
http://www.merlot.org/merlot/materials.htm?materialType=Presentation&category=2513&nosearchlanguage=&nosearchlanguage=&sort.property=dateCreated | 1,411,464,392,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657138086.23/warc/CC-MAIN-20140914011218-00040-ip-10-234-18-248.ec2.internal.warc.gz | 659,205,744 | 26,803 | Search Materials
Sort materials by
Search Materials
Get more information on the MERLOT Editors' Choice Award in a new window.
Get more information on the MERLOT Classics Award in a new window.
Go to Search Page
Go to Partner Benefits
View material results for all categories
Search all MERLOT
Select to go to your workspace
Search Terms
Please give at least one keyword of at least three characters for the search to work with. The more keywords you give, the better the search will work for you.
select OK to launch help window
cancel help
```
```
Materials 1 - 10 shown of 751 results
### Lesson 5 - Polynomials
This lesson received an honorable mention in the 2014 SoftChalk Lesson Challenge.'We have seen quadratic functions which... see more
Material Type: Presentation
Author: Phillip Clark
Date Added: May 05, 2014 Date Modified: May 05, 2014
Peer Review for material titled "Lesson 5 - Polynomials"
User Rating for material titled "Lesson 5 - Polynomials"
Member Comments for material titled "Lesson 5 - Polynomials"
Personal Collections for material titled "Lesson 5 - Polynomials"
Learning Exercises for material titled "Lesson 5 - Polynomials"
Accessibility Info for material titled "Lesson 5 - Polynomials"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Personal Collections (none)
Learning Exercises (none)
Accessibility Info (none)
### Section 3 - Introduction to Linear Functions
This lesson won an honorable mention in the 2014 SoftChalk Lesson Challenge.This lesson is an introduction to linear... see more
Material Type: Presentation
Author: Jan Spinato
Date Added: May 05, 2014 Date Modified: May 05, 2014
Peer Review for material titled "Section 3 - Introduction to Linear Functions"
User Rating for material titled "Section 3 - Introduction to Linear Functions"
Member Comments for material titled "Section 3 - Introduction to Linear Functions"
Personal Collections for material titled "Section 3 - Introduction to Linear Functions"
Learning Exercises for material titled "Section 3 - Introduction to Linear Functions"
Accessibility Info for material titled "Section 3 - Introduction to Linear Functions"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Personal Collections (none)
Learning Exercises (none)
Accessibility Info (none)
### MAT 155 Statistical Analysis - What's the Message?
This lesson received an honorable mention in the 2014 SoftChalk Lesson Challenge.'Have you ever wondered if we're really... see more
Material Type: Presentation
Author: Lisa Cooley
Date Added: May 05, 2014 Date Modified: May 05, 2014
Peer Review for material titled "MAT 155 Statistical Analysis - What's the Message?"
User Rating for material titled "MAT 155 Statistical Analysis - What's the Message?"
Member Comments for material titled "MAT 155 Statistical Analysis - What's the Message?"
Personal Collections for material titled "MAT 155 Statistical Analysis - What's the Message?"
Learning Exercises for material titled "MAT 155 Statistical Analysis - What's the Message?"
Accessibility Info for material titled "MAT 155 Statistical Analysis - What's the Message?"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Personal Collections (none)
Learning Exercises (none)
Accessibility Info (none)
### MANOVA
MANOVA is the sixth volume in a series of Advanced Educational Statistics. A straightforward, clear teaching style for... see more
Material Type: Presentation
Date Added: Jan 06, 2014 Date Modified: Mar 15, 2014
Peer Review for material titled "MANOVA"
User Rating for material titled "MANOVA"
Member Comments for material titled "MANOVA"
Personal Collections for material titled "MANOVA"
Learning Exercises for material titled "MANOVA"
Accessibility Info for material titled "MANOVA"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Learning Exercises (none)
Accessibility Info (none)
### Multiple Linear Regression
The fourth volume in a series of Advanced Educational Statistics, Multiple Linear Regression builds on the innovative... see more
Material Type: Presentation
Date Added: Jan 06, 2014 Date Modified: Mar 15, 2014
Peer Review for material titled "Multiple Linear Regression"
User Rating for material titled "Multiple Linear Regression"
Member Comments for material titled "Multiple Linear Regression"
Personal Collections for material titled "Multiple Linear Regression"
Learning Exercises for material titled "Multiple Linear Regression"
Accessibility Info for material titled "Multiple Linear Regression"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Learning Exercises (none)
Accessibility Info (none)
### Non-Parametric Analyses
Non-Parametric Analyses is the seventh volume in a series of Advanced Educational Statistics. A straightforward, clear... see more
Material Type: Presentation
Date Added: Jan 06, 2014 Date Modified: Mar 15, 2014
Peer Review for material titled "Non-Parametric Analyses"
User Rating for material titled "Non-Parametric Analyses"
Member Comments for material titled "Non-Parametric Analyses"
Personal Collections for material titled "Non-Parametric Analyses"
Learning Exercises for material titled "Non-Parametric Analyses"
Accessibility Info for material titled "Non-Parametric Analyses"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Learning Exercises (none)
Accessibility Info (none)
### t-Test & ANOVA
The fifth volume in a series of Advanced Educational Statistics, t-Test & ANOVA bulids on the innovative teaching... see more
Material Type: Presentation
Date Added: Jan 06, 2014 Date Modified: Mar 15, 2014
Peer Review for material titled "t-Test & ANOVA"
User Rating for material titled "t-Test & ANOVA"
Member Comments for material titled "t-Test & ANOVA"
Personal Collections for material titled "t-Test & ANOVA"
Learning Exercises for material titled "t-Test & ANOVA"
Accessibility Info for material titled "t-Test & ANOVA"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Learning Exercises (none)
Accessibility Info (none)
### Correlational Analysis
Finally, a clear, demystified approach to learning educational statistics. A narrative, along with videos, output data sets... see more
Material Type: Presentation
Author: Rusty Waller, Rusty Waller
Date Added: Jan 05, 2014 Date Modified: Jan 06, 2014
Peer Review for material titled "Correlational Analysis"
User Rating for material titled "Correlational Analysis"
Member Comments for material titled "Correlational Analysis"
Personal Collections for material titled "Correlational Analysis"
Learning Exercises for material titled "Correlational Analysis"
Accessibility Info for material titled "Correlational Analysis"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Learning Exercises (none)
Accessibility Info (none)
### Linear Regression
The third volume in a series of Advanced Educational Statistics, Linear Regression maintains the no nonsense teaching... see more
Material Type: Presentation
Date Added: Jan 04, 2014 Date Modified: Mar 15, 2014
Peer Review for material titled "Linear Regression"
User Rating for material titled "Linear Regression"
Member Comments for material titled "Linear Regression"
Personal Collections for material titled "Linear Regression"
Learning Exercises for material titled "Linear Regression"
Accessibility Info for material titled "Linear Regression"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Learning Exercises (none)
Accessibility Info (none)
### Introduction to Quantitative Statistics: Terms & Definitions, Vol. 2
Terms & Definitions: Part 2, Volume 2, the second volume in a series of Introduction to Quantitative Statistics, provides... see more
Material Type: Presentation
Date Added: Jan 04, 2014 Date Modified: Mar 15, 2014
Peer Review for material titled "Introduction to Quantitative Statistics: Terms & Definitions, Vol. 2"
User Rating for material titled "Introduction to Quantitative Statistics: Terms & Definitions, Vol. 2"
Member Comments for material titled "Introduction to Quantitative Statistics: Terms & Definitions, Vol. 2"
Personal Collections for material titled "Introduction to Quantitative Statistics: Terms & Definitions, Vol. 2"
Learning Exercises for material titled "Introduction to Quantitative Statistics: Terms & Definitions, Vol. 2"
Accessibility Info for material titled "Introduction to Quantitative Statistics: Terms & Definitions, Vol. 2"
Reviewed by members of Editorial board for inclusion in MERLOT.
Editor Review (not reviewed)
User Rating (not rated)
Discussion (none)
Personal Collections (none)
Learning Exercises (none)
Accessibility Info (none)
Results page 1 of 76
2 3 4 5 6 7 8 9 10 Next
``` | 2,004 | 9,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2014-41 | latest | en | 0.858 |
https://www.numbersaplenty.com/26676686730 | 1,657,100,070,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00477.warc.gz | 942,181,581 | 3,552 | Search a number
26676686730 = 2351316787961
BaseRepresentation
bin11000110110000011…
…011110111110001010
32112212010220002121110
4120312003132332022
5414113212433410
620131033551150
71633034044611
oct306603367612
975763802543
1026676686730
111034a326641
125205b754b6
132691a0c545
141410d20878
15a61ebd720
hex6360def8a
26676686730 has 32 divisors (see below), whose sum is σ = 64512790848. Its totient is φ = 7059478400.
The previous prime is 26676686723. The next prime is 26676686731. The reversal of 26676686730 is 3768667662.
It is a super-2 number, since 2×266766867302 (a number of 22 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (26676686731) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 3390051 + ... + 3397910.
It is an arithmetic number, because the mean of its divisors is an integer number (2016024714).
Almost surely, 226676686730 is an apocalyptic number.
26676686730 is an abundant number, since it is smaller than the sum of its proper divisors (37836104118).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
26676686730 is a wasteful number, since it uses less digits than its factorization.
26676686730 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 6788102.
The product of its (nonzero) digits is 18289152, while the sum is 57.
The spelling of 26676686730 in words is "twenty-six billion, six hundred seventy-six million, six hundred eighty-six thousand, seven hundred thirty". | 510 | 1,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-27 | latest | en | 0.854499 |
https://www.onlinemathlearning.com/venn-diagram-2.html | 1,721,826,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00099.warc.gz | 793,142,409 | 9,460 | # Venn Diagrams
Venn diagrams are the principal way of illustrating sets diagrammatically. The method consists primarily of entering the elements of a set into a circle or ovals.
### Venn Diagrams With Three Circles
Learn draw and interpret Venn diagrams with three circles or subsets (also called a 3-way Venn Diagram).
In a class of 50 students,
23 students speak Spanish,
25 students speak French and
20 students speak Italian.
7 students speak both French and Spanish,
9 students speak both French and Italian,
and 11 students speak both Spanish and Italian,
4 students speak neither of these languages.
### Venn Diagram GCSE Maths Revision
Exam Paper Practice & Help
(a) Use this information to fill in the Venn Diagram below.
In a class there are 32 students
23 take History.
15 take French.
6 do not take either of these subjects.
(b) How many students take French but not History?
### Solving Problems With Venn Diagrams
This video solves two problems using Venn Diagrams. One example with two sets and another example with three sets.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 267 | 1,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.920387 |
mountainflying.com | 1,501,044,366,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425766.58/warc/CC-MAIN-20170726042247-20170726062247-00362.warc.gz | 219,940,856 | 3,702 | # Proper Approach Airspeed
#### The concept of proper airspeed control for the approach to a mountain airstrip cannot be over emphasized.
The landing distance will increase by the square of the ratio of the touchdown speed to the normal touchdown speed.
In this example, we use 55 for the actual touchdown speed divided by the normal touchdown speed. This ratio is 1.1 or a 10-percent increase in speed. The landing distance increased by the square of the ratio or 21 percent increase in landing distance. You might think, that's not too bad. I only need 1,200 feet to land, so what's another 252 feet (1,200 x 1.21 = 1,452)?
Once you start believing, "What's another 252 feet," it displays the beginning of complacency. Or, maybe you decide to add an extra 5 knots for the wife, another 5 knots for the kids and another 5 knots for the dog. In this example, we change the approach speed to 70. The ratio become 1.4, and squaring the ratio means there is a 96-percent increase in landing distance. Ask yourself, do I want to nearly double my landing distance on a short mountain airstrip? If not, use exact airspeed control. | 266 | 1,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-30 | latest | en | 0.931519 |
https://www.teachoo.com/2133/579/Example-9---Let-A-=--1--2---B-=--3--4-.-Find-number-of-relations/category/Examples/ | 1,527,383,911,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00252.warc.gz | 842,369,140 | 13,243 | 1. Chapter 2 Class 11 Relations and Functions
2. Serial order wise
Transcript
Example 9, Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B. Given A = {1,2} & B = {3,4} Number of relations from A to B = 2Number of elements in A × B = 2Number of elements in set A × Number of elements in set B = 2n(A) × n(B) Number of elements in set A = 2 Number of elements in set B = 2 Number of relations from A to B = 2n(A) × n(B) = 22 × 2 = 24 = 2 × 2 × 2 × 2 = 16 | 172 | 476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-22 | longest | en | 0.881976 |
http://answers.google.com/answers/threadview/id/580852.html | 1,409,556,639,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535917463.1/warc/CC-MAIN-20140901014517-00074-ip-10-180-136-8.ec2.internal.warc.gz | 23,050,858 | 3,104 | View Question
Q: coefficient of static friction ( No Answer, 1 Comment )
Question
Subject: coefficient of static friction Category: Science Asked by: mersedeh-ga List Price: \$2.00 Posted: 16 Oct 2005 01:37 PDT Expires: 26 Oct 2005 02:05 PDT Question ID: 580852
```The coefficient of static friction between the m = 2.60 kg crate and the 35.0° incline of Figure P4.41 is 0.290. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?```
```Whew, this should be easier to imagine! (Just been to the Science-Astronomy page. Just burned me brain imagining them burstars, galactic plane, Myr-periods, Oort Cloud. Zeez, t'was Chaos Cloud, now Oort Cloud. Maybe if I did not escape from that page, I'd encounter Gungbung Clouds, whatever they are.) So Let me imagine that Figure P4.41. (It should be in a book or something in this planet of ours.) Maybe the 2.60 kg crate is not being controlled by a rope being pulled by a pulley or hand. Maybe the crate is left by itself to slide down the incline. And if the crate tends to slide down, a force F, perpendicular to the incline, must be applied on the crate to stop the motion. And you are asking what minimum F is needed. Well, without F yet, only the friction force due to the component perpendicular to the incline of the crate's weight is there to prevent the sliding down of the crate. And the force causing the motion is the component parallel to the incline of the crate's weight. Crate's weight = m*g = (2.6 kg)*(9.8 ft/sec/sec) = 25.48 newtons Weigth's component perpendicular to incline = (25.48)cos(35deg) = 20.87 newtons Available friction force = (0.290)(20.87) = 6.05 newtons. Sliding force = (25.48)sin(35deg) = 14.61 newtons Umm, the sliding force is a lot greater than the friction force fighting it. The crate will slide! A giant forefinger is needed to press the crate gently to prevent the crate from sliding down. 14.61 minus 6.05 = 8.56 newtons <---to be neutralized. friction force = (coefficient of friction)*(perpendicular force) So, 8.56 = (0.290)*(F) F = 8.56 / 0.290 = 29.52 newtons. Therefore, the giant needs to exert only enough force perpendicular to the incline of 29.52 newtons to stay the crate.``` | 617 | 2,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2014-35 | longest | en | 0.855756 |
https://www.coursehelponline.com/wk-4-signature-assignment-financial-performance-calculations-2/ | 1,643,284,674,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305260.61/warc/CC-MAIN-20220127103059-20220127133059-00512.warc.gz | 764,536,414 | 11,614 | # Wk 4 – Signature Assignment: Financial Performance Calculations
Complete the Banking Achievement Calculations worksheet. ADDITIONAL INSTRUCTION ON THIS ASSIGNMENT 1. There are a absolute of 13 problems (WORTH = 130 POINTS) and 1 Banking Achievement Arbitrary (WORTH 70 POINTS). 2. Again, anniversary botheration is account 10 credibility for a absolute of 130 credibility for all the Excel exercises. The area on the Banking Achievement Arbitrary is account 70 points. 3. Based on the calculations of the banking ratios, address a BRIEF arbitrary on the banking position of Hillside Inc. This is account 70 points. Use the afterward articulation to apprehend how to address this summary: How to use banking ratios to address banking achievement (I accept uploaded the accessories to abetment you) 4. Note that you charge to accomplish an acclimation on the Excel corpuscle so that as you blazon into the box or corpuscle the agreeable charcoal central the box. Follow the instructions below: How to blanket argument in Excel
## Order a unique copy of this paper
550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
\$26 | 266 | 1,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.878276 |
http://gmatclub.com/forum/what-is-the-range-of-numbers-in-set-s-1-the-standard-66252.html?fl=similar | 1,429,623,084,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246641468.77/warc/CC-MAIN-20150417045721-00262-ip-10-235-10-82.ec2.internal.warc.gz | 122,857,479 | 43,835 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 21 Apr 2015, 05:31
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# What is the range of numbers in Set S? (1) The standard
Author Message
TAGS:
Intern
Joined: 25 May 2008
Posts: 40
Followers: 0
Kudos [?]: 3 [0], given: 0
What is the range of numbers in Set S? (1) The standard [#permalink] 27 Jun 2008, 04:59
00:00
Difficulty:
(N/A)
Question Stats:
100% (00:00) correct 0% (00:00) wrong based on 1 sessions
What is the range of numbers in Set S?
(1) The standard deviation of the numbers is 3.
(2) The average of all numbers is 5.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Director
Joined: 01 Jan 2008
Posts: 629
Followers: 3
Kudos [?]: 138 [0], given: 1
Re: DS - Range [#permalink] 27 Jun 2008, 05:22
kapilnegi wrote:
What is the range of numbers in Set S?
(1) The standard deviation of the numbers is 3.
(2) The average of all numbers is 5.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Average and standard deviation alone don't define a distribution (unless a very special case when the kind of distribution is specified and it's a "nice" kind) -> don't define a range -> E
Re: DS - Range [#permalink] 27 Jun 2008, 05:22
Similar topics Replies Last post
Similar
Topics:
12 The range of the numbers in set S is x, and the range of the 5 14 Dec 2012, 02:47
3 The range of the numbers in set S is X, and the range of the 2 10 May 2012, 06:57
If the standard deviation of Set Y is 4, what is the range 6 14 Jul 2009, 18:48
What is the standard deviation of set S ? 1. Set S consists 6 17 Feb 2008, 05:41
Q is a set of 10 numbers. What is the standard deviation of 3 09 Jan 2008, 16:33
Display posts from previous: Sort by | 810 | 2,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2015-18 | longest | en | 0.854148 |
https://kurser.lth.se/kursplaner/arets%20eng/FMSF05.html | 1,620,839,905,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989766.27/warc/CC-MAIN-20210512162538-20210512192538-00244.warc.gz | 386,864,435 | 2,530 | Course syllabus
# Sannolikhetsteori Probability Theory
## FMSF05, 7,5 credits, G2 (First Cycle)
Valid for: 2020/21
Decided by: PLED I
Date of Decision: 2020-04-03
## General Information
Elective for: BME4, F4, F4-fm, I4, Pi4
Language of instruction: The course will be given in English
## Aim
The course gives a deaper and extended knowledge of probability theory, useful for further studies in, e.g., extreme value theory and stochastic processes and their applications.
## Learning outcomes
Knowledge and understanding
For a passing grade the student must
• be able to explain different concepts in stochastic convergence and how they relate to each other,
• be able to explain the concepts of characteristic and moment generating functions and how these functions can be used,
• be able to descibe the multi dimensional normal distribution and the invariance properties under, e.g., linear combinations and conditioning,
• be able to explain the definition and basic properties of the Poisson process.
Competences and skills
For a passing grade the student must
• show the ability to integrate knowledge from the different parts of the course when solving problems.
## Contents
The course deapens and expands the basic knowledge in probability theory. Central moments in the course are transforms of distribution, conditional expectations, multidimensional normal distribution, and stochastic convergence. Further, the concept of stochastic processes is introduced by a fairly thourough treatment of the properties of the Poisson process.
## Examination details
Grading scale: TH - (U,3,4,5) - (Fail, Three, Four, Five)
Assessment: Written exam.
The examiner, in consultation with Disability Support Services, may deviate from the regular form of examination in order to provide a permanently disabled student with a form of examination equivalent to that of a student without a disability.
• FMSF20 Mathematical Statistics, Basic Course or FMSF25 Mathematical Statistics - Complementary Project or FMSF45 Mathematical Statistics, Basic Course or FMSF50 Mathematical Statistics, Basic Course or FMSF55 Mathematical Statistics, Basic Course or FMSF70 Mathematical Statistics or FMSF75 Mathematical Statistics, Basic Course
The number of participants is limited to: No
The course overlaps following course/s: MASC01 | 496 | 2,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-21 | latest | en | 0.857893 |
https://edurev.in/studytube/Ex-1-1-NCERT-Solutions-Real-Numbers/26b8459b-1a82-43d5-90e9-5f384000d1ab_t | 1,669,682,291,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710684.84/warc/CC-MAIN-20221128235805-20221129025805-00163.warc.gz | 260,185,238 | 59,742 | NCERT Solutions: Real Numbers (Exercise 1.1)
# NCERT Solutions: Real Numbers (Exercise 1.1) Notes | Study Mathematics (Maths) Class 10 - Class 10
## Document Description: NCERT Solutions: Real Numbers (Exercise 1.1) for Class 10 2022 is part of Mathematics (Maths) Class 10 preparation. The notes and questions for NCERT Solutions: Real Numbers (Exercise 1.1) have been prepared according to the Class 10 exam syllabus. Information about NCERT Solutions: Real Numbers (Exercise 1.1) covers topics like and NCERT Solutions: Real Numbers (Exercise 1.1) Example, for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions: Real Numbers (Exercise 1.1).
Introduction of NCERT Solutions: Real Numbers (Exercise 1.1) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & NCERT Solutions: Real Numbers (Exercise 1.1) in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: NCERT Solutions: Real Numbers (Exercise 1.1) Notes | Study Mathematics (Maths) Class 10 - Class 10
1 Crore+ students have signed up on EduRev. Have you?
Q1. Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Ans:
(i) HCF of 135 and 225
Step 1: Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
⇒ 225 = 135 × 1 + 90
Step 2: Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
Step 3: We consider the new divisor 90 and new remainder 45 and apply the division lemma to obtain: 90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
(ii) HCF of 196 and 38220
Step 1: Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
⇒ 38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) HCF of 867 and 255
Step 1: Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
⇒ 867 = 255 × 3 + 102
Step 2: Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
⇒ 255 = 102 × 2 + 51
Step 3: We consider the new divisor 102 and new remainder 51 and apply the division lemma to obtain
⇒ 102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.
Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Ans: Let us consider a positive odd integer as ‘a’.
On dividing ‘a’ by 6, Let ‘a’ be the quotient and ‘r’ be the remainder.
∴ Using Euclid’s lemma, we have:
a = 6q + r where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4, or 5
Now substituting the value of r, we get:
If r = 0, then a = 6q
Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.
If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2.
A positive integer can be either even or odd.
Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.
Q3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Ans: Given,
Number of army contingent members = 616
Number of army band members = 32
If the two groups have to march in the same column, we have to find out the highest common factor between the two groups.
HCF(616, 32) gives the maximum number of columns in which they can march.
By Using Euclid’s algorithm to find their HCF, we get,
Since, 616 > 32
∴ 616 = 32 × 19 + 8
Since, 8 ≠ 0, therefore, taking 32 as the new divisor, we have,
32 = 8 × 4 + 0
Now we have got remainder as 0, therefore, HCF (616, 32) = 8.
Hence, the maximum number of columns in which they can march is 8.
Q4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in form 3m or 3m + 1.]
Ans: Let us consider an arbitrary positive integer as ‘x’ such that it is of the form 3q, (3q + 1) or (3q + 2).
Note:
For any positive integer x and b = 3,
x = 3q + r, where q is quotient and r is remainder such that 0 < r < 3.
If r = 0 then x = 3q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2
For x = 3q, we have
x2 = (3q)2
⇒ x2 = 9q2
= 3 (3q2)
= 3m ...(1) (Putting 3q2 = m where m is an integer)
For x = 3q + 1
x2 = (3q + 1)2
= 9q2 + 6q + 1
= 3 (3q2 + 2q) + 1
= 3m + 1 ...(2) (Puting 3q2 + 2q = m, where m is an integer)
For x = 3q + 2
x2 = (3r + 2)2
= 9q2 + 12q + 4
= (9q2 + 12q + 3) + 1
= 3 (3q2 + 4q + 1) + 1
= 3m + 1 ...(3) (Putting 3q2 + 4q + 1 = m, where m is an integer)
From (1), (2) and (3)
x2 = 3m or 3m + 1
Thus, the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
Q5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Ans: Let us consider an arbitrary positive integer x such that it is in the form of 3q, (3q + 1) or (3q + 2).
Note:
For any positive integer x and b = 3,
x = 3q + r, where q is quotient and r is remainder such that 0 < r < 3.
If r = 0 then x = 3q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2
∴ x is of the form 3q, (3q + 1) or (3q + 2).
For x = 3q
x3 = (3q)3
= 27q3
= 9 (3q3)
= 9m ...(1) (Putting 3q3 = m, where m is an integer)
For x = 3q + 1
x3 = (3q + 1)3
= 27q3 + 27q2 + 9q + 1
= 9 (3q3 + 3q2 + q) + 1
= 9m + 1 ...(2) (Putting (3q3 + 3q2 + 1) = m, where m is an integer)
For x = 3q + 2
x3 = (3q + 2)3
= 27q3 + 54q2 + 36q + 8
= 9 (3q3 + 6q2 + 4q) + 8
= 9m + 8 ...(3) (Putting (3q3 + 6q2 + 4q) = m, where m is an integer)
From (1), (2), (3) we have:
x3 = 9m, (9m + 1) or (9m + 8)
Thus, x3 of any positive integer can be in the form 9m, (9m + 1) or (9m + 8).
Check out the NCERT Solutions of all the exercises of Real Numbers:
Exercise 1.2 NCERT Solutions: Real Numbers
Exercise 1.3 NCERT Solutions: Real Numbers
Exercise 1.4 NCERT Solutions: Real Numbers
The document NCERT Solutions: Real Numbers (Exercise 1.1) Notes | Study Mathematics (Maths) Class 10 - Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
## Mathematics (Maths) Class 10
53 videos|395 docs|138 tests
Use Code STAYHOME200 and get INR 200 additional OFF
## Mathematics (Maths) Class 10
53 videos|395 docs|138 tests
### Top Courses for Class 10
Track your progress, build streaks, highlight & save important lessons and more!
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; | 2,507 | 6,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2022-49 | latest | en | 0.818225 |
https://gamedev.stackexchange.com/questions/tagged/lines?tab=Votes | 1,597,271,623,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738944.95/warc/CC-MAIN-20200812200445-20200812230445-00028.warc.gz | 317,815,405 | 34,175 | # Questions tagged [lines]
The tag has no usage guidance.
24 questions
Filter by
Sorted by
Tagged with
1k views
### How do you calculate the nearest point on 2 curves?
Given the points of a line and a quadratic bezier curve, how do you calculate their nearest point? .... Similarly, given the points of 2 curves, how do you get the nearest point?
379 views
### How do you calculate if 2 lines are facing toward or away?
Given the 4 points describing 2 line segments, how do you calculate if line A is towards or away from line B? The 2 lines have a fixed length, and can be measured as distance from x1/y1 to x2/y2.
15k views
### How to make Line Renderer lines stay flat?
I've noticed that as I add more vertexes to a Line Renderer line the line twists and stops being a smooth line. .GIF Here: http://i.imgur.com/hRAhCXM.gif All lines are on the same z level, even if ...
2k views
### How to find path in graph without crossing the walls?
I would like to know how to find path in a graph while having the edges "double-sided". Consider this scenario: I have list of the line segments. These represents "walls" or rather impassable edges (...
2k views
### XNA 2D Spritesheet drawing rendering problem
I'm making a tile-based game, using one spritesheet containing all tile graphics. Each tile has a size of 32x32 pixels. The main problem is: when I draw the tile to the screen, if the tile position ...
493 views
### How to create laser redirection?
Think portal 2, where a laser can go through a cube, and depending on the rotation of that cube, it will redirect it. Examples: Here is my barely functioning code. It works for one cube, I'm able to ...
2k views
### How do I draw a dotted or dashed line?
I'm trying to draw a dashed or dotted line by placing individual segments(dashes) along a path and then separating them. The only algorithm I could come up with for this gave me a dash length that was ...
724 views
### How to order points on a 3D grid such that we can connect them in a line loop correctly
I want to recreate an effect as close as possible to X-Com: Enemy Unkown's movement range feedback: (source: gameinformer.com) The blue line delimits the tiles which the soldier can reach using a ...
4k views
### Enable/ Disable Line Renderer points
I have a Line Renderer component with 4 points, I would like to disable and enable the last 2 points at a certain time. My question is, how would I in code (C#) enable and disable individual Line ...
914 views
### Drawing a lines radiating from every angle using Bresenham line algorithm
I am trying to recreate a image like this ... or just each point Using Bresenham line algorithm, but can't seem to able to do so.. The point i receive from my computation doesn't look like a like, ...
7k views
### LineRenderer sorting order - lines always in front of everything
I have 3 cameras and 3 canvases. Back Camera: depth = -5, Back Canvas: Sorting layer - BackLayer, Camera: depth = 0, Middle Canvas: Sorting layer - MiddleLayer, Camera: depth = 10, Front Canvas: ...
208 views
### How to smooth hand drawn lines?
So I am using Kinect with Unity. With the Kinect, we detect a hand gesture and when it is active we draw a line on the screen that follows where ever the hand is going. Every update the location is ...
109 views
### Why do we get artifacts on objects rendered using depth node + depth render camera on Android?
I've encountered and error I can't seem to find online. I get it on all the water shaders I test, and any thing else that I have a depth node + depth render script on the camera for. I'll share a bit ...
51 views
### Random lines drawn on screen while all vertexes are correct
I'm writing a 2D program in which a monocycle follows a Catmull-Rom spline. My problem is when I write the circle, the drawing goes crazy. There is one line on the screenshot (which seems to be 3 when ...
336 views
### Drawing many separate lines using mouse OpenGL(GLFW/glad)
So, in order to draw a line, I track the coordinates of the mouse, then I add them to the array and capture it as GL_LINE_STRIP_ADJACENCY. However, for example, I completed drawing a line1 at P1 and ...
3k views
### How to correctly draw a 2D polygon in Unity
I'm trying to draw a 2D polygon using GL.Line; however I'm not getting a polygon but separated lines instead. In the example below, I would expect a rectangle formed by the spheres: My code is the ...
963 views
### How to shade hair (line segments) correctly in GLSL
I render my hair geometry in line segments. The problem I have is with the correct shading of the hair. From what I've read I understood that I need the direction of the line segment and the ...
2k views
### LibGDX How can I draw a line at a specific angle
I'm working on a 2D platformer, shooter game and I'm attempting to draw a line to serve as a guide to indicate where the player is shooting. The line should be drawn from the player's sprite to the ...
59 views
### how can I transform a skewed line to a vertical line?
So, I have a line: And I skewed it with x += y * cos(angle) and got something like: My question is, if I have another line segment with that same skew, how can I ...
107 views
### Detect mouse over a line in Phaser
We're developing a game in Phaser as a university project and have come across an obstacle. We need to detect when the mouse is over a line between two points. There are not a lot of lines, definitely ...
119 views
### Why does my shader's Texture become unselected when I run my game?
I'd like to add a custom material to a line that I'm drawing in Unity (using Vectrosity). I'm using Unity's built-in "Unlit/Texture" shader. When I drag a texture from my Resources directory into ...
363 views
### Perspective Projection - Drawing Lines
So I sort of understand the math behind projecting 3d points onto a 2d plane. It's just some simple dividing and multiplication of distances. However I can't seem to wrap my head around projecting ... | 1,415 | 5,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-34 | latest | en | 0.927827 |
http://math.stackexchange.com/questions/189614/help-fixing-my-broken-example-of-arf-invariant | 1,469,571,573,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825124.55/warc/CC-MAIN-20160723071025-00251-ip-10-185-27-174.ec2.internal.warc.gz | 159,860,968 | 18,646 | Help fixing my broken example of Arf invariant
I need help fixing a broken example I've come up with. In particular, I wanted to use the Arf invariant to distinguish two non-homeomorphic surfaces. That's the first part that's broken since there aren't any surfaces that have the same homology groups but aren't homeomorphic. So I gain nothing because the homology groups already let me distinguish two surfaces.
Concretely, what I was trying to do was this: observe that the intersection number defines a bilinear pairing (an intersection form in this case) and then compute the Arf invariant of $q(x)=\langle x,x \rangle$. Stupidly, for orientable surfaces, this always gives me $q(x)=\langle x,x \rangle = 0$ and hence $\mathrm{Arf}(q) = 0$.
On the other hand, for $\mathbb R P^2$, there is only one basis vector hence I can't even define the Arf invariant (since I can't define an intersection form since that would need even dimension). So this is the second part that is broken.
Now I'm stuck. How can I fix this? What is the simplest example of two manifolds (or topological spaces if you like) that I can distinguish using the Arf invariant? Thanks for help.
-
Keep in mind that the Arf-Kervaire invariant is meant to be evaluated on framed manifolds. I'm pretty sure the first few nontrivial examples come from Lie groups (which are automatically frameable). – Aaron Mazel-Gee Sep 1 '12 at 22:04
@AaronMazel-Gee Lie groups? Thanks for the hint. I had read about framed manifolds but then I thought that there had to be a simpler example for this. But good to know that there are Arf examples with Lie groups! – Rudy the Reindeer Sep 2 '12 at 6:52
What makes you want to use the Arf invariant, anyways? Of course (the isomorphism class of) the intersection form is a strictly stronger invariant, so if you have it then you should use it (instead of its Arf invariant) to distinguish your manifolds. The (Arf-)Kervaire invariant problem is a global question about manifolds, i.e. it asks about all manifolds at once. Short of using HHR's theorem as an absurdly large sledgehammer, I don't know any way of constructing a manifold with nonvanishing Arf invariant without knowing its intersection form in the first place. – Aaron Mazel-Gee Sep 2 '12 at 20:28
@AaronMazel-Gee Well, I've been reading about the Arf invariant. So far I have looked at Arf invariants of knots. Next I have to look into framed manifolds. What is HHR's theorem? – Rudy the Reindeer Sep 6 '12 at 15:38
I don't see how to fix your example, surfaces aren't rich enough.
To get something a little richer, 3-manifolds have something called a torsion linking form. This is a symmetric bilinear map:
$$\tau H_1(M,\mathbb Z) \times \tau H_1(M,\mathbb Z) \to \mathbb Q / \mathbb Z$$
where $\tau H_1(M,\mathbb Z)$ is the subgroup of torsion elements of $H_1(M,\mathbb Z)$. In the lens space $L_{p,q}$ the value of this form on $(x,x)$ where $x$ generates $H_1$ is $\pm r^2 q/p$. Where $r$ is some integer.
So you could restrict the torsion linking form of 3-manifolds to the 2-torsion subgroup and compute the Arf invariant of that.
I think the Arf invariant would distinguish
$$L_{p,q} \# L_{p,q} \text{ and } L_{p,q} \# L_{p,-q}$$
Provided $p$ is even and $L_{p,q}$ does not admit an orientation-reversing homotopy-equivalence (which I think is when $q \neq -r^2 q$ modulo $p$.
Have you looked at something like that?
-
Thank you Ryan for this great answer. Yes, I'd briefly read about Lens spaces (when searching for "spaces with same homology that aren't homeomorphic") but it looked a bit less simple than I wanted it to be, since I wasn't sure what the intersection form looks like on something that is not a surface. – Rudy the Reindeer Sep 2 '12 at 6:56 | 1,003 | 3,746 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-30 | latest | en | 0.936879 |
https://www.geeksforgeeks.org/biginteger-intvalueexact-method-in-java-with-examples/?ref=rp | 1,590,738,165,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347402457.55/warc/CC-MAIN-20200529054758-20200529084758-00141.warc.gz | 739,119,516 | 25,723 | # BigInteger intValueExact() Method in Java with Examples
java.math.BigInteger.intValueExact() was introduced in Java 8. It is an inbuilt function which converts the value of BigInteger to a int and checks for lost information. If the value of BigInteger is greater than 2,147,483,647 or less than -2,147,483,648; the method will throw ArithmeticException as BigInteger doesn’t fit in int range.
Syntax:
`public int intValueExact()`
Return Value: This method returns the int value of this BigInteger.
Exception: The method throws ArithmeticException if the value of BigInteger is greater than 2,147,483,647 or less than -2,147,483,648; as this range doesn’t fit in int range.
Example:
```Input: 4561561
Output: 4561561
Explanation: 4561561 is given as input which is bigInteger
and int value of 4561561 is 4561561
Input: -8546512
Output: -8546512
Explanation: -8546512 is given as input which is bigInteger
and int value of -8546512 is -8546512
Input: 3000000000
Output: ArithmeticException
Explanation: When 3000000000 is tried to convert to int,
since 3000000000 > 2,147,483,647 (greater than a int's range),
therefore it throws an arithmetic exception.
```
Below programs illustrates intValueExact() method of BigInteger class:
Program 1: To demonstrate intValueExact() method for positive number <2,147,483,647
`// Java program to demonstrate intValueExact() ` `// method of BigInteger Class ` ` ` `import` `java.math.*; ` ` ` `public` `class` `GFG { ` ` ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``// Creating a BigInteger object ` ` ``BigInteger big; ` ` ``big = ``new` `BigInteger(``"4561561"``); ` ` ` ` ``// print value of BigInteger ` ` ``System.out.println(``"BigInteger value : "` ` ``+ big); ` ` ` ` ``// convert big to the int value ` ` ``int` `b1 = big.intValueExact(); ` ` ` ` ``// print int value ` ` ``System.out.println(``"int converted value : "` ` ``+ b1); ` ` ``} ` `} `
Output:
```BigInteger value : 4561561
int converted value : 4561561
```
Program 2: To demonstrate intValueExact() method for negative number >-2,147,483,648
`// Java program to demonstrate intValueExact() ` `// method of BigInteger Class ` ` ` `import` `java.math.*; ` ` ` `public` `class` `GFG { ` ` ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ` ` ``// Creating a BigInteger object ` ` ``BigInteger big = ``new` `BigInteger(``"-8546512"``); ` ` ` ` ``// print value of BigInteger ` ` ``System.out.println(``"BigInteger value : "` ` ``+ big); ` ` ` ` ``// convert big to the int value ` ` ``int` `b1 = big.intValueExact(); ` ` ` ` ``// print int value ` ` ``System.out.println(``"int converted value : "` ` ``+ b1); ` ` ``} ` `} `
Output:
```BigInteger value : -8546512
int converted value : -8546512
```
Program 3: To demonstrate intValueExact() method for negative number <-2,147,483,648. It will throw Arithmetic Exception
`// Java program to demonstrate intValueExact() ` `// method of BigInteger Class ` ` ` `import` `java.math.*; ` ` ` `public` `class` `GFG { ` ` ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ` ` ``// Creating a BigInteger object ` ` ``BigInteger big = ``new` `BigInteger(``"-3000000000"``); ` ` ` ` ``// print value of BigInteger ` ` ``System.out.println(``"BigInteger value : "` ` ``+ big); ` ` ` ` ``try` `{ ` ` ``// convert big to the int value ` ` ``int` `b1 = big.intValueExact(); ` ` ` ` ``// print int value ` ` ``System.out.println(``"int converted value : "` ` ``+ b1); ` ` ``} ` ` ``catch` `(Exception e) { ` ` ``System.out.println(``"Exception: "` `+ e); ` ` ``} ` ` ``} ` `} `
Output:
```BigInteger value : -3000000000
Exception: java.lang.ArithmeticException: BigInteger out of int range
```
Program 4: To demonstrate intValueExact() method for positive number >2,147,483,647. It will throw Arithmetic Exception
`// Java program to demonstrate intValueExact() ` `// method of BigInteger Class ` ` ` `import` `java.math.*; ` ` ` `public` `class` `GFG { ` ` ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ` ` ``// Creating a BigInteger object ` ` ``BigInteger big = ``new` `BigInteger(``"3000000000"``); ` ` ` ` ``// print value of BigInteger ` ` ``System.out.println(``"BigInteger value : "` ` ``+ big); ` ` ` ` ``try` `{ ` ` ``// convert big to the int value ` ` ``int` `b1 = big.intValueExact(); ` ` ` ` ``// print int value ` ` ``System.out.println(``"int converted value : "` ` ``+ b1); ` ` ``} ` ` ``catch` `(Exception e) { ` ` ``System.out.println(``"Exception: "` `+ e); ` ` ``} ` ` ``} ` `} `
Output:
```BigInteger value : 3000000000
Exception: java.lang.ArithmeticException: BigInteger out of int range
```
My Personal Notes arrow_drop_up
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Article Tags :
Practice Tags :
Be the First to upvote.
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 1,664 | 5,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-24 | longest | en | 0.687483 |
https://www.valuespreadsheet.com/calculators/annualized-return | 1,721,146,738,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00430.warc.gz | 932,766,969 | 15,923 | # Annualized Rate of Return Calculator (CAGR)
The Annualized Rate of Return Calculator helps you determine the compound annual growth rate (CAGR) of your investments. This will standardize your returns to a per year figure, which shows you your true long term average portfolio performance.
`<iframe width="500" height="350" src="https://www.valuespreadsheet.com/app/calculators/annualized-return.html?theme=dark" title="Annualized return calculator" frameborder="0" scrolling="no"></iframe><br><a href="https://www.valuespreadsheet.com" rel="noopener" target="_blank" style="color: #0095f7; font-size: 13px;">By Value Spreadsheet</a>`
## Annualized Rate of Return Calculator Explained
The annualized rate of return, also called the compound annual growth rate (CAGR), is how much your investments would have to grow each year to have the beginning value reach the ending value over the specified holding period.
So if you grow \$1000 into \$2000 over the span of 5 years, you will have made a 100% return, but your annualized return is not simply:
`100 / 5 = 20%`
Why not?
Because due to compounding, 5 years of 20% returns would result in an ending value of:
`1000 x 1.205 = 2488.32`
This result is way too high.
To calculate the correct annualized rate of return, we have to use this formula:
`CAGR = (ending value / beginning value)(1 / years held) - 1`
Using our example:
`(2000 / 1000)(1 / 5) - 1 = 14.87%`
So the annualized rate of return is in fact 14.87%.
Or in other words, if you're able to grow your investments by 14.87% per year, you will double your money in 5 years time. | 419 | 1,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-30 | longest | en | 0.847582 |
https://origin.geeksforgeeks.org/k-dimensional-tree-set-3-delete/?ref=lbp | 1,659,913,198,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570730.59/warc/CC-MAIN-20220807211157-20220808001157-00123.warc.gz | 421,937,962 | 34,636 | # K Dimensional Tree | Set 3 (Delete)
• Difficulty Level : Hard
• Last Updated : 18 Jul, 2022
We strongly recommend to refer below posts as a prerequisite of this.
K Dimensional Tree | Set 1 (Search and Insert)
K Dimensional Tree | Set 2 (Find Minimum)
In this post delete is discussed. The operation is to delete a given point from K D Tree.
Like Binary Search Tree Delete, we recursively traverse down and search for the point to be deleted. Below are steps are followed for every node visited.
1) If current node contains the point to be deleted
1. If node to be deleted is a leaf node, simply delete it (Same as BST Delete)
2. If node to be deleted has right child as not NULL (Different from BST)
1. Find minimum of current node’s dimension in right subtree.
2. Replace the node with above found minimum and recursively delete minimum in right subtree.
3. Else If node to be deleted has left child as not NULL (Different from BST)
1. Find minimum of current node’s dimension in left subtree.
2. Replace the node with above found minimum and recursively delete minimum in left subtree.
3. Make new left subtree as right child of current node.
2) If current doesn’t contain the point to be deleted
1. If node to be deleted is smaller than current node on current dimension, recur for left subtree.
2. Else recur for right subtree.
Why 1.b and 1.c are different from BST?
In BST delete, if a node’s left child is empty and right is not empty, we replace the node with right child. In K D Tree, doing this would violate the KD tree property as dimension of right child of node is different from node’s dimension. For example, if node divides point by x axis values. then its children divide by y axis, so we can’t simply replace node with right child. Same is true for the case when right child is not empty and left child is empty.
Why 1.c doesn’t find max in left subtree and recur for max like 1.b?
Doing this violates the property that all equal values are in right subtree. For example, if we delete (!0, 10) in below subtree and replace if with
```Wrong Way (Equal key in left subtree after deletion)
(5, 6) (4, 10)
/ Delete(5, 6) /
(4, 10) ------------> (4, 20)
\
(4, 20)
Right way (Equal key in right subtree after deletion)
(5, 6) (4, 10)
/ Delete(5, 6) \
(4, 10) ------------> (4, 20)
\
(4, 20) ```
Example of Delete:
Delete (30, 40): Since right child is not NULL and dimension of node is x, we find the node with minimum x value in right child. The node is (35, 45), we replace (30, 40) with (35, 45) and delete (30, 40).
Delete (70, 70): Dimension of node is y. Since right child is NULL, we find the node with minimum y value in left child. The node is (50, 30), we replace (70, 70) with (50, 30) and recursively delete (50, 30) in left subtree. Finally we make the modified left subtree as right subtree of (50, 30).
Below is C++ implementation of K D Tree delete.
## C++
`// A C++ program to demonstrate delete in K D tree ` `#include ` `using` `namespace` `std; ` ` ` `const` `int` `k = 2; ` ` ` `// A structure to represent node of kd tree ` `struct` `Node ` `{ ` ` ``int` `point[k]; ``// To store k dimensional point ` ` ``Node *left, *right; ` `}; ` ` ` `// A method to create a node of K D tree ` `struct` `Node* newNode(``int` `arr[]) ` `{ ` ` ``struct` `Node* temp = ``new` `Node; ` ` ` ` ``for` `(``int` `i=0; ipoint[i] = arr[i]; ` ` ` ` ``temp->left = temp->right = NULL; ` ` ``return` `temp; ` `} ` ` ` `// Inserts a new node and returns root of modified tree ` `// The parameter depth is used to decide axis of comparison ` `Node *insertRec(Node *root, ``int` `point[], unsigned depth) ` `{ ` ` ``// Tree is empty? ` ` ``if` `(root == NULL) ` ` ``return` `newNode(point); ` ` ` ` ``// Calculate current dimension (cd) of comparison ` ` ``unsigned cd = depth % k; ` ` ` ` ``// Compare the new point with root on current dimension 'cd' ` ` ``// and decide the left or right subtree ` ` ``if` `(point[cd] < (root->point[cd])) ` ` ``root->left = insertRec(root->left, point, depth + 1); ` ` ``else` ` ``root->right = insertRec(root->right, point, depth + 1); ` ` ` ` ``return` `root; ` `} ` ` ` `// Function to insert a new point with given point in ` `// KD Tree and return new root. It mainly uses above recursive ` `// function "insertRec()" ` `Node* insert(Node *root, ``int` `point[]) ` `{ ` ` ``return` `insertRec(root, point, 0); ` `} ` ` ` `// A utility function to find minimum of three integers ` `Node *minNode(Node *x, Node *y, Node *z, ``int` `d) ` `{ ` ` ``Node *res = x; ` ` ``if` `(y != NULL && y->point[d] < res->point[d]) ` ` ``res = y; ` ` ``if` `(z != NULL && z->point[d] < res->point[d]) ` ` ``res = z; ` ` ``return` `res; ` `} ` ` ` `// Recursively finds minimum of d'th dimension in KD tree ` `// The parameter depth is used to determine current axis. ` `Node *findMinRec(Node* root, ``int` `d, unsigned depth) ` `{ ` ` ``// Base cases ` ` ``if` `(root == NULL) ` ` ``return` `NULL; ` ` ` ` ``// Current dimension is computed using current depth and total ` ` ``// dimensions (k) ` ` ``unsigned cd = depth % k; ` ` ` ` ``// Compare point with root with respect to cd (Current dimension) ` ` ``if` `(cd == d) ` ` ``{ ` ` ``if` `(root->left == NULL) ` ` ``return` `root; ` ` ``return` `findMinRec(root->left, d, depth+1); ` ` ``} ` ` ` ` ``// If current dimension is different then minimum can be anywhere ` ` ``// in this subtree ` ` ``return` `minNode(root, ` ` ``findMinRec(root->left, d, depth+1), ` ` ``findMinRec(root->right, d, depth+1), d); ` `} ` ` ` `// A wrapper over findMinRec(). Returns minimum of d'th dimension ` `Node *findMin(Node* root, ``int` `d) ` `{ ` ` ``// Pass current level or depth as 0 ` ` ``return` `findMinRec(root, d, 0); ` `} ` ` ` `// A utility method to determine if two Points are same ` `// in K Dimensional space ` `bool` `arePointsSame(``int` `point1[], ``int` `point2[]) ` `{ ` ` ``// Compare individual pointinate values ` ` ``for` `(``int` `i = 0; i < k; ++i) ` ` ``if` `(point1[i] != point2[i]) ` ` ``return` `false``; ` ` ` ` ``return` `true``; ` `} ` ` ` `// Copies point p2 to p1 ` `void` `copyPoint(``int` `p1[], ``int` `p2[]) ` `{ ` ` ``for` `(``int` `i=0; ipoint, point)) ` ` ``{ ` ` ``// 2.b) If right child is not NULL ` ` ``if` `(root->right != NULL) ` ` ``{ ` ` ``// Find minimum of root's dimension in right subtree ` ` ``Node *min = findMin(root->right, cd); ` ` ` ` ``// Copy the minimum to root ` ` ``copyPoint(root->point, min->point); ` ` ` ` ``// Recursively delete the minimum ` ` ``root->right = deleteNodeRec(root->right, min->point, depth+1); ` ` ``} ` ` ``else` `if` `(root->left != NULL) ``// same as above ` ` ``{ ` ` ``Node *min = findMin(root->left, cd); ` ` ``copyPoint(root->point, min->point); ` ` ``root->right = deleteNodeRec(root->left, min->point, depth+1); ` ` ``} ` ` ``else` `// If node to be deleted is leaf node ` ` ``{ ` ` ``delete` `root; ` ` ``return` `NULL; ` ` ``} ` ` ``return` `root; ` ` ``} ` ` ` ` ``// 2) If current node doesn't contain point, search downward ` ` ``if` `(point[cd] < root->point[cd]) ` ` ``root->left = deleteNodeRec(root->left, point, depth+1); ` ` ``else` ` ``root->right = deleteNodeRec(root->right, point, depth+1); ` ` ``return` `root; ` `} ` ` ` `// Function to delete a given point from K D Tree with 'root' ` ` ``Node* deleteNode(Node *root, ``int` `point[]) ` `{ ` ` ``// Pass depth as 0 ` ` ``return` `deleteNodeRec(root, point, 0); ` `} ` ` ` `// Driver program to test above functions ` `int` `main() ` `{ ` ` ``struct` `Node *root = NULL; ` ` ``int` `points[][k] = {{30, 40}, {5, 25}, {70, 70}, ` ` ``{10, 12}, {50, 30}, {35, 45}}; ` ` ` ` ``int` `n = ``sizeof``(points)/``sizeof``(points[0]); ` ` ` ` ``for` `(``int` `i=0; ipoint[0] << ``", "` `<< root->point[1] << endl; ` ` ` ` ``return` `0; ` `} `
Output:
```Root after deletion of (30, 40)
35, 45```
Source:
https://www.cs.umd.edu/class/spring2008/cmsc420/L19.kd-trees.pdf | 2,686 | 8,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-33 | longest | en | 0.913146 |
https://electronics.stackexchange.com/questions/659286/is-this-way-of-maximum-current-calculation-correct-for-this-voltage-regulator | 1,721,281,674,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00180.warc.gz | 205,061,640 | 38,306 | # Is this way of maximum current calculation correct for this voltage regulator?
Regarding using the LD39050PU33R voltage regulator which is 5V to 3.3V fixed voltage regulator without any heatsink, its maximum current is given as 500mA. I was trying to figure out how they calculated this 500mA. So I tried to calculate the maximum current by myself by using the thermal resistances and maximum junction temperature as follows:
From the thermal data below: The junction to case thermal resistance R_jc is sum of those above; so R_jc = 65 °C/W.
For 5V input the voltage drop across the regulator will be Vd = 5V-3.3V = 1.7V.
Max junction temperature is 125°C. And in the worse case if we have 55°C ambient temperature the ΔT = 70°C.
That means maximum power dissipated Pd can be found from:
ΔT = Pd * R_jc
Pd = 70 / 65 = 1.077W
This makes the max current I = Pd / Vd then I = 1.077 / 1.7 = 633mA.
In my calculation I took max ambient temperature 55°C. To obtain 500mA, the ambient temperature would be taken as around 70°C. Is my conclusion correct? | 286 | 1,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.918894 |
https://www.indiabix.com/aptitude/time-and-work/discussion-401 | 1,585,819,148,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506870.41/warc/CC-MAIN-20200402080824-20200402110824-00230.warc.gz | 970,488,127 | 14,210 | # Aptitude - Time and Work - Discussion
### Discussion :: Time and Work - General Questions (Q.No.20)
20.
Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:
[A]. 15 [B]. 16 [C]. 18 [D]. 25
Explanation:
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.
Suppose Tanya takes x days to do the work.
5 : 4 :: 20 : x x = 4 x 20 5
x = 16 days.
Hence, Tanya takes 16 days to complete the work.
Kamlesh Jaiswal said: (Sep 2, 2010) The given answer is correct. Sakshi's one day work=1/20 Tanya's one day work =(1/20 + 25% of 1/20) = 1/20+1/80 = 1/16. So Tanya can complete the work in 16 days.
Siddharth said: (Sep 23, 2010) Thanks kamal you are right.
Sandeep said: (Jan 8, 2011) If tanya is 25% more efficient than sakshi (whic means tanya>sakshi) then how cum the ratio between sakshi n tanya is 125:100 it should be 100:125 right?.
Sujith said: (Jan 30, 2011) Tanya is 25% more efficient than Sakshi. Then ratio should be 3:4 rite ?.
Sundar said: (Apr 16, 2011) @Sujith and @Sandeep You are wrong. The given ratio is about "TIME TAKEN" not the ratio of work done. If Tanya is more efficient (25%) than Sakshi - means Tanya will take less time than Sakshi to complete the work. We can say the same thing in another way, i.e., Sakshi is 25% INEFFICIENT than Tanya. So, if Tanya takes 100 hours to complete a task then Sakshi will take 125 hours (25% extra than Tanya) to complete the same task. So, time taken by Tanya = 100 hours , Sakshi = 125 hours. Therefore, Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4. This is what the first line in the given explanation. Hope this will help you. Have a nice day!
Kiran Joy said: (Jun 16, 2011) Can't we take time taken by tanya and sakshi as 75:100.
Deepti said: (Jun 27, 2011) Thanks sundar that solves so many other related problems.
Chandu said: (Aug 3, 2011) 5 : 4 :: 20 : x Can you explain the meaning of the above line?
Vamshidhar Reddy said: (Sep 6, 2011) 25% efficient(so one takes 100 days means second person takes 75days[just 75%of 100]) here one takes 20 days so other takes 20*(75/100)=15 days
Preethy said: (Sep 14, 2011) I too get the answer as 15 days. Since 25% of 20 is 5 we have to subtract like 20-5=15 rite? but how the answer is 16?
Xyz said: (Oct 4, 2011) Thanks Kamlesh !!!
Samar said: (Nov 26, 2011) 20*0.25=5 subtract 5 from 20 20-5=15Ans I also get the answer 15 I did not understand how the answer is 16.
Harshit said: (Dec 14, 2011) @Vamshidar reddy, Preethy& Samar;. The answer 16 is justified. Given that Tanya is more efficient than Sakshi. So, Tanya will take less time to complete the work. If Tanya is 25% more efficient than Sakshi, it is not nessesary. That Tanya will complete the work in 25% of the days less than Sakshi.
Satya Vaag said: (Jan 18, 2012) Thanks kamlesh.
Shawisk said: (Jun 20, 2012) Read carefully for tanya & sakshi, the actual ratio is 100:75=4:3 bcz if tanya do same work complete @ 100% in x days then sakshi can complete only 75% of that work in x days so keep it in mind
Pandian said: (Aug 30, 2012) Sakshi done a work in 20 days is 100% Tanya more power is 25% So she done work in minimum days So 25% of 20 days is 4 So tanya done work in 20-4 = 16 days.
Eby said: (Sep 28, 2012) Tanya =125% of Sakshi Sakshi 1 day work =1/20 So Tanya =(125/100)*(1/20) =(5/4)*(1/20) =(1/4)*(1/4) =(1/16) =16 days
K Verma said: (Dec 11, 2012) Tanya is 25% more efficient than Sakshi. Tanya is 125. Sakshi is 100. Ratio of efficiency is 5:4. So, ratio of time will be 4:5. Given, Sakshi can do work in 20 days means 5*4=20. In the same way for Tanya 4*4=16. Answer-16 days.
Akram said: (Dec 19, 2012) x:y::p:q. Product of mean = product of extreme. ..i.e.. y*p = x*q. y*p/x*q = 1. Same way : 5:4::20:x 4*20/5 = x 4*4 = x x = 16.
Pankaj said: (Jul 20, 2013) I too get the answer as 15 days. Since 25% of 20 is 5 we have to subtract like 20-5 = 15 right? but how the answer is 16?
Soumil said: (Aug 10, 2013) The ratio should be 100:75::20:x, not 125:100::20:x.
San said: (Aug 19, 2013) Efficiency is : (1/Number of total days to complete a work). So, efficiency of sakshy = 1/20. And,efficiency of Tanya = 1/20 + 25% of (1/20). Which is, 1/16. So, Tanya will complete in 16 days.
Roji said: (Oct 3, 2013) Hi friends, Why can't we do in this manner? Sakshi can done in 20 days i.e.,100%. Where as, Tanya is x days i.e., 125%. By cross multiplication, x = (125*20)/100 = 25. Answer we get 25. So, am I wrong? Already this type of process done in 17th sum but here it is not working yar.
Dhiman Saab said: (Dec 18, 2013) @Chandu. It means 5:4 is known 125:100. :: may mean equal to 20 here is the number of days taken by Sakshi and we are supposing x to be the days taken by Tanya.
Sai said: (Jan 8, 2014) Tanya's one day work =(1/20 + 25% of 1/20) = 1/20+1/80 = 1/16 once explain the equation.
Sandeep said: (Jan 22, 2014) (1/20 + 25% of 1/20). Here 25 % of 1/20 is 1/5. Then add, (1/20+1/5), we get, (1+4/20). Then we get 1/4. Above we got 4 days, then subtract 20 from 4 20-4 = 16.
Shravani said: (Jan 25, 2014) If tanya is 25% more efficient than sakshi (which means tanya>sakshi) then how cum the ratio between sakshi n tanya is 125:100 it should be 100:125 right?.
Ramesh said: (Mar 8, 2014) Sakshi's one day work is 1/20. Tanya is 25% of sakshi it means 1/20+25%1/20. = (1/20+25/100*1/20). = (1/20+1/80). = 1/16 for one day. So 16 days to do total work.
Iqbal Sofi said: (Mar 8, 2014) It should be 15, because the 25% reduction of 20 is 15.
Manjula said: (May 27, 2014) Sakshi working days in 20 days means 100%. Tanya working days ? 100%-75% = 25% So 20*25/125 = 4. 20 days - 4 = 16 days.
Pavan said: (Aug 2, 2014) Here question starts with sakshi so we must take s=100. Then 25% eff so T=75. Then the ratio we get is 3:4. Answer will be 15. How can it be 16. Can anyone please?
Anshu said: (Aug 5, 2014) Hey, You can understand it like this. As we know working efficiency of sakshi is 1/20. And working efficiency of tanya is 25% greater. So tanya eff = 1/20*(125/100)=1/16. Hence time 16 days.
Geetha said: (Aug 10, 2014) Give a brief explain how 125:100 come?
Jasmin said: (Oct 5, 2014) Can you explain me chain rule please? 5:4 :: 20:x.
Aisha said: (Jan 27, 2015) Sakshi is doing work with x efficiency. Tanya is doing work with x+25% of x efficiency. Sakshi = x work. Tanya = x+25/100x+125x/100 = 5/4x work. ATQ, Sakshi is taking 20 days means x ----> 20 days. 5/4x ----> 20 x 4/5 = 16 days.
Deepa said: (Feb 3, 2015) 20-25%[20] = 15 days.
Jai said: (Mar 7, 2015) Hey, how come it is 16? If Tanya takes 16 days to complete, then it means Sakshi is 25% slower than Tanya. But we have calculate Tanya's time who is 25% faster than Sakshi. So I reckon it should be 15. Please correct me if I am wrong.
Kundan said: (Mar 18, 2015) Tanya does 25% efficient work. What does this mean? This means that the work that Sakshi completes is 100% and Tanya does the same work but 125%. So see efficiency as the greater percentage of work being done by Tanya 125% is greater than 100%. For example if a does 20 units of work in 5 hrs and b does 40 units of work in 5 hrs. Then 1 hr work of A and B is 4 and 8 respectively. What does this mean? It means that B is doing more work in same no of hrs. Now consider it as B was doing 40 units of work in 5 hrs. So what is his time for 1 unit of work? its 5/40. i.e 1/8! and similarly for A it is 1/4. So in respect of timing who is fast or efficient? Its certainly B because 1/8 is smaller than 1/4!
Priya said: (Mar 28, 2015) Nothing to do all this just simple calculation in question it is given Tanya is 25% more efficient than Sakshi which do in 20 days. So Tanya will do 125/100*1/20 = 4 days.
Jitendra said: (Apr 15, 2015) Let Sakshi completes the whole work = 100 days. Then Tanya will complete it = 75 days. Now, 100 = 20. 1 = 20/100 = 1/5. 75 = (1/5) *75 = 15 days. So answer should be 15 days.
Jitendra said: (Apr 15, 2015) Why we should consider 125 and 100. If we consider that Sakshi 100 and Tanya 75 then, 100 : 75 :: 20 : x. = 100/75 = 20/x. => x = 15 days.
Sanjoy said: (Apr 16, 2015) 100T = 125S, 20S = 1, consider x day required for Tania. 4T = 5S, S = 1/20, Tx =1. 4T = 5*1/20 1/16x = 1. 4T = 1/4 x = 16. T = 1/16. So Tania required 16days to complete the work.
Vijayalakshmi R said: (May 17, 2015) S can do a work in 20 days so, s=1/20; T can do a work 25% more efficient than S. So, T = 1/20 + (25/100*1/20). Finally we got T = 1/16. So 16 is the correct answer. I am sure about my answer so don't confuse, it may be 25 or other else.
Vamsi said: (Jun 26, 2015) If Tanya is 25% more efficient than Sakshi, then as 25% of 20 is 5. Tanya should complete the work in 15 days not 16 days.
Devi Prasad said: (Sep 2, 2015) Their is mixed view of this question but we got both the answer correctly.
Raghu said: (Oct 13, 2015) The answer is 15 let me tell you. Tanya is 25% efficient than Sakshi. If Sakshi do a work in 24 hours then Tanya will do in 18 hours because 12 is 50% of 24. 6 is 25% of 24. So Tanya take 6 hours less time than Sakshi, so reduce 6 hours in every day upto 20 days 6 x 20 = 120 i.e 5 days. 20 days - 5 days = 15 days.
Sonu Jangra said: (Oct 13, 2015) Sakshi do 1 day's the work = 1/20. Now, Tanya do the work 25% more efficient than Sakshi = 1/20*25/100 = 1/80. So, Tanya will do the total work = 1/20+1/80 = 5/80 = 1/16. So 1*16/1 = 16 days.
Pankaj said: (Oct 17, 2015) Answer will be 25. If the ratio taken is T:S::T:S. 125:100::X:20. 5:4::X:20 = 25. x is no. of days taken by Tanya complete a work.
Subhasis said: (Feb 4, 2016) What if I think tanaya takes 75 days and sakshi takes 100 days, then answer will be 15 days.
Suresh Ks said: (Mar 19, 2016) Hi friends I agree your answer is 16 based on your methods. So please let me know how many days Tanya takes if she is 50% more efficient than sakshi based on your methods.
Ashwini said: (Jun 7, 2016) According to me, it should be; Tanya is 25% efficient than Sakshi so that means she finishes work taking 25% less time. So, 20 - 25 * 20, then finally we get 15 days.
Tathagata said: (Jun 13, 2016) @Ashwini you are right. If you see a previous sum. If you do it in the same way it will have the same answer!
Shadab Akhtar said: (Jun 15, 2016) Sakshi = 1/20 (given). Tanya = (125/100) * Sakshi (Tanya is 25% more efficient than Sakshi). Tanya= (125/100) * 1/20. = 1/16. Hence, Tanya does the same piece of work in 16 days.
Narayan said: (Aug 10, 2016) Let the ans is 15 then, Efficiency=((20 - 15)/15) * 100 = 33.33%(wrong). Take ans as 16 then, Efficiency = ((20 - 16)/16) * 100 = 25%(correct).
Santosh said: (Sep 18, 2016) Thanks @Kamlesh.
Bulu said: (Sep 22, 2016) Shaki = 20 days, Tanya = 25% more efficient, So that 25% of 20 = 5, 20 days - 5 days = 15 days.
Aashi Mehta said: (Nov 13, 2016) If we solve in accordance with the question then the answer will be 15 I don't know why every body has taken it as 16. Please tell me.
Ayush said: (Nov 26, 2016) @All. Refer the post of Kamlesh It's quite easy to understand, Thanks @Kamlesh.
Soniya said: (Dec 17, 2016) Ratio of work done by Tanya and Sakshi = 125 : 100 = 5 : 4. Ratio of times taken by Tanya and Sakshi = 4 : 5. 5x = 20days. ( Sakshi can complete the work in 20 days). x = 4. So, Tanya can finish the work in 16 days.
Cheenu said: (Dec 22, 2016) Can anybody explain this how x = 16 has come from this equation 5:4::20:x?
Moses said: (Mar 1, 2017) 15 is the answer because we are reducing Sakshi's work time by 25% Hence multiply 20 days by 0.75 which will give 15days. 15 days is definitely the answer.
Epsilon said: (Jul 21, 2017) Sakshi does 1 unit in =20Days. Tanya does (1+1/4)=5/4 unit in = 20Days. So, Tanya does one unit in=20*4/5 = 16 days, I think efficiency here is related to the amount of work done.
Sanjay said: (Aug 16, 2017) Tanya 1 day =100 unit (make). Sakshi 1 day =75 unit. Sakshi 20 days = 1500 unit make ( so this is total work ). Then tanya make = 1500 / 100 = 15 days. 15 days 1500 unit. 15 --> Can Be Answer.
Mithun said: (Aug 31, 2017) [Tanya effective than sakshi [20days] , hence answer should be less than 20days] Sakshi 125--------->20 Tanya 100-------->? [cross multiply] 20 x 100 / 125 =16days.
Atul said: (Sep 19, 2017) Sakshi cannot be taken as 125, as Sakshi is the reference point here, she must be taken as 100. So, Tanya becomes 75 (25% more efficient). Then the ratio becomes 4:3 instated of 5:4.
Sahithikarne said: (Jan 29, 2018) 25% of 20 is 5 then the answer is 20-5=15. But how it is 16?
Mahesh Neupane said: (Feb 10, 2018) Sakshi efficiency is 1/20. Tanya Efficiency is 25% more than Sakshi. So, 1/20+(25%of1/20)=16%.
Patrick said: (Feb 11, 2018) Sakshi efficiency = 100/20 = 5; Tanya = 5 x 1.25 = 6.25; The number of days taken by Tanya to do the same piece of work is: 100/6.25 = 16 days.
Manish Niroti said: (Feb 19, 2018) Tanya & sakshi capacity =5:4. Tanya & Sakshi time taken=4:5. If Sakshi taking 20 days it means 5 is representing 20 so 20/5 =4, and multiply 4 with taken by Tanya so you will get 4*4 =16. Ratio and proportion concept.
Komal Singh said: (May 30, 2018) @Sakshi. If takes 125 times to complete her work in 20 days, Tanya takes 100. Therefore if we cross multiply to find how much days will Tanya take to complete work. i.e 125/100*20/? we get 16!
Divyansh said: (Jul 14, 2018) I am not getting it how 16? Let's think this way Sakshi take 20 days which is 240 hours. 25% of 240 = 60. 240 - 60 = 180. and 180 hours = 15days. Please can anyone help me.
Sufiyan Ansari said: (Sep 3, 2018) See This for those Who do Time and work with LCM.. It is said that Sakshi can do a piece of work in 20 days with her efficiency so we cannot take 25%of 20. Here it is given 25% more efficient. So efficiency of Sakhsi : Tanya. Assume: S:T = 100unit : 125 unit (25%more). Therefore , Total work = 100unit x 20days. :. Tanya would do the same work in: Total work/ Tanya efficiency, :. 100unit x 20/125= 16 days.
Vishal Bhadikar said: (Sep 23, 2018) Why not 15 days? Explain.. 100%=20 day. 25%=x day. X = 5 days. Now take efficient that we need to subtract 20-5 = 15 days. Explain to me if I am wrong.
Srilatha said: (Oct 8, 2018) Sakshi complete the work in 20 days . Tanya is 25% more efficient means; in 20days 25%=5 days. it means t complete the work in 5 days earlier. so 20-5=15 answer. Then how it comes as 16?
Ahsan Naveed said: (Dec 1, 2018) @All. Tania is 25% more efficient then Sakshi so we can do it like, Tania=125 units/day, Sakhi = 100 units/day. As Sakhi done it in 20 days so we can find complete work done, Total work= 20 * 100 = 2000. We also have Tania unit work so we will divide total work with Tania unit work, Tania work= 2000/125 = 16 answer.
Pitambar said: (Jan 22, 2019) A certain number of men do a piece of work in 10 days, if there are 10 men less, it will take 10 days more to finish the work. How many men were there originally. Can anyone solve this to find a solution?
Kajal said: (Sep 13, 2019) Here (125% of 1/20)=1/16 that means in 16 days.
Ryan said: (Sep 28, 2019) @All. The sentence 'Tanya is 25% more efficient than Sakshi' means Tanya will take 25% less days than Sakshi and not that Sakshi will take 25% more days than Tanya. The answer given assumes the later part which is not correct. Percents work differently for the increase and decrease. For example, 25% more than 100 is 125. But 25% less than 125 is not a hundred. The answer would be 16 if the question had been stated as Sakshi is 25% less efficient than Tanya.
Annie said: (Oct 22, 2019) Can anyone please compare it with the 17th question?
Leela said: (Nov 4, 2019) @Ashan Naveed. Consider 100 unit by Sakshi Tanya 125 unit. Y can't we consider Sakshi by 125 days. Tanya in 100 days. Mainly unitary method how can we solve.
Oakley said: (Feb 1, 2020) 25% efficient means Tanya takes 15 days. Both 1 day work=1/20+1/15=7/60. Tanya work=7/60-1/20=1/15. Why can't the answer be 15? | 5,650 | 16,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-16 | latest | en | 0.91683 |
http://slideplayer.com/slide/3162294/ | 1,503,040,562,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104612.83/warc/CC-MAIN-20170818063421-20170818083421-00085.warc.gz | 379,839,443 | 22,957 | # “velocity” is group velocity, not phase velocity
## Presentation on theme: "“velocity” is group velocity, not phase velocity"— Presentation transcript:
“velocity” is group velocity, not phase velocity
Quantum Waves Particles are actually waves Two de Broglie relations They have all the properties (and equations) that described waves before, together with some new ones Rewrite the de Broglie relations in terms of k and : We can also find a dispersion relation for these waves: From this, we can find the group and phase velocity: “velocity” is group velocity, not phase velocity
Quantum Uncertainty Two uncertainty relations for any type of wave
Multiply by h-bar Use quantum wave relations Uncertainty in Position/Momentum Classical physics: Particle is defined by position x and velocity v = (dx/dt) Or, if you prefer, exact x and exact p Quantum physics: Particle is defined by wave function (x) It cannot have a definite position and momentum – they tend to both be uncertain Uncertainty in Time/Energy Things that last eternally can have definite energy Things that last a brief time have uncertain energy Measurements of energy will have a spread in value
Time/Energy Uncertainty
The 0-meson has a rest mass of MeV/c2 and lasts an average of 8.410-17 s. What is the spread in energies of a 0-meson at rest due to its finite lifespan? E E E - mc2 (eV)
These methods give only estimates of the answer
Consequences of Position Uncertainty Position and momentum of a particle cannot be simultaneously specified or determined We can often estimate one quantity if we know the other by treating this inequality as if it were an equality Especially when the energy is being minimized L Energy of Particle in a Box By Carlson’s rule, the position is uncertain by about x = ¼L. By the uncertainty principle, momentum is uncertain by We’d like the momentum to be zero, but we can’t This causes the particle to have some energy, called Zero Point Energy ? These methods give only estimates of the answer
Solving Uncertainty/Energy Problems
A particle of mass m lies above an impenetrable barrier in a gravitational field with acceleration g. What is the minimum energy of the particle? y Write an expression for the energy: potential and kinetic Figure out what momentum (normally 0) and what position would have the “ideal” lowest kinetic and potential energies Let x = a and p = /2a Assume the momentum and position differ from ideal by about x = a and p =/2a Set derivative of energy function equal to zero, solve for a Substitute a in to determine minimum energy
Uncertainty and the Hydrogen Atom
There are two types of energy associated with the Hydrogen atom Potential Energy Kinetic Energy Classically, these two energies are at a minimum when: The electron is at rest, p = 0 The electron is at the origin, r = 0 (x = 0) In this case, the energy of the hydrogen atom E = – Quantum mechanically, you can’t control both position and momentum We will place the electron near the origin, r = 0, but there is an uncertainty x associated with it We will place the electron nearly at rest, p = 0, but there is an uncertainty p associated with it You have to compromise between these two choices to minimize E Let x = a, and let p be as small as possible by the uncertainty principle Plug these into the formula for energy
Uncertainty and the Hydrogen Atom
We need to minimize the energy Take derivative and set it to zero Substitute result back in Correct answer is 4 times smaller than this. Why so far off? It’s just an estimate There is momentum in three dimensions This makes answer 3 times smaller Then answer we get is 4/3 of correct answer
“Electric Field is potential for force at a distance on a charge”.
The Wave Function We’ve been talking about waves, which require wave functions We’d better give it a name: It would have more arguments in more dimensions It is a complex function; it has both a real and imaginary parts What does it mean? This turns out to be very hard Analogy: Electromagnetic Waves Described by electric and magnetic fields What is an electric field? It is defined in terms of a potential for something American Heritage Science Dictionary “The distribution in space of the strength and direction of forces that would be exerted on an electric charge at any point in that space”. Dr. Carlson, PHY 114 “Electric Field is potential for force at a distance on a charge”.
The Wave Function: what does it mean?
We are talking about one particle – but it is not at one location in space If we measured its position, where would we be likely to find it? The Wave Function is also called the probability amplitude Clearly, where the wave function is small (or zero), you wouldn’t expect to find the particle Where it’s negative or imaginary, wouldn’t expect to have negative or imaginary probability We’d better make darn sure that the probability is always positive For electric fields, the energy density is proportional to the field squared If working with complex waves, take amplitude first How about we make probability density proportional to wave function magnitude squared:
Sample Problem A wave in the region 0 < x < a has the wave function above. What is the probability density at all locations x at all times t?
Sample Problem A wave in the region 0 < x < a has the wave function above. What is the probability density at all locations x at all times t?
… in the region 0 < x < a …
What Does Probability Density Mean? The probability density (in 1D) has units of m-1 In a small region of size dx, the probability of finding the particle is there is given by ||2dx. To find probability over a larger region, you have to integrate it Normalization: The probability that the particle is somewhere must be 1 If we integrate over all x, we must get 1 In some cases, the problem implies that we restrict to some region … in the region 0 < x < a …
Sample Problem At t = 0, the wave function is given by the expression below. What is the normalization constant N? What is the probability that the particle is at x > ½a?
Sample Problem At t = 0, the wave function is given by the expression below. What is the normalization constant N? What is the probability that the particle is at x > ½a?
Sample Problem At t = 0, the wave function is given by the expression above. What is the most likely / least likely places to find the particle? What is the normalization constant N? What is the probability that the particle is at 0 < x < a? Least likely when function vanishes, at x = 0 Most likely when function is largest positive or negative Normalization: Let x = atan
Sample Problem At t = 0, the wave function is given by the expression above. What is the most likely / least likely places to find the particle? What is the normalization constant N? What is the probability that the particle is at 0 < x < a?
End of material for Test 2
Quantum Wave Equations You Need: End of material for Test 2 | 1,574 | 6,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-34 | longest | en | 0.898464 |
https://community.ibm.com/community/user/datascience/communities/community-home/digestviewer/viewthread?GroupId=4003&MessageKey=89182529-7e84-45be-be15-5dd35875dc8a&CommunityKey=886b6874-0fb1-402c-8243-c70ef8179a99&tab=digestviewer&ReturnUrl=%2Fcommunity%2Fuser%2Fdatascience%2Fcommunities%2Fcommunity-home%2Fdigestviewer%3FCommunityKey%3D886b6874-0fb1-402c-8243-c70ef8179a99 | 1,653,431,010,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00267.warc.gz | 226,844,416 | 274,822 | # SPSS Statistics
View Only
## Linear mixed models - interpretation of results
• #### 1. Linear mixed models - interpretation of results
Posted Mon November 08, 2021 09:59 AM
Hello, I have a mixed model wiith one categorical and one continuous predictor. When I input the continuous one as a fixed effect, both the "tests of fixed effects" and "estimates of fixed effects" table gives significant results for the continuous variable. When I input it as a covariate, suddenly there is a discrepancy between the tables - in "tests of fixed effects" it remains significant, while in "estimates of fixed effects" it becomes insignificant.
This is strange for me. If a continuous variable cannot be a fixed factor (as I understood from reading the web) - why do these results differ in both tables??
Also - there is one more difference between these two analyses - the categorical variable's estimate changes the sign. What statistics should I report to follow APA style?
------------------------------
Karolina Ziembowicz
------------------------------
• #### 2. RE: Linear mixed models - interpretation of results
Posted Mon November 08, 2021 11:25 AM
If you treat a continuous variable as a factor, you are giving it the number of degrees of freedom as the number of levels (ignoring the constant term). So that allows for a nonlinear relationship between the presumed factor and the dependent variable while as a covariate it would only pick up a linear relationship with one degree of freedom.
In other words, treating it as a factor is equivalent to creating a dummy variable for each level.
--
• #### 3. RE: Linear mixed models - interpretation of results
Posted Mon November 08, 2021 11:50 AM
Ok, I get it. Thanks for the nice clarification. So - are you saying I should not worry that the estimate and t-test are insignificant in the "tests of fixed effects" table? Should I report the value of F and ignore the "estimates of fixed effects" table?
------------------------------
Karolina Ziembowicz
------------------------------
• #### 4. RE: Linear mixed models - interpretation of results
Posted Mon November 08, 2021 04:08 PM
I am not sure which procedure you are running. Are you backtranslating table names from another language?
But if you get different significance results as a factor from as a covariate, that might suggest a nonlinearity, so you might want to try adding a quadratic term as a covariate or fitting a cubic spline, which you could create with the STATS SPLINES extension command.
--
• #### 5. RE: Linear mixed models - interpretation of results
Posted Tue November 09, 2021 04:57 AM
Hello, I tried doing as you suggested (added a cubic term as a covariate), it did not help. I am using the linear mixed models procedure and I am not back translating. I attach the output in which I conduct the normal procedure first and then go on with the cubic term. The covariate is indeed not normal, I used LN function which normalized the distribution but it also gives the same combination of significant/insignificant results between tables.
Any other ideas? I'm pretty confused.
------------------------------
Karolina Ziembowicz
------------------------------
Attachment(s)
output_SPSS.pdf 34 KB 1 version
• #### 6. RE: Linear mixed models - interpretation of results
Posted Tue November 09, 2021 09:02 AM
How very strange - I recoded my fixed factor from 01 into 10 (just swapped levels) and suddenly the tables give coherent results.
------------------------------
Karolina Ziembowicz
------------------------------
• #### 7. RE: Linear mixed models - interpretation of results
Posted Tue November 09, 2021 09:11 AM
I see that you are using MIXED, but the message is the same as with other procedures that support factors and covariates. In your case, this is more complicated, because you have an interaction term in the model as well as the main effects. You cannot judge the effect of a variable just from the estimates of individual fixed effects terms. If you need more help, I would refer you to the Linear Mixed Models Case Study available in the help system. | 907 | 4,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.875863 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/rishitc/solving-codeforces-problem-136a-presents-246g | 1,669,816,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00374.warc.gz | 502,359,440 | 32,430 | ## DEV Community π©βπ»π¨βπ» is a community of 963,503 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
Rishit Chaudhary
Posted on
# Solving Codeforces Problem: 136A - Presents
Today we are going to attempt to solve the Codeforces problem: 136A - Presents. Alright! so let's begin.
## The Question
### Problem Statement
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited n his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to n. Petya remembered that a friend number i gave a gift to a friend number pi. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend i the number of a friend who has given him a gift.
### Input
The first line contains one integer n (1ββ€βnββ€β100) β the quantity of friends Petya invited to the party. The second line contains n space-separated integers: the i-th number is pi β the number of a friend who gave a gift to friend number i. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
### Output
Print n space-separated integers: the i-th number should equal the number of the friend who gave a gift to friend number i.
## My Analysis with Sample Examples
Now, before discussing the solution, I first need to address a mistake in the description in the input of the question as given on Codeforces. In the input description, it says that:
the i-th number is pi β the number of a friend who gave a gift to friend number i.
This is not how the input is given in the various test cases. On closer inspection this is very similar to the description of the expected output of the questions:
i-th number should equal the number of the friend who gave a gift to friend number i.
So, now the natural question you'll be asking is "What exactly is the input format?"
If we re-examine the description of the Problem Statement above, we see the line:
He numbered all his friends with integers from 1 to n. Petya remembered that a friend number i gave a gift to a friend number pi.
Eureka! This is the correct description of the input format! We can verify this as well with the given sample examples. Let's try that now.
### Example-1
#### Input
4
2 3 4 1
Here, on the first line we are given that Petya invited 4 friends to the party (I sincerely hope none of his friends got left out π
), and on next line we are given a list of numbers that needs to be interpreted as follows (Notice: The numbers in bold are the ones visible in the input, in that order):
1. Friend 1 gave a present to Friend 2
2. Friend 2 gave a present to Friend 3
3. Friend 3 gave a present to Friend 4
4. Friend 4 gave a present to Friend 1
Alright, so now what we need to do is to output a list of numbers such that at the `i-th` position we give the number of the friend who gave a present to friend `i`.
So our output should be:
4 1 2 3
Which should be interpreted as (Notice: The numbers in bold are the ones visible in the output, in that order):
1. Friend 4 gave a present to Friend 1
2. Friend 1 gave a present to Friend 2
3. Friend 2 gave a present to Friend 3
4. Friend 3 gave a present to Friend 4
4 1 2 3
### Example-2
#### Input
3
1 3 2
Here, on the first line we are given that Petya invited 3 friends to the party, and on next line we are given a list of numbers that needs to be interpreted as follows (Notice: The numbers in bold are the ones visible in the input, in that order):
1. Friend 1 gave a present to Friend 1
2. Friend 2 gave a present to Friend 3
3. Friend 3 gave a present to Friend 2
Alright, so other than the fact that Friend 1 doesn't seem to have the New Year spirit, what we need to do is to output a list of numbers such that at the `i-th` position we give the number of the friend who gave a present to friend `i`.
So our output should be:
1 3 2
Which should be interpreted as (Notice: The numbers in bold are the ones visible in the output, in that order):
1. Friend 1 gave a present to Friend 1
2. Friend 3 gave a present to Friend 2
3. Friend 2 gave a present to Friend 3
1 3 2
### Example-3
#### Input
2
1 2
Here, on the first line we are given that Petya invited 2 friends to the party, and on next line we are given a list of numbers that needs to be interpreted as follows (Notice: The numbers in bold are the ones visible in the input, in that order):
1. Friend 1 gave a present to Friend 1
2. Friend 2 gave a present to Friend 2
Alright, so other than the fact that Friend 1 and Friend 2 should starting asking themselves why they came with presents to the party, what we need to do is to output a list of numbers such that at the `i-th` position we give the number of the friend who gave a present to friend `i`.
So our output should be:
1 2
Which should be interpreted as (Notice: The numbers in bold are the ones visible in the output, in that order):
1. Friend 1 gave a present to Friend 1
2. Friend 2 gave a present to Friend 2
1 2
## Code and Complexity
Alright, so now we know how to interpret out input and process it to get our output. To be extra sure we have also verified our logic over the given examples as well. Let's now translate this logic into code. For this I'll be using modern C++ (C++11 and beyond), so that it's easier to get the big picture logic which is implemented in the code.
Another interesting reason to use modern C++ is that some of its features make the code more readable and similar looking to Python code (many people consider this a plus).
``````/**
* @file 136A-Presents.cpp
* @author Rishit Chaudhary (@rishitc)
* @version 1.0
* @date 2021-06-23
*
*
*/
#include <iostream>
int main()
{
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
int n;
int p;
std::cin >> n;
int ans[n];
for (int i = 1; i < n + 1; ++i)
{
std::cin >> p;
ans[p - 1] = i; // Zero index correction for p
}
for (int i = 0; i < n; ++i)
{
std::cout << ans[i] << " ";
}
std::cout << "\n";
}
``````
### Complexity
Time Complexity O(n) n (1ββ€βnββ€β100) β the quantity of friends Petya invited to the party.
Space Complexity O(n) n (1ββ€βnββ€β100) β the quantity of friends Petya invited to the party.
Difficulty (subjective value) Easy The use of language in the question combined with the mistake in the description of the input does make the question a bit hard to grasp at first. | 1,845 | 7,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-49 | longest | en | 0.964352 |
http://sites.science.oregonstate.edu/physics/portfolioswiki/courses:hw:wvhw:wvhwsubtopic2 | 1,590,958,780,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413624.48/warc/CC-MAIN-20200531182830-20200531212830-00469.warc.gz | 110,716,931 | 6,183 | ## Homework for Waves (wave equation)
Describe the following waveforms in words (waveform, period, phase angle, direction & speed of travel … $etc$.). Demonstrate whether they are, or are not, solutions to the non-dispersive wave equation $\frac{{{\partial }^{2}}}{\partial {{t}^{2}}}\psi (x,t)={{v}^{2}}\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}\psi (x,t)$.
(a) $\psi \left( x,t \right)=4\cos \left( 4\pi x+3\pi t \right)-4\sin \left( 4\pi x+3\pi t \right)$
(b) $\psi \left( x,t \right)=3\cos \left( 2\pi x \right)\sin \left( \pi t \right)$
(c) $\psi \left( x,t \right)=3{{e}^{-\alpha x}}\cos \left( \frac{2\pi }{3}x-\pi t \right)$ , $\alpha$ is a constant
##### Views
New Users
Curriculum
Pedagogy
Institutional Change
Publications | 268 | 745 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-24 | latest | en | 0.571836 |
https://codeforces.com/blog/entry/112071 | 1,718,426,210,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00531.warc.gz | 159,903,163 | 20,930 | ### tolbi's blog
By tolbi, 16 months ago,
level: **VERY HARD*
hello today am gonna tech xor basis/ but proofs are very hard so no proof. it inituavautve
okl we hvae arary basis[37] mane basis[i] mst's i so we habvee
flolpowing codem for is to isnert x in tonhe the set;/ buist it's bveyr har adsso i wont';t exp[alaomn
# include<stdc++/bits.h>
using namespace = std; typedef int long long; void insert(x){ for(i in 30 to 0) if(x's i'th bit is not 1) : ccinonitue if(basis[i] == -1) basis[i] = x; break; else x = x $\oplus$ basis[i] } ~~~~~
isneritng is same aas qureying hwo to i mane queery if ew cna maefae the X swith gigven xorn basis ealneaments
UPD: code for query
# include<stdc++/bits.h>
using namespace = std; typedef int long long; void insert(x){ for(i in 30 to 0) if(x's i'th bit is not 1) : ccinonitue if(basis[i] == -1) return false; else x = x $\oplus$ basis[i] return true;} ~~~~~
• +1
| Write comment?
» 16 months ago, # | +6 Auto comment: topic has been updated by tolbi (previous revision, new revision, compare).
» 16 months ago, # | +31 thaknss fo4r th1s awsoeme edutiroal!
» 16 months ago, # | +17 your code didn't compile. can you fix it?
»
16 months ago, # |
Rev. 2 0
# If you enter:
cpp
#include <iostream>
using namespace std;
int main{
int array[10];
return 0;
}
# Then it will show:
#include <iostream>
int main{
int array[10];
return 0;
}
• » » 16 months ago, # ^ | 0 I thonk ne dids on purpusr
• » » » 16 months ago, # ^ | +8 Y er porbly ritgh
» 16 months ago, # | ← Rev. 3 → +3 thenk yu for the explanat1on!!! ay ken n0w AC oll ex0R pr0bbblamss
» 16 months ago, # | +11 ay dond undstersatnd pls fhelp
» 16 months ago, # | +14 Very gut tutoral. Pleas more
» 16 months ago, # | +8 n1ec bolg, nwo 1 c4n b3 a lgm in 23 s3c0nds
» 16 months ago, # | +4 Veyr ince epxlantaion, txh!
» 16 months ago, # | +3 Tghsnkcs orf isth aeswome tut6969
» 16 months ago, # | 0 r/ihadastroke
» 16 months ago, # | +11 too difficult to understand, had to go through a grad level linear algebra textbook to even understand the notation. downvoted ù_ú
» 16 months ago, # | +8 yur qurey si viod soo it caonot reutn bolo. | 785 | 2,194 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-26 | latest | en | 0.51145 |
https://www.lotterypost.com/thread/268243/162 | 1,480,760,367,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540915.89/warc/CC-MAIN-20161202170900-00454-ip-10-31-129-80.ec2.internal.warc.gz | 985,291,978 | 23,727 | Welcome Guest
You last visited December 3, 2016, 4:45 am
All times shown are
Eastern Time (GMT-5:00)
# We Are The Numeralogic Masters..
Topic closed. 5514 replies. Last post 3 years ago by Harve\$t Moon.
Page 162 of 368
nassau
Bahamas
Member #120651
December 24, 2011
3190 Posts
Offline
Posted: January 2, 2014, 7:05 pm - IP Logged
Sorry for being Late with this.....
So From now on we use this for the rest of the Year...
......................................................................................................
I will Call this ...
OUR 4 Ball Repeat Proccessor....
LCHA 448 7147 Wednesday, January 01, 2014
ENY 248 1774 Monday, December 30, 2013
......................................................................................................................................................
So we had our First Repeater...
This was its work out...
7147..=..1774..=..2908..=..7453..=..0540..=..5090..
7147..=..8371..=.3826..=..6913..=.1468..
EMIA 228 4537 Thursday, January 02, 2014
So that we never all miss Something Like this ever again....
I bring to you...
OUR 4 Ball Repeat Proccessor....
Now that is some real insight the repeaters in pick 3 also bring prs in pick 4 wow............... Thanks Mike will keep a look out for pick 4 repeaters as well going forward.......
Show me the \$\$\$\$\$money\$\$\$\$\$ ....working on my retirement plan
nassau
Bahamas
Member #120651
December 24, 2011
3190 Posts
Offline
Posted: January 2, 2014, 7:13 pm - IP Logged
Ok Lets set it out heading into Eve Draw
STEP 1
-The Due Pair List we have prs 00 49 78 so our Due Pair list looks like this
000 010 020 030 040 050 060 070 080 090
049 149 249 349 449 549 649 749 849 949
078 178 278 378 478 578 678 778 878 978
STEP 2
Combine any common prs-NIL
Strategy 1
Play back any repeats:
04-004 104 204 304 404 504 604 704 804 904
45-045 145 245 345 445 545 645 745 845 945
59-059 159 259 359 459 559 659 759 859 959
-----------------------------------------------------------------------
07-**007** **107** 207 307 407 **507** 607 707 807 907
02-002 102 202 302 402 502 602 702 802 902
25-025 125 225 325 425 525 625 725 825 925
-----------------------------------------------------------------------
47-047 147 247 347 447 547 647 747 847 947
29-029 129 229 329 429 529 629 729 829 929
79-079 179 279 379 479 579 679 779 879 979
-----------------------------------------------------------------------
12-012 112 212 312 412 512 612 **712** 812 912
17-**017** 117 **217** 317 417 517 617 717 817 917
26-026 126 226 326 426 526 626 726 826 926
67-067 167 267 367 467 567 667 767 867 967
16-016 116 216 316 416 516 616 716 816 916
11-011 111 112 113 411 511 611 711 811 911
66-066 166 266 366 466 566 666 766 866 966
284
24-024 124 224 324 424 524 624 724 824 924
79-079 179 279 379 479 579 679 779 879 979<<<<<GA>>>>
47-047 147 247 347 447 547 647 747 847 947
-----------------------------------------------------------------------------
28-028 128 228 328 428 528 628 728 828 928<<<<FL>>>>
37-037 137 237 337 437 537 637 737 837 937<<<NY>>>>>
23-023 123 223 323 423 523 623 723 823 923
-----------------------------------------------------------------------------
48-048 148 248 348 448 548 648 748 848 948
39-039 139 239 339 439 539 639 739 839 939
34-034 134 234 334 434 534 634 734 834 934<<<<<IL>>>>>
As a 123+/-
074...197...642...951...406
740...863...318...627...172
248...361...816...125...670
284...307...852...161...616
162...285...730...049...594
621...744...299...508...053
Traveller combo:
074...174...974...064...084...075...073...574...079...579...529
162...262..062...152...172...163...161...662...167...667...617
So GA 032 off the 284 and 074
Strategy 1
Play back any repeats:
04-004 104 204 304 404 504 604 704 804 904
45-045 145 245 345 445 545 645 745 845 945
59-059 159 259 359 459 559 659 759 859 959
-----------------------------------------------------------------------
07-**007** **107** 207 307 407 **507** 607 707 807 907
02-002 102 202 302 402 502 602 702 802 902
25-025 125 225 325 425 525 625 725 825 925
-----------------------------------------------------------------------
47-047 147 247 347 447 547 647 747 847 947
29-029 129 229 329 429 529 629 729 829 929
79-079 179 279 379 479 579 679 779 879 979
-----------------------------------------------------------------------
12-012 112 212 312 412 512 612 **712** 812 912
17-**017** 117 **217** 317 417 517 617 717 817 917
26-026 126 226 326 426 526 626 726 826 926
67-067 167 267 367 467 567 667 767 867 967
16-016 116 216 316 416 516 616 716 816 916
11-011 111 112 113 411 511 611 711 811 911
66-066 166 266 366 466 566 666 766 866 966
284
24-024 124 224 324 424 524 624 724 824 924
79-079 179 279 379 479 579 679 779 879 979<<<<<GA>>>>
47-047 147 247 347 447 547 647 747 847 947
-----------------------------------------------------------------------------
28-028 128 228 328 428 528 628 728 828 928<<<<FL>>>>
37-037 137 237 337 437 537 637 737 837 937<<<NY>>>>>
23-023 123 223 323 423 523 623 723 823 923
-----------------------------------------------------------------------------
48-048 148 248 348 448 548 648 748 848 948
39-039 139 239 339 439 539 639 739 839 939
34-034 134 234 334 434 534 634 734 834 934<<<<<IL>>>>>
Show me the \$\$\$\$\$money\$\$\$\$\$ ....working on my retirement plan
nassau
Bahamas
Member #120651
December 24, 2011
3190 Posts
Offline
Posted: January 2, 2014, 7:40 pm - IP Logged
So GA 032 off the 284 and 074
Strategy 1
Play back any repeats:
04-004 104 204 304 404 504 604 704 804 904
45-045 145 245 345 445 545 645 745 845 945
59-059 159 259 359 459 559 659 759 859 959
-----------------------------------------------------------------------
07-**007** **107** 207 307 407 **507** 607 707 807 907
02-002 102 202 302 402 502 602 702 802 902
25-025 125 225 325 425 525 625 725 825 925
-----------------------------------------------------------------------
47-047 147 247 347 447 547 647 747 847 947
29-029 129 229 329 429 529 629 729 829 929
79-079 179 279 379 479 579 679 779 879 979
-----------------------------------------------------------------------
12-012 112 212 312 412 512 612 **712** 812 912
17-**017** 117 **217** 317 417 517 617 717 817 917
26-026 126 226 326 426 526 626 726 826 926
67-067 167 267 367 467 567 667 767 867 967
16-016 116 216 316 416 516 616 716 816 916
11-011 111 112 113 411 511 611 711 811 911
66-066 166 266 366 466 566 666 766 866 966
284
24-024 124 224 324 424 524 624 724 824 924
79-079 179 279 379 479 579 679 779 879 979<<<<<GA>>>>
47-047 147 247 347 447 547 647 747 847 947
-----------------------------------------------------------------------------
28-028 128 228 328 428 528 628 728 828 928<<<<FL>>>>
37-037 137 237 337 437 537 637 737 837 937<<<NY>>>>>
23-023 123 223 323 423 523 623 723 823 923
-----------------------------------------------------------------------------
48-048 148 248 348 448 548 648 748 848 948
39-039 139 239 339 439 539 639 739 839 939
34-034 134 234 334 434 534 634 734 834 934<<<<<IL>>>>>
So GA 032 off the 284 and 074
TRI 985 OFF THE 407
So going with pr 29 for FLA
Strategy 1
Play back any repeats:
04-004 104 204 304 404 504 604 704 804 904
45-045 145 245 345 445 545 645 745 845 945
59-059 159 259 359 459 559 659 759 859 959 <<<<TR>>>>
-----------------------------------------------------------------------
07-**007** **107** 207 307 407 **507** 607 707 807 907
02-002 102 202 302 402 502 602 702 802 902 <<<<GA>>>>
25-025 125 225 325 425 525 625 725 825 925
-----------------------------------------------------------------------
47-047 147 247 347 447 547 647 747 847 947
29-029 129 229 329 429 529 629 729 829 929
79-079 179 279 379 479 579 679 779 879 979
-----------------------------------------------------------------------
12-012 112 212 312 412 512 612 **712** 812 912
17-**017** 117 **217** 317 417 517 617 717 817 917
26-026 126 226 326 426 526 626 726 826 926
67-067 167 267 367 467 567 667 767 867 967
16-016 116 216 316 416 516 616 716 816 916
11-011 111 112 113 411 511 611 711 811 911
66-066 166 266 366 466 566 666 766 866 966
284
24-024 124 224 324 424 524 624 724 824 924
79-079 179 279 379 479 579 679 779 879 979<<<<<GA>>>>
47-047 147 247 347 447 547 647 747 847 947
-----------------------------------------------------------------------------
28-028 128 228 328 428 528 628 728 828 928<<<<FL>>>>
37-037 137 237 337 437 537 637 737 837 937<<<NY>>>>>
23-023 123 223 323 423 523 623 723 823 923<<<<GA>>>>
-----------------------------------------------------------------------------
48-048 148 248 348 448 548 648 748 848 948
39-039 139 239 339 439 539 639 739 839 939
34-034 134 234 334 434 534 634 734 834 934<<<<<IL>>>>>
Show me the \$\$\$\$\$money\$\$\$\$\$ ....working on my retirement plan
Bahamas
Member #139033
February 12, 2013
2733 Posts
Offline
Posted: January 2, 2014, 8:05 pm - IP Logged
Sorry for being Late with this.....
So From now on we use this for the rest of the Year...
......................................................................................................
I will Call this ...
OUR 4 Ball Repeat Proccessor....
LCHA 448 7147 Wednesday, January 01, 2014
ENY 248 1774 Monday, December 30, 2013
......................................................................................................................................................
So we had our First Repeater...
This was its work out...
7147..=..1774..=..2908..=..7453..=..0540..=..5090..
7147..=..8371..=.3826..=..6913..=.1468..
EMIA 228 4537 Thursday, January 02, 2014
So that we never all miss Something Like this ever again....
I bring to you...
OUR 4 Ball Repeat Proccessor....
......................................................................................................
I will Call this ...
OUR 4 Ball Repeat Proccessor....
7147..=..1774..=..2908..=..7453..=..0540..=..5090..
7147..=..8371..=.3826..=..6913..=.1468..
1/2/2014 Late-Pennsylvania (4-Ball, PA) 1693 1/2/2014 Late-Pennsylvania (3-Ball, PA) 803 1/2/2014 Late-Pennsylvania (2-Ball, PA) 03
Bahamas
Member #139033
February 12, 2013
2733 Posts
Offline
Posted: January 2, 2014, 8:42 pm - IP Logged
......................................................................................................
I will Call this ...
OUR 4 Ball Repeat Proccessor....
7147..=..1774..=..2908..=..7453..=..0540..=..5090..
7147..=..8371..=.3826..=..6913..=.1468..
1/2/2014 Late-Pennsylvania (4-Ball, PA) 1693 1/2/2014 Late-Pennsylvania (3-Ball, PA) 803 1/2/2014 Late-Pennsylvania (2-Ball, PA) 03
OUR 4 Ball Repeat Proccessor....
7147..=..1774..=..2908..=..7453..=..0540..=..5090..
7147..=..8371..=.3826..=..6913..=.1468..
1/2/2014 Late-Pennsylvania (4-Ball, PA) 1693 1/2/2014 Late-Pennsylvania (3-Ball, PA) 803 1/2/2014 Late-Pennsylvania (2-Ball, PA) 03
So now we found a way to find our Slippery Decoherence Results..
LMIA 803 1214 Thursday, January 02, 2014
United States
Member #125799
March 31, 2012
1078 Posts
Offline
Posted: January 2, 2014, 9:31 pm - IP Logged
......................................................................................................
I will Call this ...
OUR 4 Ball Repeat Proccessor....
7147..=..1774..=..2908..=..7453..=..0540..=..5090..
7147..=..8371..=.3826..=..6913..=.1468..
1/2/2014 Late-Pennsylvania (4-Ball, PA) 1693 1/2/2014 Late-Pennsylvania (3-Ball, PA) 803 1/2/2014 Late-Pennsylvania (2-Ball, PA) 03
803 fl eve. where did this come from I did'nt see this number tonite in a lot of workouts.
1214 play 4 fl eve. how nice Jan 2, 2014 (1214) why could'nt I have thought of this? UGH!!!!!!!!!!!!!!!!!!!
Allentown, Brookyn, Pa, Ny
United States
Member #134253
October 22, 2012
5460 Posts
Offline
Posted: January 2, 2014, 10:46 pm - IP Logged
OUR 4 Ball Repeat Proccessor....
7147..=..1774..=..2908..=..7453..=..0540..=..5090..
7147..=..8371..=.3826..=..6913..=.1468..
1/2/2014 Late-Pennsylvania (4-Ball, PA) 1693 1/2/2014 Late-Pennsylvania (3-Ball, PA) 803 1/2/2014 Late-Pennsylvania (2-Ball, PA) 03
So now we found a way to find our Slippery Decoherence Results..
LMIA 803 1214 Thursday, January 02, 2014
WTG master pa tonight 803 str888!
Bahamas
Member #139033
February 12, 2013
2733 Posts
Offline
Posted: January 3, 2014, 8:05 am - IP Logged
Sorry for being Late with this.....
So From now on we use this for the rest of the Year...
......................................................................................................
I will Call this ...
OUR 4 Ball Repeat Proccessor....
LCHA 448 7147 Wednesday, January 01, 2014
ENY 248 1774 Monday, December 30, 2013
......................................................................................................................................................
So we had our First Repeater...
This was its work out...
7147..=..1774..=..2908..=..7453..=..0540..=..5090..
7147..=..8371..=.3826..=..6913..=.1468..
EMIA 228 4537 Thursday, January 02, 2014
So that we never all miss Something Like this ever again....
I bring to you...
OUR 4 Ball Repeat Proccessor....
Now We will Prove to The Lottery World that Our Logics and Witts
Are Far Superior that Anyone Else on this Planet...
...................................................................................................
...................................................................................................
...................................................................................................
This is the Chart of Ancient Secrets..
00..11..22..33..44..55..66..77..88..99.
01.02. 03. 04. 05. 06. 07. 08. 09.
12. 13. 14. 15. 16. 17. 18. 19.
23. 24. 25. 26. 27. 28. 29.
34. 35. 36. 37.38. 39.
45. 46. 47. 48. 49.
56. 57. 58. 59.
67. 68. 69.
78. 79.
89.
......................................................................................................
So are we the Master's or What.....
Our Last REAL Pairs Are..///
049..149..249..349..449..549..649..749..849..949..
078..178..278..378..478..578..678..778..878..978..
Good LUCK EVERYONE///
MEANING THESE ARE WHAT WE SPEND HEAVY ON>.
USING STRATEGY
.......................................................................................................
nassau
Bahamas
Member #120651
December 24, 2011
3190 Posts
Offline
Posted: January 3, 2014, 8:11 am - IP Logged
Morning all eve draw was not kind, no winners at all
So playing back our due pair list, o/s prs 00 49 78
000 010 020 030 040 050 060 070 080 090
049 149 249 349 449 549 649 749 849 949
078 178 278 378 478 578 678 778 878 978
Playing back our repeats 308, 612 & 245
12-012 112 212 312 412 512 612 712 812 912
26-026 126 226 326 426 526 626 726 826 926
17-017 117 217 317 417 517 617 717 817 917
67-067 167 267 367 467 567 667 767 867 967
16-016 116 216 316 416 516 616 716 816 916
11-011 111 112 113 411 511 611 711 811 911
66-066 166 266 366 466 566 666 766 866 966
03-030 103 203 303 403 503 603 703 803 903
35-035 135 235 335 435 535 635 735 835 935
58-058 158 258 358 458 558 658 758 858 958
08-080 108 208 308 408 508 608 *708* 808 908
38-038 138 238 338 438 538 638 738 838 938
33-033 133 233 333 433 533 633 733 833 933
88-088 188 288 388 488 588 688 788 888 988
24-024 124 224 324 424 524 624 724 824 924
47-047 147 247 347 447 547 647 747 847 *947*
79-079 179 279 379 *479* 579 679 779 879 979
25-025 125 225 325 425 525 625 725 825 925
02-020 120 220 320 420 520 602 702 802 902
07-070 107 207 307 407 507 607 707* 807* 907
45-045 145 245 345 445 545 645 745 845 945
04-040 104 204 304 404 504 604 704 804 *904*
09-090 109 209 309 *409* 509 609 709 809 909
Legend:
xxxx-Combos Recently Drawn
xxxx-Combos on our Due Pair List
123+/- Count
162...285...730...049...594
621...744...299...508...053
803...926...471...780...235
308...421...976...285...730
254...377...822...131...686
542...665...110...429...972
Traveller Combo
074...174...974...064...084...075...073...574...079...579...529
162...262...062...152...172...163...161...662...167...667...617
308...408...208...398...318...309...307...808...303...803...858
542...642...442...532...552...543...541...042...547...047...097
Show me the \$\$\$\$\$money\$\$\$\$\$ ....working on my retirement plan
Bahamas
Member #139033
February 12, 2013
2733 Posts
Offline
Posted: January 3, 2014, 8:17 am - IP Logged
Now We will Prove to The Lottery World that Our Logics and Witts
Are Far Superior that Anyone Else on this Planet...
...................................................................................................
...................................................................................................
...................................................................................................
This is the Chart of Ancient Secrets..
00..11..22..33..44..55..66..77..88..99.
01.02. 03. 04. 05. 06. 07. 08. 09.
12. 13. 14. 15. 16. 17. 18. 19.
23. 24. 25. 26. 27. 28. 29.
34. 35. 36. 37.38. 39.
45. 46. 47. 48. 49.
56. 57. 58. 59.
67. 68. 69.
78. 79.
89.
......................................................................................................
So are we the Master's or What.....
Our Last REAL Pairs Are..///
049..149..249..349..449..549..649..749..849..949..
078..178..278..378..478..578..678..778..878..978..
Good LUCK EVERYONE///
MEANING THESE ARE WHAT WE SPEND HEAVY ON>.
USING STRATEGY
.......................................................................................................
So From now on we use this for the rest of the Year...
......................................................................................................
This is ...
OUR 4 Ball Repeat Proccessor....
LCAL 967 5783 Thursday, January 02, 2014
ECHA 403 7853 Thursday, January 02, 2014
....................................................................................................
5783..=..7853..=..8087..=..3532..=..6629..=..1174..
5783..=..6917..=..1462..=..4559..=..9004..
LCHA 542 6266 Thursday, January 02, 2014
Well its a start....
Bahamas
Member #139033
February 12, 2013
2733 Posts
Offline
Posted: January 3, 2014, 9:17 am - IP Logged
So From now on we use this for the rest of the Year...
......................................................................................................
This is ...
OUR 4 Ball Repeat Proccessor....
LCAL 967 5783 Thursday, January 02, 2014
ECHA 403 7853 Thursday, January 02, 2014
....................................................................................................
5783..=..7853..=..8087..=..3532..=..6629..=..1174..
5783..=..6917..=..1462..=..4559..=..9004..
LCHA 542 6266 Thursday, January 02, 2014
Well its a start....
So From now on we use this for the rest of the Year...
......................................................................................................
This is ...
OUR 4 Ball Repeat Proccessor....
ETRI 405 7122 Sunday, December 29, 2013
1/2/2014 Late-Kentucky (4-Ball, KY) 7122 1/2/2014 Late-Kentucky (3-Ball, KY) 287 1/2/2014 Late-Kentucky (2-Ball, KY) 87
..........................................................................................................................................
7122..=..8356..=..3801..=..6998..=..1443..
Hmmmm...
nassau
Bahamas
Member #120651
December 24, 2011
3190 Posts
Offline
Posted: January 3, 2014, 9:17 am - IP Logged
Morning all eve draw was not kind, no winners at all
So playing back our due pair list, o/s prs 00 49 78
000 010 020 030 040 050 060 070 080 090
049 149 249 349 449 549 649 749 849 949
078 178 278 378 478 578 678 778 878 978
Playing back our repeats 308, 612 & 245
12-012 112 212 312 412 512 612 712 812 912
26-026 126 226 326 426 526 626 726 826 926
17-017 117 217 317 417 517 617 717 817 917
67-067 167 267 367 467 567 667 767 867 967
16-016 116 216 316 416 516 616 716 816 916
11-011 111 112 113 411 511 611 711 811 911
66-066 166 266 366 466 566 666 766 866 966
03-030 103 203 303 403 503 603 703 803 903
35-035 135 235 335 435 535 635 735 835 935
58-058 158 258 358 458 558 658 758 858 958
08-080 108 208 308 408 508 608 *708* 808 908
38-038 138 238 338 438 538 638 738 838 938
33-033 133 233 333 433 533 633 733 833 933
88-088 188 288 388 488 588 688 788 888 988
24-024 124 224 324 424 524 624 724 824 924
47-047 147 247 347 447 547 647 747 847 *947*
79-079 179 279 379 *479* 579 679 779 879 979
25-025 125 225 325 425 525 625 725 825 925
02-020 120 220 320 420 520 602 702 802 902
07-070 107 207 307 407 507 607 707* 807* 907
45-045 145 245 345 445 545 645 745 845 945
04-040 104 204 304 404 504 604 704 804 *904*
09-090 109 209 309 *409* 509 609 709 809 909
Legend:
xxxx-Combos Recently Drawn
xxxx-Combos on our Due Pair List
123+/- Count
162...285...730...049...594
621...744...299...508...053
803...926...471...780...235
308...421...976...285...730
254...377...822...131...686
542...665...110...429...972
Traveller Combo
074...174...974...064...084...075...073...574...079...579...529
162...262...062...152...172...163...161...662...167...667...617
308...408...208...398...318...309...307...808...303...803...858
542...642...442...532...552...543...541...042...547...047...097
Morning Mike,
Was looking back but cannot find it what do we do with the CD Series again.. as we had 576 in GA & 098 in IL
576
------------------------------
576–330,675,118 435 567–208 444 864 000 756–992 657 978 114 765-888 024 222 468 657-411 756 299 435 675-811 576 033 789
-----------------------------
098
------------------------------
089–980,225,201 447 098–111 357 555 791 809–663 908 441 687 890-531 777 197 333 908-144 809 366 012 980-744 089 522 867
-----------------------------
Show me the \$\$\$\$\$money\$\$\$\$\$ ....working on my retirement plan
Bahamas
Member #139033
February 12, 2013
2733 Posts
Offline
Posted: January 3, 2014, 9:33 am - IP Logged
Morning Mike,
Was looking back but cannot find it what do we do with the CD Series again.. as we had 576 in GA & 098 in IL
576
------------------------------
576–330,675,118 435 567–208 444 864 000 756–992 657 978 114 765-888 024 222 468 657-411 756 299 435 675-811 576 033 789
-----------------------------
098
------------------------------
089–980,225,201 447 098–111 357 555 791 809–663 908 441 687 890-531 777 197 333 908-144 809 366 012 980-744 089 522 867
-----------------------------
Good Morning Master Jeebsxx..
EGA 576 3720 Tuesday, December 31, 2013
It was like this...
When you get a series number to walk out like this in the results...
We look for the last time it came ..
LCAL 576 2829 Wednesday, December 04, 2013
Then We apply our 123Drop On It....
576..=..576..=..699..=..144..=..453..=..908..
ECHA 098 9877 Tuesday, December 31, 2013
And You see we get Our Result In Chicago>
You see this only apply to series numbers...
Respect....
nassau
Bahamas
Member #120651
December 24, 2011
3190 Posts
Offline
Posted: January 3, 2014, 9:39 am - IP Logged
Good Morning Master Jeebsxx..
EGA 576 3720 Tuesday, December 31, 2013
It was like this...
When you get a series number to walk out like this in the results...
We look for the last time it came ..
LCAL 576 2829 Wednesday, December 04, 2013
Then We apply our 123Drop On It....
576..=..576..=..699..=..144..=..453..=..908..
ECHA 098 9877 Tuesday, December 31, 2013
And You see we get Our Result In Chicago>
You see this only apply to series numbers...
Respect....
Ok Thanks got it so we still have the 098 to lookout for
098...111...666...975...420
Show me the \$\$\$\$\$money\$\$\$\$\$ ....working on my retirement plan
gay,ga
United States
Member #68810
December 30, 2008
5939 Posts
Offline
Posted: January 3, 2014, 9:48 am - IP Logged
Is 098 your mid GA choice?
Page 162 of 368 | 8,172 | 23,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-50 | longest | en | 0.558788 |
https://nptel.ac.in/srt/112104121/Lec-03.srt | 1,563,808,158,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-30/segments/1563195528037.92/warc/CC-MAIN-20190722133851-20190722155851-00060.warc.gz | 479,766,166 | 12,276 | 1 00:00:10,300 --> 00:00:16,560 In today’s lecture, we shall discuss four link planar mechanisms. 2 00:00:16,560 --> 00:00:21,650 As we shall see such mechanisms have versatile applications. 3 00:00:21,650 --> 00:00:28,930 In fact, a major portion of this course will be devoted to the study of four link planar 4 00:00:28,930 --> 00:00:49,800 mechanisms. 5 00:00:49,800 --> 00:00:56,270 Before getting into the four link planar mechanisms, let us recall the technical, that is, kinematic 6 00:00:56,270 --> 00:00:58,649 definition of a mechanism. 7 00:00:58,649 --> 00:01:16,730 We have already defined mechanism as a closed kinematic chain with one of its links fixed. 8 00:01:16,730 --> 00:01:31,620 It is obvious, that to get a closed chain, we need at least three links. 9 00:01:31,620 --> 00:01:38,970 In the next figure, we shall see a three-linked closed chain with three revolute pairs. 10 00:01:38,970 --> 00:01:45,220 In such a three-linked closed chain, it is obvious that if one of the links, say, number 11 00:01:45,220 --> 00:01:51,890 1 is held fixed, then there cannot be any relative movement between these three bodies. 12 00:01:51,890 --> 00:01:59,570 In fact, if we apply an external load there is no relative movement between various links, 13 00:01:59,570 --> 00:02:04,070 in fact this load can be supported by this assembly. 14 00:02:04,070 --> 00:02:09,259 Such an assembly with zero degree of freedom is called a structure. 15 00:02:09,259 --> 00:02:15,569 Hence, we see the four link planar mechanism is the simplest mechanism that we can think 16 00:02:15,569 --> 00:02:16,569 of. 17 00:02:16,569 --> 00:02:23,250 To start with, we consider a four link planar mechanism with four revolute pairs. 18 00:02:23,250 --> 00:02:30,530 As we see in this four-link planar mechanism, there are four revolute pairs: one at O2, 19 00:02:30,530 --> 00:02:40,610 one at O4, one at A and the other at B. The four links are numbered, 1 as the fixed link 20 00:02:40,610 --> 00:02:44,410 and 2, 3, 4 as the moving links. 21 00:02:44,410 --> 00:02:52,069 The two hinges at O2 and O4 which are connected to the fixed link are called fixed pivots 22 00:02:52,069 --> 00:02:58,900 and the revolute pairs at A and B are referred to as moving hinges. 23 00:02:58,900 --> 00:03:06,730 Link 2 and link 4, which are connected to the fixed link are normally used as the input 24 00:03:06,730 --> 00:03:08,969 and the output link. 25 00:03:08,969 --> 00:03:15,740 Link 3, which connects the links 2 and 4, is called the coupler, which is the motion 26 00:03:15,740 --> 00:03:19,780 transfer link between 2 and 4. 27 00:03:19,780 --> 00:03:32,250 Such a 4R mechanism has four kinematic dimensions namely, O2O4 which we represent by l1 and 28 00:03:32,250 --> 00:03:36,069 O2A which represent by l2. 29 00:03:36,069 --> 00:03:49,170 Similarly, AB is equal to l3 and O4B as l4. 30 00:03:49,170 --> 00:03:54,670 If one of the input links which is connected to the fixed link makes a complete rotation, 31 00:03:54,670 --> 00:04:00,420 as is normally the case, when the mechanism is driven by an electric motor, then that 32 00:04:00,420 --> 00:04:07,140 link is referred to as a crank. 33 00:04:07,140 --> 00:04:20,269 The output link 4, which is the output link, is called follower and the intermediate connecting 34 00:04:20,269 --> 00:04:29,520 link is called the coupler. 35 00:04:29,520 --> 00:04:37,039 Let us consider a 4R planar mechanism where one of these kinematic pairs at B goes to 36 00:04:37,039 --> 00:04:40,400 infinity along a vertical direction. 37 00:04:40,400 --> 00:04:47,160 Consequently, that kinematic pair is converted to a prismatic pair and what we get is a three 38 00:04:47,160 --> 00:04:57,759 revolute one prismatic pair, referred to as the 3R-1P mechanism. 39 00:04:57,759 --> 00:05:05,960 So we get a 3R-1P mechanism where there are three revolute pairs at O2, A and B and the 40 00:05:05,960 --> 00:05:13,009 kinematic pair between link 4 and link 1 is a prismatic pair. 41 00:05:13,009 --> 00:05:20,330 Such a mechanism has three kinematic dimensions namely O2A which is called the crank length 42 00:05:20,330 --> 00:05:30,310 is represented by l2, AB which is called the connecting rod whose link is represented by 43 00:05:30,310 --> 00:05:39,909 l3 and the line of reciprocation of B is at a distance from O2 and this perpendicular 44 00:05:39,909 --> 00:05:43,039 distance, which is called the offset. 45 00:05:43,039 --> 00:05:50,839 The other kinematic dimension is the offset which is equal to e. 46 00:05:50,839 --> 00:05:57,620 This is also known as an offset slider crank mechanism which is used to convert uniform 47 00:05:57,620 --> 00:06:05,439 rotation of link 2 in to rectilinear to and fro oscillation of the slider 4. 48 00:06:05,439 --> 00:06:12,729 Let us consider different kinematic inversions which can arise out of such a 3R-1P chain 49 00:06:12,729 --> 00:06:14,719 as shown in this figure. 50 00:06:14,719 --> 00:06:25,499 Here, as we see there are three revolute pairs namely at O2, O4 and at A. Comparing to the 51 00:06:25,499 --> 00:06:33,960 previous 3R-1P chain, here the link 1 which is fixed has revolute pairs at both ends, 52 00:06:33,960 --> 00:06:35,800 that is at O2 and O4. 53 00:06:35,800 --> 00:06:41,990 Whereas, in the previous mechanism, the fixed link had a revolute pair at one end and a 54 00:06:41,990 --> 00:06:44,180 prismatic pair at the other end. 55 00:06:44,180 --> 00:06:52,080 Here again, we have three kinematic dimensions namely O2O4 - that is the fixed link length, 56 00:06:52,080 --> 00:06:58,719 O2A that may be the crank length and the distance of this line of reciprocation from these fixed 57 00:06:58,719 --> 00:06:59,719 hinges. 58 00:06:59,719 --> 00:07:08,659 Such a mechanism with this offset e equal to 0 is used in a quick return mechanism that 59 00:07:08,659 --> 00:07:10,460 is used in a shaper. 60 00:07:10,460 --> 00:07:17,149 We consider again yet another kinematic inversion of the same 3R-1P chain. 61 00:07:17,149 --> 00:07:23,449 Here, we have three revolute pairs namely, at O2, A and O4. 62 00:07:23,449 --> 00:07:33,909 Here again the fixed link has revolute pairs at both ends namely O2 and O4. 63 00:07:33,909 --> 00:07:39,569 Such a mechanism is called an oscillating cylinder mechanism which is used for example 64 00:07:39,569 --> 00:07:41,930 in a bicycle foot pump. 65 00:07:41,930 --> 00:07:50,560 The thing to note is that here the input member is link 3 which is given an oscillatory motion. 66 00:07:50,560 --> 00:07:56,339 This (piston) link 3 moves within this oscillating cylinder (4) which has the same angular velocity 67 00:07:56,339 --> 00:07:57,339 as link 3. 68 00:07:57,339 --> 00:07:59,229 (Refer slide Time 07:59) 69 00:07:59,229 --> 00:08:05,349 We see another kinematic inversion from the same 3R-1P chain where we have kinematic pair 70 00:08:05,349 --> 00:08:15,279 at O2, A and B. The link 4 which has a revolute pair at B has a prismatic pair in the vertical 71 00:08:15,279 --> 00:08:18,889 direction with respect to the fixed link 1. 72 00:08:18,889 --> 00:08:24,669 Such a mechanism, as we know is used in a hand pump mechanism and here again it is link 73 00:08:24,669 --> 00:08:31,899 3 which is given the input motion such that the piston which is link 4 moves vertically 74 00:08:31,899 --> 00:08:36,580 up and down within this fixed link. 75 00:08:36,580 --> 00:08:41,310 So far, we have considered 3R-1P chain. 76 00:08:41,310 --> 00:08:47,209 Let us now have another revolute pair converted into a prismatic pair and consequently we 77 00:08:47,209 --> 00:08:51,210 get what we call a 2R-2P chain. 78 00:08:51,210 --> 00:08:59,360 However, unlike in a 3R-1P chain, there can be two varieties that is the sequential order 79 00:08:59,360 --> 00:09:01,279 of this kinematic pairs. 80 00:09:01,279 --> 00:09:08,000 For example, we can have RRPP or RPRP. 81 00:09:08,000 --> 00:09:16,800 Here, the two revolute pairs are connected by one link, then another link has a revolute 82 00:09:16,800 --> 00:09:24,140 pair at one end and a prismatic pair at the other end, another link has prismatic pair 83 00:09:24,140 --> 00:09:31,589 at both ends and this link has a P pair at one end and a revolute pair at other end. 84 00:09:31,589 --> 00:09:40,110 This is what we call RRPP chain, whereas here these kinematic pairs appear alternately as 85 00:09:40,110 --> 00:09:46,130 RPRP. 86 00:09:46,130 --> 00:09:50,709 So we see all the four links has a revolute pair at one end and a prismatic pair at the 87 00:09:50,709 --> 00:09:55,620 other end that is same for all these four links. 88 00:09:55,620 --> 00:10:02,529 It may be emphasized that in a 3R-1P chain whichever way we go, we start from one prismatic 89 00:10:02,529 --> 00:10:07,430 pair then it is followed by three sequential R pairs. 90 00:10:07,430 --> 00:10:11,550 So, PRRR is the only possibility. 91 00:10:11,550 --> 00:10:16,390 Let me now consider some inversions from this 2R-2P chain. 92 00:10:16,390 --> 00:10:24,069 Here, we consider a four link planar mechanism with a revolute pair at O2, another revolute 93 00:10:24,069 --> 00:10:32,360 pair at O, which I called as O4, that is link 2 has two revolute pairs at its either ends, 94 00:10:32,360 --> 00:10:38,830 and link number 3 has a revolute pair at O and a prismatic pair with 4 along the vertical 95 00:10:38,830 --> 00:10:39,830 direction. 96 00:10:39,830 --> 00:10:45,540 Similarly, link 1 has a revolute pair with 2 and a prismatic pair with 4 in the horizontal 97 00:10:45,540 --> 00:10:48,750 direction. 98 00:10:48,750 --> 00:10:54,230 We consider a special case when these two prismatic pairs have an angle of 90 degree 99 00:10:54,230 --> 00:10:56,290 between them. 100 00:10:56,290 --> 00:11:03,029 From such a 2R-2P, that is RRPP chain, we can have different mechanism by the process 101 00:11:03,029 --> 00:11:04,310 of kinematic inversion. 102 00:11:04,310 --> 00:11:10,019 For example, let us hold this link number 1 fixed. 103 00:11:10,019 --> 00:11:15,430 So we can see that link 4, the output link can have horizontal movement with respect 104 00:11:15,430 --> 00:11:21,640 to the fixed link as link 2 undergoes rotary motion with respect to link 1, that is the 105 00:11:21,640 --> 00:11:23,990 fixed link. 106 00:11:23,990 --> 00:11:25,160 This is the kinematic sketch. 107 00:11:25,160 --> 00:11:31,629 Let me show the physical construction of such a mechanism, which is known as the scotch 108 00:11:31,629 --> 00:11:33,370 yoke mechanism. 109 00:11:33,370 --> 00:11:41,000 As we see, such a mechanism will convert uniform rotary motion of link 2 into simple harmonic 110 00:11:41,000 --> 00:11:46,949 translatory motion in the horizontal direction for link 4. 111 00:11:46,949 --> 00:11:52,730 So again we have RRPP chain. 112 00:11:52,730 --> 00:11:58,940 There is one revolute pair at O2, one revolute pair at A, there is a vertical prismatic pair 113 00:11:58,940 --> 00:12:04,300 between 3 and 4 and a horizontal prismatic pair between 4 and 1. 114 00:12:04,300 --> 00:12:19,880 If we consider the kinematic dimension O2A as L2 and this angle if I call Theta, assuming 115 00:12:19,880 --> 00:12:26,970 at ‘t’ equal to zero is Theta is equal to 0, then I can represent Theta = Omega2t, 116 00:12:26,970 --> 00:12:34,339 where Omega2 is the constant angular speed of link 2. 117 00:12:34,339 --> 00:12:42,389 Then the position of the link 4 which can be represented by the point is given by x. 118 00:12:42,389 --> 00:12:57,090 It is easy to see that x is nothing but L2 cosTheta, that is L2 cosOmegat. 119 00:12:57,090 --> 00:13:04,839 So, we have produced a simple harmonic motion out of continuous rotation. 120 00:13:04,839 --> 00:13:11,769 It may be emphasized that continuous uniform rotary motion would have been translated into 121 00:13:11,769 --> 00:13:16,770 to and fro rectilinear oscillation by a slider crank mechanism that we have seen earlier. 122 00:13:16,770 --> 00:13:23,490 However, the to and fro rectilinear oscillation of the piston of a slider in a slider crank 123 00:13:23,490 --> 00:13:28,839 mechanism is periodic, but not purely harmonic. 124 00:13:28,839 --> 00:13:34,760 This periodic motion tends to be harmonic as the ratio of the connecting rod length 125 00:13:34,760 --> 00:13:37,670 to the crank radius keeps on increasing. 126 00:13:37,670 --> 00:13:44,690 In fact, if the connecting rod length in a slider crank mechanism tends to infinity, 127 00:13:44,690 --> 00:13:48,430 then the slider motion becomes purely harmonic. 128 00:13:48,430 --> 00:13:55,160 As soon as the connecting rod length becomes infinity, one of the kinematic pair, which 129 00:13:55,160 --> 00:14:00,689 was the revolute pair for a connecting rod, gets converted into this prismatic pair, because 130 00:14:00,689 --> 00:14:06,139 we have already seen a prismatic pair is nothing but a revolute pair at infinity. 131 00:14:06,139 --> 00:14:09,449 I will now show this through a model. 132 00:14:09,449 --> 00:14:11,110 (Refer Slide Time 14:10) 133 00:14:11,110 --> 00:14:14,699 This is the model of that scotch yoke mechanism. 134 00:14:14,699 --> 00:14:20,670 This is the link 2 that is the crank, which has a revolute pair with the fixed link at 135 00:14:20,670 --> 00:14:23,180 this point. 136 00:14:23,180 --> 00:14:29,180 Another revolute pair between link 2, that is the crank and this block is here. 137 00:14:29,180 --> 00:14:36,189 This block link 3 has a prismatic pair in the vertical direction with link 4, and link 138 00:14:36,189 --> 00:14:40,870 4 has a horizontal prismatic pair with link 1. 139 00:14:40,870 --> 00:14:49,190 So, if we rotate link 2 at a constant angular speed, as we see, the red link that is link 140 00:14:49,190 --> 00:14:55,920 4 is performing simple harmonic oscillation in the horizontal direction. 141 00:14:55,920 --> 00:15:03,110 Let us now consider another kinematic inversion of the same RRPP chain. 142 00:15:03,110 --> 00:15:10,579 As shown in this mechanism, the link which has revolute pair at its both ends at O2 and 143 00:15:10,579 --> 00:15:14,410 O4 is held fixed. 144 00:15:14,410 --> 00:15:22,370 Now link 3 has a prismatic pair in the horizontal direction with link 2 and a prismatic pair 145 00:15:22,370 --> 00:15:25,870 in a vertical direction with link 4. 146 00:15:25,870 --> 00:15:34,199 So, thus if we rotate link 2 that is in translation with link 3 which is again in translation 147 00:15:34,199 --> 00:15:41,809 with link 4, thus link 2 and link 4 has only relative translatory motion. 148 00:15:41,809 --> 00:15:45,860 In other words, they have the same angular motion. 149 00:15:45,860 --> 00:15:53,769 Thus, this mechanism known as Oldham’s coupling can be used to connect two parallel shafts: 150 00:15:53,769 --> 00:16:02,319 one at O2 and the other at O4 and transmitting unity angular velocity ratio. 151 00:16:02,319 --> 00:16:04,380 This is the kinematic diagram. 152 00:16:04,380 --> 00:16:10,160 The physical construction of this Oldham’s coupling is shown in the above slide. 153 00:16:10,160 --> 00:16:17,149 As we see, there is a revolute pair between link 1 and link 2, there is a prismatic pair 154 00:16:17,149 --> 00:16:23,509 between link 2 and link 3, another prismatic pair between link 3 and link 4. 155 00:16:23,509 --> 00:16:28,949 The direction of these two prismatic pairs are at right angles to each other. 156 00:16:28,949 --> 00:16:32,720 Link 4 again has a revolute pair with the fixed link 1. 157 00:16:32,720 --> 00:16:39,569 Thus link 4, that is, this shaft and link 2, that is the other shaft which are parallel 158 00:16:39,569 --> 00:16:45,149 can be connected by such a coupling and it will transmit unity angular velocity ratio. 159 00:16:45,149 --> 00:16:46,870 (Refer Slide Time 16:47) 160 00:16:46,870 --> 00:16:50,329 I will now show this through a model. 161 00:16:50,329 --> 00:16:53,470 This is the model of Oldham’s coupling. 162 00:16:53,470 --> 00:17:01,060 As we see, the rotation of this shaft is converted into the rotation of that shaft at the same 163 00:17:01,060 --> 00:17:07,400 speed through the intermediate member which has a prismatic pair here and a prismatic 164 00:17:07,400 --> 00:17:08,720 pair at the top. 165 00:17:08,720 --> 00:17:15,580 So, there are two prismatic pairs at 90 degrees to each other. 166 00:17:15,580 --> 00:17:23,810 This intermediate member moves in and out in this prismatic pair and also at this prismatic 167 00:17:23,810 --> 00:17:29,910 pair. 168 00:17:29,910 --> 00:17:36,070 Of course, this coupling is good enough to transmit power, only when the offset between 169 00:17:36,070 --> 00:17:42,051 these two shafts is not very large, because a lot of power is wasted in fiction at this 170 00:17:42,051 --> 00:17:46,100 two prismatic pairs. 171 00:17:46,100 --> 00:17:53,130 Now let us consider another kinematic inversion from the same RRPP chain. 172 00:17:53,130 --> 00:17:59,940 In this mechanism, we consider the link which has prismatic pair at both ends fixed. 173 00:17:59,940 --> 00:18:07,010 For example, this link 1 has a prismatic pair with link 4 and a prismatic pair with link 174 00:18:07,010 --> 00:18:10,320 2 at right angles to each other. 175 00:18:10,320 --> 00:18:13,650 This mechanism is known as elliptic trammel. 176 00:18:13,650 --> 00:18:19,630 It will be obvious now why this name elliptic trammel? 177 00:18:19,630 --> 00:18:25,310 Let us consider the physical construction of this mechanism rather than this kinematic 178 00:18:25,310 --> 00:18:27,030 representation. 179 00:18:27,030 --> 00:18:34,530 As we see, this link has two prismatic pairs, one in the horizontal direction and the other 180 00:18:34,530 --> 00:18:37,310 in the vertical direction. 181 00:18:37,310 --> 00:18:46,990 The rod ABCD has revolute pair with the block at C and another revolute pair with the block 182 00:18:46,990 --> 00:18:56,360 at A. As the rod moves, let us look at the coordinates of point D on the moving rod. 183 00:18:56,360 --> 00:19:00,740 This is x-axis and this is y-axis. 184 00:19:00,740 --> 00:19:09,900 So the x-coordinate of the point D, when the rod is making an angle Theta with the vertical 185 00:19:09,900 --> 00:19:19,450 is easily seen to be ADsinTheta. 186 00:19:19,450 --> 00:19:28,700 Similarly, the y coordinate of this moving point D is CD and this angle is Theta. 187 00:19:28,700 --> 00:19:30,530 So it is CDcosTheta. 188 00:19:30,530 --> 00:19:37,650 So, yD is CDcosTheta. 189 00:19:37,650 --> 00:19:53,620 If we eliminate Theta from these two coordinates, we can easily see that {(xD/AD)2 + (yD/CD)2 190 00:19:53,620 --> 00:19:57,470 = 1}. 191 00:19:57,470 --> 00:20:05,670 That is, as Theta changes, this point D moves on an ellipse with semi major axis given by 192 00:20:05,670 --> 00:20:07,690 AD and CD. 193 00:20:07,690 --> 00:20:10,440 That is why it is called an elliptic trammel. 194 00:20:10,440 --> 00:20:16,640 It must be pointed out that there are three points on this rod AB which are exceptions 195 00:20:16,640 --> 00:20:22,800 namely, this point A which generates a vertical straight line because of this prismatic pair. 196 00:20:22,800 --> 00:20:28,410 Similarly, this point C which generates a horizontal straight line because of this horizontal 197 00:20:28,410 --> 00:20:35,580 prismatic pair and for the mid-point B which is the midpoint of the distance AC, that is, 198 00:20:35,580 --> 00:20:38,990 AB is equal to BC. 199 00:20:38,990 --> 00:20:47,140 If AB and BC are equal, then the point B generates a circle. 200 00:20:47,140 --> 00:20:50,650 If I call this point O2, then the radius of that circle is O2B. 201 00:20:50,650 --> 00:20:58,100 We can say that these are nothing but the degenerated cases of the same ellipse. 202 00:20:58,100 --> 00:21:04,810 Now, we show you a model to generate this ellipse through an elliptic trammel. 203 00:21:04,810 --> 00:21:07,740 This is the model of the elliptic trammel. 204 00:21:07,740 --> 00:21:12,880 As we see, there are two perpendicular slots in the fixed link. 205 00:21:12,880 --> 00:21:22,540 This is rod AB, which has a revolute pair with two blocks and these two blocks move 206 00:21:22,540 --> 00:21:26,310 along these two prismatic pairs. 207 00:21:26,310 --> 00:21:34,520 If we move the rod, as we can see this particular point of the rod generates the ellipse which 208 00:21:34,520 --> 00:21:40,680 has been drawn with the green line. 209 00:21:40,680 --> 00:21:50,190 Now, if we consider the mid point of these two revolute pairs, then as we see the rod 210 00:21:50,190 --> 00:22:03,220 moves this particular mid-point generates this red circle, as mentioned earlier. 211 00:22:03,220 --> 00:22:09,960 Now that we have discussed different inversions from a RRPP chain. 212 00:22:09,960 --> 00:22:20,120 Let us go back to another 2R-2P chain namely, RPRP chain. 213 00:22:20,120 --> 00:22:26,900 We consider here only the portion covered by these dashed lines. 214 00:22:26,900 --> 00:22:32,970 This is the part of an automobile steering gear known as Davis steering gear. 215 00:22:32,970 --> 00:22:40,540 Here, as we see there is a revolute pair between link 4 and link 1, then there is a prismatic 216 00:22:40,540 --> 00:22:47,070 pair between link 4 and link 3, then there is a revolute pair between link 3 and link 217 00:22:47,070 --> 00:22:52,570 2 and then there is a prismatic pair between link 1 and link 2. 218 00:22:52,570 --> 00:22:55,610 These are the two wheels of the automobile. 219 00:22:55,610 --> 00:23:04,470 By moving the steering wheel, we move this link 2 in this prismatic pair between 1 and 220 00:23:04,470 --> 00:23:05,660 2. 221 00:23:05,660 --> 00:23:08,790 Consequently, these two wheels will rotate. 222 00:23:08,790 --> 00:23:14,890 So, here is an RPRP chain. 223 00:23:14,890 --> 00:23:21,140 Now that we have discussed 4 linked planar mechanisms which is two revolute and two prismatic 224 00:23:21,140 --> 00:23:28,380 pairs, let me now increase one more prismatic pair instead of a revolute pair, that is, 225 00:23:28,380 --> 00:23:33,480 can we talk of 3P-1R chain? 226 00:23:33,480 --> 00:23:39,900 As we see, there is a P pair and there is a link 1 connecting to these two prismatic 227 00:23:39,900 --> 00:23:40,900 pairs. 228 00:23:40,900 --> 00:23:48,050 Similarly, link 2 connects to two prismatic pairs and link 3 connects a prismatic pair 229 00:23:48,050 --> 00:23:53,080 and a revolute pair and a link 4 connects a revolute pair and a prismatic pair. 230 00:23:53,080 --> 00:24:02,300 It will be easy to show that we cannot have a mechanism with such 3P-1R chain. 231 00:24:02,300 --> 00:24:10,090 For example, we can see there is a prismatic pair between link 1 and 2, there cannot be 232 00:24:10,090 --> 00:24:11,870 any relative rotation. 233 00:24:11,870 --> 00:24:17,530 Same goes between link 2 and 3 because they are connected between a prismatic pair. 234 00:24:17,530 --> 00:24:22,540 There cannot be any relative rotation between link 2 and link 3. 235 00:24:22,540 --> 00:24:29,730 There is a prismatic pair between link 1 and link 4, so there cannot be any relative rotation 236 00:24:29,730 --> 00:24:32,030 between link 1 and link 4. 237 00:24:32,030 --> 00:24:39,320 Thus, if we follow these three prismatic pairs, we conclude there cannot be any relative rotation 238 00:24:39,320 --> 00:24:41,880 between link 4 and link 3. 239 00:24:41,880 --> 00:24:48,660 Thus, this revolute pair which allows only relative rotation between link 3 and link 240 00:24:48,660 --> 00:24:51,940 4 cannot permit any relative motion. 241 00:24:51,940 --> 00:24:59,130 Thus, we can conclude that such a 3P-1R chain cannot give rise to any mechanism. 242 00:24:59,130 --> 00:25:04,520 Now, can we have a mechanism with 4 prismatic pairs? 243 00:25:04,520 --> 00:25:08,260 A four-link planar mechanism with all pairs prismatic. 244 00:25:08,260 --> 00:25:16,470 We will see later that such a 4P mechanism is not a constrained mechanism. 245 00:25:16,470 --> 00:25:24,500 In fact, three links connected by three prismatic pairs having the relative translation at various 246 00:25:24,500 --> 00:25:32,740 angles itself constitutes a planar mechanism which is constant, but that we shall see later. 247 00:25:32,740 --> 00:25:38,780 So far we have discussed mechanisms only with revolute and prismatic pairs. 248 00:25:38,780 --> 00:25:41,000 Let me now change the topic a little bit. 249 00:25:41,000 --> 00:25:46,760 Can we consider mechanisms involving higher pairs as well? 250 00:25:46,760 --> 00:25:53,590 As we shall show, now that a mechanism with higher pair can be replaced equivalently by 251 00:25:53,590 --> 00:26:00,210 a mechanism having only lower pairs that is revolute pair and prismatic pairs. 252 00:26:00,210 --> 00:26:07,540 Of course, this equivalence is only instantaneous and holds good for velocity and acceleration 253 00:26:07,540 --> 00:26:13,950 analysis at a particular configuration. 254 00:26:13,950 --> 00:26:20,100 So we are talking of an equivalent lower pair linkage for a higher pair mechanism, i.e., 255 00:26:20,100 --> 00:26:25,490 how a higher pair can be replaced equivalently by lower pairs. 256 00:26:25,490 --> 00:26:33,110 As an example, let us look at this higher pair mechanism which consists of 3 links namely, 257 00:26:33,110 --> 00:26:38,020 one the fixed link and 2 and 3. 258 00:26:38,020 --> 00:26:46,090 There is a revolute pair between link 1 and 2 at O2 and a revolute pair between link 1 259 00:26:46,090 --> 00:26:56,560 and 3 at O3, but there is a higher pair at the point C between link 2 and link 3. 260 00:26:56,560 --> 00:27:03,590 Our objective is to replace this higher pair mechanism by a kinematically equivalent lower 261 00:27:03,590 --> 00:27:07,860 pair linkage consisting only of lower pairs. 262 00:27:07,860 --> 00:27:17,040 I repeat again that this equivalence is only instantaneous, that means, only for this particular 263 00:27:17,040 --> 00:27:18,040 configuration. 264 00:27:18,040 --> 00:27:25,350 Towards this end, let us consider that the centre of curvature of body 3 at the point 265 00:27:25,350 --> 00:27:35,020 C is at B. Similarly, the centre of curvature of the body 2 at this contact point C is at 266 00:27:35,020 --> 00:27:43,140 A. Due to the property of centre of curvature, that is circle of curvature or osculating 267 00:27:43,140 --> 00:27:52,460 circle, we can consider that for three infinitesimally separated time instances, these points A and 268 00:27:52,460 --> 00:27:54,560 B do not change. 269 00:27:54,560 --> 00:28:03,840 Thus, we can replace this higher pair by having a lower pair at the point A which is a revolute 270 00:28:03,840 --> 00:28:09,690 pair and a lower pair at the point B, which is again a revolute pair and because the distance 271 00:28:09,690 --> 00:28:17,480 between A and B are not changing for three infinitesimally separated time intervals, 272 00:28:17,480 --> 00:28:23,400 I can join these two points A and B by a rigid additional link. 273 00:28:23,400 --> 00:28:33,120 Thus, this higher pair mechanism has been replaced by a 4R-linkage which has all 4 revolute 274 00:28:33,120 --> 00:28:44,020 pairs at O2, O3, A and B. As I said earlier, because as these two bodies 2 and 3 move the 275 00:28:44,020 --> 00:28:50,450 centers of curvature also move, so for every instance, we have a different equivalent lower 276 00:28:50,450 --> 00:28:51,450 pair linkage. 277 00:28:51,450 --> 00:28:56,820 Later on, we will explain this equivalence through a model. 278 00:28:56,820 --> 00:29:05,420 But we should see that a higher pair at C has been replaced by an additional link that 279 00:29:05,420 --> 00:29:15,270 is link 4 and two additional lower pairs, one at A and the other at B, where A and B 280 00:29:15,270 --> 00:29:21,630 are respectively the centre of curvatures of the contacting surfaces between 2 and 3 281 00:29:21,630 --> 00:29:29,310 at the contact point C. Of course, this equivalence can be permanent if the centers of curvature 282 00:29:29,310 --> 00:29:30,600 do not change. 283 00:29:30,600 --> 00:29:38,040 That is, one of these contacting surfaces, say circular or cylindrical or coinstantaneous 284 00:29:38,040 --> 00:29:43,000 of curvature or flat that is again of infinite (constant) radius of curvature. 285 00:29:43,000 --> 00:29:44,120 (Refer Slide Time 29:32) 286 00:29:44,120 --> 00:29:50,210 For example, let us look at this higher pair mechanism which is a cam with a flat face 287 00:29:50,210 --> 00:29:51,380 follower. 288 00:29:51,380 --> 00:30:01,330 The cam that is this rigid link 2 has revolute pair with the fixed link at O2 and this follower 289 00:30:01,330 --> 00:30:09,540 3 has a prismatic pair with fixed link 1 and there is a higher pair at this contact point. 290 00:30:09,540 --> 00:30:19,770 So, the rotary motion of this cam 2 is converted into rectilinear motion of this follower 3. 291 00:30:19,770 --> 00:30:25,420 Let the centre of curvature of the cam surface at this contact point is at C2. 292 00:30:25,420 --> 00:30:32,141 The centre of curvature of the follower surface at the contact point is at infinity, that 293 00:30:32,141 --> 00:30:38,640 is, the revolute pair at the centre of curvature is converted to an equivalent prismatic pair 294 00:30:38,640 --> 00:30:40,960 in the horizontal direction. 295 00:30:40,960 --> 00:30:46,970 If we recall that a prismatic pair is nothing but a revolute pair at infinity. 296 00:30:46,970 --> 00:30:53,270 Consequently, this cam follower mechanism with a flat face follower is replaced by this 297 00:30:53,270 --> 00:30:55,970 equivalent lower pair linkage. 298 00:30:55,970 --> 00:31:02,890 This link 2 represents the cam, link 3 represents the follower and there is an additional link 299 00:31:02,890 --> 00:31:10,460 4 which has a revolute pair with link 2 at C2 and a prismatic pair with link 3 in the 300 00:31:10,460 --> 00:31:11,970 horizontal direction. 301 00:31:11,970 --> 00:31:17,450 This is the equivalent lower pair linkage of this cam follower mechanism. 302 00:31:17,450 --> 00:31:24,570 That means, the input-output characteristics of this cam follower mechanism can be carried 303 00:31:24,570 --> 00:31:31,900 on by analyzing this lower pair linkage, because C2 keeps on changing if the cam surface is 304 00:31:31,900 --> 00:31:33,230 not circular. 305 00:31:33,230 --> 00:31:39,370 That’s why, for every instance we have to have separate equivalent lower pair linkage, 306 00:31:39,370 --> 00:31:44,620 because this link length C2 will keep on changing. 307 00:31:44,620 --> 00:31:51,630 This equivalence is valid only up to acceleration analysis, velocity and acceleration because 308 00:31:51,630 --> 00:31:59,210 the centre of curvature or osculating circle is in contact with the surface only for three 309 00:31:59,210 --> 00:32:01,700 infinitesimally separated positions. 310 00:32:01,700 --> 00:32:06,620 Higher order derivatives do not match and consequently, higher order time derivatives 311 00:32:06,620 --> 00:32:08,360 cannot be calculated. 312 00:32:08,360 --> 00:32:15,370 Now I will show these two equivalence lower pair linkage by models. 313 00:32:15,370 --> 00:32:21,780 Let us consider this model where we have a higher pair mechanism and the link 2 and link 314 00:32:21,780 --> 00:32:25,510 3 is having a higher pair at this point. 315 00:32:25,510 --> 00:32:32,480 The centre of curvature for the body 2 at this point of contact is here at C2. 316 00:32:32,480 --> 00:32:40,630 Similarly, the centre of curvature of the surface of body 3 is at this point C3. 317 00:32:40,630 --> 00:32:47,080 Now as we said earlier, we can have an equivalent lower pair linkage by having a revolute pair 318 00:32:47,080 --> 00:33:00,990 at C2 and another at C3 and connected by an additional rigid link. 319 00:33:00,990 --> 00:33:12,700 Thus, we have a 4R mechanism which is instantaneously equivalent to the original higher pair mechanism. 320 00:33:12,700 --> 00:33:21,350 I have connected these two bodies, that is, this red link and the original link 2, rigidly. 321 00:33:21,350 --> 00:33:28,270 As we shall see that I can move this mechanism, the same motion is transmitted at least around 322 00:33:28,270 --> 00:33:35,231 this contact region between body 2 and body 3, whether, it is through the higher pair 323 00:33:35,231 --> 00:33:38,800 mechanism or through this equivalent lower pair linkage. 324 00:33:38,800 --> 00:33:45,280 To distinguish the movement of this body 3 in the lower pair linkage and this body 3 325 00:33:45,280 --> 00:33:50,890 in the higher pair mechanism, let us notice these two lines, there is a red line on this 326 00:33:50,890 --> 00:33:55,220 body and there is a blue line on this body. 327 00:33:55,220 --> 00:34:05,890 Around this point, as we see these two lines move almost the same way because velocity 328 00:34:05,890 --> 00:34:12,120 and acceleration relationship between the original higher pair mechanism and the equivalent 329 00:34:12,120 --> 00:34:15,860 lower pair linkage is just the same. 330 00:34:15,860 --> 00:34:20,700 However, when there is a lot of movement these two lines separate out. 331 00:34:20,700 --> 00:34:25,190 This blue line and this red line are not the same any more. 332 00:34:25,190 --> 00:34:37,929 But around this point, it is only here that these two lines separate out because the centers 333 00:34:37,929 --> 00:34:44,510 of curvature are very different from what it was at this configuration. 334 00:34:44,510 --> 00:34:52,600 This is what we mean by instantaneously equivalent lower pair linkage. 335 00:34:52,600 --> 00:34:56,490 We can show another model with one surface flat. 336 00:34:56,490 --> 00:35:02,650 That is, radius of curvature is infinity and the other curve is a circle, that is the radius 337 00:35:02,650 --> 00:35:04,510 of curvature is constant. 338 00:35:04,510 --> 00:35:10,330 Under such a situation of course the equivalence will hold good for the entire cycle of motion. 339 00:35:10,330 --> 00:35:20,440 A higher pair mechanism between link 1, 2 and 3 and the centre of curvature of body 340 00:35:20,440 --> 00:35:28,280 2 at the contact point is here, whereas, body 3 has a flat or a straight contact in surface. 341 00:35:28,280 --> 00:35:38,150 Consequently, the extra link that is having here gets a prismatic pair between body 3 342 00:35:38,150 --> 00:35:44,930 and that extra link and a revolute pair between body 2 and that extra link. 343 00:35:44,930 --> 00:35:52,221 If we connect these lower pair linkage link 2 with the link 2 of the original higher pair 344 00:35:52,221 --> 00:35:59,270 mechanism rigidly, then we can see that same motion is transmitted by the both lower pair 345 00:35:59,270 --> 00:36:12,760 linkage and the higher pair mechanism. 346 00:36:12,760 --> 00:36:18,200 Let me now summarize, what we have talked so far in this lecture. 347 00:36:18,200 --> 00:36:24,680 What we have seen is 4R planar mechanisms of different varieties consisting of four 348 00:36:24,680 --> 00:36:33,520 revolute pairs or three revolute and one prismatic pair or two revolute and two prismatic pairs. 349 00:36:33,520 --> 00:36:38,980 When we have two revolute and two prismatic pairs, the order of the sequence of the pairs 350 00:36:38,980 --> 00:36:40,650 becomes important. 351 00:36:40,650 --> 00:36:49,400 We have talked of two varieties namely, RRPP or RPRP. 352 00:36:49,400 --> 00:36:55,800 At the end, we have also seen we have a three-link mechanism with a higher pair and that also 353 00:36:55,800 --> 00:37:05,400 can be equivalently represented by a four-link planar linkage having only R pairs or R and 354 00:37:05,400 --> 00:37:06,400 P pairs. 355 00:37:06,400 --> 00:37:06,401 14 | 12,201 | 35,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-30 | latest | en | 0.93322 |
http://www.astrosurf.com/luxorion/qsl-antenna-comparison2.htm | 1,537,957,067,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267164750.95/warc/CC-MAIN-20180926101408-20180926121808-00245.warc.gz | 281,829,930 | 13,192 | What directive antenna to select ?
K8AJS's Cubex Skymaster III-PT-5, a 3-element 5-band quad cut for 20 to 10m bands. It offers a 7.9-9.9 dBd gain with a F/B ratio between 15-30 dB. Its weigth is 25 kg (55 lbs.) and its boom is 5.4m long (18'). It is also available in a 2-element 3-band.
Pros and Cons Quads & Yagis (II)
To close on more objective data, the table displayed below takes into account several items common to both quads and Yagis. Values of each item (4th column, left for quads, right for beams) have been balanced according their importance over the other items and over their overall importance on the final choice.
To make a choice between two or several products, there is nothing better than using a balanced table to get an objective result. The weight assigned to each item is of the uttermost importance to get significant and accurate figures. It takes more time to determine than to fill in boxes. For example, if you consider that the furtivity of your antenna is not important, you can reduce its value to very few thing or simply remove this line. If its gain is very important, gives it a maximum value. But in the same time you must also assign to each item a value compatible with the others, to which they can be somehow compared. The turning radius for example can be in your mind as important as the weigth of the antenna, thus assign to both items the same maximum value, etc.
Select also a range compatible with your estimation knowing that it is useless to use a scale between 0 and 100 for example if you are unable to fix the value of an item within 10% of accuracy; in this case use a scale between 0-10.
It is as much hard to work with decimal figures; 8.9 and 9.1 will probably give results on par in the field, but first of all that means that you can estimate each item with an accurate between 0-100... So here also I suspect that you can't do it and it is advisable to use a scale between 0-10 too. This preprocessing is mandatory because at the end, you will have to consider these terms on a equal footing to calculate them.
When your estimation is well level-headed, the result is final and irrevocable, you must select the highest value... If you are not agree with the result, that means that you did a mistake, that you underestimated one or several items or worse, that you forget to include a new variable, which weight could be predominant over all others, maybe is it your bias and subjective feeling ... But if you make this estimation seriously your subjectivity must be rejected at least 20 dB down, Hi ! Here is the table.
Think tank
Let's briefly analyze these results. According to the values assigned to each item, as amazing as it is, globally these facts incline to be in favor of quads that win with 84% of points vs 57% only for Yagis (168/200 for the quad vs. 115/200 for the beam), although we known that most amateurs prefer beams... The difference exceeds 25% and I do not see how we could reverse the balance. I suggest however to the reader having a long experience of both types of antennas to make the same or a more accurate test. Feel free to give me your feedback if you do it or if you have any other relevant information that could complete this review that visibly interest many hams.
Put back in its context, this result, based on objective technical data and field experience demonstrates already that both antenna systems have advantages and drawbacks but that quads are globally "better" than Yagis. The difference looks objective but it does not confirm the subjective feeling reported by users. That could mean that once setup in the field, amateurs changed of opinion; the weight of some items exceeded the values fixed a priori or another information omitted in this list took the step on the others.
It is not any doubt that an amateur living in a residential area, an appartment or having only at its disposal a small estate has to accept some rules of coexistence or co-ownership that will affect his or her choice.
The fact that the quad is a true 3D object, far to be stealthy, and that sometimes can be easily damaged by ice, are so many factors that impact probably negatively most amateurs wishing to install this kind of antenna on their roof or in their backyard.
If I had to make a choice, without taking into consideration the price and sizing, I should select a quad with 4 or 5 elements for its overall performances, this is an indisputable fact reading its specifications, not an a priori.
At left, a 2-element Cubex MK II PT-3 Cubical quad (\$650). This model shows a 7-9 dBi gain and a F/B ratio between 20-30 dBd. The arms are made of fiberglass 3.9 m (13ft) long with a turning radius of 3 m (10.1 ft). In option it can work on WARC band too (12 and 17 m). The boom and mast are in cast aluminum. Its weight is 16 kg. These antennas are also distributed in Europe via GB Antennes (GBanttow) in Belgium and Holland (779 €) among other dealers. At right,the impressive K2US's home-made 6-element 2-band quad cut for the 20 and 15-m bands installed at Pawleys Island, SC, USA. It was designed by Wayne Lowrance, W6ZA, and Bob Ehramjian, K2US using Nex-Wires 2. Wires are made of #12 AWG solid copper. Specifications of this unusual big gun are next : it offers respectively on 20/15 m a 14.6 / 15.7 dBd gain, F/B ratio 48.7 (max. 71) / 36.3 (max. 50), a power gain over a dipole of 29.1 / 37, and an efficiency of 91.6 / 91.8% ! Its boom is 18 m long (60'). It is installed on a 33m high (100') US Tower HDX-589MDLP. The rotator is a prop-pitch from a WWII B24 Bomber !
However, taking into account the stealth, easy of assembling, sturdiness, availability, and price, my good intentions fall back, and surely if I had to live with the opinions of a co-ownership.
On its side a beam is easy to assembly, can be stealthy if it is of small size, light and not too bulky. They are also available to many manufacturers at very attractive prices.
But in addition to financial and available space considerations, to decide on which antenna you will bear your choice, in theory you should need to test a quad in the field to appreciate its performances vs. a Yagi.
Log periodic vs. Yagi
Why don't we see much "logs" in our countries ? Most of them are used by institutions, including embassies which roof often displays a huge log periodic. But not many hams use the log periodic excepting the one maybe charmed by their design and why not... the difficulties of building.
Indeed, it is a beam hard to design and to build due to its numerous elements and wire segments. Confronted to the accuracy required to calculate and assemble all segments as well as to respect the logarithmic spacing between elements, more than one amateur moved back in front of the task. However we must also recall that today the computers and their spreasheets offer a great assistance in (re)solving such problems.
In the field some amateurs built successfully their log periodic to name K4EWG (12 elements 13-30 MHz) and K8CU (2x 13 elements stacked 14-30 MHz) who shared their experience with QST readers.
The positive thing using a log periodic is that the method of feeding the antenna is rather simple (at least compared to its design !). It consists in using a balanced feed line for each element, all adjacent elements being fed with a 180° phase shift by alternating element connections. Saying that, with all its elements and the long boom required to support them (8-12m long for a 12-element beam), the log periodic is bulkier and heavier than a Yagi offering the same performances.
At left, a 10-element Titanex log periodic in front of a low band vertical from Titanex too at LX1EA living in Beaufort. At right, an optibeam OB9-5 bander.
Contrarily to what state some authors, all elements of a log periodic are not active at all frequencies. They should say that all elements are active forward of the one most active at any given frequency. In practice only 3 elements are really active at a time (the radiator at resonance on the actual working frequency, its reflector and the first director) although all elements show some current which amplitude decreases with the distance. We can thus compare the gain offered by a log periodic with a 3-element monoband Yagi, practically as long and as wide but of course much more stealthy with its 3 elements.
Justly, Bill Jones, K8CU, works with an homemade 12-element log periodic and uses also a 3-element monoband Yagi. Where the log periodic displays a maximum gain of 5.7 dBd the beam exceeds 7 dBd, on par with a 2-element quad. We can however reach 6.9 dBd using a 4-element log periodic and optimizing the elements spacing and length in modifying the traditional log periodic design.
But worst, the maximum gain of the log periodic "falls" quite often out of the band : the 5.7 dBd gain for example is recorded on 22 MHz and it is of 5.4 dBd on 21 MHz; there is also a peak of 5.5 dBd at 27.5 MHz right in the CB band instead of 28.5 MHz (5.4 dBd gain). The 14 MHz is the only frequency where the gain is maximum (5.1 dBd gain) to fall down just after. But all theses values are ridiculously low.
The log periodic bandwidth is also broader than the one of a Yagi, and as many variables on this antenna, it varies depending on the values of the design parameter t and the relative spacing constant s.
In the same way, its gain is generally 30 to 50% lower than the one of a Yagi using the same number of elements, with a free-space forward gain that never exceed 9 dBd for a 12-element log periodic. For your information, such an lof periodic has a boom length of 8m (26.5') and with all its tubing and its large boom of 75 mm diameter (3"), its weight reaches 53 kg (116 lbs.) ! It offers a half-power bandwidth of 43°, a F/B ratio of 14.4 dB at 14 MHz and up to 21 dB on 28 MHz. Any 5-element beam exceed these values and many wire beams add to that a featherweigth. But what should be the performances of a 12-element Yagi used in the same spectrum of frequencies ?
Of course there are very few if not any Yagi of such a sizing in the field. We know however that on any kind of beam the antenna gain is proportional to the length of the array, provided the number, lengths, and spacings of the elements. We can thus estimate that a 12-element Yagi should display a 14 dBd gain; it leaves the log periodic decibels behind it, Hi!
We can also calculate the gain of long Yagis as a function of overall array length, itself being a function of the numbers of elements. So, if a 12-element Yagi gives an optimum array length of about 3.2l, the expected gain is ranging between 12.5-15.8 dBd. In all cases the Yagi wins against the log periodic.
Adding more than 15 elements on the log periodic does not significantly modify the behaviour of the circuit. However it is possible to get a higher gain tilting the elements towards the sky what increases the gain of 3 to 5 dB. But we don't reach yet the gain of the equivalent Yagi.
The fight of log periodics against its competitors is thus lost in advance.
The status of a log periodic is still more critical if you compare it with a quad. Even placed close to the earth (say the base 2m above ground), a 2-element quad will be a good DX performer and will reach far countries much easier than a 3-element Yagi for example. This comes from the quad design itself that takes advantage of both a very low takeoff angle and to its more "powerful" loop measuring a full wavelength long instead of 1/2l in the case of a beam's radiator.
A 2-element log periodic does not exist, it is called a parasitic array ! It is thus difficult to compare designs having so much differences. In all cases, erected 10m high (30') a quad will give performance to make a Yagi green. So imagine adding 10 more elements to this quad... The Yagi is under cover but the log periodic gave up the fight, Hi!
Log periodics have thus still an hard life in front of them if they want to fight on par against directive arrays like quads or even Yagis.
The ARRL Antenna Book and related books
RSGB has also published tens of books about antennas
Antenna manufacturers and dealers
Alpha DeltaAntenna.itB&WButternut Cubex Cushcraft Degen Diamond Force 12 GAP High-gain High Sierra PKW Steppir Tennadyne TGM WiMo Universal-radio eHam.net (reviews) | 2,935 | 12,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-39 | latest | en | 0.949272 |
http://www.physicsforums.com/showthread.php?t=246785 | 1,368,997,673,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698080772/warc/CC-MAIN-20130516095440-00084-ip-10-60-113-184.ec2.internal.warc.gz | 646,894,561 | 8,990 | Properties of the Absolute Value
Just wanted to say hi before I start my post!
As you may know there is a property of the absolute value that states; for $$a, b \in R$$;
$$|ab| = |a||b|$$
Well, my friend asked me if I knew a proof for this... but I don't know...
How can we prove this statement/property? I know there is a proof for the triangle inequality but for this I really don't know but I'm curious.
I'd appreciate it if anyone could show me any kind of proof or send me some links etc. Thanks!
PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
Recognitions: Gold Member Science Advisor Staff Emeritus You prove that |ab|= |a||b| the same way you prove any such elementary statement: use the definitions. The simplest definition of |x| (there are several equivalent definitions) is that |x|= x if x is positive or 0, -x if x is negative. Now break it into "cases": case 1: x and y are both positive: |x|= x and |y|= y. xy is also positive so |xy|= xy= |x||y|. case 2: x is positive while y is negative: |x|= x and |y|= -y. xy is negative so |xy|= -xy= x(-y)= |x||y|. case 3: x is negative while y is positive: |x|= -x and |y|= y. xy is negative so |xy|= -xy= (-x)y= |x||y|. case 4: x and y are both negative: |x|= -x and |y|= -y. xy is positive so |xy|= xy= (-x)(-y)= |x||y|. case 5: x= 0 and y is positive: |x|= 0 and |y|= y. xy= 0 so |xy|= 0= 0(y)= |x||y|. case 6: x= 0 and y is negative: |x|= 0 and |y|= -y. xy= 0 so |xy|= 0= (0)(-y)= |x||y|. case 7: x is positive and y is 0: |x|= x and |y|= 0. xy= 0 so |xy|= 0= x(0)= |x||y|. case 8: x is negative and y is 0: |x|= -x and |y|= 0. xy= 0 so |xy|= 0 = (-x)(0)= |x||y|. case 9: both x and y are 0: |x|= 0 and |y|= 0. xy= 0 so |xy|= 0= (0)(0)= |x||y|. There are simpler ways to prove that but I thought this would be conceptually clearest.
First, we have to understand that the absolute value is a function defined by: $$|x| = \begin{cases} x & \text{if } x\geq 0 \\ -x & \text{if } x<0 \end{cases}$$ So, $$|ab| = \begin{cases} ab & \text{if } ab\geq 0 \\ -ab & \text{if } ab<0 \end{cases}$$ Now, let's see what |a||b| is: $$|a||b| = \begin{cases} ab & \text{if } a\geq 0 \wedge b\geq0 \\ (-a)(-b) & \text{if } a\leq 0 \wedge b\leq0 \\ (-a)b & \text{if } a> 0 \wedge b<0 \\ a(-b) & \text{if } a<0 \wedge b<0 \end{cases} \Leftrightarrow$$ $$|a||b| = \begin{cases} ab & \text{if } a\geq 0 \wedge b\geq0 \\ ab & \text{if } a\leq 0 \wedge b\leq0 \\ -ab & \text{if } a> 0 \wedge b<0 \\ -ab & \text{if } a<0 \wedge b<0 \end{cases} \Leftrightarrow$$ Notice that you have ab if a and b have the same sign and that you use -ab otherwise. Now, if a and b have the same sign, $$ab\geq0$$. If they have opposite signs (and are different than zero), $$ab<0$$. Using this, $$|a||b| = \begin{cases} ab & \text{if } ab \geq 0 \\ -ab & \text{if } ab<0 \end{cases} = |ab|$$ Quod erat demonstrandum
Properties of the Absolute Value
let a = mcis(kpi) where m => 0, k is integer
and b = ncis(hpi) where n => 0, h integer
(clearly a,b are real)
|a||b|= mn
|ab|=|mcis(kpi)*ncis(hpi)| = |mncis[pi(k+h)]| = mn
as required
Recognitions:
Gold Member
Staff Emeritus
Quote by quark1005 let a = mcis(kpi) where m => 0, k is integer and b = ncis(hpi) where n => 0, h integer (clearly a,b are real) |a||b|= mn |ab|=|mcis(kpi)*ncis(hpi)| = |mncis[pi(k+h)]| = mn as required
This would make more sense if you had said that "mcis(hpi)" is
$$m (cos(h\pi)+ i sin(h\pi))$$
That is much more an "engineering notation" than mathematics.
If you really want to go to complex numbers, why not
if
$$x= r_xe^{i\theta_x}$$
and
$$y= r_ye^{i\theta_y}$$, then
$$|xy|= |r_xe^{i\theta_x}r_ye^{i\theta_y}|= |(r_xr_y)e^{i(\theta_x+\theta_y)}|$$
But for any $z= re^{i\theta}$, |z|= r, so
$$|xy|= r_x r_y= |x||y|$$
$$\Huge |ab|=\sqrt{(ab)^2}=\sqrt{a^2b^2}=\sqrt{a^2}\sqrt{b^2}=|a||b|$$ and $$\large |a+b|=\sqrt{(a+b)^2}=\sqrt{a^2+2ab+b^2}\leq \sqrt{a^2+|2ab|+b^2}=\sqrt{(|a|+|b|)^2}=||a|+|b||$$ where I utilize the following fact: |2ab|=|2(ab)|=|2||ab|=2|a||b| ----- À bientôt ?;-D | 1,585 | 4,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2013-20 | latest | en | 0.908049 |
http://lists.gnu.org/archive/html/discuss-gnuradio/2014-02/msg00129.html | 1,508,272,047,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00214.warc.gz | 205,566,869 | 3,701 | [Top][All Lists]
Re: [Discuss-gnuradio] How to specify a complex number in block: vector_
From: Activecat Subject: Re: [Discuss-gnuradio] How to specify a complex number in block: vector_source ? Date: Sat, 8 Feb 2014 17:38:48 +0800
Dear Marcus,
Now I know the reason:
Vector: ( 1 + 4j ) ==> this doesn't work
Vector: ( 1 + 4j, ) ==> this work!
So there must have a comma if it is one element.
Thanks.
On Sat, Feb 8, 2014 at 5:28 PM, Marcus Müller wrote:
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
Oh I just realized:
since python is complaining about "j" not being known:
*Don't* have a space between "2" and "j". 2j is a numeric literal,
which python understands, 2 j is a numeric literal (2) and a name (j),
which python can't understand.
Greetings,
Marcus
On 08.02.2014 08:33, Activecat wrote:
> Dear Marcus,
>
> Thanks, it accepts ( complex(1,2), complex(3,4), complex(5,6) ).
> The other two are not accepted, message: name 'j' is not defined.
>
> Nevertheless, that is good enough. Thanks.
>
> Regards, activecat
>
> On Sat, Feb 8, 2014 at 2:38 PM, Marcus Müller <address@hidden>
> wrote:
>
> Hi activecat,
>
> have you just tried the most intuitive options:
>
> Vector = ( 1 + 2j, 2 + 4.0 * 1j, complex(3,-1) )
>
> all three elements work, since this is just python :)
>
> Greetings, Marcus
>
>
> On 08.02.2014 07:28, Activecat wrote:
>>>> Dear Sir,
>>>>
>>>> I am using the built-in block of Vector Source, configured
>>>> as below: Output Type = Complex. Vector = ( 1, 2, 3 )
>>>>
>>>> Then I run the flow graph. The block sends out 1 + 0i, 2 +
>>>> 0i, 3 + 0i.
>>>>
>>>> Question: How to make it sends 1 + 2i ? (How to configure
>>>> this at the Vector field?)
>>>>
>>>> Regards, activecat
>>>>
>>>>
>>>>
>>>> _______________________________________________
>>>>
>>
>>
>
>
>
>
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1
Comment: Using GnuPG with Thunderbird - http://www.enigmail.net/
iQEcBAEBAgAGBQJS9fi0AAoJEAFxB7BbsDrLT9kH/icf6j2LbltF+M7cqX73uG5S
uG+YsMc90k3QE9wOhFQG7mDAivBOUjF2gEmxKavYLHtqXRwAJ0eu7NHif1lcHBgV
7jyrrO6Gqk3cq3pvwM+pI9/hSHyH5l1K+yLnkyNuLrWTScIJqX5y4/EWHTLZcls/
DrP9vpgnsjo0RcVtB0Tv0dWyeyAGYW9r5I8UscFtlgFQlr5XSoHQhwlZy2nPJrLt
dg4AgZuXEdhKzG8pVeJDNhXFsP2VYHThheXtBPTlRqILhtcRaQ8HHminyCp53kdA
Unm4QXUH9cZbToURySxoUd/QvohjOUEmTgGk+OoH7UW2RbCpkq0LU7E7IklDKAI=
=kFhZ
-----END PGP SIGNATURE-----
_______________________________________________ | 912 | 2,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-43 | latest | en | 0.754899 |
https://simplewebtool.com/converters/area/hectarestosquaremicrometers/hectarestosquaremicrometers.html | 1,713,129,949,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00278.warc.gz | 501,713,268 | 12,461 | # Hectares to Square Micrometers conversion
## Hectares (ha) to square Micrometers (µm2) conversion calculator and how to convert.
How to convert Hectares to Square Micrometers: 1 hectare is equal to 1e16 square micrometers. 1 ha = (1 * 1016) = 1e16 (µm2) The areaQ(µm2) in square micrometers is equal to the area Q(ha) in hectacres multiplied by 10000000000000000.
Formula: Q(µm2) = Q(ha) * 10000000000000000
Hectares to Square Micrometers conversion table
Hectares (ha) Square Micrometers (µm2)
1 ha 1e16 µm2
2 ha 2e16 µm2
3 ha 3e16 µm2
4 ha 4e16 µm2
5 ha 5e16 µm2
6 ha 6e16 µm2
7 ha 7e16 µm2
8 ha 8e16 µm2
9 ha 9e16 µm2
10 ha 1e17 µm2 | 268 | 644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.363103 |
https://www.tutorialspoint.com/p-in-the-below-figure-a-b-c-d-is-a-square-of-side-2-a-find-the-ratio-between-the-circumferences-p-p-img-src-assets-questions-media-158630-1623946692-jpg-p | 1,701,934,385,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00473.warc.gz | 1,135,461,927 | 22,474 | "
">
# In the below figure, $A B C D$ is a square of side $2 a$. Find the ratio between the circumferences."
Given:
$A B C D$ is a square of side $2 a$.
To do:
We have to find the ratio between the circumferences.
Solution:
The square $ABCD$ is inscribed a circle.
Length of the side of the square $= 2a$
From the figure,
Diameter of the outer circle $AC =$ Diagonal of the square
$=\sqrt{2} \times 2 a$
$=2 \sqrt{2} a$
This implies,
Radius of the outer circle $R=\frac{\mathrm{AC}}{2}$
$=\frac{2 \sqrt{2} a}{2}$
$=\sqrt{2} a$
Diameter of the inner circle $=2a$
Radius of the inner circle $r=\frac{2a}{2}=a$
Therefore,
The ratio between the circumferences of the circles $=\frac{\text { circumference of outer circle }}{\text { circumference of inner circle }}$
$=\frac{2 \pi \mathrm{R}}{2 \pi r}$
$=\frac{\mathrm{R}}{r}$
$=\frac{\sqrt{2} a}{a}$
$=\frac{\sqrt{2}}{1}$
The ratio between the circumferences is $\sqrt{2}:1$.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
103 Views
Get certified by completing the course | 333 | 1,059 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-50 | latest | en | 0.670383 |
https://best-excel-tutorial.com/how-to-generate-random-text-from-the-list/ | 1,695,586,990,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.30/warc/CC-MAIN-20230924191454-20230924221454-00050.warc.gz | 149,048,430 | 34,568 | #### How to generate random text from the list?
In this Excel tutorial, you will learn how to generate random text from a list.
## Random text generator formula
Suppose that your list is in the A1: A10 cell range. You have some text there from which you want to generate one in some other cell. Here is an Excel formula that solves your problem:
=INDEX(A1:A10,RAND()*ROWS(A1:A10)+1)
Drag it down to the number of cells you need it in.
Remember that every time the formula generates a new text, Each time you run the formula, it will generate a random item from a given list.
## Random generator using ROUNDUP function
Another method would be to use ROUNDUP function.
1. Prepare your list: First, create a list of the text items you want to randomly generate. This list should be in a single column, with each item in a separate cell.
2. Enter the RAND function: In an adjacent column, enter the formula “=RAND()” in the first cell. This will generate a random number between 0 and 1.
3. Copy the formula: Copy the formula down the column, so that each cell in the column contains a unique random number.
4. Enter the INDEX function: In the next column, enter the formula “=INDEX(A1:A6,ROUNDUP(B1*6,0)),” replacing “A1:A6” with the range that contains your list of text items and “6” with the number of items in your list.
5. Copy the formula: Copy the formula down the column, so that each cell in the column contains a randomly selected text item from your list.
6. Finalize the random text: You can now use the randomly generated text in your Excel worksheet as needed. If you need to generate new random text, simply copy the “RAND” and “INDEX” formulas down the column again.
See also How to Apply a Formula to an Entire Column in Excel
These steps should help you generate random text from a list in Excel. You can use this technique to randomly select items from any list, such as names, dates, or numbers, depending on the type of data you’re working with and the message you want to convey.
## How to Generate Random Text from a List Using the RANDBETWEEN Function
The RANDBETWEEN function is a built-in function in Excel that can be used to generate a random number between two values. The syntax of the RANDBETWEEN function is as follows:
`=RANDBETWEEN(start_value, end_value)`
For example, the following formula would generate a random number between 1 and 10:
`=RANDBETWEEN(1, 10)`
The RANDBETWEEN function can be used to generate random text from a list by using the INDEX function. The INDEX function returns the value from a range of cells based on a row and column number.
The following formula would generate a random text from the list in the range A1:A10:
`=INDEX(A1:A10,RANDBETWEEN(1,ROWS(A1:A10)))`
This formula first generates a random number between 1 and the number of rows in the range A1:A10. The INDEX function then returns the value from the list based on the random number. | 691 | 2,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-40 | longest | en | 0.836806 |
https://www.physicsforums.com/threads/absolute-polynomial-question-driving-me-nuts.60757/ | 1,540,293,851,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516135.92/warc/CC-MAIN-20181023111223-20181023132723-00045.warc.gz | 1,054,685,247 | 12,178 | # Absolute polynomial question (driving me nuts)
1. Jan 22, 2005
### KataKoniK
|x^2 - 2x - 3| = -(x^2 - 2x - 3), 0 <= x < 3
x^2 - 2x - 3, 3 <= x <=4
how did they get the intervals:
3 <= x <=4 and 0 <= x < 3?
I cannot, for the love of God, know how they determined it.
2. Jan 22, 2005
### Hurkyl
Staff Emeritus
Remember the definition of absolute value:
$$|q| := \left\{ \begin{array}{ll} q \quad & q \geq 0 \\ -q \quad & q \leq 0 \end{array}$$
Anyways, it sounds like there's also more to the problem than you gave us.
3. Jan 22, 2005
### KataKoniK
Thanks for the help! | 223 | 582 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-43 | latest | en | 0.870983 |
https://numbermatics.com/n/9526/ | 1,723,249,235,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782288.54/warc/CC-MAIN-20240809235615-20240810025615-00588.warc.gz | 340,112,674 | 6,594 | # 9526
## 9,526 is an even composite number composed of three prime numbers multiplied together.
What does the number 9526 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 8 divisors.
9526 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of eight divisors.
## Prime factorization of 9526:
### 2 × 11 × 433
See below for interesting mathematical facts about the number 9526 from the Numbermatics database.
### Names of 9526
• Cardinal: 9526 can be written as Nine thousand, five hundred twenty-six.
### Scientific notation
• Scientific notation: 9.526 × 103
### Factors of 9526
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 3
• Sum of prime factors: 446
### Divisors of 9526
• Number of divisors d(n): 8
• Complete list of divisors:
• Sum of all divisors σ(n): 15624
• Sum of proper divisors (its aliquot sum) s(n): 6098
• 9526 is a deficient number, because the sum of its proper divisors (6098) is less than itself. Its deficiency is 3428
### Bases of 9526
• Binary: 100101001101102
• Base-36: 7CM
### Squares and roots of 9526
• 9526 squared (95262) is 90744676
• 9526 cubed (95263) is 864433783576
• The square root of 9526 is 97.6012295005
• The cube root of 9526 is 21.1984216221
### Scales and comparisons
How big is 9526?
• 9,526 seconds is equal to 2 hours, 38 minutes, 46 seconds.
• To count from 1 to 9,526 would take you about two hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 9526 cubic inches would be around 1.8 feet tall.
### Recreational maths with 9526
• 9526 backwards is 6259
• 9526 is a Harshad number.
• The number of decimal digits it has is: 4
• The sum of 9526's digits is 22
• More coming soon!
#### Copy this link to share with anyone:
MLA style:
"Number 9526 - Facts about the integer". Numbermatics.com. 2024. Web. 10 August 2024.
APA style:
Numbermatics. (2024). Number 9526 - Facts about the integer. Retrieved 10 August 2024, from https://numbermatics.com/n/9526/
Chicago style:
Numbermatics. 2024. "Number 9526 - Facts about the integer". https://numbermatics.com/n/9526/
The information we have on file for 9526 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 9526, math, Factors of 9526, curriculum, school, college, exams, university, Prime factorization of 9526, STEM, science, technology, engineering, physics, economics, calculator, nine thousand, five hundred twenty-six.
Oh no. Javascript is switched off in your browser.
Some bits of this website may not work unless you switch it on. | 875 | 3,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-33 | latest | en | 0.888165 |
https://de.scribd.com/doc/221349868/Chap-3-HW-Assigned-Prob-1-3-4-Answers-1 | 1,585,687,631,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370503664.38/warc/CC-MAIN-20200331181930-20200331211930-00397.warc.gz | 446,449,064 | 77,628 | Sie sind auf Seite 1von 5
# Chapter 3
Problem 1
Info:
TelCo to decide if replace computer system
cost of capital
tax rate
SL depre
(000's)
Net BEFORE tax cost savings
12%
35%
Year
(\$)
2
350
3
350
4
300
## Cost of new comp
1,000
ITC
15% first year
ITC reduces Telco's taxes by an amt = to 15% of equipment purchase price
old computer can be sold for
450
cost of old computer
1,250 3 years old
life
5 yrs
both computers have -0- salvage value
no ITC on old computer
5
300
depre/yr 200
150 ITC amt yr 1
## a. What is the net investment required in the new system?
Net Investment = Cost - Salvage(old) + Tax from sale of old
cost of new machine
sale price of old machine
tax from sale of old
Net Investment
## Book value of old machine
cost
3 yrs old SL D/E
Book value of old machine
old computer can be sold for
1,000
(450)
(18)
\$ 532.50
b. Estimate the incremental operating CF's associated with the new system
Incremental CF's = after tax savings + (tax x net depreciation)
discount rate
12%
Year
1
2
Net BEFORE tax cost savings
(\$)
350
350
Tax at 35% tax rate
123
123
ITC tax credit yr 1
150
after tax savings
78
228
add: (tax x depre) of new comp
(50)
(50)
300
proceeds/loss
tax rate
tax
(50)
35%
(18)
3
300
105
4
300
105
5
300
105
195
200
195
200
195
200
Chap 3, prob 1
1,250
750
500
450
CF
Salvage value of new machine
tax on salvage value
Incremental CF's
28
28
PV =
investment
NPV =
178
178
395
395
395
395
\$959.25
(532.50)
\$426.75 The new computer should be purchased
## c. If the new computer's salvage value at the end of 5 years is projected to be
\$100,000, should TelCo purchase it?
salvage value
Book Value
cost
5 yr life D/E
Book value
Pay tax on:
tax rate
tax amt
100
1,000
1,000
100 salvage - BV
35%
35
Chap 3, prob 1
395
100 given
(35)
460
investment
NPV
## \$922.37 NPV or CF's
(532.50)
389.87
Chapter 3
Problem 3
Info:
Varico
Units sold/year
Ea. Unit needs electric motor, purchased 1 time per week
Cost per motor
Interest rate
Purchase amount
average inventory on hand (divide purchase amt by 2)
Foreign Firm to sell 100,000 motors at 9.50 ea.
Calculate the Opportunity cost of maintaining inventory =
avg. # of units on hand
price / unit
interest rate
Opportunity cost of maintaining inventory =
By buying the inventory weekly, no interest expense incurred.
The real cost of buying 100,000 motors today =
Units sold per yr x price per unit
Opportunity cost of maintaining inventory =
Result, does this exceed the cost to purchase weekly? A
Cost of purchasing motors weekly B
diff A - B
100,000
\$
10.00
15%
100,000
50,000
\$
\$
\$
\$
\$
Chap 3, prob 3
50,000
9.50
475,000.00
15%
71,250
950,000
71,250
1,021,250 A
1,000,000 B
(21,250)
## exceeds cost to purchase weekly
Chapter 3
Problem 4
Info:
Specific Foods, Inc.
i). Calculate net income & operating CF's
Discount Rate
10%
0
(1,250,000)
(25,000)
Equipment purchase
Installation costs
a). Revenue
b). COGS @ 60% of sales
GM
D/E on equipment; 10 yr life, SL
D/E on installation, 5 yr life, SL
Initial costs/expenses on equipment
EBIT
Taxes @ 35%
Net Income
Deprec
OCF's
PV =
200,000 \$
(120,000)
80,000
(10,000)
(125,000)
(5,000)
1,000,000 \$
(600,000)
400,000
(10,000)
(125,000)
(5,000)
10
## 1,150,000 \$ 1,322,500 \$ 1,520,875 \$ 1,749,006 \$ 1,749,006 \$ 1,486,655 \$ 1,263,657 \$ 1,074,108
(690,000)
(793,500)
(912,525) (1,049,404) (1,049,404)
(891,993)
(758,194)
(644,465)
460,000
529,000
608,350
699,603
699,603
594,662
505,463
429,643
(10,000)
(10,000)
(10,000)
(10,000)
(10,000)
(10,000)
(10,000)
(10,000)
(125,000)
(125,000)
(125,000)
(125,000)
(125,000)
(125,000)
(125,000)
(125,000)
(5,000)
(5,000)
(5,000)
(60,000)
21,000
(39,000)
260,000
(91,000)
169,000
320,000
(112,000)
208,000
389,000
(136,150)
252,850
468,350
(163,923)
304,428
564,603
(197,611)
366,992
564,603
(197,611)
366,992
459,662
(160,882)
298,780
370,463
(129,662)
240,801
294,643
(103,125)
191,518
130,000
91,000
130,000
299,000
130,000
338,000
130,000
382,850
130,000
434,428
125,000
491,992
125,000
491,992
125,000
423,780
125,000
365,801
125,000
316,518
\$2,120,065.46
## ii). Find NPV using 10% cost of capital
Cost today:
Equipment cost
Installation
Initial expensed costs
Tax on the exp initial costs
Net Investment
PV
Less: investment
NPV:
1,250,000
25,000
875,000 given
(306,250)
1,843,750
\$2,120,065.46
(1,843,750)
\$276,315 The project should be accepted.
## iii). Adding inflation, what is the project's NPV?
Discount Rate
Equipment purchase
Installation costs
a). Revenue
b). COGS @ 60% of sales (will grow 20%/yr from
600K level, until yr 6, remain same for yr 7, then
decline 15%/yr through yr 10)
GM
10%
0
(1,250,000)
(25,000)
200,000 \$
(120,000)
80,000
(10,000)
1,000,000 \$
(600,000)
400,000
(10,500)
10
(720,000)
430,000
(11,025)
Chap 3, prob 4
(864,000)
458,500
(11,576)
(1,036,800)
484,075
(12,155)
(1,244,160)
504,846
(12,763)
(1,244,160)
504,846
(13,401)
(1,057,536)
429,119
(14,071)
(898,906)
364,751
(14,775)
(764,070)
310,039
(15,513)
## D/E on equipment; 10 yr life, SL
D/E on installation, 5 yr life, SL
Initial costs/expenses on equipment
EBIT
Taxes @ 35%
Net Income
Deprec
OCF's
PV =
(125,000)
(5,000)
(125,000)
(5,000)
(125,000)
(5,000)
(125,000)
(5,000)
(125,000)
(5,000)
(125,000)
(125,000)
(125,000)
(125,000)
(125,000)
(60,000)
21,000
(39,000)
259,500
(90,825)
168,675
288,975
(101,141)
187,834
316,924
(110,923)
206,000
341,920
(119,672)
222,248
367,083
(128,479)
238,604
366,445
(128,256)
238,189
290,048
(101,517)
188,531
224,977
(78,742)
146,235
169,525
(59,334)
110,192
130,000
91,000
130,000
298,675
130,000
317,834
130,000
336,000
130,000
352,248
125,000
363,604
125,000
363,189
125,000
313,531
125,000
271,235
125,000
235,192
\$1,760,160.45
## ii). Find NPV using 10% cost of capital
Cost today:
Equipment
Installation
Initial expensed costs (not depre)
Tax on the exp initial costs (35%)
Investment
PV
Less: investment
NPV:
1,250,000
25,000
875,000 given
(306,250)
1,843,750
\$1,760,160.45
(1,843,750)
(\$83,590) The project should be rejected
Chap 3, prob 4 | 2,334 | 6,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-16 | latest | en | 0.72884 |
https://www.forretr.io/2011/08/physics-radioactive-decay-and-nuclear.html | 1,717,106,315,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00174.warc.gz | 664,151,932 | 16,418 | ## Saturday 6 August 2011
### Physics: radioactive decay and nuclear energy
To access the document version of this chapter of notes click the "Notes Corner" above.
Radioactive decay is the process where unstable nuclei break down and emit radiation. The nuclide before decay is the parent nuclide and that after decay is daughter nuclide.
1) α-decay: AX A-4Y + 4He (α), a helium nuclei is emitted, atomic number - 2 while mass number -4, e.g. 238U 234Th + α
2) β-decay: AX AY + e (β), the atomic number of daughter nuclei +1 while mass number unchanged, a neutron decays into a electron and a proton. (a antineutrino as well, but this is out of the discussion). Electron was written as atomic number -1 to balance the total number of protons. e.g., 14C 14N + β
3) γ-decay: AX* AX + γ, an excited nuclei emit an energetic photon and become stable. e.g., 60Ni* 60Ni + γ
Sometimes we use A-Z graph or N(number of neutrons)-Z graph to show the decay series.
Radioactive decay – a random process
Every radioactive nucleus has a certain probability to decay within a certain period of time but it’s unknown to know the time to decay. In macroscopic view the decay activity is more stable and can be calculated.
Number of disintegration of a source per second is called the activity (A) of the source, measured in (Becquerel) Bq or s-1. It’s proportional to undecayed nuclei N, i.e., A = kN where k is the decay constant in s-1.
Half life: it’s the time taken for half of the parent nuclei in any given sample to decay. i.e., N at a certain time is equal to N0(0.5)n where N0 is the original number of parent nuclei and n is the number of period of half life passed by. Similarly we have A = A0(0.5)n.
Now consider A = -dN/dt = kN, dN/N = -k dt, integrate both sides gives ln N = -kt+C, when t = 0, N = N0, so we have N = Noe-kt. Similarly A = A0e-kt.
Assume t be the half life. Then N0/2 = N0e-kt, then half life t1/2 = ln 2/k. So by plotting the graph ln A-t or ln N-t, the y-intercept is ln A0 (N0) while the slope is k. Then we can find the half life of the sample.
Nuclear energy
Binding energy of an atom is the energy to separate those neutrons and protons to infinite far. They are originally bonded by strong nuclear force (which is stronger than electrostatic force), so energy is released during the combination of nucleus. Under nuclear reaction, binding energy is provided to nucleus to separate the nucleus, and another binding energy is released through combining into a new nucleus. Higher binding energy means more energy is released in combination, hence it’s more stable. The maximum of binding energy occurs around Fe to Co, to lighter elements undergoes fusion to release energy to achieve stableness, while heavier elements undergo fission to release energy.
1) Nuclear fission: a large nucleus splits into 2 or more nuclei of comparable masses. Neutrons are used to bombard the nucleus and to supply energy to it. Proton won’t be used to prevent electrostatic repulsion between proton and the nuclei. For example consider a piece of uranium-235: 235U + n 236U (very unstable) 92Kr + 141Ba + 3n. Note that more energetic neutrons are produced, they can also used to bombard other uranium atoms and cause a chain reaction. Uncontrolled chain reaction were used in nuclear weapons like atomic bomb.
2) Nuclear fusion: two lighter nuclei unions into a heavier nucleus. For example consider a fusion occurring in stars: 2D + 3T 4He + n, where D and T are deuterium and tritium which are isotopes of hydrogen. Unfortunately, extremely high temperature is needed (107 ~ 108K) to overcome the electrostatic repulsion (in order for strong nuclear force to take place). Therefore uncontrolled fusion is invented (hydrogen bomb) but the technological level is not enough to generate electricity by fusion, through the advantages include plentiful source of hydrogen and non-radioactive waste.
Unit of smaller mass
We define the weight of 12C as 12a.m.u. or simply u. 1u = 1.66*10-27kg. However by calculation scientists find that mp = 1.0073u, mn = 1.0087u, me=0.00055u, considering 12C contains 6 protons, 6 neutrons and 6 electrons, total mass = 12.099u, however the mass of that atom is only 12u, then we say that it has a mass defect of 0.099u. In fact, the energy is released in terms of binding energy.
Mass-energy equivalence
When mass are released in forms of energy, we found that there’s a certain relationship between mass and energy. Einstein states the E = mc2 (ΔE = Δmc2), where c is the speed of light in vacuum.
For example, consider n p+ + e-, mass defect = (1.0087-1.0073-0.00055)u = 0.00085 u, then energy released = (0.00085)(1.66*10-27)(3*108)2 = 1.27 * 10-13 J. When one mole of neutrons decays, (1.27*10-13)(6.02*1023) = 7.65*1010 J of energy, so we can see nuclear reaction can give out a large amount of energy. | 1,288 | 4,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.854336 |
http://www.chegg.com/homework-help/modify-program-exercise-21-generate-display-numbers-sequence-chapter-6-problem-22re-solution-9780073523309-exc | 1,475,171,745,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661905.96/warc/CC-MAIN-20160924173741-00130-ip-10-143-35-109.ec2.internal.warc.gz | 394,678,014 | 17,438 | # An Introduction to Object-Oriented Programming with Java (5th Edition) View more editions Solutions for Chapter 6 Problem 22REProblem 22RE: Modify the program of Exercise 21 to generate and display al...
• 574 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
Modify the program of Exercise 21 to generate and display all the numbers in the sequence until a number becomes larger than the value maxNumber entered by the user.
Exercise – 21
(For Reference 3rd Chapter Exercise 10)
Write a program that does the reverse of Exercise 9, that is, input degrees Fahrenheit and prints out the temperature in degrees Celsius. The formula to convert degrees Fahrenheit to equivalent degrees Celsius is
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 5
Program Plan:
• Import the library features of classes in java.
• Define a class OutputBox to print the prime number.
• Define the main class.
• Prompt the user to input the number of Fibonacci numbers to be printed.
• Define method Recfibonacci () to print the number.
• Define method Computefibonacci() to compute the Fibonacci number.
• Display the output.
• Chapter , Problem is solved.
Corresponding Textbook
An Introduction to Object-Oriented Programming with Java | 5th Edition
9780073523309ISBN-13: 0073523305ISBN: Authors:
Alternate ISBN: 9780071283687, 9780077415686 | 349 | 1,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-40 | latest | en | 0.731169 |
http://nasawavelength.org/resource-search?facetSort=1&topicsSubjects=Physical+sciences&materialsCost=%241+-+%245&smdForumPrimary=Planetary+Science&educationalLevel=Middle+school | 1,516,294,463,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887423.43/warc/CC-MAIN-20180118151122-20180118171122-00418.warc.gz | 245,227,226 | 14,690 | ## Narrow Search
Audience
Topics
Earth and space science
Physical sciences
Resource Type [-]
View more...
Learning Time
Materials Cost
Instructional Strategies [-]
View more...
SMD Forum
Filters: Your search found 10 results.
Topics/Subjects:
Physical sciences
Materials Cost:
\$1 - \$5
SMD Forum - Primary:
Planetary Science
Educational Level:
Middle school
Sort by:
Per page:
Now showing results 1-10 of 10
# LRO Educator Resource Kit
Each lesson or activity in this toolkit is related to NASA's Lunar Reconnaissance Orbiter (LRO). The toolkit is designed so that each lesson can be done independently, or combined and taught in a sequence. The Teacher Implementation Guide provides... (View More)
# Lava Layering
After creating a model of multiple volcanic lava flows, students analyze the layers, sequence the flows, and interpret the stratigraphy. Students use that same volcanic layering model to investigate relative dating and geologic mapping principles-... (View More)
# Temperature, Pressure, and the Earth
This is a lesson where learners explore the effects of pressure on temperature and states of matter and use this information to infer the conditions of the interior of the Earth. The lesson models scientific inquiry using the 5E instructional model... (View More)
# The Flow of Matter
This is a lesson where learners are introduced to chemical change and the conservation of matter during different types of chemical changes. The lesson models scientific inquiry using the 5E instructional model and includes teacher notes,... (View More)
# Design Challenge: How to Keep Gelatin from Melting?
This is a activity about applying the scientific method to a design challenge. Learners will design and build a platform that will be placed on a heat source. The platform is expected to serve as an insulator for a cube of gelatin. The goal is to... (View More)
# My Angle on Cooling: Effect of Distance and Inclination
Learners will design and conduct experiments to answer the question, "how does distance and inclination affect the amount of heat received from a heat source?" They will measure heat change as a function of distance or viewing angle. From that... (View More)
# Sand or Rock: Finding Out from 1000 km
Learners will measure temperature of two different surfaces; sand and stone; on a sunny day, make a series of temperature measurements, and plot the results. Extensions include experimenting with different materials, using temperature sensors and... (View More)
# Impact Craters
In this activity, learners will determine the factors affecting the appearance of impact craters and ejecta on the Moon. Extensions are listed. This activity is in Unit 2 of the Exploring the Moon teachers guide, which is designed for use... (View More)
# Where Do They Come From? Looking at Asteroids
This is a lesson about the connection between meteorites and asteroids, focusing on remote-sensing techniques using light. Learners will make and record observations and measurements; analyze data and draw analogies; compare samples; measure and... (View More)
# Where Do They Come From? Impact Crater - Holes in the Ground!
This is a lesson about impact craters; the relationships between crater size, projectile size and projectile velocity; and the transfer of energy in the cratering process. Learners will create plaster of Paris or layered dry impact craters and... (View More)
1 | 708 | 3,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-05 | longest | en | 0.859252 |
http://www.scirp.org/Journal/PaperInformation.aspx?PaperID=64631 | 1,529,418,223,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863043.35/warc/CC-MAIN-20180619134548-20180619154548-00455.warc.gz | 485,256,205 | 21,524 | Global Attractors for a Class of Generalized Nonlinear Kirchhoff-Sine-Gordon Equation
In this paper, we consider a class of generalized nonlinear Kirchhoff-Sine-Gordon equation . By a priori estimation, we first prove the existence and uniqueness of solutions to the initial boundary value conditions, and then we study the global attractors of the equation.
Received 29 December 2015; accepted 14 March 2016; published 17 March 2016
1. Introduction
In 1883, Kirchhoff [1] proposed the following model in the study of elastic string free vibration:
, where is associated with the initial tension, M is related to the material
properties of the rope, and indicates the vertical displacement at the x point on the t. The equation is more accurate than the classical wave equation to describe the motion of an elastic rod.
Masamro [2] proposed the Kirchhoff equation with dissipation and damping term:
where is a bounded domain of with a smooth boundary; he uses the Galerkin method to prove the existence of the solution of the equation at the initial boundary conditions.
Sine-Gordon equation is a very useful model in physics. In 1962, Josephson [3] fist applied the Sine-Gordon equation to superconductors, where the equation:, is the two-order partial derivative of u with respect to the variable t; is the two-order partial derivative of the u about the independent variable x. Subsequently, Zhu [4] considered the following problem: (where is a bounded domain of) and he proved the existence of the global solution of the equation. For more research on the global solutions and global attractors of Kirchhoff and sine-Gordon equations, we refer the reader to [5] -[11] .
Based on Kirchhoff and Sine-Gordon model, we study the following initial boundary value problem:
(1.1)
where is a bounded domain of with a smooth boundary; is the dissipation coefficient; is a positive constant; and is the external interference. The assumptions on nonlinear terms and will be specified later.
The rest of this paper is organized as follows. In Section 2, we first obtain the basic assumption. In Section 3, we obtain a priori estimate. In Section 4, we prove the existence of the global attractors.
2. Basic Assumption
For brevity, we define the Sobolev space as follows:
In addition, we define and are the inner product and norm of H.
Nonlinear function satisfying condition (G):
(1)
(2)
(3)
Function satisfies the condition (F):
(4)
(5)
(6)
(7)
3. A Priori Estimates
Lemma 3.1. Assuming the nonlinear function satisfies the condition (G)-(F), , , then the solution of the initial boundary value problem (1.1) satisfies and
where. Thus there exists a positive constant and, such that
Proof. Let, the equation can be transformed into
(3.1)
Taking the inner product of the equations (3.1) with v in H, we find that
(3.2)
By using Holder inequality, Young’s inequality and Poincare inequality, we deal with the terms in (3.2) one by as follows
(3.3)
where is the first eigenvalue of with Dirichlet boundary conditions on.
Since and (F) (6), (7), we get
(3.4)
and
(3.5)
(3.6)
where
(3.7)
Combined (3.1)-(3.6) type, it follows from that
(3.8)
According to condition (F) (5), this will imply, then,
, and since
(3.9)
that is
(3.10)
With (3.10), (3.8) can be written as
(3.11)
Set, and, then (3.11) is equivalent to (3.12)
(3.12)
where
(3.13)
By using Gronwall inequality, we obtain
(3.14)
Let.
So, we have
(3.15)
then
(3.16)
Hence, there exists and, such that
Lemma 3.2. Assuming the nonlinear function satisfies the condition (G)-(F), , then the solution of satisfies the initial boundary value problem (1.1) satisfies and
where. Thus there exists a positive constant and, such that
Proof. The equations (3.1) in the H and have inner product, we find that
(3.17)
By using Holder inequality, Young’s inequality and Poincare inequality, we get the following results
(3.18)
(3.19)
According to condition (F) (5), (6), we obtain
(3.20)
(3.21)
where
(3.22)
By (3.18)-(3.22), (3.17) can be written
(3.23)
Noticing, this will imply
(3.24)
Substituting (3.24) into (3.23), we can get the following inequality
(3.25)
Let, and, then (3.25) type can be changed into
(3.26)
then
(3.27)
where.
By using Gronwall inequality, we obtain
(3.28)
taking, we have
(3.29)
then
(3.30)
Hence, there exists and, such that
Theorem 3.1. Assuming the nonlinear function satisfies the condition (G)-(F), , , so the initial boundary value problem (1.1) exists a unique smooth solution.
Proof. By Lemma 3.1-Lemma 3.2 and Glerkin method, we can easily obtain the existence of solutions of equ-
ation, the proof procedure is omitted. Next, we prove the uniqueness of solutions in
detail.
Assume are two solutions of equation, we denote, then, the two equations subtract and obtain
(3.31)
We take the inner product of the above equations (3.31) with in H, we have
(3.32)
We deal with the terms in (3.32) one by as follows
(3.33)
and
(3.34)
By (3.32)-(3.34), we can get the following inequality
(3.35)
Further, by mid-value theorem and Young’s inequality, we get
(3.36)
Since,
might as well set.
where.
Then, we obtain
(3.37)
Substituting (3.36), (3.37) into (3.35), we can get
(3.38)
Let, then (3.38) can be changed to
(3.39)
By using Gronwall inequality, we obtain
(3.40)
There has
(3.41)
That show that.
So as to get, the uniqueness is proved. ■
4. Global Attractor
Theorem 4.1. [12] Set be a Banach space, and are the semigroup operator on.; here I is a unit operator. Set satisfy the follow conditions.
1) is bounded, namely; it exists a constant, so that
2) It exists a bounded absorbing set, namely,; it exists a constant, so that
here and B are bounded sets.
3) When, is a completely continuous operator.
Therefore, the semigroup operators S(t) exist a compact global attractor A.
Theorem 4.2. [12] Under the assume of Theorem 3.1, equations have global attractor
where; is the bounded absorbing set of and satisfies
(1);
(2), here and it is a bounded set, .
Proof. Under the conditions of Theorem 3.1, it exists the solution semigroup S(t), here .
(1) From Lemma 3.1-Lemma 3.2, we can get that is a bounded set that includes in the ball,
This shows that is uniformly bounded in.
(2) Furthermore, for any, when, we have
So we get is the bounded absorbing set.
(3) Since is compact embedded, which means that the bounded set in is the compact set in, so the semigroup operator S(t) is completely continuous. ■
Hence, the semigroup operator S(t) exists a compact global attractor A. The proving is completed.
Acknowledgements
The authors express their sincere thanks to the anonymous reviewer for his/her careful reading of the paper, giving valuable comments and suggestions. These contributions greatly improved the paper.
Funding
This work is supported by the National Natural Sciences Foundation of People’s Republic of China under Grant 11161057.
Cite this paper
Lou, R. , Lv, P. and Lin, G. (2016) Global Attractors for a Class of Generalized Nonlinear Kirchhoff-Sine-Gordon Equation. International Journal of Modern Nonlinear Theory and Application, 5, 73-81. doi: 10.4236/ijmnta.2016.51008.
[1] Kirchhof, G. (1883) Vorlesungen fiber Mechanik. Teubner, Stuttgarty. [2] Masamro, H. and Yoshio, Y. (1991) On Some Nonlinear Wave Equations 2: Global Existence and Energy Decay of Solutions. J. Fac. Sci. Univ. Tokyo. Sect. IA, Math., 38, 239-250. [3] Josephson, B.D. (1962) Possible New Effects in Superconductive Tunneling. Physics Letters, 1, 251-253. http://dx.doi.org/10.1016/0031-9163(62)91369-0 [4] Zhu, Z.W. and Lu, Y. (2000) The Existence and Uniqueness of Solution for Generalized Sine-Gordon Equation. Chinese Quarterly Journal of Mathematics, 15, 71-77. [5] Li, Q.X. and Zhong, T. (2002) Existence of Global Solutions for Kirchhoff Type Equations with Dissipation and Damping Terms. Journal of Xiamen University: Natural Science Edition, 41, 419-422. [6] Silva, M.A.J. and Ma, T.F. (2013) Long-Time Dynamics for a Class of Kirchhoff Models with Memory. Journal of Mathematical Physics, 54, Article ID: 021505. [7] Zhang, J.W., Wang, D.X. and Wu, R.H. (2008) Global Solutions for a Class of Generalized Strongly Damped Sine-Gordon Equation. Journal of Mathematical Physics, 57, 2021-2025. [8] Guo, L., Yuan, Z.Q. and Lin, G.G. (2014) The Global Attractors for a Nonlinear Viscoelastic Wave Equation with Strong Damping and Linear Damping and Source Terms. International Journal of Modern Nonlinear Theory and Application, 4, 142-152. http://dx.doi.org/10.4236/ijmnta.2015.42010 [9] Teman, R. (1988) Infiniter-Dimensional Dynamical Systems in Mechanics and Physics. Springer-Verlag, New York, 15-26. http://dx.doi.org/10.1007/978-1-4684-0313-8_2 [10] Ma, Q.F., Wang, S.H. and Zhong, C.K. (2002) Necessary and Sufficient Congitions for the Existence of Global Attractors for Semigroup and Applications. Indiana University Mathematics Journal, 51, 1541-1559. http://dx.doi.org/10.1512/iumj.2002.51.2255 [11] Ma, Q.Z., Sun, C.Y. and Zhong, C.K. (2007) The Existence of Strong Global Attractors for Nonlinear Beam Equations. Journal of Mathematical Physics, 27A, 941-948. [12] Lin, G.G. (2011) Nonlinear Evolution Equation. Yunnan University Press, 12. | 2,597 | 9,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-26 | latest | en | 0.910496 |
https://easyteaching.net/maths-resources/number-resources/odd-and-even-numbers/ | 1,560,807,178,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998580.10/warc/CC-MAIN-20190617203228-20190617225228-00019.warc.gz | 419,864,017 | 30,977 | # Odd and Even Numbers
## Odd and Even Number Worksheets
Odd and even number worksheets to identify odd & even numbers. Teach patterns: when happens when odd & even numbers are added, multiplied or subtracted?
Even Number Maze Follow the even numbers from start to finish! Odd Number Maze Follow the odd numbers from start to finish!
Odd and Even Colour Identify odd and even numbers and colour the picture! Odd & Even Mystery Picture Colour the odd and even numbers to reveal the picture! Odd or Even? Work out whether the numbers are odd or even.
Identify Odd/Even Sheet 1 Identifying even and odd numbers. How do we tell even from odd? Identify Odd/Even Sheet 2 Identifying even and odd numbers. How do we tell even from odd? Making Odd/Even Numbers 1 Use the given digits to make odd and even numbers.
Making Odd/Even Numbers 2 Use the given digits to make odd and even numbers. Properties of Odd/Even Numbers Addition Properties of odd and even numbers. What happens when an odd number is added to an odd number? Or an even to an odd? Properties of Odd/Even Numbers Subtraction Properties of odd and even numbers. What happens when an odd number is subtracted from an odd number? Or an even from an even?
Back to Number Resources
Comments are closed | 270 | 1,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-26 | longest | en | 0.892975 |
https://www.studypool.com/questions/197971/10-if-you-do-it-bro-i-tip-25 | 1,542,728,603,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746465.88/warc/CC-MAIN-20181120150950-20181120172950-00418.warc.gz | 1,003,183,479 | 22,944 | # 10 if you do it bro i tip 25 %
Anonymous
Question description
Assignment: The Law of Large Numbers Investigation
Mrs. Hudson has made another assignment that Karen and Dakota are excited to try. To conduct your own experiment, you will need dice and a place to record your results. In this assignment, you will first calculate the theoretical probability for rolling a sum of 7. Then, you will roll the dice and add the numbers shown, recording your results as you go. Calculate the experimental probability for rolling a 7 after the 1st, 10th, and 100th rolls. Compare these results with the theoretical probability of rolling a sum of 7. How does this comparison change as the number of trials increase?
1. List out the sample space for the experiment and then calculate the theoretical probability.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
1. Roll the dice. Record your results in the table. Calculate the experimental probability for rolling a sum of 7 after the 1st, 10th, and 100th rolls.
Experimental Probability after Roll #1: ___________________________
How does the experimental probability compare to the theoretical probability after 1 roll? _________________________________________________________
_________________________________________________________
Experimental Probability after Roll #10: __________________________
How does the experimental probability compare to the theoretical probability after 10 rolls? _________________________________________________________
_________________________________________________________
Experimental Probability after Roll #100: _________________________
How does the experimental probability compare to the theoretical probability after 100 rolls?
_________________________________________________________
_________________________________________________________
Tally your results in the table.
Sum Results 2 3 4 5 6 7 8 9 10 11 12
1. Compare the results with the theoretical probability of rolling a sum of 7. How does this comparison change as the number of trials increase? Use four of your vocabulary words in your explanation.
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
(Top Tutor) Studypool Tutor
School: UIUC
Studypool has helped 1,244,100 students
flag Report DMCA
Brown University
1271 Tutors
California Institute of Technology
2131 Tutors
Carnegie Mellon University
982 Tutors
Columbia University
1256 Tutors
Dartmouth University
2113 Tutors
Emory University
2279 Tutors
Harvard University
599 Tutors
Massachusetts Institute of Technology
2319 Tutors
New York University
1645 Tutors
Notre Dam University
1911 Tutors
Oklahoma University
2122 Tutors
Pennsylvania State University
932 Tutors
Princeton University
1211 Tutors
Stanford University
983 Tutors
University of California
1282 Tutors
Oxford University
123 Tutors
Yale University
2325 Tutors | 653 | 4,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-47 | latest | en | 0.756617 |
https://community.smartsheet.com/discussion/83000/count-months-in-another-sheet-but-only-within-a-certain-year | 1,701,888,214,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00361.warc.gz | 210,772,116 | 47,524 | Count months in another sheet but only within a certain year
I have one sheet with dates listed (ex: 9/10/21) that range from 2020, 2021, and 2022. I'm using this formula to count how many are in each month:
=COUNTIF({Head And Neck_Project Overview Range 1}, IFERROR(MONTH(@cell), 0) = 1)
But I'd like to break it down by year so I'd like to find how many are in January of 2021 only.
Thanks!
Tags:
• ✭✭✭✭✭✭
Happy to help!
You'd use the COUNTIFS instead of the COUNTIF function.
Try this.
=COUNTIFS({Head And Neck_Project Overview Range 1}, IFERROR(MONTH(@cell), 0) = 1), {Head And Neck_Project Overview Range 1}, IFERROR(YEAR(@cell), 0) = 2021)
Did that work?
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:[email protected] | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭✭✭
I hope you're well and safe!
Try something like this.
```=COUNTIFS({Head And Neck_Project Overview Range 1}, IFERROR(MONTH(@cell), 0) = 1),
{Head And Neck_Project Overview Range 1}, IFERROR(YEAR(@cell), 0) = 2021)
```
Did that work/help?
I hope that helps!
Be safe and have a fantastic weekend!
Best,
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
Did my post(s) help or answer your question or solve your problem? Please support the Community by marking it Insightful/Vote Up or/and as the accepted answer. It will make it easier for others to find a solution or help to answer!
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:[email protected] | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• Hi Andree! Thanks so much!
It's coming up as #UNPARSEABLE
=COUNTIF({Head And Neck_Project Overview Range 1}, IFERROR(MONTH(@cell), 0) = 1), {Head And Neck_Project Overview Range 1}, IFERROR(YEAR(@cell), 0) = 2021)
Thoughts?!
• ✭✭✭✭✭✭
Happy to help!
You'd use the COUNTIFS instead of the COUNTIF function.
Try this.
=COUNTIFS({Head And Neck_Project Overview Range 1}, IFERROR(MONTH(@cell), 0) = 1), {Head And Neck_Project Overview Range 1}, IFERROR(YEAR(@cell), 0) = 2021)
Did that work?
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:[email protected] | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭
@Andrée Starå I'm having this same issue and my formula is coming up #UNPARSEABLE too?
=COUNTIFS({ABM Janitorial Request Tracker Range DATE}, IFERROR(MONTH(@cell), 0) = [Primary Column]@row, {ABM Janitorial Request Tracker Range DATE}, IFERROR(YEAR(@cell), 0) = 2021)
Can you see where my error might be? If I do just the Countif for the month's it works but now that we are getting into another FY I need to add the counts for 2020 Nov and 2021 Nov.
Thanks,
Michelle
• ✭✭✭✭✭✭
I hope you're well and safe!
Try something like this.
=COUNTIFS({ABM Janitorial Request Tracker Range DATE}, IFERROR(MONTH(@cell), 0) = [Primary Column]@row), {ABM Janitorial Request Tracker Range DATE}, IFERROR(YEAR(@cell), 0) = 2021)
Did that work?
I hope that helps!
Be safe and have a fantastic weekend!
Best,
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
Did my post(s) help or answer your question or solve your problem? Please support the Community by marking it Insightful/Vote Up or/and as the accepted answer. It will make it easier for others to find a solution or help to answer!
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:[email protected] | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭
No that didn't work, but thanks for responding.
• ✭✭✭✭✭✭
Happy to help!
I'd be happy to take a quick look.
Can you describe your process in more detail and maybe share the sheet(s)/copies of the sheet(s) or some screenshots? (Delete/replace any confidential/sensitive information before sharing) That would make it easier to help. (share too, [email protected])
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:[email protected] | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭
@Andrée Starå - Great news I found my error! -
=COUNTIFS({ABM Janitorial Request Tracker Range DATE}, IFERROR(MONTH(@cell), 0) = [Primary Column]@row)👈️🤦♀️, {ABM Janitorial Request Tracker Range DATE}, IFERROR(YEAR(@cell), 0) = 2021)
No parenthesis after =[Primary Column]@row,
This one is working!! - =COUNTIFS({ABM Janitorial Request Tracker Range DATE}, IFERROR(MONTH(@cell), 0) = [Primary Column]@row, {ABM Janitorial Request Tracker Range DATE}, IFERROR(YEAR(@cell), 0) = 2021)
Thanks again for your assistance I'm good to go at the moment!
• ✭✭✭✭✭✭
Excellent!
You're more than welcome!
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:[email protected] | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• I'm using the same sort of formula counting how many items are in Jan of 2022 and my formula works, what I'm wondering is it somehow possible to not have to update the year? I'll be coping his summary sheet each year and I'm hoping there's a way that each Dec when setting up the new year's sheet I won't need to manually change the 2022 to 2023 (then 2023 to 2024, etc.) several dozen times.
edited 12/02/22
Hi @CaseyC
What about referencing a Cell in the formula instead of typing in the number? For example a cell in a Sheet Summary field which identifies the Year.
Then you would update the formula to look there:
=COUNTIFS({Reference}, IFERROR(YEAR(@cell), 0) = [Year Reference]#)
This means you'd only have to update one cell when you copy the sheet and it will update all of your formulas.
Cheers,
Genevieve
• Great suggestion @Genevieve P. thanks!
Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 1,903 | 6,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-50 | latest | en | 0.84947 |
https://www.physicsforums.com/threads/how-do-you-derive-a-range-prediction-equation.301854/ | 1,527,329,920,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867416.82/warc/CC-MAIN-20180526092847-20180526112847-00073.warc.gz | 817,220,504 | 14,221 | # Homework Help: How do you derive a Range prediction equation?
1. Mar 23, 2009
### genu
1. The problem statement, all variables and given/known data
Derive the equations for v0 and R, and show how you made your prediction.
2. Relevant equations
d = v0t +1/2at^2
3. The attempt at a solution
The acceleration was zero, so I rewrote the equation:
R(range) = v0t
but then they substituted t with$$\sqrt{\frac{2Y}{g}}$$ to get
$$R=V{0}\sqrt{\frac{2Y}{g}$$
How did they get that for t?
also note: that Y is the vertical displacement of the ball to the floor.
let me know if I need to attach the diagram that they provided
2. Mar 23, 2009
### alphysicist
Hi genu,
Think about what the range R means in this problem. In other words, as the object moves, the x-displacement keeps getting larger and larger until some event occurs; what event is that? The time when that occurs is what tells you how large your range is, and so the time when that occurs has to be used in your range equation. Does that help? | 275 | 1,016 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-22 | latest | en | 0.946436 |
https://au.mathworks.com/matlabcentral/cody/problems/45881-reindex-a-vector/solutions/2804824 | 1,601,462,633,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00774.warc.gz | 288,166,227 | 16,611 | Cody
# Problem 45881. Reindex a vector (★★)
Solution 2804824
Submitted on 7 Aug 2020 by Rafael S.T. Vieira
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
N = [1 3 2 7]; IDX= [2 3 4 6]; y_correct = [0 1 3 2 0 7]; assert(isequal(reindex_a_vector(N,IDX),y_correct))
2 Pass
N=[ -1 0 4 2]; IDX=[3 8 1 6]; y_correct = [4 0 -1 0 0 2 0 0]; assert(isequal(reindex_a_vector(N,IDX),y_correct))
3 Pass
N=[-1 0 4 2]'; IDX=[3 8 1 6]; y_correct = [4 0 -1 0 0 2 0 0]'; assert(isequal(reindex_a_vector(N,IDX),y_correct))
4 Pass
N=[1 2 3 4 5 6 7 8 9 10]; IDX=[3 4 5 6 7 8 9 10 11 13]; y_correct=[0 0 1 2 3 4 5 6 7 8 9 0 10]; assert(isequal(reindex_a_vector(N,IDX),y_correct)) | 344 | 790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-40 | latest | en | 0.455171 |
https://finance.yahoo.com/news/cautious-siegfried-holding-ags-vtx-064912618.html | 1,566,811,125,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331485.43/warc/CC-MAIN-20190826085356-20190826111356-00222.warc.gz | 448,609,264 | 105,276 | U.S. Markets open in 4 hrs 12 mins
# Should We Be Cautious About Siegfried Holding AG's (VTX:SFZN) ROE Of 8.3%?
Many investors are still learning about the various metrics that can be useful when analysing a stock. This article is for those who would like to learn about Return On Equity (ROE). By way of learning-by-doing, we'll look at ROE to gain a better understanding of Siegfried Holding AG (VTX:SFZN).
Over the last twelve months Siegfried Holding has recorded a ROE of 8.3%. Another way to think of that is that for every CHF1 worth of equity in the company, it was able to earn CHF0.083.
### How Do You Calculate ROE?
The formula for return on equity is:
Return on Equity = Net Profit ÷ Shareholders' Equity
Or for Siegfried Holding:
8.3% = CHF56m ÷ CHF679m (Based on the trailing twelve months to December 2018.)
Most know that net profit is the total earnings after all expenses, but the concept of shareholders' equity is a little more complicated. It is all the money paid into the company from shareholders, plus any earnings retained. You can calculate shareholders' equity by subtracting the company's total liabilities from its total assets.
### What Does ROE Mean?
Return on Equity measures a company's profitability against the profit it has kept for the business (plus any capital injections). The 'return' is the amount earned after tax over the last twelve months. That means that the higher the ROE, the more profitable the company is. So, as a general rule, a high ROE is a good thing. That means it can be interesting to compare the ROE of different companies.
### Does Siegfried Holding Have A Good ROE?
One simple way to determine if a company has a good return on equity is to compare it to the average for its industry. The limitation of this approach is that some companies are quite different from others, even within the same industry classification. As is clear from the image below, Siegfried Holding has a lower ROE than the average (12%) in the Life Sciences industry.
That certainly isn't ideal. We'd prefer see an ROE above the industry average, but it might not matter if the company is undervalued. Nonetheless, it might be wise to check if insiders have been selling.
### How Does Debt Impact Return On Equity?
Companies usually need to invest money to grow their profits. The cash for investment can come from prior year profits (retained earnings), issuing new shares, or borrowing. In the first and second cases, the ROE will reflect this use of cash for investment in the business. In the latter case, the debt used for growth will improve returns, but won't affect the total equity. Thus the use of debt can improve ROE, albeit along with extra risk in the case of stormy weather, metaphorically speaking.
### Combining Siegfried Holding's Debt And Its 8.3% Return On Equity
Siegfried Holding has a debt to equity ratio of 0.16, which is far from excessive. Although the ROE isn't overly impressive, the debt load is modest, suggesting the business has potential. Careful use of debt to boost returns is often very good for shareholders. However, it could reduce the company's ability to take advantage of future opportunities.
### But It's Just One Metric
Return on equity is one way we can compare the business quality of different companies. Companies that can achieve high returns on equity without too much debt are generally of good quality. If two companies have around the same level of debt to equity, and one has a higher ROE, I'd generally prefer the one with higher ROE.
But when a business is high quality, the market often bids it up to a price that reflects this. Profit growth rates, versus the expectations reflected in the price of the stock, are a particularly important to consider. So you might want to check this FREE visualization of analyst forecasts for the company.
Of course Siegfried Holding may not be the best stock to buy. So you may wish to see this free collection of other companies that have high ROE and low debt.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 1,002 | 4,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-35 | latest | en | 0.948124 |
https://www.neetprep.com/questions/55-Physics/701-Dual-Nature-Radiation-Matter?courseId=8&testId=1183774-Recommended-MCQs---Questions&subtopicId=471-Particle-Nature-Light | 1,675,829,057,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500671.13/warc/CC-MAIN-20230208024856-20230208054856-00747.warc.gz | 928,931,727 | 48,272 | The energy of a quanta of frequency ${10}^{15}$ Hz and will be:
1.
$\mathrm{}$2.
3.
$\mathrm{}$4.
Subtopic: Particle Nature of Light |
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
Launched MCQ Practice Books
Prefer Books for Question Practice? Get NEETprep's Unique MCQ Books with Online Audio/Video/Text Solutions via Telegram Bot
The spectrum of radiation 1.0 x 1014 Hz is in the infrared region. The energy of one photon of this in joules will be:
1. $6.62×{10}^{-48}$
2. $6.62×{10}^{-20}$
3. $\frac{6.62}{3}×{10}^{-28}$
4. $3×6.62×{10}^{-28}$
Subtopic: Particle Nature of Light |
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
Launched MCQ Practice Books
Prefer Books for Question Practice? Get NEETprep's Unique MCQ Books with Online Audio/Video/Text Solutions via Telegram Bot
A helium-neon laser produces monochromatic light of a wavelength of 667 nm. The power emitted is 9 mW. The average number of photons arriving per second on average at a target irradiated by this beam is:
1. $9×{10}^{17}$
2. $3×{10}^{16}$
3. $9×{10}^{15}$
4. $3×{10}^{19}$
Subtopic: Particle Nature of Light |
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
Launched MCQ Practice Books
Prefer Books for Question Practice? Get NEETprep's Unique MCQ Books with Online Audio/Video/Text Solutions via Telegram Bot
A source S1 is producing, 1015 photons per sec of wavelength 5000 Å. Another source S2 is producing 1.02×1015 photons per second of wavelength 5100 Å. Then the ratio of the power of S2 to the power of S1 is equal to:
1. 1.00
2. 1.02
3. 1.04
4. 0.98
Subtopic: Particle Nature of Light |
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
Launched MCQ Practice Books
Prefer Books for Question Practice? Get NEETprep's Unique MCQ Books with Online Audio/Video/Text Solutions via Telegram Bot
A 200 W sodium street lamp emits yellow light of wavelength 0.6 $\mathrm{\mu m}$. If it is 25% efficient in converting electrical energy to light, how many photons of yellow light does it emit per second?
1. $1.5×{10}^{20}$
2. $6×{10}^{18}$
3. $62×{10}^{20}$
4. $3×{10}^{19}$
Subtopic: Particle Nature of Light |
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
Launched MCQ Practice Books
Prefer Books for Question Practice? Get NEETprep's Unique MCQ Books with Online Audio/Video/Text Solutions via Telegram Bot
Monochromatic light of frequency 6.0×1014 Hz is produced by a laser. The power emitted is 2×10-3 W. What will be the average number of photons emitted by the source per second?
1. $5×{10}^{15}$
2. $5×{10}^{16}$
3. $5×{10}^{17}$
4. $5×{10}^{14}$
Subtopic: Particle Nature of Light |
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh | 1,079 | 3,616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.755639 |
https://www.cazoommaths.com/maths-worksheet/tiling-problems-a-worksheet/ | 1,725,859,977,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00296.warc.gz | 642,278,721 | 68,963 | ## Tiling Problems (A) worksheet
Suitable for Year groups: 7
GCSE Tier:
Foundation/Higher
0.0/5
## Total Reviews: (0)
Prerequisite knowledge: Metric measures, area, compound shapes.
0.0/5
## Tiling Problems (A) worksheet description
This activity is designed to promote thinking about areas, factors and metric measures. Learners will need to work out the number of tiles, of a given size, that can fit in a rectangular wall or a compound shape made from rectangles. Sections A, B and C ask learners to use square tiles of size, 20cm, 50cm and 25cm respectively. Section D then extends the situation to involve rectangular tiles. Section E then provides two real-life problems to solve.
## WE LOVE TO GET YOUR FEEDBACK:
{{ reviewsTotal }}{{ options.labels.singularReviewCountLabel }}
{{ reviewsTotal }}{{ options.labels.pluralReviewCountLabel }}
{{ userData.canReview.message }}
X
#### Get 20 FREE MATHS WORKSHEETS
Fill out the form below to get 20 FREE maths worksheets! | 235 | 980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.753306 |
https://oeis.org/A097700 | 1,628,001,180,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00065.warc.gz | 442,837,295 | 3,710 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A097700 Numbers not of form x^2 + 2y^2. 10
5, 7, 10, 13, 14, 15, 20, 21, 23, 26, 28, 29, 30, 31, 35, 37, 39, 40, 42, 45, 46, 47, 52, 53, 55, 56, 58, 60, 61, 62, 63, 65, 69, 70, 71, 74, 77, 78, 79, 80, 84, 85, 87, 90, 91, 92, 93, 94, 95, 101, 103, 104, 105, 106, 109, 110, 111, 112, 115, 116, 117, 119, 120, 122 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Is lim_{n->inf} a(n)/n = 3/4? LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..6800 MATHEMATICA formQ[n_]:=Reduce[x>=0&&y>=0&&n==x^2 + 2 y^2, {x, y}, Integers]==False; Select[Range[0, 200], formQ] (* Vincenzo Librandi, Jan 15 2017 *) PROG (MAGMA) [n: n in [0..160] | NormEquation(2, n) eq false]; // Vincenzo Librandi, Jan 15 2017 CROSSREFS Complement of A002479. Sequence in context: A066513 A218852 A028810 * A172321 A154689 A175766 Adjacent sequences: A097697 A097698 A097699 * A097701 A097702 A097703 KEYWORD nonn AUTHOR Ralf Stephan, Aug 23 2004 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 3 08:21 EDT 2021. Contains 346435 sequences. (Running on oeis4.) | 589 | 1,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-31 | latest | en | 0.598302 |
https://likegeeks.com/filter-colors-numpy-python/ | 1,709,531,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00185.warc.gz | 350,842,619 | 53,195 | # Filter Colors Using NumPy in Python
Color filtering is an essential aspect of image processing, it involves isolating specific colors or color ranges in images, allowing for a range of applications from simple photo enhancements to object detection and real-time video analysis.
In this tutorial, you’ll learn how to filter colors using NumPy in Python.
## Understanding Image Color Models
In this section, we’ll dive into the RGB color model, grayscale conversion, and HSV/HSL color models, highlighting their significance in color filtering.
### The RGB Color Model
The RGB color model represents images using the colors Red, Green, and Blue.
We can create a simple RGB color using NumPy:
```import numpy as np
red_color = np.array([255, 0, 0])
print(red_color)
```
Output:
```[255 0 0]
```
This output shows an array representing the color red in the RGB model, with maximum red (255) and no green or blue.
### Grayscale Conversion
Converting an image to grayscale simplifies it by reducing the color information, which is useful for certain processing tasks. Here’s how you convert an RGB color to grayscale:
```# Convert the red color to grayscale
gray_scale_value = np.dot(red_color, [0.2989, 0.5870, 0.1140])
print(gray_scale_value)
```
Output:
```76.2195
```
The grayscale value is a weighted sum of the RGB components, reflecting the perceived brightness of the color.
### HSV/HSL Color Models
HSV (Hue, Saturation, Value) and HSL (Hue, Saturation, Lightness) are alternative representations of the color model.
They are useful in filtering because they separate color information (hue) from lighting (value/lightness).
Here’s how you can create a simple HSV color:
```hsv_color = np.array([30, 100, 100]) # Hue: 30, Saturation: 100, Value: 100
print(hsv_color)
```
Output:
```[ 30 100 100]
```
This represents an HSV color with specific hue, saturation, and value levels.
## Isolating Single Color Channels
By focusing on individual Red, Green, or Blue (RGB) channels, you can gain insights into specific aspects of an image.
Here, we’ll explore how to isolate these color channels using NumPy in Python.
### Isolating the Red Channel
To isolate the red channel, you’ll extract the red component from each pixel while setting the green and blue components to zero. Here’s how to do it:
```import numpy as np
# Sample RGB image represented as a 3x3x3 NumPy array
sample_image = np.array([[[255, 0, 0], [0, 255, 0], [0, 0, 255]],
[[0, 255, 255], [255, 0, 255], [255, 255, 0]],
[[0, 0, 0], [255, 255, 255], [128, 128, 128]]])
red_channel = sample_image.copy()
red_channel[:,:,[1, 2]] = 0
print(red_channel)
```
Output:
```[[[255 0 0]
[ 0 0 0]
[ 0 0 0]]
[[ 0 0 0]
[255 0 0]
[255 0 0]]
[[ 0 0 0]
[255 0 0]
[128 0 0]]]
```
In this output, only the red values are retained, while green and blue are set to zero.
### Isolating the Green Channel
Similarly, to isolate the green channel, retain only the green values:
```green_channel = sample_image.copy()
green_channel[:,:,[0, 2]] = 0
print(green_channel)
```
Output:
```[[[ 0 0 0]
[ 0 255 0]
[ 0 0 0]]
[[ 0 255 0]
[ 0 0 0]
[ 0 255 0]]
[[ 0 0 0]
[ 0 255 0]
[ 0 128 0]]]
```
Here, only the green components are visible.
### Isolating the Blue Channel
For the blue channel:
```blue_channel = sample_image.copy()
blue_channel[:,:,[0, 1]] = 0
print(blue_channel)
```
Output:
```[[[ 0 0 0]
[ 0 0 0]
[ 0 0 255]]
[[ 0 0 255]
[ 0 0 255]
[ 0 0 0]]
[[ 0 0 0]
[ 0 0 255]
[ 0 0 128]]]
```
Only blue values are present in this output.
Now, Let’s isolate and display the Red, Green, and Blue channels separately using Matplotlib.
This code will create a randomly generated image and then isolate each color channel for display:
```import numpy as np
import matplotlib.pyplot as plt
# Sample image array
image = np.random.rand(100, 100, 3) # Randomly generated sample image
# Function to isolate a color channel
def isolate_color_channel(image, channel_index):
# Creating a copy of the image
filtered_image = image.copy()
# Setting the other two channels to 0
zero_indices = [i for i in range(3) if i != channel_index]
filtered_image[:, :, zero_indices] = 0
return filtered_image
# Isolating each color channel
red_filtered = isolate_color_channel(image, 0)
green_filtered = isolate_color_channel(image, 1)
blue_filtered = isolate_color_channel(image, 2)
plt.figure(figsize=(15, 5))
# Displaying original image
plt.subplot(1, 4, 1)
plt.imshow(image)
plt.title("Original Image")
# Displaying red channel image
plt.subplot(1, 4, 2)
plt.imshow(red_filtered)
plt.title("Red Color Filtered Image")
# Displaying green channel image
plt.subplot(1, 4, 3)
plt.imshow(green_filtered)
plt.title("Green Color Filtered Image")
# Displaying blue channel image
plt.subplot(1, 4, 4)
plt.imshow(blue_filtered)
plt.title("Blue Color Filtered Image")
plt.show()
```
Output:
This code creates a sample image and then isolates each of the RGB channels by setting the other two channels to zero in each case using `isolate_color_channel` function.
## Creating Binary Masks Based on Color Thresholds
Creating binary masks based on color thresholds involves setting a threshold value to isolate specific colors or color ranges in an image.
These masks are useful for various applications like object detection, background removal, and more.
Let’s create a threshold to isolate a specific color range.
```import numpy as np
import matplotlib.pyplot as plt
# Generating a random image for demonstration
image = np.random.rand(100, 100, 3)
# Defining color thresholds for red
lower_red = np.array([0.5, 0, 0]) # Lower bound for red
upper_red = np.array([1, 0.2, 0.2]) # Upper bound for red
plt.imshow(image)
plt.title("Sample Image")
plt.show()
```
Output:
Now, create a binary mask that identifies pixels within the red color range:
```# Creating a binary mask
mask = np.all(np.logical_and(lower_red <= image, image <= upper_red), axis=-1)
plt.title("Binary Mask for Red Color Range")
plt.show()
```
Output:
This code generates a binary mask where white areas (True) indicate pixels that fall within the defined red color range, and black areas (False) represent other colors.
Finally, apply this mask to the original image to see the result:
```masked_image = np.where(mask[..., None], image, 0)
plt.title("Image After Applying Red Color Mask")
plt.show()
```
Output:
The masked image highlights only the areas that meet the red color criteria, effectively filtering the image based on the color threshold.
## Combining and Splitting Color Channels
First, let’s split an image into its individual Red, Green, and Blue (RGB) channels.
```import numpy as np
import matplotlib.pyplot as plt
# Generating a random image
image = np.random.rand(100, 100, 3)
# Splitting the color channels
red_channel = image[:, :, 0]
green_channel = image[:, :, 1]
blue_channel = image[:, :, 2]
# Displaying the color channels
plt.figure(figsize=(15, 5))
plt.subplot(1, 3, 1)
plt.imshow(red_channel, cmap='Reds')
plt.title("Red Channel")
plt.subplot(1, 3, 2)
plt.imshow(green_channel, cmap='Greens')
plt.title("Green Channel")
plt.subplot(1, 3, 3)
plt.imshow(blue_channel, cmap='Blues')
plt.title("Blue Channel")
plt.show()
```
Output:
This code separates the image into its red, green, and blue components, displaying each channel individually.
The `cmap` parameter in `imshow` ensures that each channel is visualized in its respective color.
Now, let’s combine different color channels to form a new image. For instance, we can combine the red and green channels to create a yellowish image:
```# Combining red and green channels to form yellow
yellow_image = np.stack([red_channel, green_channel, np.zeros_like(blue_channel)], axis=-1)
plt.imshow(yellow_image)
plt.title("Combined Yellow Image (Red + Green)")
plt.show()
```
Output:
This code combines the red and green channels while setting the blue channel to zero, resulting in a yellow-colored image.
This demonstrates how combining different color channels can create various color effects.
## Detecting and Isolating Specific Color Ranges
Detecting and isolating specific color ranges in images allows for the identification and extraction of objects or features based on their color.
It’s useful in applications like object tracking, image segmentation, and automated quality control.
First, define the color range you want to detect. For this example, let’s focus on isolating a specific shade of blue. We start by setting up the color thresholds:
```import numpy as np
import matplotlib.pyplot as plt
# Generating a random image
image = np.random.rand(100, 100, 3)
# Defining color thresholds for blue
lower_blue = np.array([0, 0, 0.5]) # Lower bound for blue
upper_blue = np.array([0.3, 0.3, 1]) # Upper bound for blue
plt.imshow(image)
plt.title("Sample Image")
plt.show()
```
Output:
Next, create a mask that detects the pixels within the specified blue color range:
```# Creating a binary mask for the blue color range
blue_mask = np.all(np.logical_and(lower_blue <= image, image <= upper_blue), axis=-1)
plt.title("Binary Mask for Blue Color Range")
plt.show()
```
Output:
This binary mask shows the areas in the image that fall within the specified blue color range.
Now, apply this mask to the original image to isolate the blue color range:
```# Applying the mask to the original image
isolated_blue = np.where(blue_mask[..., None], image, 0)
plt.imshow(isolated_blue)
plt.title("Image After Isolating Blue Color Range")
plt.show()
```
Output:
The resulting image highlights the regions that match the defined blue color range. Other areas are set to black, effectively isolating the color of interest.
## Dealing with Varying Lighting Conditions
One of the primary challenges in color filtering is the variation in how colors appear under different lighting conditions.
For example, an object that appears red in daylight might look quite different under artificial light.
This variation can significantly affect the accuracy of color-based image processing.
To deal with varying lighting conditions, consider the following strategies:
Adjust color thresholds dynamically based on the lighting conditions of the image. This can be done through real-time analysis of the image’s overall brightness and contrast.
```average_brightness = np.mean(image)
adjusted_lower_blue = lower_blue * average_brightness
adjusted_upper_blue = upper_blue * average_brightness
```
In this example, the blue color thresholds are adjusted based on the average brightness of the image.
Using HSV/HSL Color Models:
The HSV (Hue, Saturation, Value) and HSL (Hue, Saturation, Lightness) models are less sensitive to lighting variations compared to RGB.
Converting your image to one of these models before applying color filters can improve consistency.
```from matplotlib.colors import rgb_to_hsv
hsv_image = rgb_to_hsv(image)
hsv_mask = np.all(np.logical_and(lower_hsv <= hsv_image, hsv_image <= upper_hsv), axis=-1)
```
This code converts an RGB image to HSV and applies a mask based on HSV thresholds.
## Real-Time Color Filtering in Video Streams
Processing a video stream is different from processing a static image. A video is a sequence of images (frames), and real-time processing involves applying color filtering to each frame as it is received.
### Essential Steps for Real-Time Color Filtering
Frame Capture:
Continuously capture frames from the video stream. Python libraries like OpenCV are commonly used for this purpose.
```import cv2
import numpy as np
video_capture = cv2.VideoCapture(0)
```
This code initializes video capture from the default webcam.
Applying Color Filters:
Apply the desired color filter to each frame. This involves selecting a color range and creating a mask, similar to image processing.
```while True:
ret, frame = video_capture.read() # Read a frame
if not ret:
break
# Convert frame to HSV (better for color filtering)
hsv_frame = cv2.cvtColor(frame, cv2.COLOR_BGR2HSV)
# Example: Detecting red color
lower_red = np.array([160, 100, 100])
upper_red = np.array([180, 255, 255])
mask = cv2.inRange(hsv_frame, lower_red, upper_red)
# Apply the mask to the frame | 3,083 | 12,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-10 | latest | en | 0.750629 |
https://chat.stackexchange.com/transcript/100450?m=59411124 | 1,652,977,773,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529538.2/warc/CC-MAIN-20220519141152-20220519171152-00434.warc.gz | 219,005,771 | 6,909 | 5:10 AM
@JohnRennie hi
I need ur help in a simple question
how the trees are so tall?
@JackRod Hi :-)
I mean I was observing a tree which was roughly 115m
Do you mean how do they have the mechanical strength to stay upright?
no,to survive they need water right at stem of the the top?
for water to travel 115 m from ground is looking totally unreasonable
Ah, you mean how do they suck water up higher than 30m i.e. one atmosphere?
5:13 AM
yes excatly
I did some calculation and some physical experiment
and maximum height I got is 10m from a very thin tube
obviously more thinner the tube more the hight
Yes, because in simple column taller than 10m the pressure difference exceeds 1 atm at the top and the water the top would boil and turn to water vapour.
yes
But in trees water transport involves osmosis and it isn't just a simple column of water. In effect cells pump water between them.
5:18 AM
but can U give what should be pressure difference between ground and top(115m)?
if assume ground to be at 1 atm
The pressure difference in the tree is not the same as the pressure difference would be ina column of water.
That's because the cells of the tree have mechanical strength and hence they can exert an upward force on the water.
Suppose you had a chain of buckets 1m apart and extending up to an arbitrary height. To pump water up you only need a pressure equal to 1m height i.e. just enough to get to the next bucket.
ah
do there exist a chance of air bubble inside the tube of the tree
I don't know to be honest. I suspect the bubble would dissolve i.e. the gas inside would go into solution in the fluids around it.
my friend says it does not convert water into vapour
because of activation energy
9 mins ago, by John Rennie
Yes, because in simple column taller than 10m the pressure difference exceeds 1 atm at the top and the water the top would boil and turn to water vapour.
5:26 AM
cool thanks sir
I would have guessed that inside a tree there were lots of things that could act as nuclei for bubble formation so I doubt bubble nucleation would be a barrier.
one more question I have
Yes ... ?
related to mechanical tv
I forgot the name of the scientist who discover the mechanical tv
sir how to they convert a image into beam of electrons
John Logie Baird FRSE (; 13 August 1888 – 14 June 1946) was a Scottish inventor, electrical engineer, and innovator who demonstrated the world's first live working television system on 26 January 1926. He went on to invent the first publicly demonstrated colour television system, and the first viable purely electronic colour television picture tube.In 1928 the Baird Television Development Company achieved the first transatlantic television transmission. Baird's early technological successes and his role in the practical introduction of broadcast television for home entertainment have earned him...
5:30 AM
I did read the pendulum experiment to duplicate an image but how they used a mechanical tv two show a real-time image on tv
Baird's TV did use an electron beam. The "mechanical" bit was just how the image was scanned.
sir I cannot imagine how they convert an image into the beam of electron how this beam of electron form exactly the same image
I don't know how Baird's TV worked. | 745 | 3,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-21 | latest | en | 0.969154 |
http://grinebiter.com/Numbers/Cardinal/197-one-hundred-ninety-seven.html | 1,510,940,087,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803848.60/warc/CC-MAIN-20171117170336-20171117190336-00410.warc.gz | 137,096,765 | 3,433 | One hundred ninety-seven
Here is information about "one hundred ninety-seven" that you may find useful and interesting. Number Systems One hundred ninety-seven is a decimal number and can be written with numbers: 197 Binary is a number system with only 0s and 1s. One hundred ninety-seven in binary form is displayed below: 11000101 A Hexadecimal number has a base of 16 which means it includes the numbers 0 to 9 and A through F. One hundred ninety-seven converted to hexadecimal is: C5 Roman Numerals is another number system. Below is one hundred ninety-seven in roman numerals: CXCVII Scientific Notation Sometimes calculators and scientists shorten numbers using scientific notation. Here is one hundred ninety-seven as a scientific notation: 1.97E+02 Math Here are some math facts about One hundred ninety-seven: One hundred ninety-seven is a rational number and an integer. One hundred ninety-seven is an odd number because it is not divisible by two. One hundred ninety-seven is divisible by the following numbers: 1, 197 One hundred ninety-seven is not a square number because no number multiplied by itself will equal one hundred ninety-seven. Number Lookup One hundred ninety-seven is not the only number we have information about. Go here to look up other numbers.
Translated Here we have translated one hundred ninety-seven into some of the most commonly used languages: Chinese: 一百九十七 French: cent quatre-vingt-dix-sept German: hundert siebenundneunzig Italian: cento novantasette Spanish: ciento noventa y siete
Currency Here is one hundred ninety-seven written in different currencies: US Dollars: \$197 Canadian Dollars: CA\$197 Australian Dollars: A\$197 British Pounds: £197 Indian Rupee: ₹197 Euros: €197
Ordinal The cardinal number one hundred ninety-seven can also be written as an ordinal number: 197th Or if you want to write it with letters only: one hundred ninety-seventh.
One hundred ninety-eight Go here for the next number on our list that we have information about. | 435 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-47 | longest | en | 0.861253 |
https://devhubby.com/thread/how-to-check-if-a-number-is-prime-using-javascript | 1,701,955,237,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100674.56/warc/CC-MAIN-20231207121942-20231207151942-00005.warc.gz | 228,171,642 | 27,744 | # How to check if a number is prime using JavaScript?
, in category: Other , 8 months ago
How to check if a number is prime using JavaScript?
Member
by wiley , 8 months ago
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ``` ```function isPrime(num) { // If number is less than 2, it's not a prime number if (num < 2) { return false; } // Loop from 2 to the square root of the number for (let i = 2; i <= Math.sqrt(num); i++) { // If any number between 2 and the square root of the number divides // the number without a remainder, it's not a prime number if (num % i === 0) { return false; } } // If the number is not divisible by any number between 2 and the // square root of the number, it is a prime number return true; } ```
```1 2 3 ``` ```console.log(isPrime(5)); // true console.log(isPrime(10)); // false console.log(isPrime(13)); // true ``` | 260 | 863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | longest | en | 0.543594 |
https://fr.scribd.com/document/252966969/%D9%83%D9%84-%D9%82%D9%88%D8%A7%D9%86%D9%8A%D9%86-%D8%A7%D9%84%D9%83%D9%87%D8%B1%D8%A8%D8%A7%D8%A1-goooood | 1,571,714,713,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987798619.84/warc/CC-MAIN-20191022030805-20191022054305-00521.warc.gz | 504,851,134 | 94,933 | Vous êtes sur la page 1sur 37
# A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 1
FORMULAS
AND THEIR VALUES
Q1 (coul)
k (cnstnt)
Formula 1-1
F=k x Q1 x Q2/d2
CHAPTER 1
Formula 1-2
I=Q/T
=D16/E16
Q (coul)
CHAPTER 2
d (mils)
Formula 2-1
A = d2
CHAPTER 2
Formula 2-2
R = l/A
=D22*E22/F22
(CM-/ft)
CHAPTER 3
Formula 3-1
I = V/R
=D25/E25
V (volts)
CHAPTER 3
Formula 3-2
V=IxR
=D28*E28
I (amps)
CHAPTER 3
Formula 3-3
R = V/I
=D31/E31
V (volts)
CHAPTER 3
Formula 3-4
## To find work in foot pounds
Work (ft lbs)=Force x Distance
=D34*E34
F (lbs)
CHAPTER 3
Formula 3-5
## To find power in watts
Power = Energy / Time
=D37/E37
E (joules)
CHAPTER 3
Formula 3-6
## To find energy in watthours
Energy (Watthours)= Power x Time
=D40*E40
P (watts)
CHAPTER 3
Formula 3-7
P=VxI
=D43*E43
V (volts)
CHAPTER 3
## Excel formula in column H is:
I (amps)
Formula 3-8
P = I2x R
=D13*E13*F13/G13^2
Q2 (coul)
2
d (mtr)
3
1.5
T (sec)
5
=D19^2
=D46^2*E46
From Use of Formulas
(Use Approp. Units)
2.5
16
l (feet)
10.4
A (CM)
250
39.75
65.41
R (ohms)
10
10
100
10000
10
10
200
600
10
I (amps)
0.05
0.5
R (ohms)
2
I (amps)
10
D (feet)
100
T (sec)
100
T (hours)
R (ohms)
0.05
200
0.5
A
1
2
3
4
5
6
7
8
9
10
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
CHAPTER 3
Formula 3-9
P = V2/R
CHAPTER 4
RT = R1+R2+Rn
RT = VT / IT
V1 = IR1 x R1
Formula 4-3
CHAPTER 4
Formula 4-4
VX = RX/RT x VT
CHAPTER 4
Formula 4-5
VT = V1 + V2 +Vn
=SUM(D64:F64)
CHAPTER 4
## Excel formula in column H is:
Formula 4-6
PT = P1 + P2 + Pn
=SUM(D67:F67)
CHAPTER 5
## Excel formula in column H is:
Formula 5-1
IT = I1 + I2 + I3 +In
=SUM(D70:F70)
CHAPTER 5
Formula 5-2
## GT = 1/R1 + 1/R2 +1/Rn
=1/D73+1/E73+1/F73
CHAPTER 5
## Excel formula in column H is:
Formula 5-3
RT=1/(1/R1+1/R2+1/R3...1/Rn)
=1/(1/D76+1/E76+1/F76)
CHAPTER 5
## Excel formula in column H is:
Formula 5-4
GT = G1 + G2 + G3
CHAPTER 5
## Excel formula in column H is:
Formula 5-5
RT = R1 x R2/R1 + R2
=D82*E82/(D82+E82)
Formula 4-1
CHAPTER 4
Formula 4-2
CHAPTER 4
## Excel formula in column H is:
AND THEIR VALUES
V (volts)
R (ohms)
=D49^2/E49
10
R1 (ohms)
10
VT (volts)
R3 (ohms)
10
70
IR1 (amps)
10
30
35
100,000
200
R1 (ohms)
0.002
RX (ohms)
=D61/E61*F61
0.5
IT (amps)
=D55/E55
=SUM(D79:F79)
200
R2 (ohms)
=SUM(D52:F52)
=D58*E58
From Use of Formulas
(Use Approp. Units)
RT (ohms)
4700
V1 (volts)
V2 (volts)
20
PR1 (watts)
PR2 (watts)
I2 (amps)
R1 (ohms)
R2 (ohms)
G1 (S)
R1 (ohms)
3900
5.5
47
0.097
20000
5346.53
0.1
0.4
R3 (ohms)
R3 (ohms)
27000
G2 (S)
0.1
400
33
10000
60
I3 (amps)
R2 (ohms)
R1 (ohms)
10
160
1.5
22
13.51
PR3 (watts)
200
50
V3 (volts)
30
40
I1 (amps)
VT (volts)
17400
G3 (S)
0.2
R2 (ohms)
6800
2478.50
A
1
2
3
4
5
6
7
8
9
10
83
84
85
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
CHAPTER 5
## Excel formula in column H is:
Formula 5-6
Ru = Rk x Re/Rk -Re
=D85*E85/(D85-E85)
AND THEIR VALUES
Rk (ohms)
From Use of Formulas
(Use Approp. Units)
Re (ohms)
10
15
A
1
2
3
4
5
6
7
8
9
10
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
CHAPTER 5
Formula 5-7
IX = RT/RX x IT
CHAPTER 5
## Excel formula in column H is:
Formula 5-8
I1 = R2/R1 + R2 x IT
=E91/(D91+E91)*F91
CHAPTER 5
## Excel formula in column H is:
Formula 5-9
I2 = R1/R1 + R2 x IT
CHAPTER 6
R1/R2 = R3/R4
Formula 6-1
CHAPTER 6
Formula 6-2
CHAPTER 7
## To find efficiency of power transfer
Efficiency (%) = Pout/Pin x 100
## To find I load using Thevenin's Th.
IL = VTH / RTH + RL
## To find V load using Thevenin's Th.
VL = RL/RL + RTH x VTH
Formula 7-5
## To find I load using Norton's Th.
IL = RN/RN + RL x IN
CHAPTER 8
Formula 8-1
## RA= (R1xR2) + (R2xR3) + (R3xR1)/R1
=((D118*E118)+(E118*F118)+(F118*D118))/D118
CHAPTER 8
## Excel formul in column H is:
Formula 8-2
RB = (R1xR2)+(R2xR3)+(R3xR1)/R2
=((D121*E121)+(E121*F121)+(F121*D121))/E121
Formula 7-1
CHAPTER 7
Formula 7-2
CHAPTER 7
Formula 7-3
CHAPTER 7
Formula 7-4
CHAPTER 7
## Excel formula in column H is:
=D88/E88*F88
=D94/(D94+E94)*F94
AND THEIR VALUES
RT (ohms)
RX (ohms)
550
R1 (ohms)
R1 (ohms)
R1 (ohms)
R1 (ohms)
10
Pout (watts)
0.0055
0.003025
IT (amps)
0.05
0.02619
IT (amps)
0.05
R3 (ohms)
0.02381
R4 (ohms)
30
## 60 Ans. here is 600=600
R3 (ohms)
20
30
60
Pin (watts)
8.84
13
VTH (volts)
26.6
12
RTH (ohms)
50
RL (ohms)
RN (ohms)
R1 (ohms)
R1 (ohms)
4.7
R2 (ohms)
3.3
100
4.7
0.01
RL (ohms)
25
RTH (ohms)
1
VTH (v)
25
RL (ohms)
25
R2 (ohms)
3.3
25
RN (ohms)
25
33.23
25
RL (ohms)
25
=D115/(E115+F115)*G115
IT (amps)
20
R2 (ohms)
=E100*F100/D100
=D112/(E112+F112)*G112
11000
R2 (ohms)
10
=D109/(E109+F109)
11000
R2 (ohms)
10000
=(D106-E106)/F106
1000
R2 (ohms)
10000
D97*G97=E97*F97
=D103/E103*100
From Use of Formulas
(Use Approp. Units)
50
25
0.5
IN (amps)
75
R3 (ohms)
5.6
18.28
R3 (ohms)
5.6
12.83
A
1
2
3
4
5
6
7
8
9
10
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
122
123 CHAPTER 8
124 Formula 8-3
FORMULAS
## Excel formula in column H is:
RC = (R1xR2)+(R2xR3)+(R3xR1)/R3
=((D124*E124)+(E124*F124)+(F124*D124))/F124
AND THEIR VALUES
R1 (ohms)
R2 (ohms)
3.3
4.7
From Use of Formulas
(Use Approp. Units)
R3 (ohms)
5.6
10.77
A
1
2
3
4
5
6
7
8
9
10
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
AND THEIR VALUES
CHAPTER 8
## Excel formula in column H is:
Formula 8-4
R1 = RB x RC/RA + RB + RC
=E127*F127/(D127+E127+F127)
CHAPTER 8
## Excel formula in column H is:
Formula 8-5
R2 = RC x RA/RA + RB + RC
=F130*D130/(D130+E130+F130)
CHAPTER 8
## Excel formula in column H is:
Formula 8-6
R3 = RA x RB/RA + RB + RC
=D133*E133/(D133+E133+F133)
CHAPTER 9
Formula 9-1
## Force = Pole 1 Strng. x Pole 2 Strng./d2
=D136*E136/F136^2
CHAPTER 9
Formula 9-2
mmf = NI
## Excel formula in column H is:
=D139*E139
N (# of trns) I (amps)
100
CHAPTER 9
Formula 9-3
H = AT/meter
=D142*E142/F142
A (amps)
CHAPTER 9
## To find induced voltage per turn
Vind = # of lines of flux cut/per unit time
flux (Wb)
## To find induced voltage in coil
Vind = N (# of turns) x flux/time
Formula 9-5
CHAPTER 10
CHAPTER 11
Formula 11-1
## To find time of one period in ac
T (sec) = 1/f (Hz)
=1/D153
f (Hz)
CHAPTER 11
Formula 11-2
## To find frequency from time of 1 per.
f (Hz) = 1/T (sec)
=1/D156
T (time in sec)
0.0001
CHAPTER 11
Formula 11-3
## To find effective value from peak
Effective value (rms) = 0.707 x pk. val.
=D159*E159
Formula 9-4
CHAPTER 9
RA (ohms)
RB (ohms)
18.28
RA (ohms)
RB (ohms)
RA (ohms)
10.77
5.60
d (meters)
2
0.05
800
500
T (turns)
meters
100
0.1
5000
time( sec)
2
## N (# of turns) flux (Wb)
=D148*E148/F148
4.70
RC (ohms)
P2 Strength
10.77
12.83
3.30
RC (ohms)
RB (ohms)
P1 Strength
10.77
12.83
18.28
=D145/E145
RC (ohms)
12.83
18.28
From Use of Formulas
(Use Approp. Units)
100
time( sec)
3
1000
0.707
300
0.001
V pk. (volts)
100
10000
70.7
A
1
2
3
4
5
6
7
8
9
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
161 CHAPTER 11
162 Formula 11-4
10
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
FORMULAS
To find avg. value (1/2 cycle) from pk.
Avg. value (1/2 cycle) = 0.637 x pk. val.
## EXCEL EQUIVALENT FORMULAS
Excel formula in column H is:
=D162*E162
AND THEIR VALUES
V pk. (volts)
0.637
100
From Use of Formulas
(Use Approp. Units)
63.7
A
1
2
3
4
5
6
7
8
9
10
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
AND THEIR VALUES
## To find instant. V of sine wave
v inst = Vp x sin of angle value
CHAPTER 11
Formula 11-6
=D168/E168*100
tw (sec)
CHAPTER 11
Baseln value
Formula 11-7
## Vavg = Baseln val. + (duty cycle x p-p amp.)
=D171+(E171*F171)
CHAPTER 12
CHAPTER 13
## To find V induced in a coil
Vind = N (# of trns) x d /dt
Formula 13-1
CHAPTER 13
Formula 13-2
VL = L x di/dt
CHAPTER 13
## Excel formula in column H is:
Formula 13-3
L = VL / di/dt
=D182/(E182/F182)
CHAPTER 13
Formula 13-4
## To find L from physical parameters.
L=12.57 x 10-7x r x N2 x A / l
CHAPTER 13
## Excel formula in column H is:
Formula 13-5
L = 12.57 x 10-7 x N2 x A / l
=0.000001257*D188^2*E188/F188
CHAPTER 13
## Excel formula in column H is:
Formula 13-6
LT = L1 + L2 . . . +Ln
= SUM(D191:F191)
CHAPTER 13
Formula 13-7
## 1/LT = 1/L1 + 1/L2 + 1/L3+1/Ln
=1/D194+1/E194+1/F194
CHAPTER 13
Formula 13-8
## LT = 1 / 1/L1 + 1/L2 + 1/L3+1/Ln
=1/(1/D197+1/E197+1/F197)
CHAPTER 11
Formula 11-5
From Use of Formulas
(Use Approp. Units)
V pk. (volts)
=D165*F165
sin
angle
141
30
20
duty cycle
pk-pk volts
500
L (H)
di (amps)
di (amps)
r
1
N (# trns)
1000
50
2.5
dt (secs)
12
N (# trns)
=0.000001257*D185*E185^2*F185/G185
2.5
dt (secs)
5
dt (secs)
2
10
VL (volts)
20
0.1
d(Wb)
N (# trns)
=D179*E179/F179
70.5
T (sec)
4
=D176*E176/F176
0.5
A (sq mtrs)
2000
0.001
l (mtrs)
0.01
0.5028
2000
L1 (H)
0.001
L2 (H)
10
L1 (H)
L2 (H)
L1 (H)
2.5
17.5
10
0.4
10
2.5
L3 (H)
5
L2 (H)
10
0.5028
L3 (H)
5
10
0.01
L3 (H)
5
A
1
2
3
4
5
6
7
8
9
10
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
199 CHAPTER 13
200 Formula 13-9
## EXCEL EQUIVALENT FORMULAS
FORMULAS
To find LT of 2 parallel L's
## Excel formula in column H is:
LT = L1 x L2 / L1 + L2
=D200*E200/(D200+E200)
AND THEIR VALUES
L2 (H)
L1 (H)
10
10
From Use of Formulas
(Use Approp. Units)
A
1
2
3
4
5
6
7
8
9
10
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
## EXCEL EQUIVALENT FORMULAS
From Use of Formulas
(Use Approp. Units)
AND THEIR VALUES
L (H)
CHAPTER 13
Formula 13-10
## Energy = 0.5 LI2
=D203*E203*F203^2
CHAPTER 13
Formula 13-11
=D206/E206
L (H)
CHAPTER 13
## Excel formula in column H is:
Vs (volts)
Formula 13-12
i = Vs/R x (1 - -Rt/L)
=D209/E209*((1-2.71828^ (E209*F209/G209)))
CHAPTER 13
## Excel formula in column H is:
Formula 13-13
i = Vs/R x ( -Rt/L)
=D212/E212*((2.71828^-(E212*F212/G212)))
CHAPTER 14
Formula 14-1
XL = 2fL
CHAPTER 14
## Excel formula in column H is:
Formula 14-2
L= XL/2f
=D218/(E218*F218)
CHAPTER 14
## Excel formula in column H is:
Formula 14-3
f = XL/2L
=D221/(E221*F221)
CHAPTER 14
## Excel formula in column H is:
Formula 14-4
1/XLT=1/XL1+1/XL2+1/XL3
=1/(D224)+1/(E224)+1/(F224)
CHAPTER 14
## Excel formula in column H is:
Formula 14-5
XLT=1/(1/XL1+1/XL2+)
=1/(1/D227+1/E227)
CHAPTER 14
Formula 14-6
## XLT = XL1 x XL2/XL1 + XL2
=D230*E230/(D230+E230)
CHAPTER 14
To find Q in an RL ac circuit
Q = XL/R
## Excel formula in column H is:
Formula 14-7
CHAPTER 15
Pythagorean Theorem
Formula 15-1
c2 = a2 + b2
=D236^2+E236^2
I (amps)
0.5
20
12
R (ohms)
10
Vs (volts)
t (sec)
R (ohms)
f (Hz)
XL1 (ohms)
XL2 (ohms)
4000
5000
1884
1500000
0.000212
0.00003
21231423
XL3 (ohms)
6000
0.000917
2500
XL2 (ohms)
5000
XL (ohms)
XL2 (ohms)
5000
XL1 (ohms)
0.000137
L (H)
6.28
2000
0.05
f (Hz)
6.28
XL1 (ohms)
0.001
L (H)
4000
10
L (H)
0.000068
60
2000
XL (ohms)
L (H)
0.001
t (sec)
2200
6.28
XL (ohms)
0.83
10000
=D233/E233
250
R (ohms)
10
=D215*E215*F215
5000
2500
R (ohms)
3140
a
50
62.8
10
200
b
10
A
1
2
3
4
5
6
7
8
9
10
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
AND THEIR VALUES
CHAPTER 15
## Excel formula in column H is:
Formula 15-2
VT = VR2 + VL2
=SQRT(D239^2+E239^2)
CHAPTER 15
Z = R2 + XL2
Formula 15-3
CHAPTER 15
## Excel formula in column H is:
Formula 15-4
IT = R2 + IL2
=SQRT(D245^2+E245^2)
CHAPTER 15
Formula 15-5
## To find Z in a parallel RL circuit
Z = R x XL / R2 + XL2
CHAPTER 15
Formula 15-6
G = 1/R
=D251/E251
CHAPTER 15
BL = 1/XL
Formula 15-7
=D254/E254
1005
0.000995
CHAPTER 15
Formula 15-8
Y = 1/Z
=D257/E257
Z (ohms)
835
0.001198
CHAPTER 15
YT = G2 + BL2
## To find magnitude of phase angle
(magnitude) = tan-1 R / XL
M = k L1 x L2
Formula 16-1
CHAPTER 16
## Excel formula in column H is:
Formula 16-2
LT = L1 + L2 + 2M
=D269+E269+2*F269
CHAPTER 16
## Excel formula in column H is:
Formula 16-3
LT = L1 + L2 - 2M
=D272+E272-2*F272
Formula 15-9
CHAPTER 15
Formula 15-10
CHAPTER 16
VR (volts)
From Use of Formulas
(Use Approp. Units)
VL (volts)
10
5
IR (amps)
8.66
10.00
IL (amps)
3
5.83
50
30
25.72
R (ohms)
1500
0.000667
XL (ohms)
R (ohms)
=D248*E248/SQRT(D248^2+E248^2)
14.14
XL (ohms)
R (ohms)
=SQRT(D242^2+E242^2)
10
One
XL (ohms)
BL (S)
G (S)
=SQRT(D260^2+E260^2)
0.000667
1500
1005
L1 (H)
=D266*SQRT(E266*F266)
0.001198
XL (ohms)
R (ohms)
=DEGREES(ATAN(D263/E263))
0.000995
1
L1 (H)
L2 (H)
3
L2 (H)
8
L1 (H)
24
M (H)
8
L2 (H)
8
56.178
M (H)
8
A
1
2
3
4
5
6
7
8
9
10
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
CHAPTER 16
Formula 16-4
LT = L1 + L2 2M
=D275+E2752*F275
CHAPTER 16
## Excel formula in column H is:
Formula 16-5
LT = 1/1/L1 + M + 1/L2 + M
=1/((1/(D278+F278)+1/(E278+F278)))
CHAPTER 16
## Excel formula in column H is:
Formula 16-6
LT = 1/1/L1 - M + 1/L2 - M
=1/((1/(D281-F281)+1/(E281-F281)))
CHAPTER 16
k = M/ L1 x L2
VP/VS = NP/NS
NP/NS = VP/VS
NP/NS = IS/IP
VP/VS = IS/IP
Formula 16-12
(NS/NP)2 = ZS/ZP
CHAPTER 16
## Excel formula in column H is:
Formula 16-13
ZP = ZS(NP/NS)2
=D302*(E302/F302)^2
CHAPTER 16
NP/NS = ZP/ZS
## To find transformer efficiency
Efficiency % = Pout/Pin x 100
## Excel formula in column H is:
Formula 16-7
CHAPTER 16
Formula 16-8
CHAPTER 16
Formula 16-9
CHAPTER 16
Formula 16-10
CHAPTER 16
Formula 16-11
CHAPTER 16
Formula 16-14
CHAPTER 16
Formula 16-15
AND THEIR VALUES
L1 (H)
L2 (H)
L1 (H)
10
L2 (H)
L1 (H)
8
L2 (H)
8
L2 (H)
8
VS (volts)
20
NP (trns)
NP (trns)
VP (volts)
VS (volts)
14
0.1
IP (amps)
0.5
ZS (ohms)
4
0.1
ZP(ohms)
8
128
NS (trns)
2000
ZP (ohms)
IP (amps)
IS (amps)
NP (trns)
14
VS(volts)
0.5
1
ZS (ohms)
0.5
IS (amps)
NP (trns)
or =(D299/E299)^2=F299/G299
5
NS (trns)
VP (volts)
or =D296*G296=E296*F296
280
NS (trns)
or = D293*G293 = E293*F293
NS(trns)
280
20
10
NP (trns)
NS (trns)
or = D290*G290=E290*F290
30
M (H)
8
VP (volts)
(aid)
(opp)
M (H)
8
L1 (H)
5
M (H)
or = D287*G287=E287*F287
=D308/E308*100
M (H)
10
=F284/SQRT(D284*E284)
=SQRT(D305/E305)
From Use of Formulas
(Use Approp. Units)
8000
ZS (ohms)
1000
10
10
86
100
86
A
1
2
3
4
5
6
7
8
9
10
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
## EXCEL EQUIVALENT FORMULAS
From Use of Formulas
(Use Approp. Units)
AND THEIR VALUES
CHAPTER 17
Formula 17-1
C = Q/V
=D311/E311
Q (coul)
V (volts)
CHAPTER 17
Formula 17-2
Q=CxV
=D314*E314
CHAPTER 17
Formula 17-3
V = Q/C
=D317/E317
Q (coul)
CHAPTER 17
Formula 17-4
=0.5*D320*E320^2
CHAPTER 17
Formula 17-5
k = o/ v
CHAPTER 17
## Excel formula in column H is:
Formula 17-6
C = 8.85 x A/1012s
=8.85*D326/(10^12*E326)
CHAPTER 17
## Excel formula in column H is:
Formula 17-7
C = 8.85 x k x A/1012 x s
=8.85*D329*E329/(10^12*F329)
CHAPTER 17
## Excel formula in column H is:
Formula 17-8
C = 2.25*k*A/107 x s x (n-1)
=2.25*D332*E332/10^7*F332*(G332-1)
CHAPTER 17
## Excel formula in column H is:
Formula 17-9
CT = C1 x C2 / C1 + C2
=D335*E335/(D335+E335)
CHAPTER 17
Formula 17-10
CT = C/N
=D338/G338
CHAPTER 17
## Excel formula in column H is:
Formula 17-11
CT = 1/1/C1+1/C2+1/C3...+1/Cn
=1/(1/D341+1/E341+1/F341)
CHAPTER 17
Formula 17-12
VX = VS x CT/CX
=D344*E344/F344
10
10
250
0.15625
0.001
8.85E-011
V (volts)
5
10
V (volts)
0.000005
o (abs)
v (vac)
=D323/E323
6
A (sq mtrs)
s (mtrs)
0.01
k (number)
3
k (number)
0.01
A (sq in.)
100
C1 (F)
0.001
n(# plates)
s (in.)
2
2.655E-010
0.005
0.009
C2 (F)
3
C1 (F)
6
C2 (F)
10
C1 (F)
C3 (F)
10
C2 (F)
10
VS (volts)
n(# plates)
10
3.33
C3 (F)
5
CT (uF)
60
10
2.5
15
30
CX (F)
7.5
A
1
2
3
4
5
6
7
8
9
10
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
## EXCEL EQUIVALENT FORMULAS
From Use of Formulas
(Use Approp. Units)
AND THEIR VALUES
CHAPTER 17
## Excel formula in column H is:
Formula 17-13
CT = C1 + C2 + C3+Cn
C1 (F)
C2 (F)
C3 (F)
=D347+E347+F347+G347
CHAPTER 17
Formula 17-14
1 TC (T) = R x C
=D350*E350
R (ohms)
1000
0.000001
CHAPTER 17
## Excel formula in column H is:
VS (volts)
Formula 17-15
vR = VS x ( -t/T)
=D353*((E353^-(F353/G353)))
CHAPTER 17
## Excel formula in column H is:
Formula 17-16
vC = VS x (1 - -t/T)
=D356*(((1-(E356^-(F356/G356)))))
CHAPTER 18
## Excel formula in column H is:
Formula 18-1
iC = C x dv/dt
=(D359*10^-12)*E359/(F359*10^-6)
CHAPTER 18
## Excel formula in column H is:
Formula 18-2
XC = 1/2x f x C
=1/(D362*E362*F362)
CHAPTER 18
## Excel formula in column H is:
Formula 18-3
f = 1/2 x C x XC
=1/(D365*E365*F365)
CHAPTER 18
## Excel formula in column H is:
Formula 18-4
C = 1/2 x f x XC
=1/(D368*E368*F368)
CHAPTER 18
Formula 18-5
## XCT = 1/1/XC1 + 1/XC2 + 1/XC31/XCn
=1/(1/D371+1/E371+1/F371)
CHAPTER 18
Formula 18-6
## XCT = XC1 x XC2 / XC1 + XC2
=D374*E374/(D374+E374)
CHAPTER 18
Formula 18-7
XCT = XC1 / N
=D377/E377
CHAPTER 19
## Excel formula in column H is:
Formula 19-1
VT = VR2 + VC2
=SQRT(D380^2+E380^2)
15
10
50
VS (volts)
C (value)
2.71828
dv (volts)
f (Hz)
XC1 (ohms)
XC1 (ohms)
10000
18.39
(sec)
18000
10000
41.74
100
0.0001
0.000002
663.48
1000
318471
XC (ohms)
2000
XC2 (ohms)
1000
10000
XC (ohms)
5E-010
f (Hz)
6.28
(sec)
120
6.28
38
dt (sec)
100
6.28
0.001
t (sec)
2.71828
100
10
t (sec)
50
C4 (F)
3180
0.000000025
XC3 (ohms)
2000
3000
545.45
XC2 (ohms)
1000
XC1 (ohms)
2000
666.67
N (number)
1000
VR (volts)
200
50
70.71
VC (volts)
50
A
1
2
3
4
5
6
7
8
9
10
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
AND THEIR VALUES
XC (ohms)
Formula 19-2
Z = R2 + XC 2
CHAPTER 19
## Excel formula in column H is:
Formula 19-3
IT = IR2 + IC2
=SQRT(D386^2+E386^2)
CHAPTER 19
## To find Z in a parallel RC circuit
Z = R x XC / R2 + XC 2
Formula 19-4
CHAPTER 19
Formula 19-5
G = 1/R
=D392/E392
CHAPTER 19
BC = 1/XC or 2fC
YT = G2 + BC2
## Excel formula in column H is:
G (S)
BC (S)
=SQRT(D398^2+E398^2)
0.000667
0.001
## To find magnitude of phase angle
(magnitude) = tan-1 R / XC
## Excel formula in column H is:
R (ohms)
XC (ohms)
=DEGREES(ATAN(D401/E401))
1500
Z = R2 + Xnet2
## Excel formula in column H is:
R (ohms)
CHAPTER 19
Formula 19-6
CHAPTER 19
Formula 19-7
CHAPTER 19
Formula 19-8
CHAPTER 20
Formula 20-1
R (ohms)
=SQRT(D383^2+E383^2)
From Use of Formulas
(Use Approp. Units)
50
IR (amps)
50
70.71
20
25
IC (amps)
15
XC (ohms)
R (ohms)
=D389*E389/SQRT(D389^2+E389^2)
2.4
R (ohms)
1500
0.000667
One
Xc (ohms)
=D395/E395
=SQRT(D404^2+E404^2)
995
0.001
0.0012
995
56.44
Xnet (ohms)
60
80
100
## (where Xnet = XL- XC if XL>XC
or Xnet = XC- XL if XC>XL)
CHAPTER 20
## Excel formula in column H is:
Formula 20-2
VT = VR2 + VXnet2
=SQRT(D409^2+E409^2)
VR (volts)
Vxnet(volts)
60
80
100.00
60
1.33
## (where Vxnet=VL- VC if VL>VC
or Vxnet=VC- VL if VC>VL)
CHAPTER 20
Formula 20-3
Tang of = Xnet/R
=D414/E414
Xnet (ohms)
80
R (ohms)
A
1
2
3
4
5
6
7
8
9
10
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
## EXCEL EQUIVALENT FORMULAS
CHAPTER 20
FORMULAS
To find Tan knowing VXnet and VR
## Excel formula in column H is:
Formula 20-4
Tang of = VXnet / VR
=D417/E417
CHAPTER 20
## Excel formula in column H is:
Formula 20-5
IT = IR2 + IXnet2
=SQRT(D420^2+E420^2)
AND THEIR VALUES
VXnet (volts) VR (volts)
80
IR (amps)
60
From Use of Formulas
(Use Approp. Units)
1.33
IXnet (amps)
2
2.24
## (where Ixnet = IL- IC if IL>IC
or Ixnet = IC- IL if IC>IL)
CHAPTER 20
Formula 20-6
## To find p.f. knowing phase angle
Power factor (p.f.) = cos
## Excel formula in column H is:
=COS(D425*PI()/180)
(deg)
CHAPTER 20
## To find app. power knowing V and I
Apparent power (S) = Vrms x Irms
V (volts)
Formula 20-7
CHAPTER 20
Formula 20-8
=D432^2*E432
=D428*E428
45
0.707
I (amps)
141.5
I (amps)
141.5
100
100
R (ohms)
1
A
1
2
3
4
5
6
7
8
9
10
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
AND THEIR VALUES
CHAPTER 20
Formula 20-9
## To find true power knowing S and p.f.
P = Apparent power x power factor (p.f.)
=D435*E435
S (VA)
CHAPTER 20
Formula 20-10
## To find p.f. knowing P and S
p.f. = True power (P)/App. power (S)
=D438/E438
Ptrue (W)
CHAPTER 20
S = P2 + Q2
Ptrue (W)
## To convert Z polar to rect. data
R factor(Z cos )
j factor(Z sin )
45 (Z cos (jZ sin )
## Excel formulas in column H are:
=(D444*COS(E444*PI()/180))
=(D444*SIN(E444*PI()/180))
CHAPTER 20
Formula 20-13
## Real axis(VT x cos )
=(D450*COS(E450*PI()/180))
(Series circuit)
## Imag. axis(jVT x sin )
=(D450*SIN(E450*PI()/180))
Formula 20-11
CHAPTER 20
Formula 20-12
(Series circuit)
141.5
p.f. (numb.)
0.707
100
S (VA)
100
=SQRT(D441^2+E441^2)
141.5
0.707
Q (VAR)
100
100
141.42
30
38.97
22.5
(deg)
Z (ohms)
45
45 30 = 38.97 + j22.5
VT (volts)
(deg)
66
60
33
57.16
## (VT x cos (jVT x sin )
6645 = 33 + j57.16
CHAPTER 20
Formula 20-14
## Real axis(IT x cos )
=(D456*COS(E456*PI()/180))
(Parallel circuit)
## Imag. axis(jIT x sin )
=(D456*SIN(E456*PI()/180))
IT (amps)
(deg)
15
30
12.99
7.5
## (IT x cos (jIT x sin )
CHAPTER 20
Formula 20-15
(Parallel circuit)
CHAPTER 20
Formula 20-16
(Parallel circuit)
From Use of Formulas
(Use Approp. Units)
15 = 12.99 + j7.5
BL = 1/2 x f x L
BC = 2 x f x C
## Excel formula in column H is:
=1/(D462*E462*F462)
=D466*E466*F466
f (Hz)
6.28
L (H)
60
f (Hz)
6.28
0.00265
0.000004
0.00151
60
A
1
2
3
4
5
6
7
8
9
10
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 20
Formula 20-17
(Parallel circuit)
## EXCEL EQUIVALENT FORMULAS
FORMULAS
Y = G2 + jBnet2
(Note: tan of B/G)
(Note: of Y arctan B/G)
## Excel formulas in column H are:
AND THEIR VALUES
Bnet (S)
G (S)
=SQRT(D470^2+E470^2)
=E470/D470
=ATAN(H471)*180/pi()
From Use of Formulas
(Use Approp. Units)
0.002
0.00114
0.0023
0.57
29.7
Y = 0.0023 29.7S
CHAPTER 21
## Excel formula in column H is:
Formula 21-1
fr = 1/2 x L x C
=1/(D477*SQRT(E477*F477))
L (H)
6.28
C (Fd)
0.0002
4E-010
562983
CHAPTER 21
## Excel formula in column H is:
Formula 21-2
L = 0.02533/fr2 x C
=0.02533/(D481^2*E481)
fr (Hz)
C (Fd)
1000000 2.533E-010
0.0001
CHAPTER 21
## Excel formula in column H is:
Formula 21-3
C = 0.02533/fr2 x L
=0.02533/(D485^2*E485)
fr (Hz)
L (H)
1000000
0.0001
2.533E-010
CHAPTER 21
## Excel formula in column H is:
Formula 21-4
L = 25,330/fr2 x C
=25330/(D489^2*E489)
fr (MHz)
C (pF)
1
253.3
100
100
253.3
990000
20000
Where: L= H; C=pF;f=MHz
CHAPTER 21
## Excel formula in column H is:
Formula 21-5
C = 25,330/fr2 x L
=25330/(D493^2*E493)
fr (MHz)
L ( H)
1
Where: L= H; C=pF;f=MHz
CHAPTER 21
Formula 21-6
Information only!
No computations for this formula are required in text!
CHAPTER 21
BW = f2 - f1
Formula 21-7
## (Where f2= higher f 0.707 pt
and f1 = lower f 0.707 pt)
=D500 - E500
f2 (Hz)
1010000
f1 (Hz)
A
1
2
3
4
5
6
7
8
9
10
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
504 CHAPTER 21
505 Formula 21-8
## EXCEL EQUIVALENT FORMULAS
FORMULAS
To find BW knowing fr and Q
## Excel formula in column H is:
BW = fr/Q
=D505/E505
AND THEIR VALUES
fr (Hz)
From Use of Formulas
(Use Approp. Units)
Q (number)
5000
25
200
A
1
2
3
4
5
6
7
8
9
10
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
CHAPTER 21
Formula 21-9
Q = fr/BW
=D508/E508
CHAPTER 21
fc = 1/2 x R x C
## Excel formula in column H is:
Formula 21-10
AND THEIR VALUES
fr (Hz)
From Use of Formulas
(Use Approp. Units)
BW (Hz)
5000
=1/(D511*E511*F511)
200
R (ohms)
6.28
1000
25
0.0000001
1592.36
CHAPTER 21
Formula 21-11
fc = 1/2 x (L/R)
## Excel formula in column H is:
=1/(D515*(E515/F515))
L (H)
R (ohms)
6.28
0.1
1000
1592.36
CHAPTER 22
## Excel formula in column H is:
Formula 22-1
Max # of electrons = 2 x n2
=2*D519^2
CHAPTER 23
## To find I in circ. w/series diode and R
I = (VS - VD) / R
P = I x VD
P = Iz x Vz
## To find regulation transition point
Vs = VzT x (R1 + RL) / RL
## To find output V w/source <trans. pt.
VL = Vs x RL / (RL + R1)
## Excel formula in column H is:
Formula 23-5
CHAPTER 23
To find V to be dropped by VR
Formula 23-6
VR = Vs - VD
=D537-E537
CHAPTER 24
Formula 24-1
## Vdc = 0.45 x Vrms
=0.45*D540
Formula 23-1
CHAPTER 23
Formula 23-2
CHAPTER 23
Formula 23-3
CHAPTER 23
Formula 23-4
CHAPTER 23
n
3
VS (volts)
18
VD (volts)
=(D522-E522)/F522
10
0.01
Iz (amps)
=D528*E528
150
0.062
0.7
0.007
5.1
0.255
Vz (volts)
0.05
R1 (ohms)
VzT
=D531*(E531+F531)/F531
R (ohms)
VD
I (amps)
=D525*E525
0.7
9.1
Vs (volts)
100
RL (ohms)
=D534*E534/(E534+F534)
9
Vs (volts)
750
RL (ohms)
750
10.31
R1 (ohms)
100
7.94
VD (volts)
12
1.5
10.5
Vrms (volts)
70.7
31.8
A
1
2
3
4
5
6
7
8
9
10
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
542 CHAPTER 24
543 Formula 24-2
## EXCEL EQUIVALENT FORMULAS
FORMULAS
To find Vdc out of unfilt. H.W. rect.
## Vdc = 0.318 x VPK
=0.318*D543
AND THEIR VALUES
From Use of Formulas
(Use Approp. Units)
VPK (volts)
100
31.8
A
1
2
3
4
5
6
7
8
9
10
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 24
FORMULAS
## EXCEL EQUIVALENT FORMULAS
AND THEIR VALUES
From Use of Formulas
(Use Approp. Units)
Vrms (V)
Formula 24-3
## To find PIV of unfilt. H.W. rect.
PIV (H.W.) = 1.414 x Vrms
CHAPTER 24
Formula 24-4
## To find ripple freq. H.W. rect.
Ripple freq. = ac input frequency
=1*D549
ac input f
CHAPTER 24
V (1/2 sec.)
Formula 24-5
=0.9*D552
CHAPTER 24
Formula 24-6
=0.45*D555
CHAPTER 24
Formula 24-7
=0.637*D558
CHAPTER 24
Formula 24-8
=0.318*D561
CHAPTER 24
## To find PIV of unfilt. F.W. rect.
PIV (F.W.) = 1.414 x Vrms full secondary
Formula 24-9
CHAPTER 24
Formula 24-10
## To find ripple freq. F.W. rect.
Ripple freq. = 2 x ac input frequency
=2*D567
ac input f
CHAPTER 24
Vrms input
Formula 24-11
=0.9*D570
CHAPTER 24
Formula 24-12
=0.637*D573
CHAPTER 24
Formula 24-13
## To find PIV of unfilt. bridge rect.
PIV (Bridge) = 1.414 x Vrms input
CHAPTER 24
Formula 24-14
## To find ripple freq. (bridge rect.)
Ripple frequency (bridge) = 2 x ac in f
=2*D579
=1.414*D546
200
282.8
60
60
100
90
V (full sec.)
200
90
141.4
90.1
282.8
89.9
V (Full sec.)
=1.414*D564
200
282.8
60
120
12.6
11.34
17.8
11.34
200
282.8
60
120
Vpk input
Vrms input
=1.414*D576
ac input f
A
1
2
3
4
5
6
7
8
9
10
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
580
581 CHAPTER 24
582 Formula 24-15
FORMULAS
To find power supply out % ripple
% Ripple = ac(ripple p-p)/dc out x 100
## EXCEL EQUIVALENT FORMULAS
Excel formula in column H is:
=D582/E582*100
AND THEIR VALUES
ac ripple p-p
From Use of Formulas
(Use Approp. Units)
dc out
1
48
2.08
A
1
2
3
4
5
6
7
8
9
10
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 24
FORMULAS
## To find power supply % regulation
% Reg. = (VNL- VFL)/ VFL x 100%
## To find power diss. by IC regulator
PREG = IB x VIN + IL x (VIN -VOUT )
## To find efficiency of IC regulator
= Pout / (Pout + Preg) x 100%
## To find regulated dc output voltage
Vout = Vref x (1+R2 / R1 )
Formula 24-20
## To find D factor for "buck" reg.
D = Ton / (Ton + Toff)
CHAPTER 24
Formula 24-21
Vout = Vin x D
=D600*E600
CHAPTER 24
Formula 24-22
=D603/(1-E603)
CHAPTER 24
Formula 24-23
## Vout = -Vin x D/(1-D)
=-D606*E606/(1-E606)
CHAPTER 25
IE = IC + IB
dc = IC/IB
IC = dc x IB
IB = IC / dc
## Excel formula in column H is:
Formula 24-16
CHAPTER 24
Formula 24-17
CHAPTER 24
Formula 24-18
CHAPTER 24
Formula 24-19
CHAPTER 24
Formula 25-1
CHAPTER 25
Formula 25-2
CHAPTER 25
Formula 25-3
CHAPTER 25
Formula 25-4
AND THEIR VALUES
VNL(volts)
=(D585-E585)/E585*100
VFL(volts)
12.2
IB (amps)
=(D588*E588)+F588*(E588-G588)
11.8
VIN (V)
Pout (W)
24
Vref (V)
1.446
R2 (ohms)
1.25
Ton (msec)
=D597/(D597+E597)
240
Vin (volts)
1.46
60.88
R1 (ohms)
240
2.5
0.5
0.5
12
0.5
48.00
0.5
-24
0.001
18.001
D value
Vin (volts)
D value
24
Vin (volts)
D value
24
IC (mA)
IB (mA)
18
IC (mA)
IB (mA)
1.2
dc
=D615*E615
15
50
24
=D612/E612
VOUT (V)
Toff (msec)
50
=D609+E609
0.15
Preg (W)
2.25
=D594*(1+E594/F594)
3.39
IL (amps)
0.0044
=D591/(D591+E591)*100
=D618/E618
From Use of Formulas
(Use Approp. Units)
0.024
50
IB (mA)
50
IC (mA)
0.024
1.2
50
0.024
dc
1.2
A
1
2
3
4
5
6
7
8
9
10
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 25
Formula 25-5
CHAPTER 25
Formula 25-6
CHAPTER 26
Formula 26-1
CHAPTER 26
Formula 26-2
CHAPTER 26
Formula 26-3
CHAPTER 26
Formula 26-4
CHAPTER 26
Formula 26-5
CHAPTER 26
Formula 26-6
CHAPTER 26
Formula 26-7
CHAPTER 26
Formula 26-8
CHAPTER 26
Formula 26-9
CHAPTER 26
Formula 26-10
## EXCEL EQUIVALENT FORMULAS
FORMULAS
To find Alpha of transistor
= IC / IE
ac = IC / IB
AV = Vout / Vin
AI = Iout / Iin
AP = AV x AI
## To find base V with V divider bias
VB VTH = Vcc x R2 / (R1 + R2)
## To find emitter curr. with V div. bias
IE = (VB - VBE) / RE
## To find collector-emitter voltage
VCE = VCC - IE x (RC + RE)
re' = 25 mV / IE
## To find input impedance
ZIN = R1|| R2|| (Beta +1) x re'
ZOUT = RC
AV = -RC / re'
## Excel formula in column H is:
AND THEIR VALUES
IC (mA)
IE (mA)
=D621/E621
9.8
IC (mA)
=D624/E624
Vout (Vpp)
Iout (Ipp)
10
AV
50
0.05
24
0.5
20
20
480
AI
=D633*E633
24
VCC (volts)
=D636*F636/(E636+F636)
R1 (ohms)
12
VB (volts)
VBE (volts)
VCC (volts)
2000
1.71
RE (ohms)
0.7
1000
RC (k )
IE (mA)
12
R2 (ohms)
12000
1.714
=D642-(E642*(F642+G642))
1.014
0.001014
RE (k )
3.9
7.03
IE (mA)
(mV)
=D645/E645
=-D654/E654
0.04
Iin (Ipp)
=D630/E630
=D651
0.98
Vin (Vpp)
1.2
=(D639-E639)/F639
10
IB (mA)
=D627/E627
=1/(1/D648+1/E648+(1/(F648+1)))*G648
From Use of Formulas
(Use Approp. Units)
25
R1 (ohms)
12000
1.014
R2 (ohms)
2000
24.65
re' (ohms)
Beta
100
24.65
2351.13
RC (ohms)
3000
RC (ohms)
3900
3000
re' (ohms)
24.65
-158.22
A
1
2
3
4
5
6
7
8
9
10
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
655
656 CHAPTER 26
657 Formula 26-11
658
FORMULAS
Approximation of current gain
(R1||R2)/(R1||R2)+( +1)re'
## EXCEL EQUIVALENT FORMULAS
Excel formula in column H is:
=D657*(E657*F657/(E657+F657))/
((E657*F657/(E657+F657))+(D657+1)*G657
From Use of Formulas
(Use Approp. Units)
AND THEIR VALUES
R1 (ohms)
100
12000
R2 (ohms)
2000
re' (ohms)
24.65
40.78
A
1
2
3
4
5
6
7
8
9
10
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 26
Formula 26-12
CHAPTER 26
Formula 26-13
CHAPTER 26
Formula 26-14
CHAPTER 26
Formula 26-15
CHAPTER 26
Formula 26-16
where:
## EXCEL EQUIVALENT FORMULAS
FORMULAS
Approximation of input impedance
ZIN = R1|| R2|| (Beta +1) x (re'+RE1))
ZOUT = RC
## Approximation of voltage gain
AV -RC / (re' + RE1 )
## Approximation of current gain
(R1||R2)/(R1||R2)+( +1)x(re'+RE1)
## Approx. gain of multistage amplifier
AV = AV1 x AV2 x AV3
## Excel formula in column H is:
From Use of Formulas
(Use Approp. Units)
AND THEIR VALUES
R1 (ohms)
=1/(1/E661+1/F661+(1/((D661+1)*(G661))))
100
R2 (ohms)
12000
2000
(re' +RE1)
1024.65
1686.35
RC (ohms)
=D664
3900
RC (ohms)
=-D667/(E667+F667)
3900
re' (ohms)
3900
24.65
R1 (ohms)
=D670*(E670*F670/(E670+F670))/
((E670*F670/(E670+F670))+(D670+1)*(G670))
100
AV1
10
1000
R2 (ohms)
12000
AV2
=D674*E674*F674
RE1 (ohms)
2000
AV3
-3.81
(re' +RE1)
1024.65
1.63
---
50
50
25000
## AV1 = Zin1/(Zin1 + Zs)
AV2 is the gain of first amplifier
AV3 = Zin2/(Zin2 + Zout1)
AV4 is gain of 2nd amp
CHAPTER 26
Formula 26-17
## ZIN = R1||R2|| ((x (re' +RE))
=1/(1/D681+1/E681+(1/((F681+1)*G681)))
CHAPTER 26
## Excel formula in column H is:
Formula 26-18
ZOUT = RE || re'
=1/(1/D684+1/E684)
CHAPTER 26
## To approx. v. gain of CC amplifier
AV = RE/(re' + RE)
## Excel formula in column H is:
Formula 26-19
R1 (ohms)
12000
RE (ohms)
500
RE (ohms)
=D687/(E687+F687)
1000
R2 (ohms)
2000
re' + RE
100
1024
1686.33
re' (ohms)
24.65
re' (ohms)
24.65
23.49
RE (ohms)
1000
0.98
Note: AV 1
CHAPTER 26
Ai -
Formula 26-20
CHAPTER 26
=-D691
100
RE (ohms)
-100
re' (ohms)
A
1
2
3
4
5
6
7
8
9
10
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
## EXCEL EQUIVALENT FORMULAS
FORMULAS
Formula 26-21
ZIN = RE || re'
=D694*E694/(D694+E694)
CHAPTER 26
Formula 26-22
ZOUT = RC
=D697
CHAPTER 26
AV = RC / re'
Formula 26-24
Ai -1
CHAPTER 26
Formula 26-25
VCE = VCC
=D706
CHAPTER 26
Formula 26-26
## IC(sat) = VCC/(RC + RE)
=D709/(E709+F709)
CHAPTER 27
## To find drain curr JFETs & MOSFETs
ID=IDSS(1-(VGS/VGS(off) ))2
Formula 26-23
CHAPTER 26
Formula 27-1
AND THEIR VALUES
1000
24.65
From Use of Formulas
(Use Approp. Units)
24.06
RC (ohms)
3900
RC (ohms)
=D700/E700
3900
re' (ohms)
3900
24.65
158.22
AI
=D703
-1
-1
12
12
VCC (volts)
VCC (volts)
RC (ohms)
12
RE (ohms)
3900
1000
0.00245
CHAPTER 27
Formula 27-2
CHAPTER 27
Formula 27-3
gm = ID / VGS
VGS = VGG
=D716/E716
ID (amps)
VGS(V)
0.002
0.000667
VGG (volts)
=D719
CHAPTER 27
Formula 27-4
## To find JFET drain to source V
VDS = VDD - ID x RD
## Excel formula in column H is:
VDD (volts)
=D723-(E723*F723)
ID (amps)
12
0.005
RD (ohms)
1000
CHAPTER 27
Formula 27-5
## To find JFET gate to source V
VGS = - ID x RS
(Note: Using self bias)
=-D727*E727
ID (amps)
0.004
RS (ohms)
1000
-4
A
1
2
3
4
5
6
7
8
9
10
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 27
Formula 27-6
FORMULAS
To find JFET drain to source V
VDS = VDD - ID x RD - ID x RS
## EXCEL EQUIVALENT FORMULAS
Excel formula in column H is:
AND THEIR VALUES
ID (amps)
RD (ohms)
RS (ohms)
VDD (volts)
=D731-(E731*F731)-(E731*G731)
18
0.01
1000
390
From Use of Formulas
(Use Approp. Units)
4.1
CHAPTER 27
Formula 27-7
## To find JFET gate voltage
VG = VDD x R2 / (R1 + R2)
## Excel formula in column H is:
VDD (volts)
=D735*F735/(E735+F735)
R1 (ohms)
12
R2 (ohms)
100000
33000
2.98
CHAPTER 27
Formula 27-8
## To find JFET gate to source V
VGS = VG - ID x RS
## Excel formula in column H is:
VG (volts)
=D739-(E739*F739)
ID (amps)
2.98
RS (ohms)
0.0085
470
-1.02
CHAPTER 27
Formula 27-9
## To find E-MOSFET drain curr.
ID = k x (VGS - VT)2
(NOTE: Shockley's equation, shown here for information only!)
CHAPTER 27
Formula 27-10
CHAPTER 28
Formula 28-1
(open loop)
CHAPTER 28
Formula 28-2
(open loop)
CHAPTER 28
Formula 28-3
(open loop)
CHAPTER 28
Formula 28-4
PFET = VDS x ID
## To find op-amp output V
Vout = - AVOL x Vin
VDS (volts)
=D747*E747
ID (amps)
0.12
50000
0.0002
-10
## (Note: Inverting input)
To find op-amp output V
Vout = AVOL x Vin
Vin (volts)
AVOL
=D754*E754
50000
0.0002
10
## (Note: Non-inverting input)
To find op-amp output V
Vout = AVOL x (V1 - V2)
## Excel formula in column H is:
=D758*(E758-F758)
V1 (volts)
AVOL
50000
V2 (volts)
0.002
0.0015
25
## (Note: Differential amp circuit config.)
To find inv. op-amp V gain
AV = Vout / Vin
Vout (volts)
=D762/E762
Vin (volts)
-15
CHAPTER 28
0.156
Vin (volts)
AVOL
=-D750*E750
1.3
Rf (ohms)
Ri (ohms)
-7.5
A
1
2
3
4
5
6
7
8
9
10
766
767
768
769
770
771
772
773
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
Formula 28-5
FORMULAS
AV = -Rf / Ri
## EXCEL EQUIVALENT FORMULAS
=-(D766/E766)
From Use of Formulas
(Use Approp. Units)
AND THEIR VALUES
27000
1800
-15
CHAPTER 28
Formula 28-6
CHAPTER 28
Formula 28-7
## To find non-inv. op-amp V gain
AV = (Rf / Ri) + 1
## To find V gain of V div. non-inv. amp.
AV=[(R2/R1)+1] x [R4/(R4 + R3)]
## Excel formula in column H is:
=(D770/E770)+1
=(E773/D773+1)*((G773/(G773+F773)))
Rf (ohms)
13000
R1 (ohms)
10000
Ri (ohms)
1000
R2 (ohms)
50000
14
R3 (ohms)
20000
R4 (ohms)
10000
A
1
2
3
4
5
6
7
8
9
10
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 28
Formula 28-8
CHAPTER 28
Formula 28-9
FORMULAS
## To find V out of subtractor circuit
Vout=Vin+x((R2/R1)+1)x(R4/(R4+R3))+
## Excel formula in column H is:
Vin- x (-R2/R1)
G776*(-F776/E776)
## To find V out of subtractor circuit
Vout = Vin+ - Vin-
AND THEIR VALUES
Vin+ (volts)
R1 (ohms)
=D776*(F776/E776+1)*(E778/(E778+D778))+
## Excel formula in column H is:
From Use of Formulas
(Use Approp. Units)
R2 (ohms)
10000
R4 (ohms)
5000
5000
R3 (ohms)
Vin+ (volts)
Vin- (volts)
10000
Vin- (volts)
=D781-E781
-5
-7
R1=R2=R3=R4)
CHAPTER 28
Formula 28-10
CHAPTER 28
Formula 28-11
## To find V out of summing amplifier
Vout=Vin1 x (-Rf/R1)+Vin2 x (-Rf/R2)
## To find V out of summing amplifier
Vout=Vin1 x (-Rf/R1)+Vin2 x (-Rf/R2) +
Formula 28-12
Rf&R1&R2
2
Vin1 (volts)
5
R2 (ohms)
R3 (ohms)
## To find V out of summing amplifier
Vout = (1+R2/R1) x [Vin1xR4/(R3+R4)] +
## Excel formula in column H is:
R1 (ohms)
=(1+E795/D795)*(D797*(G795/(F795+G795))+
## Vin2 x R3/(R3 + R4)]
E797*F795/(F795+G795))
10000
20000
Vin1 (volts)
Vin2 (volts)
10000
## To find V out. change for integ. circ.
Vout-Vin / (R x C)] x t
## Excel formula in column H is:
=D800/(E800*F800)*G800
- V in
CHAPTER 28
## To find V out. for differentiator circ.
Vout = (-R x C) x (Vint)
R (ohms)
## To find V output of instrum. amp.
Vout = (1 + 2 x R/RG) x (V2 - V1)
## To find upper thresh. pt. of Sch. trig.
UTP = Vmax x [R1/(R1 + R2)]
## Excel formula in column H is:
CHAPTER 28
Formula 28-15
CHAPTER 28
Formula 28-16
Vin3 (volts)
(-F792/E792)
(Note: 3-input inverting amp with R1 and R3 of equal value)
CHAPTER 28
Formula 28-13
=(D803*E803)*(F803/G803)
=(1+2*D806/E806)*(F806-G806)
=D809*(E809/(E809+F809))
10000
Vmax (volts)
12
-2.5
1000
R3 (ohms)
R4 (ohms)
10000
C (F)
R (ohms)
1000
20000
5.5
t (sec)
0.01
Vin (V)
0.01
RG (ohms)
R (ohms)
5000
Rf (ohms)
2.5
C (F)
-1000
-5
R1 (ohms)
5
5000
R2 (ohms)
-3
1.5
Formula 28-14
Vin2 (volts)
10000
Vin2 (volts)
D790*(-F792/G790)+(E790)*(-F792/D792)+(F790)*
Vin3 x (-Rf/R3)
CHAPTER 28
Vin1 (volts)
=D786*(-E786/E786)+F786*(-E786/E786)
(Note: For inverting amplifier with all R's equal value)
10000
-1.5
-5
0.5
7.5
t (sec)
0.5
V2 (volts)
5000
R1 (ohms)
V1 (volts)
2
R2 (ohms)
100000
1.09
A
1
2
3
4
5
6
7
8
9
10
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 28
Formula 28-17
CHAPTER 28
Formula 28-18
FORMULAS
Formula 28-19
Vmin (volts)
R1 (ohms)
## To find output V of IC audio amplifier
Vout = Vin x [R1B / (R1A+R1B-jXC1)]
## Excel formula in column H is:
Vin (volts)
=(D815*E815)/SQRT((F815+E815)^2+(G815)^2)*
0.1
2000
RSPK (ohms) jXC2 (ohms)
=D812*(E812/(E812+F812))
(20*D817)/SQRT(D817^2+E817^2)
(Note: R1A and R1B = input signal potentiometer)
CHAPTER 28
AND THEIR VALUES
## To find lower thresh. pt. of Sch. trig.
LTP = Vmin x [R1/(R1 + R2)]
x 20 x [RSPK) / (RSPK-jXC2 )]
## To find Vout of OTA diff. amplifier
Vout = (V1 - V2) x k x IBIAS x RL
## Excel formula in column H is:
From Use of Formulas
(Use Approp. Units)
-12
10000
R1B (ohms)
8
V1 (volts)
=(D820-E820)*F820*(G820*D822)
R2 (ohms)
100000
R1A (ohms)
8000
-1.09
jXC1 (ohms)
723.4
0.397
0.7234
V2 (volts)
0.2
IBIAS (amps)
0.15
20
0.000486
4.86
RL(ohms)
10000
CHAPTER 28
Formula 28-20
UTP = IBIAS x R1
## Excel formula in column H is:
IBIAS (amps)
=D825*E825
R1 (ohms)
0.000486
10000
4.86
where:
IBIAS = VBIAS - V - - 0.7/RBIAS
To find LTP V for Sch. trig. comp.
LTP = -IBIAS x R1
Formula 28-21
CHAPTER 28
## Excel formula in column H is:
Formula 28-22
fc = 1/(2 x R x C)
=1/(D833*E833*F833)
CHAPTER 28
Formula 28-23
## Vout = Vin x [-jXC/(R-jXC)] x (1+R2/R1)
=D836*E836/SQRT(F836^2+E836^2)*
CHAPTER 28
IBIAS (amps)
=D830*E830
(1+G836/D838)
R1 (ohms)
-0.000486
2
10000
R (ohms)
6.28
Vin (volts)
1000
- j XC
-4.86
0.000001
R (ohms)
10000
10000
159.24
R2 (ohms)
10000
1.414
R1 (ohms)
10000
CHAPTER 28
## Excel formula in column H is:
Formula 28-24
fc = 1/(2 x R x C)
=1/(D841*E841*F841)
CHAPTER 28
Formula 28-25
## Vout = Vin x [R/(R - jXC)] x (1+R2/R1)
=D844*(E844/SQRT(E844^2+F844^2))
*(1+G844/D846))
R (ohms)
6.28
Vin (volts)
R (ohms)
1
R1 (ohms)
10000
10000
2.2E-009
7238
R2 (ohms)
- j XC
10000
10000
1.414
A
1
2
3
4
5
6
7
8
9
10
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
## EXCEL EQUIVALENT FORMULAS
From Use of Formulas
(Use Approp. Units)
AND THEIR VALUES
10000
CHAPTER 28
Formula 28-26
## To find center f of bandpass filter
fo=1/2 x[R1xR2/(R1+R2)]xR3xC1xC2)
## Excel formula in column H is:
R1 (ohms)
=1/(6.28*SQRT(D849*E849/(D849+E849)*
1000
F849*G849*D851))
R2 (ohms)
R3 (ohms)
1000
20000 0.00000001
7345
4.7E-009
CHAPTER 28
BW = fo/Q
fc1 = fo - BW/2
fc2 = fo + BW/2
## To find center f of bandstop filter
fo = 1/(2 x R1 x R2 x C1 x C2 )
## To find digit. dec. value for ADC circ.
DATA = Vin1 x (255/5 V) dec. rep.
Formula 28-31
CHAPTER 28
Formula 28-32
## Vout = (-VREF/R14) x [D 7 /2)+(D6/4)+D5/8)
=(-D869/E869)*((F869/2)+(G869/4)+(D871/8)
+(D4/16)+(D3/32)+(D2/64)+(D1/128)+
+(E871/16)+(F871/32)+(G871/64)+(D873/128)
(D0/256)] x (- R1)
+(E873/256))*(-F873)
Formula 28-27
CHAPTER 28
Formula 28-28
CHAPTER 28
Formula 28-29
CHAPTER 28
Formula 28-30
CHAPTER 28
fo (Hz)
=D854/E854
Q
7345
fo (Hz)
=D857-(E857/2)
3.16
BW (Hz)
7345
fo (Hz)
=D860+(E860/2)
2324
R1 (ohms)
2324
R2 (ohms)
1000
## Excel formula in column H is:
fr = 1/2 x LC
=1/(D876*SQRT(E876*F876))
CHAPTER 29
## Excel formula in column H is:
Formula 29-2
fr = 1/2 x LtC
=1/(D879*SQRT(E879*F879))
CHAPTER 29
## Excel formula in column H is:
0.00000001 0.00000001
3561
VREF (volts)
102
R14 (ohms)
10
D5
D4
0.00004
Lt (H)
2
6.28
5.390625
R1 (ohms)
L (H)
6.28
0
D2
0
D0
D6
1
D3
L1A (H)
D7
10000
Formula 29-1
20000
D1
## To find fr of LC tank circuit
8507
Vin1 (volts)
=D866*(255/5)
CHAPTER 29
6183
BW (Hz)
7345
=1/(6.28*SQRT(D863*E863*F863*G863))
2324
0.00004
L1B (H)
10000
1E-010
2517737
1E-010
2517737
A
1
2
3
4
5
6
7
8
9
10
882
883
884
885
886
887
888
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
AND THEIR VALUES
Formula 29-3
Lt = L1A + L1B
=D882+E882
CHAPTER 29
## Excel formula in column H is:
Formula 29-4
fr = 1/2 x L1CT
=1/(D885*SQRT(E885*F885))
CHAPTER 29
## Excel formula in column H is:
Formula 29-5
CT = C1 x C2/(C1 + C2)
=D888*E888/(D888+E888)
0.000000001
0.00002
2
0.00002
0.00004
L (H)
6.28
From Use of Formulas
(Use Approp. Units)
0.0001
3.33E-010
872606
5E-010
0.000000000333
A
1
2
3
4
5
6
7
8
9
10
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
FORMULAS
AND THEIR VALUES
CHAPTER 29
Formula 29-6
## CT = 1/(1/C1 + 1/C2 + 1/C3)
=1/(1/D891+1/E891+1/F891)
0.000000001
CHAPTER 29
## Excel formula in column H is:
Formula 29-7
fr = 1/2 x R x C x 6
=1/((D894*E894*F894*SQRT(6)))
0.0000001
R (ohms)
6.28
From Use of Formulas
(Use Approp. Units)
0.0000001
0.000000000980
3300
0.00000001
1970
CHAPTER 29
## Excel formula in column H is:
Formula 29-8
fr = 1/2 x R x C
=1/(D898*E898*F898)
CHAPTER 29
## Estimated timing for monostable MV
ton = 0.69 x RC x C
## Est. timing for 555 type mono. MV
tw = 1.1 x RC x C
Formula 29-10
CHAPTER 29
Formula 29-11
t = 0.7 x R x C
=0.7*D907*E907
RC1 (ohms)
CHAPTER 29
## Operating freq. of an astable MV
f = 1/1.38 x RC x C
## Excel formula in column H is:
RC (ohms)
Formula 29-9
CHAPTER 29
Formula 29-12
=0.69*D901*E901
=1.1*D904*E904
=1/(1.38*D910*E910)
R (ohms)
6.28
RC (ohms)
100000
0.0000001
15.92
100000
RC (ohms)
0.0000001
0.0069
1000000
0.00001
11
CC1 (Fd)
10000 0.00000001
0.00007
47000 0.00000001
1542
CHAPTER 29
Formula 29-13
## Operating freq. of an IC astable MV
f = 1/0.69 x C x (RA + 2 x RB)
CHAPTER 30
CHAPTER 31
VR = VS - VD
Formula 31-1
CHAPTER 31
Formula 31-2
## To find R value for given LED curr.
R = VR / ID
(Note: Knowing that VR = VS-VD)
RA (ohms)
## Excel formula in column H is:
=1/(0.69*D914*(E914+2*F914))
0.000000001
## Excel formula in column H is:
VS (volts)
=D919-E919
Excel formula in column H is:
=D922/E922
10000
RB (ohms)
47000
13935
VD (volts)
12
VR (volts)
1.5
10.5
ID (amps)
7.5
0.015
500
A
1
2
3
4
5
6
7
8
9
10
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
925 CHAPTER 31
926 Formula 31-3
FORMULAS
To find R value for given LED curr.
R = (VS - VD) / ID
## EXCEL EQUIVALENT FORMULAS
Excel formula in column H is:
=(D926-E926)/F926
AND THEIR VALUES
VD (volts)
ID (amps)
VS (volts)
9
1.5
0.015
From Use of Formulas
(Use Approp. Units)
500
A
1
2
3
4
5
6
7
8
9
10
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
## CHAPTER-BY-CHAPTER FORMULAS AND SAMPLE EXCEL EQUIVALENT SPREADSHEET FORMULAS
NOTE: Some of the most-used, more-complex and useful formulas and Excel equivalents are highlighted in yellow.
The chapters' text-style formulas in Column B are shown in Excel formula format in Column C.
These same formulas are used as operative formulas in Column H, where the formula-computed answers are shown.
DIRECTIONS: Replace sample parameter values shown in columns D through E, F, or G with the values for your problem, and your answer appears in Column H.
CHAPTER # &
FORMULA #
CHAPTER 31
Formula 31-4
FORMULAS
To find I in photodiode circuit
I = R x PRCVD
## EXCEL EQUIVALENT FORMULAS
AND THEIR VALUES
PRCVD
R (ohms)
=D929*E929
From Use of Formulas
(Use Approp. Units)
100000 (unknown)
## (Note: Knowing device spectral response
and amount of received optical power)
CHAPTER 31
Formula 31-5
IC = CTR x ID
ID (amps)
CTR
=D934*E934
0.01
0.01
333
-20
CHAPTER 31
Formula 31-6
CHAPTER 31
Formula 31-7
## Power gain/loss in fiber optic cable
A = 10 log10 (Pout / Pin)
## Find output power of fiber optic cbl.
Pout = Pin x ( 10A/10 )
## Excel formula in column H is:
Pout (W)
=10*LOG(D938/E938,10)
=D941*(10^-20/10)
Pin (W)
3.33
Pin (W)
333
3.33E-019 | 25,669 | 68,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-43 | latest | en | 0.539032 |
http://theperihelioneffect.com/how-far-is-the-earth-from-the-sun/ | 1,669,659,920,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00393.warc.gz | 52,733,610 | 33,790 | The Perihelion Effect
# How Far is the Earth from the Sun?
The AU is the average distance from the Earth to the sun. Earth makes a complete revolution around the sun every 365.25 days — one year. However, Earth’s orbit is not a perfect circle; it is shaped more like an oval, or an ellipse. Over the course of a year, Earth moves sometimes closer to the sun and sometimes farther away from the sun. Earth’s closest approach to the sun, called perihelion, comes in early January and is about 91 million miles (146 million km), just shy of 1 AU. The farthest from the sun Earth gets is called aphelion. It comes in early July and is about 94.5 million miles (152 million km), just over 1 AU.
Historically, the first person to measure the distance to the sun was the Greek astronomer Aristarchus around the year 250 B.C. He used the phases of the moon to measure the sizes and distances of the sun and moon. During a half moon, the three celestial bodies should form a right angle. By measuring the angle at Earth between the sun and moon, he determined the sun was 19 times as far from the planet as the moon, and thus 19 times as big. In fact, the sun is about 400 times larger than the moon.
The sun is at the heart of the solar system. All of the bodies in the solar system — planets, asteroids, comets, etc. — revolve around it at various distances. Mercury, the planet closest to the sun, gets as close as 29 million miles (47 million kilometers) in its elliptical orbit, while objects in the Oort Cloud, the solar system’s icy shell, are thought to lie as far as 9.3 trillion miles (15 trillion km).
Earth orbits the sun 100,000 times closer than the Oort Cloud, at an average of 92,955,807 miles (149,597,870,700 meters). The distance from Earth to the sun is called an astronomical unit, or AU, which is used to measure distances throughout the solar system.
Jupiter, for example, is 5.2 AU from the sun. Neptune is 30.07 AU from the sun. The distance to the nearest star, Proxima Centauri, is about 268,770 AU, according to NASA. However, to measure longer distances, astronomers use light-years, or the distance that light travels in a single Earth-year, which is equal to 63,239 AU. So Proxima Centauri is about 4.25 light-years away
The AU is the average distance from the Earth to the sun. Earth makes a complete revolution around the sun every 365.25 days — one year. However, Earth’s orbit is not a perfect circle; it is shaped more like an oval, or an ellipse. Over the course of a year, Earth moves sometimes closer to the sun and sometimes farther away from the sun. Earth’s closest approach to the sun, called perihelion, comes in early January and is about 91 million miles (146 million km), just shy of 1 AU. The farthest from the sun Earth gets is called aphelion. It comes in early July and is about 94.5 million miles (152 million km), just over 1 AU.
Historically, the first person to measure the distance to the sun was the Greek astronomer Aristarchus around the year 250 B.C. He used the phases of the moon to measure the sizes and distances of the sun and moon. During a half moon, the three celestial bodies should form a right angle. By measuring the angle at Earth between the sun and moon, he determined the sun was 19 times as far from the planet as the moon, and thus 19 times as big. In fact, the sun is about 400 times larger than the moon.
“Aristarchus’ measurement was probably off because first, it is hard to determine the exact centers of the sun and the moon and second, it is hard to know exactly when the moon is half full,” says Cornell University’s astronomy website.
Although imprecise, Aristarchus provided a simple understanding of the sizes and distances of the three bodies, which led him to conclude that the Earth goes around the sun, about 1,700 years before Nicolaus Copernicusproposed his heliocentric model of the solar system.
In 1653, astronomer Christiaan Huygens calculated the distance from Earth to the sun. He used the phases of Venus to find the angles in a Venus-Earth-sun triangle. For example, when Venus appears half illuminated by the sun, the three bodies form a right triangle from Earth’s perspective. Guessing (correctly, by chance) the size of Venus, Huygens was able to determine the distance from Venus to Earth, and knowing that distance, plus the angles made by the triangle, he was able to measure the distance to the sun. However, because Huygens’ method was partly guesswork and not completely scientifically grounded, he usually doesn’t get the credit.
In 1672, Giovanni Cassini used a method involving parallax, or angular difference, to find the distance to Mars and at the same time figured out the distance to the sun. He sent a colleague, Jean Richer, to French Guiana while he stayed in Paris. They took measurements of the position of Mars relative to background stars, and triangulated those measurements with the known distance between Paris and French Guiana. Once they had the distance to Mars, they could also calculate the distance to the sun. Since his methods were more scientific, he usually gets the credit.
“Expressing distances in the astronomical unit allowed astronomers to overcome the difficulty of measuring distances in some physical unit,” astronomer Nicole Capitaine of Paris University told Space.com. “Such a practice was useful for many years, because astronomers were not able to make distance measurements in the solar system as precisely as they could measure angles.”
With the advent of spacecraft and radar, more precise methods emerged for making a direct measure of the distance between the Earth and the sun. The definition of AU had been “the radius of an unperturbed circular Newtonian orbit about the sun of a particle having infinitesimal mass, moving with a mean motion of 0.01720209895 radians per day (known as the Gaussian constant).”
Along with making things unnecessarily difficult for astronomy professors, that definition actually didn’t jibe with general relativity. Using the old definition, the value of AU would change depending on an observer’s location in the solar system. If an observer on Jupiter used the old definition to calculate the distance between the Earth and the sun, the measurement would vary from one made on Earth by about 1,000 meters (3,280 feet).
Moreover, the Gaussian constant depends on the mass of the sun, and because the sun loses mass as it radiates energy, the value of AU was changing along with it.
The International Astronomical Union voted in August 2012 to change the definition of the astronomical unit to a plain old number: 149,597,870,700 meters. The measurement is based on the speed of light, a fixed distance that has nothing to do with the sun’s mass. A meter is defined as the distance traveled by light in a vacuum in 1 / 299,792,458 of a second.
“The new definition is much simpler than the old one,” said astronomer Sergei Klioner of the Technical University of Dresden in Germany. Both Klioner and Capitaine were both part of the International Astronomical Union group that worked to refine the definition.
https://www.space.com/17081-how-far-is-earth-from-the-sun.html | 1,651 | 7,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | longest | en | 0.94135 |
https://us.metamath.org/mpeuni/limsssuc.html | 1,726,590,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00355.warc.gz | 536,974,897 | 6,328 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > limsssuc Structured version Visualization version GIF version
Theorem limsssuc 7570
Description: A class includes a limit ordinal iff the successor of the class includes it. (Contributed by NM, 30-Oct-2003.)
Assertion
Ref Expression
limsssuc (Lim 𝐴 → (𝐴𝐵𝐴 ⊆ suc 𝐵))
Proof of Theorem limsssuc
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 sssucid 6251 . . 3 𝐵 ⊆ suc 𝐵
2 sstr2 3901 . . 3 (𝐴𝐵 → (𝐵 ⊆ suc 𝐵𝐴 ⊆ suc 𝐵))
31, 2mpi 20 . 2 (𝐴𝐵𝐴 ⊆ suc 𝐵)
4 eleq1 2839 . . . . . . . . . . . 12 (𝑥 = 𝐵 → (𝑥𝐴𝐵𝐴))
54biimpcd 252 . . . . . . . . . . 11 (𝑥𝐴 → (𝑥 = 𝐵𝐵𝐴))
6 limsuc 7569 . . . . . . . . . . . . . 14 (Lim 𝐴 → (𝐵𝐴 ↔ suc 𝐵𝐴))
76biimpa 480 . . . . . . . . . . . . 13 ((Lim 𝐴𝐵𝐴) → suc 𝐵𝐴)
8 limord 6233 . . . . . . . . . . . . . . 15 (Lim 𝐴 → Ord 𝐴)
9 ordelord 6196 . . . . . . . . . . . . . . . . 17 ((Ord 𝐴𝐵𝐴) → Ord 𝐵)
108, 9sylan 583 . . . . . . . . . . . . . . . 16 ((Lim 𝐴𝐵𝐴) → Ord 𝐵)
11 ordsuc 7534 . . . . . . . . . . . . . . . 16 (Ord 𝐵 ↔ Ord suc 𝐵)
1210, 11sylib 221 . . . . . . . . . . . . . . 15 ((Lim 𝐴𝐵𝐴) → Ord suc 𝐵)
13 ordtri1 6207 . . . . . . . . . . . . . . 15 ((Ord 𝐴 ∧ Ord suc 𝐵) → (𝐴 ⊆ suc 𝐵 ↔ ¬ suc 𝐵𝐴))
148, 12, 13syl2an2r 684 . . . . . . . . . . . . . 14 ((Lim 𝐴𝐵𝐴) → (𝐴 ⊆ suc 𝐵 ↔ ¬ suc 𝐵𝐴))
1514con2bid 358 . . . . . . . . . . . . 13 ((Lim 𝐴𝐵𝐴) → (suc 𝐵𝐴 ↔ ¬ 𝐴 ⊆ suc 𝐵))
167, 15mpbid 235 . . . . . . . . . . . 12 ((Lim 𝐴𝐵𝐴) → ¬ 𝐴 ⊆ suc 𝐵)
1716ex 416 . . . . . . . . . . 11 (Lim 𝐴 → (𝐵𝐴 → ¬ 𝐴 ⊆ suc 𝐵))
185, 17sylan9r 512 . . . . . . . . . 10 ((Lim 𝐴𝑥𝐴) → (𝑥 = 𝐵 → ¬ 𝐴 ⊆ suc 𝐵))
1918con2d 136 . . . . . . . . 9 ((Lim 𝐴𝑥𝐴) → (𝐴 ⊆ suc 𝐵 → ¬ 𝑥 = 𝐵))
2019ex 416 . . . . . . . 8 (Lim 𝐴 → (𝑥𝐴 → (𝐴 ⊆ suc 𝐵 → ¬ 𝑥 = 𝐵)))
2120com23 86 . . . . . . 7 (Lim 𝐴 → (𝐴 ⊆ suc 𝐵 → (𝑥𝐴 → ¬ 𝑥 = 𝐵)))
2221imp31 421 . . . . . 6 (((Lim 𝐴𝐴 ⊆ suc 𝐵) ∧ 𝑥𝐴) → ¬ 𝑥 = 𝐵)
23 ssel2 3889 . . . . . . . . . 10 ((𝐴 ⊆ suc 𝐵𝑥𝐴) → 𝑥 ∈ suc 𝐵)
24 vex 3413 . . . . . . . . . . 11 𝑥 ∈ V
2524elsuc 6243 . . . . . . . . . 10 (𝑥 ∈ suc 𝐵 ↔ (𝑥𝐵𝑥 = 𝐵))
2623, 25sylib 221 . . . . . . . . 9 ((𝐴 ⊆ suc 𝐵𝑥𝐴) → (𝑥𝐵𝑥 = 𝐵))
2726ord 861 . . . . . . . 8 ((𝐴 ⊆ suc 𝐵𝑥𝐴) → (¬ 𝑥𝐵𝑥 = 𝐵))
2827con1d 147 . . . . . . 7 ((𝐴 ⊆ suc 𝐵𝑥𝐴) → (¬ 𝑥 = 𝐵𝑥𝐵))
2928adantll 713 . . . . . 6 (((Lim 𝐴𝐴 ⊆ suc 𝐵) ∧ 𝑥𝐴) → (¬ 𝑥 = 𝐵𝑥𝐵))
3022, 29mpd 15 . . . . 5 (((Lim 𝐴𝐴 ⊆ suc 𝐵) ∧ 𝑥𝐴) → 𝑥𝐵)
3130ex 416 . . . 4 ((Lim 𝐴𝐴 ⊆ suc 𝐵) → (𝑥𝐴𝑥𝐵))
3231ssrdv 3900 . . 3 ((Lim 𝐴𝐴 ⊆ suc 𝐵) → 𝐴𝐵)
3332ex 416 . 2 (Lim 𝐴 → (𝐴 ⊆ suc 𝐵𝐴𝐵))
343, 33impbid2 229 1 (Lim 𝐴 → (𝐴𝐵𝐴 ⊆ suc 𝐵))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 209 ∧ wa 399 ∨ wo 844 = wceq 1538 ∈ wcel 2111 ⊆ wss 3860 Ord word 6173 Lim wlim 6175 suc csuc 6176 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2729 ax-sep 5173 ax-nul 5180 ax-pr 5302 ax-un 7465 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-clab 2736 df-cleq 2750 df-clel 2830 df-nfc 2901 df-ne 2952 df-ral 3075 df-rex 3076 df-rab 3079 df-v 3411 df-sbc 3699 df-dif 3863 df-un 3865 df-in 3867 df-ss 3877 df-pss 3879 df-nul 4228 df-if 4424 df-pw 4499 df-sn 4526 df-pr 4528 df-tp 4530 df-op 4532 df-uni 4802 df-br 5037 df-opab 5099 df-tr 5143 df-eprel 5439 df-po 5447 df-so 5448 df-fr 5487 df-we 5489 df-ord 6177 df-on 6178 df-lim 6179 df-suc 6180 This theorem is referenced by: cardlim 9447
Copyright terms: Public domain W3C validator | 2,227 | 3,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.159389 |
https://blog.csdn.net/qq_38650028/article/details/107776127 | 1,600,493,157,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00706.warc.gz | 298,446,476 | 19,859 | # leetcode 230. 二叉搜索树中第K小的元素
230. 二叉搜索树中第K小的元素
3
/ \
1 4
\
2
5
/ \
3 6
/ \
2 4
/
1
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def kthSmallest(self, root: TreeNode, k: int) -> int:
res = []
def solve(root):
if not root: return
if solve(root.left): return True
res.append(root.val)
if len(res) == k:
return True
if solve(root.right): return True
solve(root)
return res[-1]
11-28
04-24 291
07-29 117
04-04 517 | 213 | 596 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-40 | latest | en | 0.183467 |
https://www.tutorialsteacher.com/python/recursion-in-python | 1,685,780,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00622.warc.gz | 1,136,768,855 | 14,485 | Recursion in Python
A function that calls itself is a recursive function. This method is used when a certain problem is defined in terms of itself. Although this involves iteration, using an iterative approach to solve such a problem can be tedious. The recursive approach provides a very concise solution to a seemingly complex problem. It looks glamorous but can be difficult to comprehend!
The most popular example of recursion is the calculation of the factorial. Mathematically the factorial is defined as: n! = n * (n-1)!
We use the factorial itself to define the factorial. Hence, this is a suitable case to write a recursive function. Let us expand the above definition for the calculation of the factorial value of 5.
``````5! = 5 X 4!
5 X4 X 3!
5 X4 X 3 X 2!
5 X4 X 3 X 2 X 1!
5 X4 X 3 X 2 X 1
= 120
``````
While we can perform this calculation using a loop, its recursive function involves successively calling it by decrementing the number until it reaches 1. The following is a recursive function to calculate the factorial.
Example: Recursive Function
``````def factorial(n):
if n == 1:
print(n)
return 1
else:
print (n,'*', end=' ')
return n * factorial(n-1)
factorial(5) #calling recursive function
``````
The above recursive function will produce the following output:
Output
``````5 * 4 * 3 * 2 * 1
``````
When the factorial function is called with 5 as argument, successive calls to the same function are placed, while reducing the value of 5. Functions start returning to their earlier call after the argument reaches 1. The return value of the first call is a cumulative product of the return values of all calls.
Related Articles | 405 | 1,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-23 | longest | en | 0.870554 |
https://cs.brown.edu/courses/cs1951x/docs/algebra/algebra/basic.html | 1,709,277,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00616.warc.gz | 193,639,560 | 43,685 | # mathlibdocumentation
algebra.algebra.basic
# Algebras over commutative semirings #
In this file we define associative unital algebras over commutative (semi)rings, algebra homomorphisms alg_hom, and algebra equivalences alg_equiv.
subalgebras are defined in algebra.algebra.subalgebra.
For the category of R-algebras, denoted Algebra R, see the file algebra/category/Algebra/basic.lean.
See the implementation notes for remarks about non-associative and non-unital algebras.
## Main definitions: #
• algebra R A: the algebra typeclass.
• alg_hom R A B: the type of R-algebra morphisms from A to B.
• alg_equiv R A B: the type of R-algebra isomorphisms between A to B.
• algebra_map R A : R →+* A: the canonical map from R to A, as a ring_hom. This is the preferred spelling of this map.
• algebra.linear_map R A : R →ₗ[R] A: the canonical map from R to A, as a linear_map.
• algebra.of_id R A : R →ₐ[R] A: the canonical map from R to A, as n alg_hom.
• Instances of algebra in this file:
## Notations #
• A →ₐ[R] B : R-algebra homomorphism from A to B.
• A ≃ₐ[R] B : R-algebra equivalence from A to B.
## Implementation notes #
Given a commutative (semi)ring R, there are two ways to define an R-algebra structure on a (possibly noncommutative) (semi)ring A:
• By endowing A with a morphism of rings R →+* A denoted algebra_map R A which lands in the center of A.
• By requiring A be an R-module such that the action associates and commutes with multiplication as r • (a₁ * a₂) = (r • a₁) * a₂ = a₁ * (r • a₂).
We define algebra R A in a way that subsumes both definitions, by extending has_smul R A and requiring that this scalar action r • x must agree with left multiplication by the image of the structure morphism algebra_map R A r * x.
As a result, there are two ways to talk about an R-algebra A when A is a semiring:
1. variables [comm_semiring R] [semiring A]
variables [algebra R A]
2. variables [comm_semiring R] [semiring A]
variables [module R A] [smul_comm_class R A A] [is_scalar_tower R A A]
The first approach implies the second via typeclass search; so any lemma stated with the second set of arguments will automatically apply to the first set. Typeclass search does not know that the second approach implies the first, but this can be shown with:
The advantage of the first approach is that algebra_map R A is available, and alg_hom R A B and subalgebra R A can be used. For concrete R and A, algebra_map R A is often definitionally convenient.
The advantage of the second approach is that comm_semiring R, semiring A, and module R A can all be relaxed independently; for instance, this allows us to:
While alg_hom R A B cannot be used in the second approach, non_unital_alg_hom R A B still can.
You should always use the first approach when working with associative unital algebras, and mimic the second approach only when you need to weaken a condition on either R or A.
@[nolint, class]
structure algebra (R : Type u) (A : Type v) [semiring A] :
Type (max u v)
• to_has_smul : A
• to_ring_hom : R →+* A
• commutes' : ∀ (r : R) (x : A),
• smul_def' : ∀ (r : R) (x : A), r x =
An associative unital R-algebra is a semiring A equipped with a map into its center R → A.
See the implementation notes in this file for discussion of the details of this definition.
Instances of this typeclass
Instances of other typeclasses for algebra
def algebra_map (R : Type u) (A : Type v) [semiring A] [ A] :
R →+* A
Embedding R →+* A given by algebra structure.
Equations
def ring_hom.to_algebra' {R : Type u_1} {S : Type u_2} [semiring S] (i : R →+* S) (h : ∀ (c : R) (x : S), i c * x = x * i c) :
S
Creating an algebra from a morphism to the center of a semiring.
Equations
def ring_hom.to_algebra {R : Type u_1} {S : Type u_2} (i : R →+* S) :
S
Creating an algebra from a morphism to a commutative semiring.
Equations
theorem ring_hom.algebra_map_to_algebra {R : Type u_1} {S : Type u_2} (i : R →+* S) :
S = i
@[reducible]
def algebra.of_module' {R : Type u} {A : Type w} [semiring A] [ A] (h₁ : ∀ (r : R) (x : A), r 1 * x = r x) (h₂ : ∀ (r : R) (x : A), x * r 1 = r x) :
A
Let R be a commutative semiring, let A be a semiring with a module R structure. If (r • 1) * x = x * (r • 1) = r • x for all r : R and x : A, then A is an algebra over R.
Equations
@[reducible]
def algebra.of_module {R : Type u} {A : Type w} [semiring A] [ A] (h₁ : ∀ (r : R) (x y : A), r x * y = r (x * y)) (h₂ : ∀ (r : R) (x y : A), x * r y = r (x * y)) :
A
Let R be a commutative semiring, let A be a semiring with a module R structure. If (r • x) * y = x * (r • y) = r • (x * y) for all r : R and x y : A, then A is an algebra over R.
Equations
• h₂ =
@[ext]
theorem algebra.algebra_ext {R : Type u_1} {A : Type u_2} [semiring A] (P Q : A) (w : ∀ (r : R), A) r = A) r) :
P = Q
To prove two algebra structures on a fixed [comm_semiring R] [semiring A] agree, it suffices to check the algebra_maps agree.
@[protected, instance]
def algebra.to_module {R : Type u} {A : Type w} [semiring A] [ A] :
A
Equations
theorem algebra.smul_def {R : Type u} {A : Type w} [semiring A] [ A] (r : R) (x : A) :
r x = A) r * x
theorem algebra.algebra_map_eq_smul_one {R : Type u} {A : Type w} [semiring A] [ A] (r : R) :
A) r = r 1
theorem algebra.algebra_map_eq_smul_one' {R : Type u} {A : Type w} [semiring A] [ A] :
A) = λ (r : R), r 1
theorem algebra.commutes {R : Type u} {A : Type w} [semiring A] [ A] (r : R) (x : A) :
A) r * x = x * A) r
mul_comm for algebras when one element is from the base ring.
theorem algebra.left_comm {R : Type u} {A : Type w} [semiring A] [ A] (x : A) (r : R) (y : A) :
x * ( A) r * y) = A) r * (x * y)
mul_left_comm for algebras when one element is from the base ring.
theorem algebra.right_comm {R : Type u} {A : Type w} [semiring A] [ A] (x : A) (r : R) (y : A) :
x * A) r * y = x * y * A) r
mul_right_comm for algebras when one element is from the base ring.
@[protected, instance]
def is_scalar_tower.right {R : Type u} {A : Type w} [semiring A] [ A] :
A
@[protected, simp]
theorem algebra.mul_smul_comm {R : Type u} {A : Type w} [semiring A] [ A] (s : R) (x y : A) :
x * s y = s (x * y)
This is just a special case of the global mul_smul_comm lemma that requires less typeclass search (and was here first).
@[protected, simp]
theorem algebra.smul_mul_assoc {R : Type u} {A : Type w} [semiring A] [ A] (r : R) (x y : A) :
r x * y = r (x * y)
This is just a special case of the global smul_mul_assoc lemma that requires less typeclass search (and was here first).
@[simp]
theorem smul_algebra_map {R : Type u} {A : Type w} [semiring A] [ A] {α : Type u_1} [monoid α] [ A] (a : α) (r : R) :
a A) r = A) r
@[simp]
theorem algebra.bit0_smul_one {R : Type u} {A : Type w} [semiring A] [ A] {r : R} :
bit0 r 1 = bit0 (r 1)
theorem algebra.bit0_smul_one' {R : Type u} {A : Type w} [semiring A] [ A] {r : R} :
bit0 r 1 = r 2
@[simp]
theorem algebra.bit0_smul_bit0 {R : Type u} {A : Type w} [semiring A] [ A] {r : R} {a : A} :
bit0 r bit0 a = r bit0 (bit0 a)
@[simp]
theorem algebra.bit0_smul_bit1 {R : Type u} {A : Type w} [semiring A] [ A] {r : R} {a : A} :
bit0 r bit1 a = r bit0 (bit1 a)
@[simp]
theorem algebra.bit1_smul_one {R : Type u} {A : Type w} [semiring A] [ A] {r : R} :
bit1 r 1 = bit1 (r 1)
theorem algebra.bit1_smul_one' {R : Type u} {A : Type w} [semiring A] [ A] {r : R} :
bit1 r 1 = r 2 + 1
@[simp]
theorem algebra.bit1_smul_bit0 {R : Type u} {A : Type w} [semiring A] [ A] {r : R} {a : A} :
bit1 r bit0 a = r bit0 (bit0 a) + bit0 a
@[simp]
theorem algebra.bit1_smul_bit1 {R : Type u} {A : Type w} [semiring A] [ A] {r : R} {a : A} :
bit1 r bit1 a = r bit0 (bit1 a) + bit1 a
@[protected]
def algebra.linear_map (R : Type u) (A : Type w) [semiring A] [ A] :
The canonical ring homomorphism algebra_map R A : R →* A for any R-algebra A, packaged as an R-linear map.
Equations
@[simp]
theorem algebra.linear_map_apply (R : Type u) (A : Type w) [semiring A] [ A] (r : R) :
A) r = A) r
theorem algebra.coe_linear_map (R : Type u) (A : Type w) [semiring A] [ A] :
A) = A)
@[protected, instance]
def algebra.id (R : Type u) :
R
Equations
@[simp]
theorem algebra.id.map_eq_id {R : Type u} :
R =
theorem algebra.id.map_eq_self {R : Type u} (x : R) :
R) x = x
@[simp]
theorem algebra.id.smul_eq_mul {R : Type u} (x y : R) :
x y = x * y
@[protected, instance]
def punit.algebra {R : Type u} :
Equations
@[simp]
theorem algebra.algebra_map_punit {R : Type u} (r : R) :
punit) r = punit.star
@[protected, instance]
def ulift.algebra {R : Type u} {A : Type w} [semiring A] [ A] :
(ulift A)
Equations
theorem ulift.algebra_map_eq {R : Type u} {A : Type w} [semiring A] [ A] (r : R) :
(ulift A)) r = {down := A) r}
@[simp]
theorem ulift.down_algebra_map {R : Type u} {A : Type w} [semiring A] [ A] (r : R) :
( (ulift A)) r).down = A) r
@[protected, instance]
def prod.algebra (R : Type u) (A : Type w) (B : Type u_1) [semiring A] [ A] [semiring B] [ B] :
(A × B)
Equations
@[simp]
theorem algebra.algebra_map_prod_apply {R : Type u} {A : Type w} {B : Type u_1} [semiring A] [ A] [semiring B] [ B] (r : R) :
(A × B)) r = ( A) r, B) r)
@[protected, instance]
def algebra.of_subsemiring {R : Type u} {A : Type w} [semiring A] [ A] (S : subsemiring R) :
A
Algebra over a subsemiring. This builds upon subsemiring.module.
Equations
theorem algebra.algebra_map_of_subsemiring {R : Type u} (S : subsemiring R) :
R = S.subtype
theorem algebra.algebra_map_of_subsemiring_apply {R : Type u} (S : subsemiring R) (x : S) :
R) x = x
@[protected, instance]
def algebra.of_subring {R : Type u_1} {A : Type u_2} [comm_ring R] [ring A] [ A] (S : subring R) :
A
Algebra over a subring. This builds upon subring.module.
Equations
theorem algebra.algebra_map_of_subring {R : Type u_1} [comm_ring R] (S : subring R) :
R = S.subtype
theorem algebra.coe_algebra_map_of_subring {R : Type u_1} [comm_ring R] (S : subring R) :
theorem algebra.algebra_map_of_subring_apply {R : Type u_1} [comm_ring R] (S : subring R) (x : S) :
R) x = x
def algebra.algebra_map_submonoid {R : Type u} (S : Type u_1) [semiring S] [ S] (M : submonoid R) :
Explicit characterization of the submonoid map in the case of an algebra. S is made explicit to help with type inference
Equations
theorem algebra.mem_algebra_map_submonoid_of_mem {R : Type u} {S : Type u_1} [semiring S] [ S] {M : submonoid R} (x : M) :
S) x
theorem algebra.mul_sub_algebra_map_commutes {R : Type u} {A : Type w} [ring A] [ A] (x : A) (r : R) :
x * (x - A) r) = (x - A) r) * x
theorem algebra.mul_sub_algebra_map_pow_commutes {R : Type u} {A : Type w} [ring A] [ A] (x : A) (r : R) (n : ) :
x * (x - A) r) ^ n = (x - A) r) ^ n * x
@[reducible]
def algebra.semiring_to_ring (R : Type u) {A : Type w} [comm_ring R] [semiring A] [ A] :
A semiring that is an algebra over a commutative ring carries a natural ring structure. See note [reducible non-instances].
Equations
@[protected, instance]
def mul_opposite.algebra {R : Type u_1} {A : Type u_2} [semiring A] [ A] :
Equations
@[simp]
theorem mul_opposite.algebra_map_apply {R : Type u_1} {A : Type u_2} [semiring A] [ A] (c : R) :
@[protected, instance]
def module.End.algebra (R : Type u) (M : Type v) [ M] :
M)
Equations
theorem module.algebra_map_End_eq_smul_id (R : Type u) (M : Type v) [ M] (a : R) :
M)) a =
@[simp]
theorem module.algebra_map_End_apply (R : Type u) (M : Type v) [ M] (a : R) (m : M) :
( M)) a) m = a m
@[simp]
theorem module.ker_algebra_map_End (K : Type u) (V : Type v) [field K] [ V] (a : K) (ha : a 0) :
theorem linear_map.map_algebra_map_mul {R : Type u_1} {A : Type u_2} {B : Type u_3} [semiring A] [semiring B] [ A] [ B] (f : A →ₗ[R] B) (a : A) (r : R) :
f ( A) r * a) = B) r * f a
An alternate statement of linear_map.map_smul for when algebra_map is more convenient to work with than .
theorem linear_map.map_mul_algebra_map {R : Type u_1} {A : Type u_2} {B : Type u_3} [semiring A] [semiring B] [ A] [ B] (f : A →ₗ[R] B) (a : A) (r : R) :
f (a * A) r) = f a * B) r
def alg_hom.to_ring_hom {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (self : A →ₐ[R] B) :
A →+* B
Reinterpret an alg_hom as a ring_hom
@[nolint]
structure alg_hom (R : Type u) (A : Type v) (B : Type w) [semiring A] [semiring B] [ A] [ B] :
Type (max v w)
Defining the homomorphism in the category R-Alg.
Instances for alg_hom
@[nolint, instance]
def alg_hom_class.to_ring_hom_class (F : Type u_1) (R : out_param (Type u_2)) (A : out_param (Type u_3)) (B : out_param (Type u_4)) [semiring A] [semiring B] [ A] [ B] [self : A B] :
B
@[class]
structure alg_hom_class (F : Type u_1) (R : out_param (Type u_2)) (A : out_param (Type u_3)) (B : out_param (Type u_4)) [semiring A] [semiring B] [ A] [ B] :
Type (max u_1 u_3 u_4)
• coe : F → Π (a : A), (λ (_x : A), B) a
• coe_injective' :
• map_mul : ∀ (f : F) (x y : A), f (x * y) = f x * f y
• map_one : ∀ (f : F), f 1 = 1
• map_add : ∀ (f : F) (x y : A), f (x + y) = f x + f y
• map_zero : ∀ (f : F), f 0 = 0
• commutes : ∀ (f : F) (r : R), f ( A) r) = B) r
alg_hom_class F R A B asserts F is a type of bundled algebra homomorphisms from A to B.
Instances of this typeclass
Instances of other typeclasses for alg_hom_class
• alg_hom_class.has_sizeof_inst
@[protected, instance]
def alg_hom_class.linear_map_class {R : Type u_1} {A : Type u_2} {B : Type u_3} [semiring A] [semiring B] [ A] [ B] {F : Type u_4} [ A B] :
A B
Equations
@[protected, instance]
def alg_hom.has_coe_to_fun {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
has_coe_to_fun (A →ₐ[R] B) (λ (_x : A →ₐ[R] B), A → B)
Equations
@[simp]
theorem alg_hom.to_fun_eq_coe {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₐ[R] B) :
@[protected, instance]
def alg_hom.alg_hom_class {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
alg_hom_class (A →ₐ[R] B) R A B
Equations
@[protected, instance]
def alg_hom.coe_ring_hom {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
has_coe (A →ₐ[R] B) (A →+* B)
Equations
@[protected, instance]
def alg_hom.coe_monoid_hom {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
has_coe (A →ₐ[R] B) (A →* B)
Equations
@[protected, instance]
def alg_hom.coe_add_monoid_hom {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
has_coe (A →ₐ[R] B) (A →+ B)
Equations
@[simp, norm_cast]
theorem alg_hom.coe_mk {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] {f : A → B} (h₁ : f 1 = 1) (h₂ : ∀ (x y : A), f (x * y) = f x * f y) (h₃ : f 0 = 0) (h₄ : ∀ (x y : A), f (x + y) = f x + f y) (h₅ : ∀ (r : R), f ( A) r) = B) r) :
{to_fun := f, map_one' := h₁, map_mul' := h₂, map_zero' := h₃, map_add' := h₄, commutes' := h₅} = f
@[simp]
theorem alg_hom.to_ring_hom_eq_coe {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₐ[R] B) :
@[simp, norm_cast]
theorem alg_hom.coe_to_ring_hom {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₐ[R] B) :
@[simp, norm_cast]
theorem alg_hom.coe_to_monoid_hom {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₐ[R] B) :
@[simp, norm_cast]
theorem alg_hom.coe_to_add_monoid_hom {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₐ[R] B) :
theorem alg_hom.coe_fn_injective {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
theorem alg_hom.coe_fn_inj {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] {φ₁ φ₂ : A →ₐ[R] B} :
φ₁ = φ₂ φ₁ = φ₂
theorem alg_hom.coe_ring_hom_injective {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
theorem alg_hom.coe_monoid_hom_injective {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
theorem alg_hom.coe_add_monoid_hom_injective {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
@[protected]
theorem alg_hom.congr_fun {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] {φ₁ φ₂ : A →ₐ[R] B} (H : φ₁ = φ₂) (x : A) :
φ₁ x = φ₂ x
@[protected]
theorem alg_hom.congr_arg {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) {x y : A} (h : x = y) :
φ x = φ y
@[ext]
theorem alg_hom.ext {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] {φ₁ φ₂ : A →ₐ[R] B} (H : ∀ (x : A), φ₁ x = φ₂ x) :
φ₁ = φ₂
theorem alg_hom.ext_iff {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] {φ₁ φ₂ : A →ₐ[R] B} :
φ₁ = φ₂ ∀ (x : A), φ₁ x = φ₂ x
@[simp]
theorem alg_hom.mk_coe {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] {f : A →ₐ[R] B} (h₁ : f 1 = 1) (h₂ : ∀ (x y : A), f (x * y) = f x * f y) (h₃ : f 0 = 0) (h₄ : ∀ (x y : A), f (x + y) = f x + f y) (h₅ : ∀ (r : R), f ( A) r) = B) r) :
{to_fun := f, map_one' := h₁, map_mul' := h₂, map_zero' := h₃, map_add' := h₄, commutes' := h₅} = f
@[simp]
theorem alg_hom.commutes {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (r : R) :
φ ( A) r) = B) r
theorem alg_hom.comp_algebra_map {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) :
φ.comp A) = B
@[protected]
theorem alg_hom.map_add {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (r s : A) :
φ (r + s) = φ r + φ s
@[protected]
theorem alg_hom.map_zero {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) :
φ 0 = 0
@[protected]
theorem alg_hom.map_mul {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (x y : A) :
φ (x * y) = φ x * φ y
@[protected]
theorem alg_hom.map_one {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) :
φ 1 = 1
@[protected]
theorem alg_hom.map_pow {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (x : A) (n : ) :
φ (x ^ n) = φ x ^ n
@[protected, simp]
theorem alg_hom.map_smul {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (r : R) (x : A) :
φ (r x) = r φ x
@[protected]
theorem alg_hom.map_sum {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) {ι : Type u_1} (f : ι → A) (s : finset ι) :
φ (s.sum (λ (x : ι), f x)) = s.sum (λ (x : ι), φ (f x))
@[protected]
theorem alg_hom.map_finsupp_sum {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) {α : Type u_1} [has_zero α] {ι : Type u_2} (f : ι →₀ α) (g : ι → α → A) :
φ (f.sum g) = f.sum (λ (i : ι) (a : α), φ (g i a))
@[protected]
theorem alg_hom.map_bit0 {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (x : A) :
φ (bit0 x) = bit0 (φ x)
@[protected]
theorem alg_hom.map_bit1 {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (x : A) :
φ (bit1 x) = bit1 (φ x)
def alg_hom.mk' {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →+* B) (h : ∀ (c : R) (x : A), f (c x) = c f x) :
If a ring_hom is R-linear, then it is an alg_hom.
Equations
@[simp]
theorem alg_hom.coe_mk' {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →+* B) (h : ∀ (c : R) (x : A), f (c x) = c f x) :
h) = f
@[protected]
def alg_hom.id (R : Type u) (A : Type v) [semiring A] [ A] :
Identity map as an alg_hom.
Equations
@[simp]
theorem alg_hom.coe_id (R : Type u) (A : Type v) [semiring A] [ A] :
A) = id
@[simp]
theorem alg_hom.id_to_ring_hom (R : Type u) (A : Type v) [semiring A] [ A] :
A) =
theorem alg_hom.id_apply {R : Type u} {A : Type v} [semiring A] [ A] (p : A) :
A) p = p
def alg_hom.comp {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (φ₁ : B →ₐ[R] C) (φ₂ : A →ₐ[R] B) :
Composition of algebra homeomorphisms.
Equations
@[simp]
theorem alg_hom.coe_comp {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (φ₁ : B →ₐ[R] C) (φ₂ : A →ₐ[R] B) :
(φ₁.comp φ₂) = φ₁ φ₂
theorem alg_hom.comp_apply {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (φ₁ : B →ₐ[R] C) (φ₂ : A →ₐ[R] B) (p : A) :
(φ₁.comp φ₂) p = φ₁ (φ₂ p)
theorem alg_hom.comp_to_ring_hom {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (φ₁ : B →ₐ[R] C) (φ₂ : A →ₐ[R] B) :
(φ₁.comp φ₂) = (φ₁.comp φ₂)
@[simp]
theorem alg_hom.comp_id {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) :
φ.comp A) = φ
@[simp]
theorem alg_hom.id_comp {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) :
B).comp φ = φ
theorem alg_hom.comp_assoc {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} {D : Type v₁} [semiring A] [semiring B] [semiring C] [semiring D] [ A] [ B] [ C] [ D] (φ₁ : C →ₐ[R] D) (φ₂ : B →ₐ[R] C) (φ₃ : A →ₐ[R] B) :
(φ₁.comp φ₂).comp φ₃ = φ₁.comp (φ₂.comp φ₃)
def alg_hom.to_linear_map {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) :
R-Alg ⥤ R-Mod
Equations
@[simp]
theorem alg_hom.to_linear_map_apply {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (p : A) :
theorem alg_hom.to_linear_map_injective {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
@[simp]
theorem alg_hom.comp_to_linear_map {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (f : A →ₐ[R] B) (g : B →ₐ[R] C) :
@[simp]
theorem alg_hom.to_linear_map_id {R : Type u} {A : Type v} [semiring A] [ A] :
def alg_hom.of_linear_map {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₗ[R] B) (map_one : f 1 = 1) (map_mul : ∀ (x y : A), f (x * y) = f x * f y) :
Promote a linear_map to an alg_hom by supplying proofs about the behavior on 1 and *.
Equations
@[simp]
theorem alg_hom.of_linear_map_apply {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₗ[R] B) (map_one : f 1 = 1) (map_mul : ∀ (x y : A), f (x * y) = f x * f y) (ᾰ : A) :
map_one map_mul) = f ᾰ
@[simp]
theorem alg_hom.of_linear_map_to_linear_map {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (map_one : (φ.to_linear_map) 1 = 1) (map_mul : ∀ (x y : A), (φ.to_linear_map) (x * y) = (φ.to_linear_map) x * (φ.to_linear_map) y) :
map_one map_mul = φ
@[simp]
theorem alg_hom.to_linear_map_of_linear_map {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₗ[R] B) (map_one : f 1 = 1) (map_mul : ∀ (x y : A), f (x * y) = f x * f y) :
map_one map_mul).to_linear_map = f
@[simp]
theorem alg_hom.of_linear_map_id {R : Type u} {A : Type v} [semiring A] [ A] (map_one : = 1) (map_mul : ∀ (x y : A), linear_map.id (x * y) = ) :
map_mul = A
theorem alg_hom.map_smul_of_tower {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) {R' : Type u_1} [has_smul R' A] [has_smul R' B] [ R] (r : R') (x : A) :
φ (r x) = r φ x
theorem alg_hom.map_list_prod {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (φ : A →ₐ[R] B) (s : list A) :
φ s.prod = s).prod
theorem alg_hom.End_mul {R : Type u} {A : Type v} [semiring A] [ A] (φ₁ φ₂ : A →ₐ[R] A) :
φ₁ * φ₂ = φ₁.comp φ₂
theorem alg_hom.End_one {R : Type u} {A : Type v} [semiring A] [ A] :
1 = A
@[protected, instance]
def alg_hom.End {R : Type u} {A : Type v} [semiring A] [ A] :
Equations
@[simp]
theorem alg_hom.one_apply {R : Type u} {A : Type v} [semiring A] [ A] (x : A) :
1 x = x
@[simp]
theorem alg_hom.mul_apply {R : Type u} {A : Type v} [semiring A] [ A] (φ ψ : A →ₐ[R] A) (x : A) :
* ψ) x = φ (ψ x)
def alg_hom.fst (R : Type u) (A : Type v) (B : Type w) [semiring A] [semiring B] [ A] [ B] :
A × B →ₐ[R] A
First projection as alg_hom.
Equations
def alg_hom.snd (R : Type u) (A : Type v) (B : Type w) [semiring A] [semiring B] [ A] [ B] :
A × B →ₐ[R] B
Second projection as alg_hom.
Equations
def alg_hom.prod {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (f : A →ₐ[R] B) (g : A →ₐ[R] C) :
A →ₐ[R] B × C
The pi.prod of two morphisms is a morphism.
Equations
@[simp]
theorem alg_hom.prod_apply {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (f : A →ₐ[R] B) (g : A →ₐ[R] C) (ᾰ : A) :
theorem alg_hom.coe_prod {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (f : A →ₐ[R] B) (g : A →ₐ[R] C) :
(f.prod g) = g
@[simp]
theorem alg_hom.fst_prod {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (f : A →ₐ[R] B) (g : A →ₐ[R] C) :
B C).comp (f.prod g) = f
@[simp]
theorem alg_hom.snd_prod {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (f : A →ₐ[R] B) (g : A →ₐ[R] C) :
B C).comp (f.prod g) = g
@[simp]
theorem alg_hom.prod_fst_snd {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] :
A B).prod A B) = 1
@[simp]
theorem alg_hom.prod_equiv_apply {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (f : (A →ₐ[R] B) × (A →ₐ[R] C)) :
= f.fst.prod f.snd
@[simp]
theorem alg_hom.prod_equiv_symm_apply {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] (f : A →ₐ[R] B × C) :
= ((alg_hom.fst R B C).comp f, B C).comp f)
def alg_hom.prod_equiv {R : Type u} {A : Type v} {B : Type w} {C : Type u₁} [semiring A] [semiring B] [semiring C] [ A] [ B] [ C] :
(A →ₐ[R] B) × (A →ₐ[R] C) (A →ₐ[R] B × C)
Taking the product of two maps with the same domain is equivalent to taking the product of their codomains.
Equations
theorem alg_hom.algebra_map_eq_apply {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (f : A →ₐ[R] B) {y : R} {x : A} (h : A) y = x) :
B) y = f x
@[protected]
theorem alg_hom.map_multiset_prod {R : Type u} {A : Type v} {B : Type w} [ A] [ B] (φ : A →ₐ[R] B) (s : multiset A) :
φ s.prod = s).prod
@[protected]
theorem alg_hom.map_prod {R : Type u} {A : Type v} {B : Type w} [ A] [ B] (φ : A →ₐ[R] B) {ι : Type u_1} (f : ι → A) (s : finset ι) :
φ (s.prod (λ (x : ι), f x)) = s.prod (λ (x : ι), φ (f x))
@[protected]
theorem alg_hom.map_finsupp_prod {R : Type u} {A : Type v} {B : Type w} [ A] [ B] (φ : A →ₐ[R] B) {α : Type u_1} [has_zero α] {ι : Type u_2} (f : ι →₀ α) (g : ι → α → A) :
φ (f.prod g) = f.prod (λ (i : ι) (a : α), φ (g i a))
@[protected]
theorem alg_hom.map_neg {R : Type u} {A : Type v} {B : Type w} [ring A] [ring B] [ A] [ B] (φ : A →ₐ[R] B) (x : A) :
φ (-x) = -φ x
@[protected]
theorem alg_hom.map_sub {R : Type u} {A : Type v} {B : Type w} [ring A] [ring B] [ A] [ B] (φ : A →ₐ[R] B) (x y : A) :
φ (x - y) = φ x - φ y
@[simp]
theorem rat.smul_one_eq_coe {A : Type u_1} [ A] (m : ) :
m 1 = m
@[nolint]
def alg_equiv.to_add_equiv {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (self : A ≃ₐ[R] B) :
A ≃+ B
@[nolint]
def alg_equiv.to_equiv {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (self : A ≃ₐ[R] B) :
A B
@[nolint]
def alg_equiv.to_ring_equiv {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (self : A ≃ₐ[R] B) :
A ≃+* B
structure alg_equiv (R : Type u) (A : Type v) (B : Type w) [semiring A] [semiring B] [ A] [ B] :
Type (max v w)
An equivalence of algebras is an equivalence of rings commuting with the actions of scalars.
Instances for alg_equiv
@[nolint]
def alg_equiv.to_mul_equiv {R : Type u} {A : Type v} {B : Type w} [semiring A] [semiring B] [ A] [ B] (self : A ≃ₐ[R] B) :
A ≃* B
@[class]
structure alg_equiv_class (F : Type u_1) (R : out_param (Type u_2)) (A : out_param (Type u_3)) (B : out_param (Type u_4)) [semiring A] [semiring B] [ A] [ B] :
Type (max u_1 u_3 u_4)
• coe : F → A → B
• inv : F → B → A
• left_inv : ∀ (e : F),
• right_inv : ∀ (e : F),
• coe_injective' : ∀ (e g : F), e = g
• map_mul : ∀ (f : F) (a b : A), f (a * b) = f a * f b
• map_add : ∀ (f : F) (a b : A), f (a + b) = f a + f b
• commutes : ∀ (f : F) (r : R), f ( A) r) = B) r
alg_equiv_class F R A B states that F is a type of algebra structure preserving equivalences. You should extend this class when you extend alg_equiv.
Instances of this typeclass
Instances of other typeclasses for alg_equiv_class
• alg_equiv_class.has_sizeof_inst
@[nolint, instance]
def alg_equiv_class.to_ring_equiv_class (F : Type u_1) (R : out_param (Type u_2)) (A : out_param (Type u_3)) (B : out_param (Type u_4)) [semiring A] [semiring B] [ A] [ B] [self : A B] :
B
@[protected, instance]
def alg_equiv_class.to_alg_hom_class (F : Type u_1) (R : Type u_2) (A : Type u_3) (B : Type u_4) [semiring A] [semiring B] [ A] [ B] [h : A B] :
A B
Equations
@[protected, instance]
def alg_equiv_class.to_linear_equiv_class (F : Type u_1) (R : Type u_2) (A : Type u_3) (B : Type u_4) [semiring A] [semiring B] [ A] [ B] [h : A B] :
A B
Equations
@[protected, instance]
def alg_equiv.alg_equiv_class {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
alg_equiv_class (A₁ ≃ₐ[R] A₂) R A₁ A₂
Equations
@[protected, instance]
def alg_equiv.has_coe_to_fun {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
has_coe_to_fun (A₁ ≃ₐ[R] A₂) (λ (_x : A₁ ≃ₐ[R] A₂), A₁ → A₂)
Helper instance for when there's too many metavariables to apply fun_like.has_coe_to_fun directly.
Equations
@[ext]
theorem alg_equiv.ext {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {f g : A₁ ≃ₐ[R] A₂} (h : ∀ (a : A₁), f a = g a) :
f = g
@[protected]
theorem alg_equiv.congr_arg {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {f : A₁ ≃ₐ[R] A₂} {x x' : A₁} :
x = x'f x = f x'
@[protected]
theorem alg_equiv.congr_fun {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {f g : A₁ ≃ₐ[R] A₂} (h : f = g) (x : A₁) :
f x = g x
@[protected]
theorem alg_equiv.ext_iff {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {f g : A₁ ≃ₐ[R] A₂} :
f = g ∀ (x : A₁), f x = g x
theorem alg_equiv.coe_fun_injective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
function.injective (λ (e : A₁ ≃ₐ[R] A₂), e)
@[protected, instance]
def alg_equiv.has_coe_to_ring_equiv {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
has_coe (A₁ ≃ₐ[R] A₂) (A₁ ≃+* A₂)
Equations
@[simp]
theorem alg_equiv.coe_mk {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {to_fun : A₁ → A₂} {inv_fun : A₂ → A₁} {left_inv : function.left_inverse inv_fun to_fun} {right_inv : function.right_inverse inv_fun to_fun} {map_mul : ∀ (x y : A₁), to_fun (x * y) = to_fun x * to_fun y} {map_add : ∀ (x y : A₁), to_fun (x + y) = to_fun x + to_fun y} {commutes : ∀ (r : R), to_fun ( A₁) r) = A₂) r} :
{to_fun := to_fun, inv_fun := inv_fun, left_inv := left_inv, right_inv := right_inv, map_mul' := map_mul, map_add' := map_add, commutes' := commutes} = to_fun
@[simp]
theorem alg_equiv.mk_coe {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (e' : A₂ → A₁) (h₁ : e) (h₂ : e) (h₃ : ∀ (x y : A₁), e (x * y) = e x * e y) (h₄ : ∀ (x y : A₁), e (x + y) = e x + e y) (h₅ : ∀ (r : R), e ( A₁) r) = A₂) r) :
{to_fun := e, inv_fun := e', left_inv := h₁, right_inv := h₂, map_mul' := h₃, map_add' := h₄, commutes' := h₅} = e
@[simp]
theorem alg_equiv.to_fun_eq_coe {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[simp]
theorem alg_equiv.to_equiv_eq_coe {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[simp]
theorem alg_equiv.to_ring_equiv_eq_coe {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[simp, norm_cast]
theorem alg_equiv.coe_ring_equiv {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
theorem alg_equiv.coe_ring_equiv' {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
theorem alg_equiv.coe_ring_equiv_injective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
@[protected]
theorem alg_equiv.map_add {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x y : A₁) :
e (x + y) = e x + e y
@[protected]
theorem alg_equiv.map_zero {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
e 0 = 0
@[protected]
theorem alg_equiv.map_mul {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x y : A₁) :
e (x * y) = e x * e y
@[protected]
theorem alg_equiv.map_one {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
e 1 = 1
@[simp]
theorem alg_equiv.commutes {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (r : R) :
e ( A₁) r) = A₂) r
@[simp]
theorem alg_equiv.map_smul {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (r : R) (x : A₁) :
e (r x) = r e x
theorem alg_equiv.map_sum {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) {ι : Type u_1} (f : ι → A₁) (s : finset ι) :
e (s.sum (λ (x : ι), f x)) = s.sum (λ (x : ι), e (f x))
theorem alg_equiv.map_finsupp_sum {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) {α : Type u_1} [has_zero α] {ι : Type u_2} (f : ι →₀ α) (g : ι → α → A₁) :
e (f.sum g) = f.sum (λ (i : ι) (b : α), e (g i b))
def alg_equiv.to_alg_hom {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
A₁ →ₐ[R] A₂
Interpret an algebra equivalence as an algebra homomorphism.
This definition is included for symmetry with the other to_*_hom projections. The simp normal form is to use the coercion of the has_coe_to_alg_hom instance.
Equations
@[protected, instance]
def alg_equiv.has_coe_to_alg_hom {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
has_coe (A₁ ≃ₐ[R] A₂) (A₁ →ₐ[R] A₂)
Equations
@[simp]
theorem alg_equiv.to_alg_hom_eq_coe {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[simp, norm_cast]
theorem alg_equiv.coe_alg_hom {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
theorem alg_equiv.coe_alg_hom_injective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
theorem alg_equiv.coe_ring_hom_commutes {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
The two paths coercion can take to a ring_hom are equivalent
@[protected]
theorem alg_equiv.map_pow {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x : A₁) (n : ) :
e (x ^ n) = e x ^ n
@[protected]
theorem alg_equiv.injective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[protected]
theorem alg_equiv.surjective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[protected]
theorem alg_equiv.bijective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[refl]
def alg_equiv.refl {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
A₁ ≃ₐ[R] A₁
Algebra equivalences are reflexive.
Equations
@[protected, instance]
def alg_equiv.inhabited {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
inhabited (A₁ ≃ₐ[R] A₁)
Equations
@[simp]
theorem alg_equiv.refl_to_alg_hom {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
@[simp]
theorem alg_equiv.coe_refl {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
@[symm]
def alg_equiv.symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
A₂ ≃ₐ[R] A₁
Algebra equivalences are symmetric.
Equations
def alg_equiv.simps.symm_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
A₂ → A₁
Equations
@[simp]
theorem alg_equiv.inv_fun_eq_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {e : A₁ ≃ₐ[R] A₂} :
@[simp]
theorem alg_equiv.symm_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
e.symm.symm = e
theorem alg_equiv.symm_bijective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
@[simp]
theorem alg_equiv.mk_coe' {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (f : A₂ → A₁) (h₁ : f) (h₂ : f) (h₃ : ∀ (x y : A₂), f (x * y) = f x * f y) (h₄ : ∀ (x y : A₂), f (x + y) = f x + f y) (h₅ : ∀ (r : R), f ( A₂) r) = A₁) r) :
{to_fun := f, inv_fun := e, left_inv := h₁, right_inv := h₂, map_mul' := h₃, map_add' := h₄, commutes' := h₅} = e.symm
@[simp]
theorem alg_equiv.symm_mk {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (f : A₁ → A₂) (f' : A₂ → A₁) (h₁ : f) (h₂ : f) (h₃ : ∀ (x y : A₁), f (x * y) = f x * f y) (h₄ : ∀ (x y : A₁), f (x + y) = f x + f y) (h₅ : ∀ (r : R), f ( A₁) r) = A₂) r) :
{to_fun := f, inv_fun := f', left_inv := h₁, right_inv := h₂, map_mul' := h₃, map_add' := h₄, commutes' := h₅}.symm = {to_fun := f', inv_fun := f, left_inv := _, right_inv := _, map_mul' := _, map_add' := _, commutes' := _}
@[simp]
theorem alg_equiv.refl_symm {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
@[trans]
def alg_equiv.trans {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] (e₁ : A₁ ≃ₐ[R] A₂) (e₂ : A₂ ≃ₐ[R] A₃) :
A₁ ≃ₐ[R] A₃
Algebra equivalences are transitive.
Equations
@[simp]
theorem alg_equiv.apply_symm_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x : A₂) :
e ((e.symm) x) = x
@[simp]
theorem alg_equiv.symm_apply_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x : A₁) :
(e.symm) (e x) = x
@[simp]
theorem alg_equiv.symm_trans_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] (e₁ : A₁ ≃ₐ[R] A₂) (e₂ : A₂ ≃ₐ[R] A₃) (x : A₃) :
((e₁.trans e₂).symm) x = (e₁.symm) ((e₂.symm) x)
@[simp]
theorem alg_equiv.coe_trans {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] (e₁ : A₁ ≃ₐ[R] A₂) (e₂ : A₂ ≃ₐ[R] A₃) :
(e₁.trans e₂) = e₂ e₁
@[simp]
theorem alg_equiv.trans_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] (e₁ : A₁ ≃ₐ[R] A₂) (e₂ : A₂ ≃ₐ[R] A₃) (x : A₁) :
(e₁.trans e₂) x = e₂ (e₁ x)
@[simp]
theorem alg_equiv.comp_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
e.comp (e.symm) = A₂
@[simp]
theorem alg_equiv.symm_comp {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
(e.symm).comp e = A₁
theorem alg_equiv.left_inverse_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
theorem alg_equiv.right_inverse_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
def alg_equiv.arrow_congr {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {A₁' : Type u_1} {A₂' : Type u_2} [semiring A₁'] [semiring A₂'] [ A₁'] [ A₂'] (e₁ : A₁ ≃ₐ[R] A₁') (e₂ : A₂ ≃ₐ[R] A₂') :
(A₁ →ₐ[R] A₂) (A₁' →ₐ[R] A₂')
If A₁ is equivalent to A₁' and A₂ is equivalent to A₂', then the type of maps A₁ →ₐ[R] A₂ is equivalent to the type of maps A₁' →ₐ[R] A₂'.
Equations
theorem alg_equiv.arrow_congr_comp {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] {A₁' : Type u_1} {A₂' : Type u_2} {A₃' : Type u_3} [semiring A₁'] [semiring A₂'] [semiring A₃'] [ A₁'] [ A₂'] [ A₃'] (e₁ : A₁ ≃ₐ[R] A₁') (e₂ : A₂ ≃ₐ[R] A₂') (e₃ : A₃ ≃ₐ[R] A₃') (f : A₁ →ₐ[R] A₂) (g : A₂ →ₐ[R] A₃) :
(e₁.arrow_congr e₃) (g.comp f) = ((e₂.arrow_congr e₃) g).comp ((e₁.arrow_congr e₂) f)
@[simp]
theorem alg_equiv.arrow_congr_refl {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
@[simp]
theorem alg_equiv.arrow_congr_trans {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] {A₁' : Type u_1} {A₂' : Type u_2} {A₃' : Type u_3} [semiring A₁'] [semiring A₂'] [semiring A₃'] [ A₁'] [ A₂'] [ A₃'] (e₁ : A₁ ≃ₐ[R] A₂) (e₁' : A₁' ≃ₐ[R] A₂') (e₂ : A₂ ≃ₐ[R] A₃) (e₂' : A₂' ≃ₐ[R] A₃') :
(e₁.trans e₂).arrow_congr (e₁'.trans e₂') = (e₁.arrow_congr e₁').trans (e₂.arrow_congr e₂')
@[simp]
theorem alg_equiv.arrow_congr_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {A₁' : Type u_1} {A₂' : Type u_2} [semiring A₁'] [semiring A₂'] [ A₁'] [ A₂'] (e₁ : A₁ ≃ₐ[R] A₁') (e₂ : A₂ ≃ₐ[R] A₂') :
(e₁.arrow_congr e₂).symm = e₁.symm.arrow_congr e₂.symm
def alg_equiv.of_alg_hom {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (f : A₁ →ₐ[R] A₂) (g : A₂ →ₐ[R] A₁) (h₁ : f.comp g = A₂) (h₂ : g.comp f = A₁) :
A₁ ≃ₐ[R] A₂
If an algebra morphism has an inverse, it is a algebra isomorphism.
Equations
theorem alg_equiv.coe_alg_hom_of_alg_hom {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (f : A₁ →ₐ[R] A₂) (g : A₂ →ₐ[R] A₁) (h₁ : f.comp g = A₂) (h₂ : g.comp f = A₁) :
h₁ h₂) = f
@[simp]
theorem alg_equiv.of_alg_hom_coe_alg_hom {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (f : A₁ ≃ₐ[R] A₂) (g : A₂ →ₐ[R] A₁) (h₁ : f.comp g = A₂) (h₂ : g.comp f = A₁) :
h₁ h₂ = f
theorem alg_equiv.of_alg_hom_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (f : A₁ →ₐ[R] A₂) (g : A₂ →ₐ[R] A₁) (h₁ : f.comp g = A₂) (h₂ : g.comp f = A₁) :
h₁ h₂).symm = h₂ h₁
noncomputable def alg_equiv.of_bijective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (f : A₁ →ₐ[R] A₂) (hf : function.bijective f) :
A₁ ≃ₐ[R] A₂
Promotes a bijective algebra homomorphism to an algebra equivalence.
Equations
@[simp]
theorem alg_equiv.coe_of_bijective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {f : A₁ →ₐ[R] A₂} {hf : function.bijective f} :
hf) = f
theorem alg_equiv.of_bijective_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] {f : A₁ →ₐ[R] A₂} {hf : function.bijective f} (a : A₁) :
hf) a = f a
@[simp]
theorem alg_equiv.to_linear_equiv_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (ᾰ : A₁) :
(e.to_linear_equiv) ᾰ = e ᾰ
def alg_equiv.to_linear_equiv {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
A₁ ≃ₗ[R] A₂
Forgetting the multiplicative structures, an equivalence of algebras is a linear equivalence.
Equations
@[simp]
theorem alg_equiv.to_linear_equiv_refl {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
@[simp]
theorem alg_equiv.to_linear_equiv_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[simp]
theorem alg_equiv.to_linear_equiv_trans {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] (e₁ : A₁ ≃ₐ[R] A₂) (e₂ : A₂ ≃ₐ[R] A₃) :
(e₁.trans e₂).to_linear_equiv =
theorem alg_equiv.to_linear_equiv_injective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
def alg_equiv.to_linear_map {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
A₁ →ₗ[R] A₂
Interpret an algebra equivalence as a linear map.
Equations
@[simp]
theorem alg_equiv.to_alg_hom_to_linear_map {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[simp]
theorem alg_equiv.to_linear_equiv_to_linear_map {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) :
@[simp]
theorem alg_equiv.to_linear_map_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x : A₁) :
theorem alg_equiv.to_linear_map_injective {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] :
@[simp]
theorem alg_equiv.trans_to_linear_map {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] (f : A₁ ≃ₐ[R] A₂) (g : A₂ ≃ₐ[R] A₃) :
@[simp]
theorem alg_equiv.of_linear_equiv_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (l : A₁ ≃ₗ[R] A₂) (map_mul : ∀ (x y : A₁), l (x * y) = l x * l y) (commutes : ∀ (r : R), l ( A₁) r) = A₂) r) (ᾰ : A₁) :
map_mul commutes) = l ᾰ
def alg_equiv.of_linear_equiv {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (l : A₁ ≃ₗ[R] A₂) (map_mul : ∀ (x y : A₁), l (x * y) = l x * l y) (commutes : ∀ (r : R), l ( A₁) r) = A₂) r) :
A₁ ≃ₐ[R] A₂
Upgrade a linear equivalence to an algebra equivalence, given that it distributes over multiplication and action of scalars.
Equations
@[simp]
theorem alg_equiv.of_linear_equiv_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (l : A₁ ≃ₗ[R] A₂) (map_mul : ∀ (x y : A₁), l (x * y) = l x * l y) (commutes : ∀ (r : R), l ( A₁) r) = A₂) r) :
map_mul commutes).symm =
@[simp]
theorem alg_equiv.of_linear_equiv_to_linear_equiv {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (map_mul : ∀ (x y : A₁), (e.to_linear_equiv) (x * y) = (e.to_linear_equiv) x * (e.to_linear_equiv) y) (commutes : ∀ (r : R), (e.to_linear_equiv) ( A₁) r) = A₂) r) :
map_mul commutes = e
@[simp]
theorem alg_equiv.to_linear_equiv_of_linear_equiv {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (l : A₁ ≃ₗ[R] A₂) (map_mul : ∀ (x y : A₁), l (x * y) = l x * l y) (commutes : ∀ (r : R), l ( A₁) r) = A₂) r) :
map_mul commutes).to_linear_equiv = l
theorem alg_equiv.aut_one {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
@[protected, instance]
def alg_equiv.aut {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
group (A₁ ≃ₐ[R] A₁)
Equations
theorem alg_equiv.aut_mul {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] (ϕ ψ : A₁ ≃ₐ[R] A₁) :
ϕ * ψ = ψ.trans ϕ
@[simp]
theorem alg_equiv.one_apply {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] (x : A₁) :
1 x = x
@[simp]
theorem alg_equiv.mul_apply {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] (e₁ e₂ : A₁ ≃ₐ[R] A₁) (x : A₁) :
(e₁ * e₂) x = e₁ (e₂ x)
def alg_equiv.aut_congr {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (ϕ : A₁ ≃ₐ[R] A₂) :
(A₁ ≃ₐ[R] A₁) ≃* A₂ ≃ₐ[R] A₂
An algebra isomorphism induces a group isomorphism between automorphism groups
Equations
@[simp]
theorem alg_equiv.aut_congr_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (ϕ : A₁ ≃ₐ[R] A₂) (ψ : A₁ ≃ₐ[R] A₁) :
(ϕ.aut_congr) ψ = ϕ.symm.trans (ψ.trans ϕ)
@[simp]
theorem alg_equiv.aut_congr_refl {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
@[simp]
theorem alg_equiv.aut_congr_symm {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (ϕ : A₁ ≃ₐ[R] A₂) :
@[simp]
theorem alg_equiv.aut_congr_trans {R : Type u} {A₁ : Type v} {A₂ : Type w} {A₃ : Type u₁} [semiring A₁] [semiring A₂] [semiring A₃] [ A₁] [ A₂] [ A₃] (ϕ : A₁ ≃ₐ[R] A₂) (ψ : A₂ ≃ₐ[R] A₃) :
= (ϕ.trans ψ).aut_congr
@[protected, instance]
def alg_equiv.apply_mul_semiring_action {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
mul_semiring_action (A₁ ≃ₐ[R] A₁) A₁
The tautological action by A₁ ≃ₐ[R] A₁ on A₁.
This generalizes function.End.apply_mul_action.
Equations
@[protected, simp]
theorem alg_equiv.smul_def {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] (f : A₁ ≃ₐ[R] A₁) (a : A₁) :
f a = f a
@[protected, instance]
def alg_equiv.apply_has_faithful_smul {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
has_faithful_smul (A₁ ≃ₐ[R] A₁) A₁
@[protected, instance]
def alg_equiv.apply_smul_comm_class {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
(A₁ ≃ₐ[R] A₁) A₁
@[protected, instance]
def alg_equiv.apply_smul_comm_class' {R : Type u} {A₁ : Type v} [semiring A₁] [ A₁] :
smul_comm_class (A₁ ≃ₐ[R] A₁) R A₁
@[simp]
theorem alg_equiv.algebra_map_eq_apply {R : Type u} {A₁ : Type v} {A₂ : Type w} [semiring A₁] [semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) {y : R} {x : A₁} :
A₂) y = e x A₁) y = x
theorem alg_equiv.map_prod {R : Type u} {A₁ : Type v} {A₂ : Type w} [comm_semiring A₁] [comm_semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) {ι : Type u_1} (f : ι → A₁) (s : finset ι) :
e (s.prod (λ (x : ι), f x)) = s.prod (λ (x : ι), e (f x))
theorem alg_equiv.map_finsupp_prod {R : Type u} {A₁ : Type v} {A₂ : Type w} [comm_semiring A₁] [comm_semiring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) {α : Type u_1} [has_zero α] {ι : Type u_2} (f : ι →₀ α) (g : ι → α → A₁) :
e (f.prod g) = f.prod (λ (i : ι) (a : α), e (g i a))
@[protected]
theorem alg_equiv.map_neg {R : Type u} {A₁ : Type v} {A₂ : Type w} [ring A₁] [ring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x : A₁) :
e (-x) = -e x
@[protected]
theorem alg_equiv.map_sub {R : Type u} {A₁ : Type v} {A₂ : Type w} [ring A₁] [ring A₂] [ A₁] [ A₂] (e : A₁ ≃ₐ[R] A₂) (x y : A₁) :
e (x - y) = e x - e y
def mul_semiring_action.to_alg_hom {M : Type u_1} (R : Type u_3) (A : Type u_4) [semiring A] [ A] [monoid M] [ A] [ A] (m : M) :
Each element of the monoid defines a algebra homomorphism.
This is a stronger version of mul_semiring_action.to_ring_hom and distrib_mul_action.to_linear_map.
Equations
@[simp]
theorem mul_semiring_action.to_alg_hom_apply {M : Type u_1} (R : Type u_3) (A : Type u_4) [semiring A] [ A] [monoid M] [ A] [ A] (m : M) (a : A) :
a = m a
theorem mul_semiring_action.to_alg_hom_injective {M : Type u_1} (R : Type u_3) (A : Type u_4) [semiring A] [ A] [monoid M] [ A] [ A] [ A] :
def mul_semiring_action.to_alg_equiv {G : Type u_2} (R : Type u_3) (A : Type u_4) [semiring A] [ A] [group G] [ A] [ A] (g : G) :
Each element of the group defines a algebra equivalence.
This is a stronger version of mul_semiring_action.to_ring_equiv and distrib_mul_action.to_linear_equiv.
Equations
@[simp]
theorem mul_semiring_action.to_alg_equiv_symm_apply {G : Type u_2} (R : Type u_3) (A : Type u_4) [semiring A] [ A] [group G] [ A] [ A] (g : G) (ᾰ : A) :
g).symm) =
@[simp]
theorem mul_semiring_action.to_alg_equiv_apply {G : Type u_2} (R : Type u_3) (A : Type u_4) [semiring A] [ A] [group G] [ A] [ A] (g : G) (ᾰ : A) :
=
theorem mul_semiring_action.to_alg_equiv_injective {G : Type u_2} (R : Type u_3) (A : Type u_4) [semiring A] [ A] [group G] [ A] [ A] [ A] :
@[protected, instance]
def algebra_nat {R : Type u_1} [semiring R] :
R
Semiring ⥤ ℕ-Alg
Equations
@[protected, instance]
def nat_algebra_subsingleton {R : Type u_1} [semiring R] :
def ring_hom.to_nat_alg_hom {R : Type u_1} {S : Type u_2} [semiring R] [semiring S] (f : R →+* S) :
Reinterpret a ring_hom as an -algebra homomorphism.
Equations
def ring_hom.to_int_alg_hom {R : Type u_1} {S : Type u_2} [ring R] [ring S] [ R] [ S] (f : R →+* S) :
Reinterpret a ring_hom as a -algebra homomorphism.
Equations
@[simp]
theorem ring_hom.map_rat_algebra_map {R : Type u_1} {S : Type u_2} [ring R] [ring S] [ R] [ S] (f : R →+* S) (r : ) :
f ( R) r) = S) r
def ring_hom.to_rat_alg_hom {R : Type u_1} {S : Type u_2} [ring R] [ring S] [ R] [ S] (f : R →+* S) :
Reinterpret a ring_hom as a -algebra homomorphism. This actually yields an equivalence, see ring_hom.equiv_rat_alg_hom.
Equations
@[simp]
theorem ring_hom.to_rat_alg_hom_to_ring_hom {R : Type u_1} {S : Type u_2} [ring R] [ring S] [ R] [ S] (f : R →+* S) : | 22,213 | 52,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-10 | latest | en | 0.821566 |
https://www.coursehero.com/file/5874189/styexam1ceg416wi10/ | 1,526,834,837,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863626.14/warc/CC-MAIN-20180520151124-20180520171124-00224.warc.gz | 710,567,967 | 51,457 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
styexam1ceg416wi10
# styexam1ceg416wi10 - CEG/MTH/416/416 Mid-Exam Study...
This preview shows page 1. Sign up to view the full content.
Page 1 of 1 2/2/2010 styexam1ceg416wi10.docx CEG/MTH/416/616 Winter 2010 CEG/MTH/416/416 Mid-Exam - Study Topics/Procedures February 4, 2010 Study Topics Some topics listed will be covered in lecture on February 2, 2010 Study class notes. Study Datta text: 1. Chapter 0 (all sections), 2. Chapter 1 (topics covered in lectures), 3. Chapter 3 (sect 3.1), 4. Chapter 5 (sect 5.1, 5.2), 5. Chapter 6 ( sect 6.1, 6.2, 6.4.1-6.4.4, 6.5.1, 6.5.2, 6.7.1-6.7.4, parts of 6.11) Be able to prove matrix identities using basic mathematics Solving linear equations - basic mathematics. Properties of determinants and matrices. Solving eigenvalue problems - basic mathematics. Trace and determinant properties of eigenvalues. Gerschgorin circle theorem for eigenvalue localization. Orthogonality of eigenvectors for real, symmetric matrices. Properties of special matrices such as the Hilbert matrix. Properties and application of permutation matrices. Properties of orthogonal and unitary matrices. Geometric interpretation of eigenvalues and eigenvectors.
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 592 | 2,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-22 | latest | en | 0.824912 |
https://www.jiskha.com/questions/22320/The-Randolphs-used-12-more-gal-of-fuel-oil-in-October-than-in-September-and-twice | 1,571,285,709,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00403.warc.gz | 956,317,523 | 4,903 | # math
The Randolphs used 12 more gal of fuel oil in October
than in September and twice as much oil in November as in September. If they
used 132 gal for the 3 months, how much was used during each month?
Let x = the gallons of fuel oil used in September, and solve this equation.
x + (x + 12) + 2x = 132
1. 👍 0
2. 👎 0
3. 👁 68
## Similar Questions
1. ### math
The Randolph’s used 12 more gal of fuel oil in October than in September and twice as much oil in November as in September. If they used 132 gal for the 3 months, how much was used during each month? I can solve the problem but
asked by Millah on September 29, 2007
2. ### math
The Randolph’s used 12 more gal of fuel oil in October than in September and twice as much oil in November as in September. If they used 132 gal for 3 months, how much was used during each month? Let the usage in the three
asked by cheryl on March 17, 2007
3. ### MATH, ALGEBRA
cAN SOMEONE JUST SHOW ME HOW TO GET THE EQUATION IN ORDER TO SOLVE IT MYSELF. i'LL APPRECIATED. bUSINESS AND FINANCE. The Randolphs used 12 more gal of fuel oil in October than in September and twice as much oil in November as in
asked by jasmine20 on December 18, 2006
4. ### calculus
An automobile computer gives a digital readout of fuel consumption in gallons per hous. During a trip, a passenger recorded the fuel consumption every 5 minutes for a full hour of travel. time: 0 / gal/h: 2.5 time: 5 / gal/h: 2.4
asked by Anonymous on October 26, 2007
5. ### math
LPG (liquedfied petroleum gas) is stored in highly insulated spherical vessels in most oil refineries. If the inside diameter of one of these tanks is 11.4ft, find it's volume. Report your answer to the nearest one-tenth of a gal
asked by Anonymous on October 6, 2010
6. ### College Math
LPG is stored in highly insulated spherical vessels in most oil refineries. If the inside diameter of one of these tanks is 11.4 ft, find it's volume. Report your answer to the nearest one-tenth of a gal. Use pie-3014, and 1cubic
asked by Carol on August 31, 2012
7. ### accounting
Cindy Lou makes \$30,000 per year as an exempt employee. She worked 15 hours of overtime in the month of September. Payday for wages earned in the month of September is October 7. Ignoring Federal and State income tax withholdings,
asked by Anonymous on April 21, 2013
8. ### accounting
2. Cindy Lou makes \$30,000 per year as an exempt employee. She worked 15 hours of overtime in the month of September. Payday for wages earned in the month of September is October 7. Ignoring Federal and State income tax
asked by Anonymous on April 20, 2013
9. ### math
An oil tank has a capacity of 45 gallons.At the beginning of October it is 80% full.At the end of October it is 1/3 full.How many gallons of oil were used in October? a. 21 b. 25 c. 41 d. 27 e. 30 my answer is (e)30, 45 x 1/3 =15
asked by vincent on October 1, 2014
10. ### algebra
given two containers, one holding 4 gal. and the other holding 3 gal. how does one measure out two gal.? Fill the 3 gal, pour it in the 4 gal. Fill the 3 gal, then pout it int the 4 gal until filled. How much is left in the 3 gal?
asked by Ethan on April 26, 2007
More Similar Questions | 914 | 3,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.97683 |
https://www.proprofs.com/quiz-school/story.php?title=quiz-newtons-laws-motion-fbds | 1,722,952,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00843.warc.gz | 758,064,182 | 108,683 | # Newton's Laws Of Motion And FBDs Quiz
Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By Junho Song
Junho Song, Physics/Science Teacher
Junho Song, a dedicated educator, specializes in teaching physics and science. Passionate about inspiring students in the fascinating world of scientific discovery and understanding.
Quizzes Created: 11 | Total Attempts: 93,426
Questions: 18 | Attempts: 7,656
Settings
Have you studied Newton's laws of motion? If yes, then it's time for you to test your knowledge with this sample Newton's Laws of Motion and FBDs (free-body diagrams) Quiz. Here, we'll ask you questions related to Newton's theory and see how well you understand this topic in this quiz. If you score more than 80% marks on this quiz, it means that you have passed the test. So, do you think you can crack this quiz? Let's find out!
• 1.
### When analyzing dynamics problems, free-body diagrams
• A.
Should always be used
• B.
Are more useful when analyzing horizontal forces than when analyzing vertical forces
• C.
Should include only the forces that are directly responsible for the acceleration
• D.
Should be used only when objects are accelerating
• E.
Only apply to objects in equilibrium
A. Should always be used
Explanation
Free-body diagrams should always be used when analyzing dynamics problems because they provide a visual representation of all the forces acting on an object. By including all the forces, both horizontal and vertical, the diagram helps to accurately analyze the motion and acceleration of the object. It is important to consider all the forces that are directly responsible for the acceleration, rather than including unrelated forces. Free-body diagrams are not limited to objects that are accelerating or in equilibrium, but are used in all dynamic situations.
Rate this question:
• 2.
### A force of 12 N acting in a direction [30ºE of S] is equivalent to which of the following pairs of forces acting simultaneously?
• A.
24 N [S], 14 N [E]
• B.
14 N [S], 24 N [E]
• C.
12 N [S], 12 N [E]
• D.
6 N [S], 10 N [E]
• E.
10 N [S], 6 N [E]
E. 10 N [S], 6 N [E]
Explanation
The correct answer is 10 N [S], 6 N [E]. This is because the given force of 12 N acting in a direction [30ºE of S] can be resolved into two components: one along the south direction and one along the east direction. By using trigonometry, the south component is found to be 10 N and the east component is found to be 6 N. Therefore, the equivalent pair of forces is 10 N [S], 6 N [E].
Rate this question:
• 3.
### A hockey puck slides along an ice surface shortly after it has left the hockey stick that propelled it. Which of the following free-body diagrams best represents the hockey puck?
• A.
A
• B.
B
• C.
C
• D.
D
• E.
E
D. D
Explanation
The correct free-body diagram for the hockey puck would show two forces acting on it: the force of gravity pulling it downwards and the force of friction opposing its motion along the ice surface. Diagram D accurately represents these two forces, with the force of gravity acting downwards and the force of friction acting opposite to the direction of the puck's motion.
Rate this question:
• 4.
### Study the free-body diagram below and choose the statement that best describes the object's motion.
• A.
The object will accelerate north.
• B.
The object will be motionless.
• C.
The object will accelerate northeast.
• D.
The object will travel with uniform motion.
• E.
The object will travel north with a constant velocity.
C. The object will accelerate northeast.
Explanation
The free-body diagram shows that there is a net force acting on the object in the northeast direction. Since there is a net force, Newton's second law states that the object will accelerate in the direction of the net force. Therefore, the object will accelerate northeast.
Rate this question:
• 5.
### Study the free-body diagram below and determine what additional force(s) would be required for the object to achieve uniform motion.
• A.
1 N [W]
• B.
1 N [E]
• C.
2 N [N] and 1 N [W]
• D.
2 N [S] and 1 N [E]
• E.
2 N [S] and 1 N [W]
E. 2 N [S] and 1 N [W]
Explanation
The free-body diagram shows a single force of 1 N in the west direction acting on the object. In order for the object to achieve uniform motion, there must be an additional force acting on it. The additional force should have a magnitude of 2 N and should be directed towards the south (S) direction. Additionally, there should be another force of 1 N directed towards the west (W) direction. These two additional forces, 2 N [S] and 1 N [W], when combined with the existing 1 N [W] force, will result in a net force of zero, allowing the object to achieve uniform motion.
Rate this question:
• 6.
### The free-body diagram below illustrates the forces acting on a 2.0-kg object as it is being pushed along a horizontal surface. What is the motion of the object?
• A.
Moving at 5.3 m/s2
• B.
Moving at 3.1 m/s
• C.
Moving at 5.3 m/s
• D.
Moving at 3.1 m/s2
• E.
Not moving
D. Moving at 3.1 m/s2
Explanation
The object is moving at 3.1 m/s^2 because there is an unbalanced force acting on it. The force of friction is opposing the force applied to the object, causing it to accelerate in the direction of the applied force. The acceleration can be calculated using Newton's second law, F = ma, where F is the net force and m is the mass of the object. Since the net force is non-zero, the object is accelerating and therefore moving.
Rate this question:
• 7.
### The free-body diagram of a 4.0-kg object is shown below. What additional force must act so that the object has an acceleration of 2.5 m/s2 [W]?
• A.
14.0 N [W]
• B.
10.0 N [W]
• C.
8.0 N [W]
• D.
6.0 N [W]
• E.
4.0 N [W]
C. 8.0 N [W]
Explanation
The free-body diagram shows that the object has a weight force of 14.0 N acting downwards. In order for the object to have an acceleration of 2.5 m/s2 in the westward direction, an additional force of equal magnitude but in the westward direction is required. The only option that matches this requirement is 8.0 N [W]. This force will counteract the friction and other forces acting on the object, allowing it to accelerate in the desired direction.
Rate this question:
• 8.
### Two identical arrows, A and B, are fired with different bows. The bow that fires arrow A exerts twice the average force as the bow that fires arrow B. Compare the accelerations of the two arrows.
• A.
Arrow B will have twice the acceleration of arrow A.
• B.
Arrow A will have twice the acceleration of arrow B.
• C.
Arrow A and arrow B will have the same acceleration.
• D.
Arrow B will have four times the acceleration of arrow A.
• E.
Arrow A will have four times the acceleration of arrow B.
B. Arrow A will have twice the acceleration of arrow B.
Explanation
The acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. Since both arrows have the same mass, the arrow fired with the bow exerting twice the average force will experience twice the acceleration. Therefore, Arrow A will have twice the acceleration of Arrow B.
Rate this question:
• 9.
### A hockey puck of mass 150 g is sliding south along the ice and slows at a rate of 1.2 m/s2. What is the net force acting on the puck?
• A.
1.8 x 102 N [N]
• B.
1.8 x 102 N [S]
• C.
1.8 x 101 N [N]
• D.
1.8 x 10-1 N [N]
• E.
1.9 x 10-1 N [S]
D. 1.8 x 10-1 N [N]
Explanation
The net force acting on the puck can be calculated using Newton's second law, which states that force is equal to mass multiplied by acceleration. In this case, the mass of the puck is given as 150 g (or 0.15 kg) and the acceleration is given as 1.2 m/s^2. By multiplying these values together, we get a net force of 0.18 N. The direction of the force is north, as indicated by the [N] in the answer. Therefore, the correct answer is 1.8 x 10-1 N [N].
Rate this question:
• 10.
### Newton's third law essentially states
• A.
Objects won't move unless pushed
• B.
Acceleration only occurs if there is net force
• C.
The acceleration of an object depends on its mass and the net force acting on it
• D.
Objects which are moving tend to stay moving
• E.
Forces always occur in pairs
E. Forces always occur in pairs
Explanation
Newton's third law states that forces always occur in pairs. This means that for every action, there is an equal and opposite reaction. When one object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object. This principle is fundamental in understanding the interactions between objects and is applicable in various scenarios, such as the recoil of a gun when a bullet is fired or the propulsion of a rocket through the expulsion of exhaust gases.
Rate this question:
• 11.
### Which of the following observations would be adequately explained by Newton's third law?
• A.
When turning a corner, a passenger in a car is pushed against the door.
• B.
A skater stands on a frictionless pond, tosses her bag in one direction and she accelerates in the opposite direction.
• C.
The normal force on an object is always equal to the force of gravity acting on the same object.
• D.
The friction of a surface causes a sliding object to come to rest.
• E.
When firing a rifle, the force of the rifle on the bullet, is greater than the force of the bullet back on the rifle.
B. A skater stands on a frictionless pond, tosses her bag in one direction and she accelerates in the opposite direction.
Explanation
When the skater tosses her bag in one direction, according to Newton's third law, there will be an equal and opposite reaction. This means that the skater will experience a force in the opposite direction, causing her to accelerate in the opposite direction. This observation can be adequately explained by Newton's third law, which states that for every action, there is an equal and opposite reaction.
Rate this question:
• 12.
### An object is pushed horizontally at a constant velocity. What can correctly be said about the forces acting on the object?
• A.
The force(s) acting forward is/are greater than the force(s) acting backward.
• B.
The sum of all forces is zero.
• C.
The sum of all forces has a value directed forward.
• D.
The forces acting on the object can be said to be “unbalanced.”
• E.
Newton’s second law best summarizes the effect of the forces acting on the object.
B. The sum of all forces is zero.
Explanation
When an object is pushed horizontally at a constant velocity, it means that the object is experiencing a balanced force. According to Newton's first law of motion, an object at rest or in motion will remain in that state unless acted upon by an external force. In this case, the constant velocity indicates that the forces acting forward and backward are equal in magnitude and opposite in direction, resulting in a net force of zero. Therefore, the correct answer is that the sum of all forces is zero.
Rate this question:
• 13.
### Which of the following units is equivalent to a newton (N)?
• A.
Kg·m/s2
• B.
G·cm/s
• C.
Kg·s2/m
• D.
Kg·m/s
• E.
Kg·cm/s2
A. Kg·m/s2
Explanation
The unit kg·m/s2 is equivalent to a newton (N) because it represents the formula for force, which is mass (kg) multiplied by acceleration (m/s2). A newton is defined as the force required to accelerate a mass of one kilogram at a rate of one meter per second squared.
Rate this question:
• 14.
### Three masses are suspended vertically as shown in the diagram below. The system is accelerating upward. What is the relationship among the forces of tension?
• A.
|FT1| = |FT2| = |FT3|
• B.
|FT1| > |FT2| > |FT3|
• C.
|FT1| < |FT2| < |FT3|
• D.
|FT1| > |FT2| < |FT3|
• E.
|FT1| > |FT3| < |FT2|
B. |FT1| > |FT2| > |FT3|
Explanation
The correct answer is |FT1| > |FT2| > |FT3|. In a system with multiple masses suspended vertically, the tension force decreases as you move down the system. This is because the force of gravity acting on each mass increases as you move down, and the tension force must counteract this force. Therefore, the top mass (FT1) experiences the greatest tension force, followed by the middle mass (FT2), and finally the bottom mass (FT3).
Rate this question:
• 15.
### The physical quantity, which is the measure of inertia, is
• A.
Density
• B.
Weight
• C.
Force
• D.
Acceleration
• E.
Mass
E. Mass
Explanation
Mass is the physical quantity that measures inertia. Inertia refers to an object's resistance to changes in its motion. The greater the mass of an object, the greater its inertia, meaning it is more difficult to change its state of motion. Therefore, mass is the correct answer as it directly relates to the measure of inertia.
Rate this question:
• 16.
### Which law is in control of a spaceship which cruises through space at constant speed without using any fuel?
• A.
Newton's first law
• B.
Newton's second law
• C.
Newton's third law
• D.
Both A and B
• E.
All of the above
A. Newton's first law
Explanation
Newton's first law, also known as the law of inertia, states that an object at rest will stay at rest and an object in motion will stay in motion with the same speed and direction, unless acted upon by an external force. In the case of a spaceship cruising through space at a constant speed without using any fuel, there are no external forces acting on it to change its state of motion. Therefore, the spaceship follows Newton's first law of motion.
Rate this question:
• 17.
### An object is hanging by two strings, A and B, anchored to the ceiling as in the diagram below. Each string holds the object in place with a force of 40 N at 30 degrees to the vertical. Determine the force of gravity required to keep the object motionless.
• A.
35 N [down]
• B.
46 N [down]
• C.
69 N [down]
• D.
80 N [down]
• E.
92 N [down]
C. 69 N [down]
Explanation
The object is being held in place by two strings, each exerting a force of 40 N at 30 degrees to the vertical. To keep the object motionless, the force of gravity must be equal to the sum of the forces exerted by the strings. Since the forces are at an angle, we can use trigonometry to find the vertical component of each force. The vertical component of each force is 40 N * sin(30°) = 20 N. Therefore, the total force of gravity required to keep the object motionless is 20 N + 20 N = 40 N. However, the answer choices are given in the downward direction, so the correct answer is 69 N [down].
Rate this question:
• 18.
### Describe the motion of the 2.0 kg mass below in the horizontal plane.
• A.
The object accelerates at 1.5 m/s2 [right]
• B.
The object accelerates at 11 m/s2 [right]
• C.
The object accelerates at 12 m/s2 [right]
• D.
The object accelerates at 13 m/s2 [right]
• E.
The object does not accelerate
A. The object accelerates at 1.5 m/s2 [right]
Explanation
The object accelerates at 1.5 m/s2 [right], indicating that there is a net force acting on the object in the right direction. This means that the object's velocity is increasing at a rate of 1.5 m/s2 in the right direction.
Rate this question:
Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
• Current Version
• Dec 05, 2023
Quiz Edited by
ProProfs Editorial Team
• Oct 03, 2009
Quiz Created by
Junho Song
Related Topics | 3,957 | 16,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-33 | latest | en | 0.915316 |
https://www.askiitians.com/forums/Algebra/22/4273/complex-numbers.htm | 1,708,898,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474643.29/warc/CC-MAIN-20240225203035-20240225233035-00347.warc.gz | 652,245,803 | 42,969 | # solve for x :x7 = 1{ the answer is not only = 1, it probably has 6 other values }
24 Points
14 years ago
Hi Anthony
1 = cos2pi + isin2pi
11/7 = cos(2rpi/7) + isin(2rpi/7)
where
r = 1,2,3,4,5,6,7
This was derived using Euler's Formula
If you want you can join Askiitians.com's Online IITJEE Coaching Sessions.
For more details, visit the Packages Section of Askiitians.com
Regards,
Rajat
anthony rebello
8 Points
14 years ago
thnx for the reply, but i hav a doubt :
why cant 1=cos (0) + i sin (0) | 180 | 514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.79807 |
https://electricalengineeringmcq.com/the-value-of-the-demand-factor-is/ | 1,723,709,049,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00615.warc.gz | 170,117,336 | 16,519 | # The value of the Demand Factor is – Power Generation MCQ#59
The value of the Demand Factor is generally:
1. Less than 1
2. Equal to 1
3. More than 1
4. None of these
Correct answer: The value of the Demand Factor is less than 1.
The demand factor is mathematically defined as the ratio of maximum demand and connected load.
Demand Factor = Maximum demand/Connected
Since the maximum demand is always smaller than the connected load in practical cases, the value of the demand factor is always less than 1. | 122 | 513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-33 | latest | en | 0.900834 |
http://openstudy.com/updates/560e90a6e4b0fa7de699690b | 1,495,654,900,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607860.7/warc/CC-MAIN-20170524192431-20170524212431-00491.warc.gz | 265,666,567 | 9,381 | • Virnios
Brad wants to buy flowers for his friend with $33. The daises are$1 each and the roses are \$2 each. He buys 3 more daises than roses. How much did the roses cost? (I just need the equation and variable to solve for)
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Looking for something else?
Not the answer you are looking for? Search for more explanations. | 305 | 1,156 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-22 | longest | en | 0.415952 |
https://www.jiskha.com/display.cgi?id=1371910792 | 1,503,114,862,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105297.38/warc/CC-MAIN-20170819031734-20170819051734-00187.warc.gz | 916,610,010 | 3,838 | # math
posted by .
Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a discount rate of 14.5%. The note was made on March 21. What was the maturity date of the note?
• math -
14. Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the Grove Isle Bank at a discount rate of 14.5%. The note was made on March 21. What was the maturity date of the note? (Points : 2)
15. Using the scenario from the previous question, calculate the maturity value of the note. (Points : 2)
16. What was the discount date of the note from the previous question? (Points : 2)
• math -
may 20th
## Similar Questions
1. ### Math
Lisa borrowed 8,000 on an 8% , 60 day note after 15 days she paid 2,000 on the note on day 45 she paid 1,000 on the note what is the total interest and ending balance due by the us rule of ordinary interest.
2. ### math
Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the …
3. ### math
Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the …
Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the …
5. ### contemporary math
Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the …
6. ### choice
Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the …
7. ### finance math
Schell Publishing received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers. After 20 days, Schell discounted the note at the bank at a discount rate of 14.5%. The note was made on March 21. …
8. ### math
Varsity Press, a publisher of college textbooks, received a \$70,000 promissory note at 12% ordinary interest for 60 days from one of its customers, Reader’s Choice Bookstores. After 20 days, Varsity Press discounted the note at the …
9. ### math
Use the ordinary interest method, 360 days, to solve the following word problem. Round to the nearest cent when necessary. Pinnacle Manufacturing received a \$40,000 promissory note at 12% simple interest for 95 days from one of its …
10. ### math
Webster Digital received a promissory note of \$8,000 for 9 months at 7% simple interest from one of its customers. After 4 months, the note was discounted at Bank of Aventura at a discount rate of 10%. What are the proceeds Webster …
More Similar Questions | 866 | 3,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-34 | latest | en | 0.920619 |
https://www.scienceforums.net/topic/118740-black-hole-why-do-we-believe-that-matter-could-be-such-dense/?page=2&tab=comments | 1,558,769,078,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257920.67/warc/CC-MAIN-20190525064654-20190525090654-00102.warc.gz | 931,493,089 | 45,725 | # Black Hole: Why do we believe that matter could be such dense?
## Recommended Posts
22 hours ago, MaximT said:
Is a sum of mass orbiting a Black Hole, could contribute to the gravity force maintaining the rotation in orbit around the center of the galaxy, to make that galaxy objects spin around the center?
Is that sum of mass (all orbiting objects) could reduced the theoretical mass of the Black Hole itself, creating a empty, cannot be seen, zone around it, at the place of dilating it's volume to achieved the same density?
Can objects orbit a common center of gravity? Yes. Can this explain or effectively reduce the predicted mass of the central black hole? No. It's not just the orbital speeds involved, but the shapes of the orbits. When you look at the plot of the orbits of those stars you will note that since they are very elliptical, sometimes a given star will be closer to the center than other stars and sometimes further away. Any star that is further from the center than you are will not contribute to the force you feel pulling you towards the center.* By plotting the shape of the orbit as well as its speed at different points, you can calculate just how much of the total mass of the entire system has to be actually be located at the center. This is what gives you the mass of the BH.
* An extreme example of this would be a spherical cloud of stars with a hollow at it center. For any star in that cloud its orbit is determined by the stars as close or close to the center than it is, the other stars have no effect. An object that wandered into the central hollow, would be behave as if there were no stars surrounding the hollow at all.
##### Share on other sites
Posted (edited)
On 4/13/2019 at 4:46 PM, MigL said:
People are often skeptical of Black Holes because they cannot understand how matter can be compressed to such a high degree in the central singularity ( near zero size ) to reach ( almost ? ) infinite density.
One has to remember that elementary particles, like leptons, are treated as if they have zero size, and only other properties, like the exclusion principle and the statistics they obey, keep an infinity of them from being stacked on top of each other.
As to mass, it is a property of these leptons, not a thing onto itself, and in a Black Hole, this property is conveniently left with the Event Horizon ( along with charge, angular momentum and in a non-classical treatment, entropy ). The EH is the mathematical construct where pre-collapse properties are stored, not the central singularity.
So we have a dimensionless singularity where a large number of dimensionless particles, that no longer have the property of mass, reside.
Is there actually a problem there ?
If the particles have lost the property of mass, where does gravity come from? Doesn't that mean that gravity is something transcendental?
And second question: if space is so much curved in the BH that straight lines become curved, how can we deduce volume & density? Isn't it an error to measure it from our FoR as it is was a ball in regular space?
And 3rd question: from the diagrams as explained in the video I understand that an observer that is inside the BH must have be able to observe events happening outside the BH (since the rays of light are entering the BH) In this case, how can he figure out (measuring) that he is inside a BH or not?
Edited by michel123456
##### Share on other sites
42 minutes ago, michel123456 said:
If the particles have lost the property of mass, where does gravity come from?
They have probably lost the property of being particles, as well.
The mass of the black hole, like anything else, is defined by the space-time curvature associated with it.
43 minutes ago, michel123456 said:
And second question: if space is so much curved in the BH that straight lines become curved, how can we deduce volume & density? Isn't it an error to measure it from our FoR as it is was a ball in regular space?
Yes. The volume, and even the radius, of a black hole is not a well defined property. We know the surface area. From that, we can use normal geometry to deduce values we call radius and volume. But they don't mean anything when it comes to the interior of a black hole. For example, what we call the radius is time.
However, as there are no other words (beyond the mathematics of GR) to describe it, these are still useful analogies.
45 minutes ago, michel123456 said:
And 3rd question: from the diagrams as explained in the video I understand that an observer that is inside the BH must have be able to observe events happening outside the BH (since the rays of light are entering the BH) In this case, how can he figure out (measuring) that he is inside a BH or not?
They can't. If you were to fall into a black hole, the only way you would know you had crossed the event horizon would be if you could do the necessary calculations to know when you would pass it.
##### Share on other sites
Posted (edited)
1 hour ago, michel123456 said:
If the particles have lost the property of mass, where does gravity come from? Doesn't that mean that gravity is something transcendental?
The gravity is a fossil field from the star whence the BH formed from, so is already there..
Quote
And second question: if space is so much curved in the BH that straight lines become curved, how can we deduce volume & density? Isn't it an error to measure it from our FoR as it is was a ball in regular space?
The volume is deduced from the EH and the density is actually meaningless.
Quote
And 3rd question: from the diagrams as explained in the video I understand that an observer that is inside the BH must have be able to observe events happening outside the BH (since the rays of light are entering the BH) In this case, how can he figure out (measuring) that he is inside a BH or not?
In fact any observer inside the BH, as long as tidal gravitational effects have not torn the observer asunder, would actually probably see the whole universe in the form of a circle above his head, due to lensing. I dare say he would be ripped apart before he had any inkling he had crossed a point of no return.
Edited by beecee
##### Share on other sites
2 hours ago, beecee said:
In fact any observer inside the BH, as long as tidal gravitational effects have not torn the observer asunder, would actually probably see the whole universe in the form of a circle above his head, due to lensing. I dare say he would be ripped apart before he had any inkling he had crossed a point of no return.
Fact?
##### Share on other sites
Posted (edited)
5 hours ago, Strange said:
They can't. If you were to fall into a black hole, the only way you would know you had crossed the event horizon would be if you could do the necessary calculations to know when you would pass it.
But from this diagram below I understand that the rays of light enter inside the BH.
And in the BH, space is intensely curved, but for any observer living in it I suppose Space is straight, light rays are not curved, and time is different. I suppose that for an internal observer things are just as usual. Why would it be different? With light going at C as for us.
Edited by michel123456
##### Share on other sites
13 minutes ago, michel123456 said:
But from this diagram below I understand that the rays of light enter inside the BH.
And in the BH, space is intensely curved, but for any observer living in it I suppose Space is straight, light rays are not curved, and time is different. I suppose that for an internal observer things are just as usual. Why would it be different? With light going at C as for us.
We already live in curved spacetime. The main effect we notice is that thing we call "gravity". With sensitive instruments we can measure some other effects such as gravitational red-shift or lensing.
These effects would all be greater as we approach (and fall into) a black hole: the force of gravity (and tidal forces) would be greater, gravitational red shift would be greater (the rest of the universe would look increasingly blue-shifted), gravitational lensing would be obvious (the event horizon would appear to be 2.6 times larger than expected; the entire accretion disk would be visible, including that on the far-side of the event horizon; our view of the universe would narrow; etc)
##### Share on other sites
1 hour ago, michel123456 said:
But from this diagram below I understand that the rays of light enter inside the BH.
And in the BH, space is intensely curved, but for any observer living in it I suppose Space is straight, light rays are not curved, and time is different. I suppose that for an internal observer things are just as usual. Why would it be different? With light going at C as for us.
One of the bizarre things that occurs after crossing the the event horizon is that time and space switch roles. So describing what you would "see" is a bit difficult. For example, outside of a black hole, if we are looking at a point 1 light hr away, we can only see, at any given moment, events that occurred 1 hr ago. Inside the event horizon, If you are looking at a point further out from the center than you are, at the same 1 light hr away, you would see everything that occurs at that point between 1 hr in the past to 1 hr in the future, all at once.
##### Share on other sites
4 hours ago, Janus said:
One of the bizarre things that occurs after crossing the the event horizon is that time and space switch roles. So describing what you would "see" is a bit difficult.
Yes, most descriptions gloss over this. For perhaps obvious reasons. It is hard to comprehend how that might appear to our senses and we will probably never know.
##### Share on other sites
Posted (edited)
9 hours ago, et pet said:
Fact?
7 hours ago, michel123456 said:
But from this diagram below I understand that the rays of light enter inside the BH.
And in the BH, space is intensely curved, but for any observer living in it I suppose Space is straight, light rays are not curved, and time is different. I suppose that for an internal observer things are just as usual. Why would it be different? With light going at C as for us.
The EH is not any physical barrier, but simply a point/circumference in spacetime around the BH, where escape velocity equals and then exceeds "c". As mentioned before, other the tidal gravitational effects [that depend on the size of the BH] we have no indication of crossing anything.
4 hours ago, Strange said:
Yes, most descriptions gloss over this. For perhaps obvious reasons. It is hard to comprehend how that might appear to our senses and we will probably never know.
I'll drink to that.
Edited by beecee
##### Share on other sites
5 hours ago, Strange said:
Yes, most descriptions gloss over this. For perhaps obvious reasons. It is hard to comprehend how that might appear to our senses and we will probably never know.
This seems to conflict with this "fact" that was Posted earlier?
12 hours ago, beecee said:
In fact any observer inside the BH, as long as tidal gravitational effects have not torn the observer asunder, would actually probably see the whole universe in the form of a circle above his head, due to lensing. I dare say he would be ripped apart before he had any inkling he had crossed a point of no return
##### Share on other sites
2 minutes ago, et pet said:
This seems to conflict with this "fact" that was Posted earlier?
I don't think so. We do "know" (based on what GR tells us) how things would look to an observer falling in. We can even run simulations, such as raytracing, to generate images that represent what we see.
Whether the change of the radial direction from being spatial to being temporal is something we would be aware of or not, is not really known. Simulating what something looks like doesn't tell us anything about how it would feel
But it does explain why you still would not be able to see the singularity (or whatever is at the centre of the black hole) even after you pass the event horizon: because it is in your future and, as we all know, you can't see the future!
BTW. People often use words like "know" or "fact" to describe our best current understanding based on theory and available evidence. But, as with all science, this knowledge and these facts are, of course, provisional and subject to change with further evidence or new theories.
##### Share on other sites
Posted (edited)
27 minutes ago, et pet said:
This seems to conflict with this "fact" that was Posted earlier?
You fail to recognise that what I have posted aligns with GR predictions, which obviously we all agree has a pretty good track record. You also fail to see that in my statement I also used the word "probably" thus, "In fact any observer inside the BH, as long as tidal gravitational effects have not torn the observer asunder, would actually "probably" see the whole universe in the form of a circle above his head, due to lensing. I dare say he would be ripped apart before he had any inkling he had crossed a point of no return"
Plus of course the obvious that you seemed to have missed, in that Strange's statement was referring to the time and space changing places, while my statement was in relation to what anyone inside the EH would see outside the EH. And even in that scenario, facts may aspire to prevent you seeing anything outside....[1]Surviving tidal gravity effects, [2] Surviving any infalling matter/energy which would probably be in the form of lethal radiation, [3] and simply surviving longe enough, eg: Even our own SMBH, one would only have around 12 seconds or so, from the trip from the EH to the Singularity.
Couple all that with other contributions of mine and others, that clearly state that "while we certainly can never observe or have direct evidence of anything about what happens inside a BH, we do have the overwhelmingly supportive and incredibly correct predicitive powers of GR, to at least form a reasonable picture of that which will never be seen.
Hope that helps.
Edited by beecee
##### Share on other sites
24 minutes ago, beecee said:
You fail to recognise that what I have posted aligns with GR predictions, which obviously we all agree has a pretty good track record. You also fail to see that in my statement I also used the word "probably" thus, "In fact any observer inside the BH, as long as tidal gravitational effects have not torn the observer asunder, would actually "probably" see the whole universe in the form of a circle above his head, due to lensing. I dare say he would be ripped apart before he had any inkling he had crossed a point of no return"
Plus of course the obvious that you seemed to have missed, in that Strange's statement was referring to the time and space changing places, while my statement was in relation to what anyone inside the EH would see outside the EH. And even in that scenario, facts may aspire to prevent you seeing anything outside....[1]Surviving tidal gravity effects, [2] Surviving any infalling matter/energy which would probably be in the form of lethal radiation, [3] and simply surviving longe enough, eg: Even our own SMBH, one would only have around 12 seconds or so, from the trip from the EH to the Singularity.
Couple all that with other contributions of mine and others, that clearly state that "while we certainly can never observe or have direct evidence of anything about what happens inside a BH, we do have the overwhelmingly supportive and incredibly correct predicitive powers of GR, to at least form a reasonable picture of that which will never be seen.
Hope that helps.
Digging yourself deeper.
Maybe you should have began your statement : According to Theory, or my interpretation of the Theory of what one might see
BTW : the Link you provided : https://jila.colorado.edu/~ajsh/insidebh/index.html , contains this disclaimer " The creation of this website was supported by the National Science Foundation. Statements made herein are those of the author, and do not necessarily reflect the views of the National Science Foundation or of any other entity."
re, your : "Hope that helps." , "facts may aspire" means exactly what?
##### Share on other sites
Posted (edited)
5 minutes ago, et pet said:
Digging yourself deeper.
Maybe you should have began your statement : According to Theory, or my interpretation of the Theory of what one might see
BTW : the Link you provided : https://jila.colorado.edu/~ajsh/insidebh/index.html , contains this disclaimer " The creation of this website was supported by the National Science Foundation. Statements made herein are those of the author, and do not necessarily reflect the views of the National Science Foundation or of any other entity."
re, your : "Hope that helps." , "facts may aspire" means exactly what?
I'll let you do the digging, OK? At this time, I'm happy with what I have said and ignore your pedant and ignorance, but if any of our on line professionals object or find fault in what I have said, then I'll humbly listen as always. And "theory"of course is our best estimation as to what is happening and what we see and interpret at any particular time.
19 minutes ago, et pet said:
re, your : "Hope that helps." , "facts may aspire" means exactly what?
Well looking at the whole context of what I said, and ignoring pedantic nonsense, I believe it entirely comprehensible "And even in that scenario, facts may aspire to prevent you seeing anything outside....[1]Surviving tidal gravity effects, [2] Surviving any infalling matter/energy which would probably be in the form of lethal radiation, [3] and simply surviving longe enough, eg: Even our own SMBH, one would only have around 12 seconds or so, from the trip from the EH to the Singularity". Self explanatory I suggest.
Edited by beecee
##### Share on other sites
54 minutes ago, beecee said:
I'll let you do the digging, OK? At this time, I'm happy with what I have said and ignore your pedant and ignorance, but if any of our on line professionals object or find fault in what I have said, then I'll humbly listen as always. And "theory"of course is our best estimation as to what is happening and what we see and interpret at any particular time.
Well looking at the whole context of what I said, and ignoring pedantic nonsense, I believe it entirely comprehensible "And even in that scenario, facts may aspire to prevent you seeing anything outside....[1]Surviving tidal gravity effects, [2] Surviving any infalling matter/energy which would probably be in the form of lethal radiation, [3] and simply surviving longe enough, eg: Even our own SMBH, one would only have around 12 seconds or so, from the trip from the EH to the Singularity". Self explanatory I suggest.
" facts may aspire ", So, are you trying to attribute a human trait, the conscious ability to aspire, to a fact?
##### Share on other sites
Posted (edited)
You seem to have a problem with Beecee and words and phrases he has used ( fact, hope that helps, facts may aspire ), yet you have not posted any meaningful contribution to this topic.
What you are doing is not discussion.
Grow up, before someone decides to report you.
Edited by MigL
##### Share on other sites
Getting back to the subject of the thread, as entitled.....The first BH suspect was "Cygnus X1" https://www.nasa.gov/mission_pages/chandra/multimedia/cygnusx1.html
"On the left, an optical image from the Digitized Sky Survey shows Cygnus X-1, outlined in a red box. Cygnus X-1 is located near large active regions of star formation in the Milky Way, as seen in this image that spans some 700 light years across. An artist's illustration on the right depicts what astronomers think is happening within the Cygnus X-1 system. Cygnus X-1 is a so-called stellar-mass black hole, a class of black holes that comes from the collapse of a massive star. The black hole pulls material from a massive, blue companion star toward it. This material forms a disk (shown in red and orange) that rotates around the black hole before falling into it or being redirected away from the black hole in the form of powerful jets.
A trio of papers with data from radio, optical and X-ray telescopes, including NASA's Chandra X-ray Observatory, has revealed new details about the birth of this famous black hole that took place millions of years ago. Using X-ray data from Chandra, the Rossi X-ray Timing Explorer, and the Advanced Satellite for Cosmology and Astrophysics, scientists were able to determine the spin of Cygnus X-1 with unprecedented accuracy, showing that the black hole is spinning at very close to its maximum rate. Its event horizon -- the point of no return for material falling towards a black hole -- is spinning around more than 800 times a second.
Using optical observations of the companion star and its motion around its unseen companion, the team also made the most precise determination ever for the mass of Cygnus X-1, of 14.8 times the mass of the Sun. It was likely to have been almost this massive at birth, because of lack of time for it to grow appreciably.
The researchers also announced that they have made the most accurate distance estimate yet of Cygnus X-1 using the National Radio Observatory's Very Long Baseline Array (VLBA). The new distance is about 6,070 light years from Earth. This accurate distance was a crucial ingredient for making the precise mass and spin determinations".
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Other such observations have been made. In recent times the discovery of gravitational waves that just happen to match templates as determined by GR, and of course the new image of a BH at m87, or at least the shadow of the BH.
I believe we can be as sure as any scientific theory allows us to be, that BH's are real entities in this big wide wonderful universe we inhabit.
Plenty more info and data in the "First real Black Hole image - 10 April 2019" thread.
##### Share on other sites
Posted (edited)
12 hours ago, Strange said:
We already live in curved spacetime. The main effect we notice is that thing we call "gravity". With sensitive instruments we can measure some other effects such as gravitational red-shift or lensing.
These effects would all be greater as we approach (and fall into) a black hole: the force of gravity (and tidal forces) would be greater, gravitational red shift would be greater (the rest of the universe would look increasingly blue-shifted), gravitational lensing would be obvious (the event horizon would appear to be 2.6 times larger than expected; the entire accretion disk would be visible, including that on the far-side of the event horizon; our view of the universe would narrow; etc)
That in red is wrong. It's well known that GR predicts the opposite - rest of universe will appear redshifted, not blueshifted. See e.g.:
Some 'authorities' do get it wrong, e.g. here (and reference therein):
https://physics.stackexchange.com/questions/26185/what-will-the-universe-look-like-for-anyone-falling-into-a-black-hole
Edited by Q-reeus
##### Share on other sites
1 hour ago, MigL said:
You seem to have a problem with Beecee and words and phrases he has used ( fact, hope that helps, facts may aspire ), yet you have not posted any meaningful contribution to this topic.
What you are doing is not discussion.
Grow up, before someone decides to report you.
fact : beecee said:
"In fact any observer inside the BH, as long as tidal gravitational effects have not torn the observer asunder, would actually probably see the whole universe in the form of a circle above his head, due to lensing. I dare say he would be ripped apart before he had any inkling he had crossed a point of no return."
I asked if this was fact.
beecee replied with a Link https://jila.colorado.edu/~ajsh/ , nothing at that Link corroborated nothing that beecee Posted as fact. Indeed, the Linked site contains this disclaimer " The creation of this website was supported by the National Science Foundation. Statements made herein are those of the author, and do not necessarily reflect the views of the National Science Foundation or of any other entity."
hope that helps, said beecee.
How could that Link possibly help?
facts may aspire ?
Yes, MigL, the problem is that what beecee Posted is NOT A FACT!
It is no more than conjecture, or speculation, or a subjective interpretation of a somewhat misunderstood understanding of a Theoretical construct.
From my understanding of the Theory of Relativity, everything inside the Black Hole - including Light - is moving at the very least, at the speed of Light towards some "center". There would be no light or photons that could escape the intense pull of gravity so as to be able to be seen by any eye or receptor.
What Janus Posted : "One of the bizarre things that occurs after crossing the the event horizon is that time and space switch roles. So describing what you would "see" is a bit difficult. For example, outside of a black hole, if we are looking at a point 1 light hr away, we can only see, at any given moment, events that occurred 1 hr ago. Inside the event horizon, If you are looking at a point further out from the center than you are, at the same 1 light hr away, you would see everything that occurs at that point between 1 hr in the past to 1 hr in the future, all at once."
And also alluded to by Strange : "Yes, most descriptions gloss over this. For perhaps obvious reasons. It is hard to comprehend how that might appear to our senses and we will probably never know."
Both of these views are much more in line with the current interpretation of Relativity.
MigL, there may indeed be students visiting this site to actually learn some science.
MigL, there is not currently, and most likely never will be any "fact" concerning what one would "see" inside a Black Hole.
MigL, the only thing close to being attributed as fact would be what Janus, Strange and myself surmise - We Do Not Know And Might Never Know - what one would "see" inside a Black Hole!
##### Share on other sites
4 minutes ago, et pet said:
MigL, the only thing close to being attributed as fact would be what Janus, Strange and myself surmise - We Do Not Know And Might Never Know - what one would "see" inside a Black Hole!
Fact: While I have always agreed we will never observationally verify anything within a BH's EH, I also stipulate that we can reasonably use GR as a guide.
Hope that helps.
##### Share on other sites
1 hour ago, et pet said:
fact : beecee said:
"In fact any observer inside the BH, as long as tidal gravitational effects have not torn the observer asunder, would actually probably see the whole universe in the form of a circle above his head, due to lensing. I dare say he would be ripped apart before he had any inkling he had crossed a point of no return."
I asked if this was fact.
beecee replied with a Link https://jila.colorado.edu/~ajsh/ , nothing at that Link corroborated nothing that beecee Posted as fact. Indeed, the Linked site contains this disclaimer " The creation of this website was supported by the National Science Foundation. Statements made herein are those of the author, and do not necessarily reflect the views of the National Science Foundation or of any other entity."
hope that helps, said beecee.
beecee also said in direct reply to you...
6 hours ago, beecee said:
You fail to recognise that what I have posted aligns with GR predictions, which obviously we all agree has a pretty good track record. You also fail to see that in my statement I also used the word "probably" thus, "In fact any observer inside the BH, as long as tidal gravitational effects have not torn the observer asunder, would actually "probably" see the whole universe in the form of a circle above his head, due to lensing. I dare say he would be ripped apart before he had any inkling he had crossed a point of no return"
Plus of course the obvious that you seemed to have missed, in that Strange's statement was referring to the time and space changing places, while my statement was in relation to what anyone inside the EH would see outside the EH. And even in that scenario, facts may aspire to prevent you seeing anything outside....[1]Surviving tidal gravity effects, [2] Surviving any infalling matter/energy which would probably be in the form of lethal radiation, [3] and simply surviving longe enough, eg: Even our own SMBH, one would only have around 12 seconds or so, from the trip from the EH to the Singularity.
Couple all that with other contributions of mine and others, that clearly state that "while we certainly can never observe or have direct evidence of anything about what happens inside a BH, we do have the overwhelmingly supportive and incredibly correct predicitive powers of GR, to at least form a reasonable picture of that which will never be seen.
Hope that helps.
So one must ask, why do you keep misrepresenting what I said?
Strange also said in direct reply to you......
6 hours ago, Strange said:
I don't think so. We do "know" (based on what GR tells us) how things would look to an observer falling in. We can even run simulations, such as raytracing, to generate images that represent what we see.
Whether the change of the radial direction from being spatial to being temporal is something we would be aware of or not, is not really known. Simulating what something looks like doesn't tell us anything about how it would feel
But it does explain why you still would not be able to see the singularity (or whatever is at the centre of the black hole) even after you pass the event horizon: because it is in your future and, as we all know, you can't see the future!
BTW. People often use words like "know" or "fact" to describe our best current understanding based on theory and available evidence. But, as with all science, this knowledge and these facts are, of course, provisional and subject to change with further evidence or new theories.
And obviously, GR is by far our best understanding of BH's considering its incredible predictive powers, so obvious in recent times, so again its reasonable to assume or speculate what is inside. Or do you have a better answer to add to the knowledge and data in this and/or other BH' threads?
##### Share on other sites
1 hour ago, beecee said:
Fact: While I have always agreed we will never observationally verify anything within a BH's EH, I also stipulate that we can reasonably use GR as a guide.
Hope that helps.
So, at the point that GR fails, and even beyond that point, you choose to "stipulate that we can reasonably use GR as a guide" ?
Is there any possibility that you can explain why you choose to make such a stipulation, or that you may be able to supply any evidence to support such a stipulation?
##### Share on other sites
Posted (edited)
27 minutes ago, et pet said:
So, at the point that GR fails, and even beyond that point, you choose to "stipulate that we can reasonably use GR as a guide" ?
Is there any possibility that you can explain why you choose to make such a stipulation, or that you may be able to supply any evidence to support such a stipulation?
No, I stipulate that GR tells us collapse is compulsory once the EH/Schwarzchild radius is reached. But since it fails at the quantum/Planck level, it is then reasonable to assume that instead of the highly unlikely so called singularity of infinite density and spacetime curvature [which mainstream cosmologists now reject] there probably exists a surface of sorts at or below this quantum/Planck level, which as yet we are unable to really speculate on with any confidence, the likes that GR give us for the rest of the volume that makes up the BH. So any singularity, is simply a singularity defined by the fact that GR and all the other laws of physics fail us there.
Just as I have said many many times before with the confidence of the professionals we do have on this forum
2 hours ago, Q-reeus said:
That in red is wrong. It's well known that GR predicts the opposite - rest of universe will appear redshifted, not blueshifted. See e.g.:
Some 'authorities' do get it wrong, e.g. here (and reference therein):
https://physics.stackexchange.com/questions/26185/what-will-the-universe-look-like-for-anyone-falling-into-a-black-hole
Image distortion inside the black hole At 0.35 Schwarzschild radii. Compare this view to the unconventional view you would see if the Schwarzschild surface were attached to another Universe via a wormhole. Images are being distorted by two effects: a tidal distortion from the gravity of the black hole, and a special relativistic beaming from our near light speed motion. Just as the tidal distortion redshifts images from above and below, and blueshifts them about your middle, so also it tends to repel images from above and below, and concentrate them about your middle. At first, images appear distorted into a kidney shape. As the distortion grows, images become stretched and squashed into a doughnut shape about your waist. Our near light speed motion concentrates our view ahead, by special relativistic beaming. Relative to observers freely falling radially from rest at infinity, our velocity increases towards the speed of light: the relativistic Lorentz gamma factor at radius rr is 1+2rs/r1+2rs/r.
The distortions grow At 0.01 Schwarzschild radii. The tidal force continues to concentrate our view into a ‘horizon’ shape, while our near light speed motion further concentrates the view ahead. The tidal force and our motion blueshifts photons from the outside world eventually to very high energies, which we would see as x and gamma rays.
Edited by beecee
##### Share on other sites
31 minutes ago, beecee said:
Quote
2 hours ago, Q-reeus said:
That in red is wrong. It's well known that GR predicts the opposite - rest of universe will appear redshifted, not blueshifted. See e.g.:
Some 'authorities' do get it wrong, e.g. here (and reference therein):
https://physics.stackexchange.com/questions/26185/what-will-the-universe-look-like-for-anyone-falling-into-a-black-hole
Image distortion inside the black hole At 0.35 Schwarzschild radii. Compare this view to the unconventional view you would see if the Schwarzschild surface were attached to another Universe via a wormhole. Images are being distorted by two effects: a tidal distortion from the gravity of the black hole, and a special relativistic beaming from our near light speed motion. Just as the tidal distortion redshifts images from above and below, and blueshifts them about your middle, so also it tends to repel images from above and below, and concentrate them about your middle. At first, images appear distorted into a kidney shape. As the distortion grows, images become stretched and squashed into a doughnut shape about your waist. Our near light speed motion concentrates our view ahead, by special relativistic beaming. Relative to observers freely falling radially from rest at infinity, our velocity increases towards the speed of light: the relativistic Lorentz gamma factor at radius rr is 1+2rs/r1+2rs/r.
The distortions grow At 0.01 Schwarzschild radii. The tidal force continues to concentrate our view into a ‘horizon’ shape, while our near light speed motion further concentrates the view ahead. The tidal force and our motion blueshifts photons from the outside world eventually to very high energies, which we would see as x and gamma rays.
It's typical of you to hunt for a contrary finding and just cut & paste it without acknowledging it represents conflicting opinions among assumed GR experts, or explaining your own pov and giving detailed reasoning why. So which of those two conflicting positions do you support, and why exactly? And btw, there is a link from my 2nd linked ref:
, that gives detailed calculations for (notionally)you, or Strange, to follow. Feel free to 'spot the fatal error'. And note I made it clear in earlier post this is all reasoned on the assumption GR is true hence EH's actually exist. I don't believe in either, but nevertheless expect a better, horizonless theory will share to some degree some of those general tendencies. | 8,055 | 36,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-22 | latest | en | 0.949073 |
https://studyadda.com/question-bank/critical-thinking-questions_q33/1449/106991 | 1,600,598,516,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400197946.27/warc/CC-MAIN-20200920094130-20200920124130-00403.warc.gz | 664,039,654 | 18,431 | • # question_answer If $x,\ y$ and $r$ are positive integers, then $^{x}{{C}_{r}}{{+}^{x}}{{C}_{r-1}}^{y}{{C}_{1}}{{+}^{x}}{{C}_{r-2}}^{y}{{C}_{2}}+.......{{+}^{y}}{{C}_{r}}=$ [Karnataka CET 1993; RPET 2001] A) $\frac{x\ !\ y\ !}{r\ !}$ B) $\frac{(x+y)\ !}{r\ !}$ C) $^{x+y}{{C}_{r}}$ D) $^{xy}{{C}_{r}}$
The result is trivially true for$r=1,\ 2$. It can be easily proved by the principle of mathematical induction that the result is true for $r$ also. | 186 | 453 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-40 | latest | en | 0.512218 |
https://wisdomessays.com/econ-213/ | 1,719,231,611,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00654.warc.gz | 533,975,859 | 12,407 | # Econ 213
Econ 213
Problem Set 4
Name: ______________________________________________
Problem Set 4 is to be completed by 11:59 p.m. (ET) on Friday of Module/Week 8.
1. Movies are distributed in a variety of forms, not just first run theatrical presentations. What other ways are movies distributed? What are the different price points? Using this information, draw a fully labeled graph of the market for movies in which the distributor of the film price discriminates. (NOTE: This should not be perfect price discrimination.)
2. Assume the following game is played one time only. Based on the information in the payoff matrix, PNC Bank and Citizens Bank are considering an implicit collusive agreement on interest rates. Payoffs to the two firms are represented in terms of profits in thousands of dollars:
Citizens Bank Collude: Raise Rates Defect: Keep Rates where they are PNC Collude: Raise Rates (900, 600) (700, 800) Defect: Keep Rates where they are (1100, 300) (800,400)
a. Does PNC have a dominant strategy? What is it? Does Citizens have a dominant strategy? What is it?
b. Does the result of your answer change if the game is played an infinite number of times? Why or why not. Properly use game theoretic terminology in your answer.
3. What is the profit maximizing output of the monopolist shown below? _____________
What price do they set? _______________________
What is the mark up over cost? _______________________
Why will this price not fall?
4. Draw the cheese market for the United States showing the world price as the price for this market. How much cheese does the U.S. import at the world price? Now assume that the cheese lobby promotes and successfully gains a tariff on cheese. What happens to the price paid by cheese lovers in the U.S.? How does this change the value generated by the market? Why do you say this? Where does this appear in your graph? | 425 | 1,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.922551 |
http://math.mit.edu/~djk/18.310/18.310F04/23_finite_fourier.html | 1,386,265,194,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163046954/warc/CC-MAIN-20131204131726-00051-ip-10-33-133-15.ec2.internal.warc.gz | 117,898,174 | 6,850 | 23. The Finite Fourier Transform and the Fast Fourier Transform Algorithm
1. Introduction: Fourier Series
Early in the Nineteenth Century, Fourier, in studying sound and oscillatory motion conceived of the idea of representing periodic functions by their coefficients in an expansion as a sum of sines and cosines rather than their values.
He noticed that if, for example, you represented the shape of a vibrating string of length L, fixed at its ends as
y(x) = ak sin 2kx/L,
the coefficients, ak, contained important and useful information about the quality of the sound that the string produces that was not easily accessible from the ordinary y=f(x) representation of the shape of the string.
This kind of representation of a function is called a Fourier Series, and there is a tremendous amount of mathematical lore about properties of such series and for what classes of functions they can be shown to exist.
One particularly useful fact about them is how we can obtain the coefficients ak from the function.
This follows from the orthogonality property of sines:
sin 2kx/L sin 2jx/L dx
if the integral has limits 0 and L, is 0 if k is different from j and is when k is j.
(To see this notice that the product of these sines can be written as a constant multiple of the difference between cosines of 2p(k+j)x/L and 2p(k-j)x/L, and each of these cosines has 0 integral over this range.)
By multiplying the expression for y(x) above by 2pjx/L and integrating the result from 0 to L we get then the expression
aj = (1/p)ò f(x) sin 2pjx/L dx.
Fourier series represent only one of many alternate ways we can represent a function. Whenever we can, by introducing an appropriate weight function in the integral, obtain a similar orthogonality relation among functions, we can derive similar formulae for coefficients in a series.
2. The Fourier Transform
Given a function f defined for all real arguments, we can give an alternative representation to it as an integral rather than as an infinite series, as follows.
f(x) = ò exp(ikx) g(k) dk
where the integral is over all real values of k.
The representation of f by the function g is called a Fourier transform of f, and it is very important tool in physics.
One reason for this is that exponential functions eikx, which f is written as an integral which is a sort of a sum of are eigenfunctions of the derivative. That is, the derivative, acting on an exponential, merely multiplies the exponential by ik. This makes the Fourier transform a useful tool in investigating differential equations.
Another example of it’s application: In quantum mechanics, we represent the state of a particle in a physical system has wave function, (x), and |(x)|2 dx represents the probability that the particle in this state has position that lies between x and x+dx.
The same state can also be represented by its wave function in momentum space, and that wave function of the variable p, is a constant multiple of the Fourier transform of j(x):
(p) = c ò exp(ipx)j(x)dx.
We can invert the Fourier Transform in much the same way that we can invert Fourier Series. The resulting formula is
g(k) = (1/2p) ò exp(-ikx) f(x) dx
again the integration is over all real values of x.
3. The Finite Fourier Transform
Given a finite sequence consisting of n numbers, for example the ccoefficients of a polynomial of degree n-1, we can define a Finite Fourier Transform that produces a different set of n numbers, in a way that has a close relationship to the Fourier Transform just mentioned.
I like to look at it backwards.
Suppose we have a polynomial p of degree n-1. It can be described by its coefficients, {aj}: with
p(x) = j= 0 to n-1 aj xj .
We can also represent p by giving its values at any n arguments {p(xk)}.
This can be done as follows. Observe first that the polynomial of degree n-1
f(xj)((x-x1) /(xj-x1))*. . .(exclude (x-xj) /(xj-xj)) . . . *((x-xn)/ (xj-xn))
takes the value f(xj) at x=xj and is 0 at all other of our arguments xk.
We can recover p from its values by summing similar terms over all j.
To evaluate a polynomial of degree n-1 at n values appears to require n2 actions: n evaluations each of n terms. Similarly the procedure just described for recovering p from its values requires at least n2 operations to obtain all the coefficients of p.
When you evaluate a polynomial at an argument x whose magnitude is not close to 1, the powers of x that are big dominate those that are small by so much that you have to worry about losing the smaller terms entirely from round off errors.
The finite Fourier transform can be defined as the act of evaluating a polynomial of degree n-1 at n roots of unity, that is, at n solutions to the equation xn=1.
This transform can be performed upon polynomials with coefficients in any field in which this equation has n solutions, which will happen when there is a primitive n-th root of unity in the field. (This means a number such that xn=1 but xk is not 1 for any k between 1 and n-1.) The n roots of unity are then the various powers of the primitive root.
When does this happen? It does for complex numbers, in which case we have
exp(2pi/n) which is cos(2p/n)+isin(2p/n).
as primitive n-th root of unity.
But it also happens for remainders on dividing by a prime number of the form kn+1. In such fields there is a primitive kn-th root of unity and hence a primitive n-th root of unity (such as the k-th power of the former.)
The analogy between this finite transform and the Fourier transform is mnost apparent when we use complex numbers. Then, if the coefficients of the polynomial are {aj}, the evaluations become
p(exp(2pik/n)) = j aj (cos(2pjk/n)) + ij aj (sin(2pjk/n)).
(This is why we say that we are doing things backwards. It is the aj which are analogous to the Fourier coefficients for the function p.)
In general, if we let z be our primitive n-th root of unity, the same expression becomes
Transforms of this kind can be defined for any value of n. And there is a symmetric form for the inverse transformation which takes the values
{p(zk)}, which we shall abbreviate as {pk}, and produces the aj, so there is no significant difference between defining this transform forward or backward.
We can obtain the inverse transformation by multiplying each pk by z-sk,
and summing over the n values of k.
We get j,k aj zjk*z-sk or j aj (Sk z(j-s)k) or Sj aj ts-j, where tr is our old friend the sum of the r-th powers of the n roots, zk, of the equation zn –1.
Recall please, that this equation, zn – 1 = 0 has the form Sk zksk = 0, where sk is the k-th elementary symmetric function of the roots of the equation.
This implies that the sk are all 0, for k=1 up to n-1 for our equation, while s0 is 1.
Recall also that the the t’s and the s’s are linearly dependent according to the relations, for each k
Sj=0 to k-1 sjtk-j(- 1)j + ksk(-1)k =0,
from which we can deduce that the ts-j here are all 0, unless s=j.
When s=j, the sum that forms t0, or tn, which is n, so that we get
Sk pk z-k = Sj,k aj zjk*z-sk = nas,
and
as = (1/n) S k pkz-k.
Since z-1 is another primitive n-th root of 1 the only difference in this equation, and the inverse equation for evaluations is in the factor 1/n.
4. The Cooley-Tukey Fast Fourier Transform Algorithm
Suppose n is even, so that n can be written as 2s. Then this algorithm is a procedure for reducing 2s evaluations of polynomials of degree 2s to 2s evaluations of polynomials of degree s, upon making a total of s additions, s subtractions and s multiplications. Moreover the evaluations consist of evaluating the FFT’s of two polynomials, each of degree s-1, at primitive s-th roots of unity.
To keep things straight let us describe the evaluations of a polynomial of degree at most n-1 at n n-th roots of unity.as an nFFT.
If n is a power of 2, we can iterate this procedure n times, until we reduce the problem to n evaluations of polynomials of degree 0, which is a weird way to say that we will have obtained our n evaluations.
The reduction that is the heart of this algorithm is based upon the following observations. To perform an nFFT require evaluating our polynomial of degree up to n-1 at the n powers of a primitive n-th root of unity, z.
1. If we consider the evaluations we seek at the even powers of z, (1, z2, z4, . . . ), these powers are the powers 0 to s-1 of the s-th root of unity z2.
Thus, these evaluations are exactly what is involved in an sFFT; the only difference being that we are here evaluating a polynomial of degree up to n-1, not up to s-1.
2. In every even power evaluation, say at z2k, the term in our polynomial ajxj contributes ajz2kj. Thus, the contribution from the j-th and (s+j)-th terms together are
ajz2kj + aj+sz2kj+2sk.
But, z2sk is znk which is 1, so that the aj+s contribution here is multiplied by the same power of z as the aj contribution, and can be added to it instead of being treated as a separate term.
3. But this means that the z2k evaluations here are exactly those of sFFT({aj+aj+s}).
4. The odd power evaluations, (at z, z3, . . .) each gets a contribution from aj of the form ajz(2k+1)j which we can write as (ajzj)z2jk. Notice that these evaluations can be considered evaluations at even roots of unity of a polynomial whose coefficients are given by the products (ajzj).
5. In every odd power evalution, say at z2k+1, the term in our polynomial ajxj contributes ajzjz2kj while the term aj+sxj+s contributes aj+szjzsz2kj +2ks. As in our second observation we have z2ks=1, but now we have an additional factor of zs, which is a primitive square root of 1, which is –1. In other words the contributions from aj and (–aj+s) are the same here.
6. We conclude that we can combine the j and s+j terms by subtraction in the odd power evaluations and these become exactly those of sFFT({(aj-aj+s)zj}).
You will notice that we make an addition for each of our even power evaluations in making these reductions, and a subtraction and a multiplication for each of the odd powers reductions, and this means that we must perform s of each of these operations to reduce the problem by a factor of two.
The remaining task in completing our evaluations after this reduction consists of repeating it in parallel for the even and odd evaluations. After a second reduction, we perform 4 reductions in parallel for the even-even, odd-even even-odd and odd-odd evaluations (starting in positions z0,z1, z2,z4), and so on.
And that is the algorithm.
We illustrate it by starting with 4 coefficients 1,3,2,5 (of powers 0 through 3) and do our calculations mod 17. 4 is a primitive 4th root of 1 whose inverse is –4 or 13.
In the first step we replace the even power entries, which are the first and third here, by 1+2 and 3+5 respectively, and the second and fourth entries by (1-2)*40 and (3-5)*41 which produces the sequence 3,16,8,9 mod 17.
In the second step we replace the first and second entries by 3+8 or 11 and 16+9 or 8, and the third and fourth by 3-8 or 12 and 16-9 or 7, for an answer of (11,8,12,7) as the result of evaluating the polynomial 1+3x+2x2+5x3 at x= 1,4,16 and 13 mod 17.
It is instructive to see what happens when we apply the same procedure again to the sequence obtained here, namely (11,8,12,7).
In the first step this becomes (6,16,15,4) (the last value comes from (8-7)*4 mod 17). And in the last step we get (4,3,8,12).
Notice that if we divide this result by 4 we get (1,5,2,3) (to divide 3 by 4 you can add 17 to the 3, divide 20 by 4 and get 5).
These are the original coefficients of our polynomial, in the order 0,3,2,1.
The reason for this is that the formula for the inverse of our transformation requires dividing by n and also using z-1 in place of z as the primitive n-th root of unity at whose powers the evaluations are made.
And of course evaluating at z-k is the same thing as evaluating at z(n-k), which means the k-th power evaluation for z-1 is the n-k-th for z. | 3,035 | 11,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2013-48 | latest | en | 0.914697 |
https://ch.mathworks.com/matlabcentral/profile/authors/14108066 | 1,713,510,658,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00789.warc.gz | 152,340,299 | 19,857 | Community Profile
# jie hu
Last seen: 3 months ago Active since 2019
Followers: 0 Following: 0
#### Feeds
View by
Question
create animation from several figures
how to create an animation from several figures?
3 months ago | 1 answer | 0
### 1
Question
I have created a .csv file with date and value by fid=fopen([file(1:4) '.csv'],'w'); fprintf(fid,'%s,%s,%s\n', 'Time...
4 months ago | 1 answer | 0
### 1
how can I calculate the nautical direction angle from Cartesian x- and y-component of velocity
thanks, But I am talking about the direction in Nautical convention. for y1 = atan2d(10, 10) = 45, shall that angle be 225 in Na...
4 months ago | 0
Question
how can I calculate the nautical direction angle from Cartesian x- and y-component of velocity
I have a set of Cartesian x- (u) and y-component (v) of velocity. But I want the velocity to be Nautical direction based, which ...
4 months ago | 1 answer | 0
### 1
Question
semilog plot with a self defined x-axis range
I have a time series data and I would like to plot them in the semilogx ranging from [10^(-4) to 10^2] as the attached figure. M...
5 months ago | 1 answer | 0
### 1
Question
problem of using findpeaks
I have two vectors of time (datetime formate) and elevation, but there is an error to use findpeaks(elev,time), saying "Expected...
5 months ago | 2 answers | 0
### 2
Question
The sample points must be finite.
I like to interpolate a sample data on a new timeline by useing interp1, saying "The sample points must be finite." data_origin...
10 months ago | 1 answer | 0
### 1
Question
"griddata" results in much smaller values than original data
I am using "griddata" function to interpolate original data on query points. However, the interpolation value on the query point...
1 year ago | 2 answers | 0
### 2
Question
formulation of square matrix
Hello everyone, I want to formulate a very big square matrix, in which every element is again a square matrix. I want the sqaur...
5 years ago | 1 answer | 0 | 543 | 2,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-18 | latest | en | 0.886134 |
https://pythondict.com/python-tutorials/python-return-value-function/ | 1,726,120,540,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00874.warc.gz | 444,277,007 | 45,674 | # Python 教程 6 — 有返回值的函数
## 1.返回值 #
```e = math.exp(1.0)
```def area(radius):
return a```
```def area(radius):
```def absolute_value(x):
if x < 0:
return -x
else:
return x```
```def absolute_value(x):
if x < 0:
return -x
if x > 0:
return x```
```>>> absolute_value(0)
None```
## 2.增量式开发 #
```def distance(x1, y1, x2, y2):
return 0.0```
```>>> distance(1, 2, 4, 6)
0.0```
```def distance(x1, y1, x2, y2):
dx = x2 - x1
dy = y2 - y1
print('dx is', dx)
print('dy is', dy)
return 0.0```
```def distance(x1, y1, x2, y2):
dx = x2 - x1
dy = y2 - y1
dsquared = dx**2 + dy**2
print('dsquared is: ', dsquared)
return 0.0```
```def distance(x1, y1, x2, y2):
dx = x2 - x1
dy = y2 - y1
dsquared = dx**2 + dy**2
result = math.sqrt(dsquared)
return result```
1. 从一个能运行的程序开始，并且每次只增加少量改动。无论你何时遇到错误，都能够清楚定位错误的源头。
2. 用临时变量存储中间值，这样你就能显示并检查它们。
3. 一旦程序正确运行，你要删除一些脚手架代码，或者将多条语句组成复合表达式，但是前提是不会影响程序的可读性。
## 3.组合 #
```radius = distance(xc, yc, xp, yp)
```
```result = area(radius)
```
```def circle_area(xc, yc, xp, yp):
radius = distance(xc, yc, xp, yp)
return result```
```def circle_area(xc, yc, xp, yp):
return area(distance(xc, yc, xp, yp))```
## 4.布尔函数 #
```def is_divisible(x, y):
if x % y == 0:
return True
else:
return False```
```>>> is_divisible(6, 4)
False
>>> is_divisible(6, 3)
True```
`==`运算符的结果是布尔值，因此我们直接返回它，让代码变得更简洁。
```def is_divisible(x, y):
return x % y == 0```
```if is_divisible(x, y):
print('x is divisible by y')```
```if is_divisible(x, y) == True:
print('x is divisible by y')```
## 5.再谈递归 #
`def factorial(n):`
```def factorial(n):
if n == 0:
return 1```
```def factorial(n):
if n == 0:
return 1
else:
recurse = factorial(n-1)
result = n * recurse
return result```
## 7.再举一例 #
F(0)=0，F(1)=1, F(n)=F(n – 1)+F(n – 2)（≥ 2，∈ N*）
```def fibonacci (n):
if n == 0:
return 0
elif n == 1:
return 1
else:
return fibonacci(n-1) + fibonacci(n-2)```
## 8.检查类型 #
```>>> factorial(1.5)
RuntimeError: Maximum recursion depth exceeded```
```def factorial (n):
if not isinstance(n, int):
print('Factorial is only defined for integers.')
return None
elif n < 0:
print('Factorial is not defined for negative integers.')
return None
elif n == 0:
return 1
else:
return n * factorial(n-1)```
```>>> factorial('fred')
Factorial is only defined for integers.
None
>>> factorial(-2)
Factorial is not defined for negative integers.
None```
## 9.调试 #
• 该函数获得的实参有些问题，违反先决条件。
• 该函数有些问题，违反后置条件。
• 返回值或者它的使用方法有问题。
```def factorial(n):
space = ' ' * (4 * n)
print(space, 'factorial', n)
if n == 0:
print(space, 'returning 1')
return 1
else:
recurse = factorial(n-1)
result = n * recurse
print(space, 'returning', result)
return result```
`space`是一个空格字符的字符串，用来控制输出的缩进。 下面是 `factorial(4)` 的输出结果：
``` factorial 4
factorial 3
factorial 2
factorial 1
factorial 0
returning 1
returning 1
returning 2
returning 6
returning 24```
## 11.练习题 #
### 习题6-1 #
```def b(z):
prod = a(z, z)
print(z, prod)
return prod
def a(x, y):
x = x + 1
return x * y
def c(x, y, z):
total = x + y + z
square = b(total)**2
return square
x = 1
y = x + 1
print(c(x, y+3, x+y))```
### 习题6-2 #
Ackermann函数定义如下：
```"""This module contains a code example related to
Think Python, 2nd Edition
by Allen Downey
http://thinkpython2.com
"""
from __future__ import print_function, division
def ackermann(m, n):
"""Computes the Ackermann function A(m, n)
See http://en.wikipedia.org/wiki/Ackermann_function
n, m: non-negative integers
"""
if m == 0:
return n+1
if n == 0:
return ackermann(m-1, 1)
return ackermann(m-1, ackermann(m, n-1))
print(ackermann(3, 4))```
### 习题6-3 #
```def first(word):
return word[0]
def last(word):
return word[-1]
def middle(word):
return word[1:-1]```
1. 将它们录入到文件 `palindrome.py` 中并测试。当你用一个两个字母的字符串调用 `middle` 时会发生什么？一个字母的呢？空字符串呢？空字符串这样 `''` 表示，中间不含任何字母。
2. 编写一个叫 `is_palindrome` 的函数，接受一个字符串作为实参。如果是回文词，就返回 `True` ，反之则返回 `False`。记住，你可以使用内建函数 `len` 来检查字符串的长度。
```"""This module contains a code example related to
Think Python, 2nd Edition
by Allen Downey
http://thinkpython2.com
"""
from __future__ import print_function, division
def first(word):
"""Returns the first character of a string."""
return word[0]
def last(word):
"""Returns the last of a string."""
return word[-1]
def middle(word):
"""Returns all but the first and last characters of a string."""
return word[1:-1]
def is_palindrome(word):
"""Returns True if word is a palindrome."""
if len(word) <= 1:
return True
if first(word) != last(word):
return False
return is_palindrome(middle(word))
print(is_palindrome('allen'))
print(is_palindrome('bob'))
print(is_palindrome('otto'))
print(is_palindrome('redivider'))```
### 习题6-5 #
aa和 bb 的最大公约数（GCD）是能被二者整除的最大数。
1. 翻译：@theJian
2. 校对：@bingjin
3. 参考：@carfly
## 推荐阅读#
##### bert-serving 模型及源代码#
bert-serving 模型及源代码 提供百度网盘下载。
##### 推荐一个非常优惠的QMT开户渠道#
Python实用宝典 (pythondict.com) | 1,854 | 4,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-38 | latest | en | 0.259708 |
http://aperiodical.com/2018/04/the-chromatic-number-of-the-plane-is-at-least-5/ | 1,529,701,305,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00236.warc.gz | 22,848,763 | 8,106 | # The chromatic number of the plane is at least 5
A long-standing mathematical problem has had a recent breakthrough – scientist Aubrey de Grey has proved that the chromatic number of the plane is at least 5.
In geometric graph theory, the Hadwiger–Nelson problem poses the question: what is the minimum number of colours required to colour the 2-dimensional plane, so that no two points that are exactly one unit of distance apart are coloured the same colour? The answer to this question is called the chromatic number of the plane.
A 7-colouring of the plane, with four-chromatic unit distance graph (image from Wikipedia)
Up until now, we knew the answer was one of 4, 5, 6 or 7. The diagram shown here contains proofs of the upper and lower bounds: for the upper bound, you can assume the hexagons are each less than one unit in diameter, and so each point inside a hexagon is one unit away from a circle of points which use only the other six colours, as the colouring has no adjacent or adjacent-but-one hexagons the same colour.
The graph in the diagram also proves the minimum is four: known as the Moser Spindle, the graph consists of two pairs of equilateral triangles, and all the arcs are one unit long. It can be used to prove a three-colouring is not possible, as follows. The colour of the top left vertex forces the other two vertices of each of the two attached triangles to use the other two colours, but this in turn forces the two lower right vertices to be the same colour, and they are a unit apart. Contradiction!
Aubrey de Grey’s graph, from the paper
Aubrey de Grey’s proof, uploaded to the arXiv earlier this month, uses a similar line of reasoning to prove it’s not possible to colour the plane with four colours in this way – only this graph has around 1600 vertices (see diagram, if you can), and the construction and checking of the graph is computer-assisted. The paper itself is still just on the arXiv for now as far as I can tell, but if we spot it being peer-reviewed and published anywhere we’ll make an update here.
In the wake of this announcement, a 16th Polymath project has been proposed to try to find a simpler graph, and improve on this result – some have conjectured it might not be possible to reduce the size of this graph further without redesigning it completely, as de Grey’s method already reduces it to as minimal as possible from the initial approach used (it involves taking smaller graphs and rotating them to align on top, much like the Moser Spindle).
Some blog posts on the topic, including one which makes attempts to verify the proof, and the paper itself, are listed below.
The chromatic number of the plane is at least 5, de Grey’s paper on the ArXiv
The chromatic number of the plane is at least 5
, on Jordan Ellenberg’s blog
Amazing progress on longstanding open problems, on Scott Aaronson’s blog
The chromatic number of the plane is at least 5, on Dustin Mixon’s blog
Aubrey de Grey: The chromatic number of the plane is at least 5, on Gil Kalai’s blog | 696 | 3,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-26 | latest | en | 0.943835 |
https://finance.yahoo.com/news/chow-tai-fook-jewellery-group-001544624.html | 1,571,029,010,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00160.warc.gz | 579,948,632 | 110,091 | U.S. Markets open in 8 hrs 34 mins
# Chow Tai Fook Jewellery Group Limited (HKG:1929) Delivered A Better ROE Than Its Industry
One of the best investments we can make is in our own knowledge and skill set. With that in mind, this article will work through how we can use Return On Equity (ROE) to better understand a business. To keep the lesson grounded in practicality, we'll use ROE to better understand Chow Tai Fook Jewellery Group Limited (HKG:1929).
Over the last twelve months Chow Tai Fook Jewellery Group has recorded a ROE of 15%. One way to conceptualize this, is that for each HK\$1 of shareholders' equity it has, the company made HK\$0.15 in profit.
### How Do I Calculate ROE?
The formula for ROE is:
Return on Equity = Net Profit ÷ Shareholders' Equity
Or for Chow Tai Fook Jewellery Group:
15% = HK\$4.6b ÷ HK\$31b (Based on the trailing twelve months to March 2019.)
It's easy to understand the 'net profit' part of that equation, but 'shareholders' equity' requires further explanation. It is the capital paid in by shareholders, plus any retained earnings. You can calculate shareholders' equity by subtracting the company's total liabilities from its total assets.
### What Does Return On Equity Mean?
ROE looks at the amount a company earns relative to the money it has kept within the business. The 'return' is the amount earned after tax over the last twelve months. That means that the higher the ROE, the more profitable the company is. So, all else being equal, a high ROE is better than a low one. That means it can be interesting to compare the ROE of different companies.
### Does Chow Tai Fook Jewellery Group Have A Good ROE?
One simple way to determine if a company has a good return on equity is to compare it to the average for its industry. The limitation of this approach is that some companies are quite different from others, even within the same industry classification. As is clear from the image below, Chow Tai Fook Jewellery Group has a better ROE than the average (9.3%) in the Specialty Retail industry.
That is a good sign. In my book, a high ROE almost always warrants a closer look. One data point to check is if insiders have bought shares recently.
### The Importance Of Debt To Return On Equity
Most companies need money -- from somewhere -- to grow their profits. That cash can come from issuing shares, retained earnings, or debt. In the first and second cases, the ROE will reflect this use of cash for investment in the business. In the latter case, the debt required for growth will boost returns, but will not impact the shareholders' equity. That will make the ROE look better than if no debt was used.
### Chow Tai Fook Jewellery Group's Debt And Its 15% ROE
Although Chow Tai Fook Jewellery Group does use debt, its debt to equity ratio of 0.58 is still low. The combination of modest debt and a very respectable ROE suggests this is a business worth watching. Judicious use of debt to improve returns can certainly be a good thing, although it does elevate risk slightly and reduce future optionality.
### The Key Takeaway
Return on equity is a useful indicator of the ability of a business to generate profits and return them to shareholders. In my book the highest quality companies have high return on equity, despite low debt. If two companies have the same ROE, then I would generally prefer the one with less debt.
But ROE is just one piece of a bigger puzzle, since high quality businesses often trade on high multiples of earnings. Profit growth rates, versus the expectations reflected in the price of the stock, are a particularly important to consider. So you might want to take a peek at this data-rich interactive graph of forecasts for the company.
Of course, you might find a fantastic investment by looking elsewhere. So take a peek at this free list of interesting companies.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 970 | 4,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-43 | latest | en | 0.940995 |
https://books.google.com.jm/books?qtid=ed7e789c&lr=&id=9e4DAAAAQAAJ&sa=N&start=40 | 1,695,372,241,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00554.warc.gz | 170,980,786 | 5,884 | Books Books
If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle ; the angles which this line makes with the line touching the circle, shall be equal to the angles which are in the alternate segments of the...
Euclid in Paragraphs: The Elements of Euclid: Containing the First Six Books ... - Page 69
by Euclid - 1845 - 199 pages
## The first three books of Euclid's Elements of geometry, with theorems and ...
Euclid, Thomas Tate - 1849 - 120 pages
...the adjacent angles are equal, they are right angles. PROP. XXXH. THEOK. If a straight line touches a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle, shall be equal...
## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
Euclid, John Playfair - Geometry - 1853 - 336 pages
...contact a straight line be drawn at right angles to the touching line, the centre of the circle is in that line. Let the straight line DE touch the circle ABC, in C, and from C let CA be drawn Łt right angles to DE ; the centre of the circle is in CA. For, if not, let F be the centre, if possible,...
## The synoptical Euclid; being the first four books of Euclid's Elements of ...
Euclides - 1853 - 146 pages
...and when the adjacent angles are equal, they are (I. Def. 10.) right angles. PROP. XXXII. THEOREM. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to...
## The Elements of Euclid, books i-vi; xi. 1-21; xii. 1,2; ed. by H.J. Hose, Book 1
Euclides - Geometry - 1853 - 334 pages
...the first part of the prop" that BAC is a right angle. Which was to be proved. PEOP. XXXII. THEOE. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle : then the angles which this line makes with the touching line shall be respectively...
## The Elements of Euclid: With Many Additional Propositions ..., Part 1
Euclid - Geometry - 1853 - 176 pages
...part ff the circumference for its base as an angle at the circumference. If a straight line touches a circle, and from the point of contact a straight line be drawn cutting the circle. They shall stand upon equal parts of the circumference, whether they be at the...
## The first six books of the Elements of Euclid, with numerous exercises
Euclides - 1853 - 176 pages
...angles are equal they are right angles. PROPOSITION XXXII. — THEOREM. If a straight line touches a circle, and from the point of contact a straight line be drawn cutting the circle, the angles moule bу this Une with the line touching the circle shall be equal...
## Orr's Circle of the Sciences: Organic nature, vols. 1-3 (1854-1856)
William Somerville Orr - Science - 1854 - 532 pages
...here only to avoid interruption in the numbering of the propositions. PROPOSITION XIX.— THEOREM. If a straight line touch a circle, and from the point of contact a straight line be draw» at right angles to the touching line, the centre of the circle shall be in that. lim. PROPOSITION...
## Elements of Geometry: Containing the First Six Books of Euclid with a ...
John Playfair - Geometry - 1855 - 336 pages
...; and when the adjacent angles are equal, they are right anles. •T*"*'V i^ PROP. XXXII. THEOR. Jf a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the lute which touches the circle, shall be equal... | 903 | 3,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-40 | latest | en | 0.85055 |
https://www.physicsforums.com/threads/work-by-a-gas.373287/ | 1,521,855,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649508.48/warc/CC-MAIN-20180323235620-20180324015620-00770.warc.gz | 816,772,538 | 14,456 | # Work by a gas
1. Jan 27, 2010
### ronaldor9
1. The problem statement, all variables and given/known data
Consider 5.5L of a gas at a pressure of 3 atm in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5L. Calculate the work done.
Calculate the work if the process is done in two steps. At the end of the first step the volume is 7L and after the second step the volume is 10.5L
2. Relevant equations
W=PV?
3. The attempt at a solution
Not sure since the external pressure is not constant I don't think I can use W=PV to calculate work done.
Last edited: Jan 27, 2010 | 173 | 630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-13 | longest | en | 0.914409 |
https://www.teacherspayteachers.com/Product/USA-Mystery-Picture-695156 | 1,537,581,090,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158001.44/warc/CC-MAIN-20180922005340-20180922025740-00546.warc.gz | 879,397,240 | 19,272 | # USA Mystery Picture
Subject
Resource Type
Product Rating
File Type
PDF (Acrobat) Document File
418 KB|4 pages
Share
Product Description
Great for Veteran's Day
This project can be used for Memorial Day, Veteran’s Day, Election Day or any time you wish to honor the USA. Students will put their knowledge of coordinate systems, coordinate graphing and ordered pairs to work creating a drawing of the letters USA with a star over each letter. This will be created by plotting ordered pairs and then connecting them with straight lines.
This activity can be a class project or something to be worked on independently when time allows. It also works as an extra credit assignment or when there is a substitute. Looks great on a bulletin board, too.
This project is for the beginner graphing student who is familiar with fractions but can also be used by the intermediate student. The graph consists of points in the first quadrant using ordered pairs with whole numbers and a few fractional points using the fractions 1/3, 1/2 and 2/3 for detail. When finished, they can use markers, colored pencils or other medium to enhance the project.
This lesson includes coordinate graph paper for plotting, a coordinate list, a full sized answer key and a sample of the completed project in color.
If you like this project please let me know. There are over 50 projects covering the entire school year and some projects are cross curricular (Sports, Patriotism, Seasons, Holidays, History, and Science). Lessons begin at 4th grade and are appropriate for the Middle School and some for High School students. These projects have been used successfully at both the Elementary and Middle School levels.
http://www.teacherspayteachers.com/Store/Anthony-And-Linda-Iorlano
Anthony Iorlano and Linda Iorlano
NCTM Standards
Specify locations and describe spatial relationships using coordinate geometry and other representational systems.
Common Core State Standards
CCSS.Math.Content.5.G.A.1 Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate).
Keywords: Olympics, Memorial Day, Veteran's Day, USA, Coordinate Drawing, Coordinate Graphing
Total Pages
4 pages
Included
Teaching Duration
N/A
Report this Resource
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 592 | 2,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-39 | latest | en | 0.889299 |
https://metanumbers.com/1071791 | 1,638,419,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00399.warc.gz | 475,408,732 | 7,386 | # 1071791 (number)
1,071,791 (one million seventy-one thousand seven hundred ninety-one) is an odd seven-digits composite number following 1071790 and preceding 1071792. In scientific notation, it is written as 1.071791 × 106. The sum of its digits is 26. It has a total of 2 prime factors and 4 positive divisors. There are 918,672 positive integers (up to 1071791) that are relatively prime to 1071791.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 26
• Digital Root 8
## Name
Short name 1 million 71 thousand 791 one million seventy-one thousand seven hundred ninety-one
## Notation
Scientific notation 1.071791 × 106 1.071791 × 106
## Prime Factorization of 1071791
Prime Factorization 7 × 153113
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1071791 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,071,791 is 7 × 153113. Since it has a total of 2 prime factors, 1,071,791 is a composite number.
## Divisors of 1071791
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 1.22491e+06 Sum of all the positive divisors of n s(n) 153121 Sum of the proper positive divisors of n A(n) 306228 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1035.27 Returns the nth root of the product of n divisors H(n) 3.49998 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,071,791 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 1,071,791) is 1,224,912, the average is 306,228.
## Other Arithmetic Functions (n = 1071791)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 918672 Total number of positive integers not greater than n that are coprime to n λ(n) 459336 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 83515 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 918,672 positive integers (less than 1,071,791) that are coprime with 1,071,791. And there are approximately 83,515 prime numbers less than or equal to 1,071,791.
## Divisibility of 1071791
m n mod m 2 3 4 5 6 7 8 9 1 2 3 1 5 0 7 8
The number 1,071,791 is divisible by 7.
## Classification of 1071791
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (1071791)
Base System Value
2 Binary 100000101101010101111
3 Ternary 2000110012222
4 Quaternary 10011222233
5 Quinary 233244131
6 Senary 34545555
8 Octal 4055257
10 Decimal 1071791
12 Duodecimal 4382bb
20 Vigesimal 6dj9b
36 Base36 myzz
## Basic calculations (n = 1071791)
### Multiplication
n×y
n×2 2143582 3215373 4287164 5358955
### Division
n÷y
n÷2 535896 357264 267948 214358
### Exponentiation
ny
n2 1148735947681 1231204850100966671 1319594277494565169277761 1414329270270177497345380739951
### Nth Root
y√n
2√n 1035.27 102.338 32.1757 16.0702
## 1071791 as geometric shapes
### Circle
Diameter 2.14358e+06 6.73426e+06 3.60886e+12
### Sphere
Volume 5.15726e+18 1.44354e+13 6.73426e+06
### Square
Length = n
Perimeter 4.28716e+06 1.14874e+12 1.51574e+06
### Cube
Length = n
Surface area 6.89242e+12 1.2312e+18 1.8564e+06
### Equilateral Triangle
Length = n
Perimeter 3.21537e+06 4.97417e+11 928198
### Triangular Pyramid
Length = n
Surface area 1.98967e+12 1.45099e+17 875114
## Cryptographic Hash Functions
md5 2e0ffc188f5e5ee81f83b1aab4fe1823 d303aa6af677ede70d9e9917a9c8ff8edef6cbd9 fd2b1fa66f78f77243dc1e0cef2f301fa9571ea25f62a45b8c928d0180c520dc 9d3194befa96f14020f01877e86ec7b596e84625d24aca3f73f65d6e340ed4863de68e601f404736e8c7f98ed3547c77c622ace2824bee065fdcbc69ef276c42 06a049e5c3b090a999881f471a611678f8796f15 | 1,526 | 4,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-49 | latest | en | 0.811844 |
http://www.statemaster.com/correlations/edu_ele_sec_tot_num_of_sch_percap-total-number-schools-per-capita | 1,579,695,026,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606975.49/warc/CC-MAIN-20200122101729-20200122130729-00528.warc.gz | 284,571,427 | 14,540 | FACTOID # 5: Cheap sloppy joes: Looking for reduced-price lunches for schoolchildren? Head for Oklahoma!
Home Encyclopedia Statistics States A-Z Flags Maps FAQ About
WHAT'S NEW TOP STATS Which countries have the most: More Top Stats »
SEARCH ALL
Search encyclopedia, statistics and forums:
(* = Graphable)
## Correlations > Education Statistics > Elementary & Secondary > Total Number of Schools (per capita)
DEFINITION: The total number of schools as reported by the district. Per capita figures expressed per 1,000 population.
VIEW DATA: Totals Per capita
Definition Source Printable version
Bar Graph Map Correlations
Showing latest available data.
### Correlations between Education > Elementary & Secondary > Total Number of Schools (per capita) ...
A correlation is a statistical measure of similarity between at least two given sets of data. StateMaster's correlations compare two variables from our database and reveal statistical relationships between them. The percentages you see represent the strength (or likelihood) that a change in the topic variable is matched by a change in the listed variables below it. But remember: These correlations do not imply causation, that is, one does not necessarily cause the other. Also, not all variables contain all states, rather subsets of states matched together.
#### NOTES:
• Outliers have been removed only where they are outside 3 standard deviations of the mean.
• Only variable pairs where at least 15 states match for each have been considered.
• Strength is given by the correlation coefficient (R squared). It is the fraction of variation in Y that can be attributed to the variation in X. 100% signifies a perfect fit (R squared of 1). The top 50 such stats are displayed | 351 | 1,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-05 | latest | en | 0.862634 |
https://math.stackexchange.com/questions/2817124/is-the-definability-axiom-schema-consistent-with-zf | 1,571,021,994,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00312.warc.gz | 683,843,339 | 33,182 | # Is the definability axiom schema consistent with ZF?
Let us construct an axiom schema that declares that every set is definable from previous sets. For any formulas $\phi$, $\pi$, $\tau$, the following is instance of this axiom schema:
$$\forall x\forall y.\left( \begin{array} {rl} & \phi(\emptyset) \\ \land & \phi(\mathbb N) \\ \land & (\phi(x) \land \phi(y) \implies \phi(\{x,y\}) \\ \land & (\phi(x) \implies \phi (\bigcup x)) \\ \land & (\phi(x) \implies \phi (P(x))) \\ \land & (\phi(x) \implies \phi(\{z \in x: \pi(z)\})) \\ \land & (\text{\tau is a function formula} \land \phi(x) \implies \phi(\tau[x]))\end{array} \right) \implies \forall x. \phi(x)$$
It essentially goes through each of the ZF axioms, constructing a statement saying that $\phi$ satisfies the set is defines. For example, $(\text{$\tau$is a function} \land \phi(x) \implies \phi(pi[x]))$ represents the axiom schema of replacement. The axiom then states that this statement implies that all sets satisfy $\phi$.
Note in particular that this is more restrictive than $V = L$, since it asserts that ordinals need to be definable as well.
In particular, it implies there are no inaccessible cardinals.
Is this axiom schema consistient with ZF? (Also does it have a name?)
EDIT: The motivation for this axiom is to be comparable to the induction scheme on peano arithmetic. Essentially, its asking if we make ZF more like peano arithmetic, is it still consistient? We could even replace this with a second order axiom, and then ask if like peano arithmetic, it becomes categorical! Of course, this would be a different set theory then the normally envisioned one, but it is still of interest.
• Con(ZFC) does not imply there are inaccessible cardinals – Carl Mummert Jun 12 '18 at 15:55
• Rather than just posting a problem, you could improve the post by adding more context. What is the motivation for the problem? What question does it address? What attempts have you made to solve it? Posts that merely state a problem are generally discouraged, but this might appear to be a pattern; here is a a similarly unmotivated post from just a few minutes ago math.stackexchange.com/questions/2817100/… – Carl Mummert Jun 12 '18 at 15:58
• @CarlMummert Okay, I tried to add more motivation. – PyRulez Jun 12 '18 at 16:05
• There is already an induction schema for ZF, and it is equivalent to Replacement (over ZF-Replacement, of course). Or $\in$-induction, which is equivalent to Regularity (over ZF-Regularity, of course). – Asaf Karagila Jun 12 '18 at 16:05
• I don't see why this implies "only definable sets exist". How do you even define "definable set" internally? – Asaf Karagila Jun 12 '18 at 16:08
Your proposed axiom schema is not consistent with ZF:
ZF proves that $V_{\omega+\omega}$ exists, and if we instantiate your axiom schema with
• $\phi(x)$ meaning $x\in V_{\omega+\omega}$
• $\pi(x)$ meaning some tautology
• $\tau(x,y)$ meaning $x=y$
then it is easy to establish the premises of your schema. (The two last conjuncts on the LHS become trivial with this $\pi$ and $\tau$). It then concludes $\forall x: x\in V_{\omega+\omega}$, or in other words $V_{\omega+\omega}$ is a universal set, which is known to be inconsistent with Selection.
The fact that there are instances with more complex $\pi$ and $\tau$ that are not as obviously inconsistent with ZF doesn't prevent this instance from being inconsistent.
You might be able to say something closer to what you intend if you work in something like NBG instead of ZF, so you can quantify directly over your $\pi$ and $\tau$. | 950 | 3,580 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-43 | latest | en | 0.87403 |
http://www.slate.com/articles/briefing/articles/2000/07/math_for_jeopardy_players.html | 1,411,063,332,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657128337.85/warc/CC-MAIN-20140914011208-00162-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 787,860,494 | 81,987 | # Math for Jeopardy! Players
articles
July 20 2000 3:00 AM
# Math for Jeopardy! Players
## How contestants regularly blow their final bet.
Why do so many Jeopardy! contestants blow it in the final round? Picture this scene, from the March 21, 2000 show: Going into "Final Jeopardy!" Andrew was in the lead with \$8,000, Haley was in second with \$5,700, and Dave was back in third with \$2,700.
If you're Andrew in this situation, deciding on your bet is simple, assuming for argument's sake that the Final Jeopardy! category is neutral, i.e., one you know neither particularly well nor particularly poorly. Andrew's rational path is to wager the minimum he needs to put himself out of Haley's reach—that is, enough to give him twice her current score, plus \$1. That's \$3,401 in this case, which is precisely what Andrew wound up betting.
For Haley, betting is more complicated. Before I tell you how she should have bet, consider how she did bet. Like most contestants, she took a deep final-scene-of-Thelma-&-Louise breath, bet \$5,600, got the final question wrong, and lost. Andrew got it right, won \$11,401, and went back the next day. Dave, if anyone cares, bet the house, got Final Jeopardy! wrong, and wound up with nothing.
Here's what Haley should have bet: \$299. Notice that the way she actually bet, the only way she could have won is if she'd gotten Final Jeopardy! right and Andrew had gotten it wrong. Obviously, if Andrew answers correctly, the game's over, no matter what Haley does.
By betting \$299, Haley gives herself an extra chance. If Andrew gets it right, he still wins, as before. And, as before, if Haley gets it right and Andrew misses it, Haley wins. Here's the difference: If Haley bets \$299 and they both miss Final Jeopardy! Haley wins. Her final total would be \$5,401, while Andrew would be down at \$4,599.
Why can't Haley bet more than \$299? Because she has to guard against Dave, whose maximum score, if he bets everything and gets Final Jeopardy! right, would be \$5,400. Note that, with correct wagering, Dave is a non-factor in this Final Jeopardy! equation. Even if he bet it all and got it right, he still wouldn't be able to overcome Haley, even if she answered incorrectly.
All this wouldn't have helped Haley in this case, since Andrew answered Final Jeopardy! correctly. But had he missed it, she would've won.
For the player in second place, this all boils down to betting an amount that still gives you the win if both you and the player in the lead miss Final Jeopardy! A wagering-savvy former Jeopardy! champ has labeled this "The Two-Thirds Rule," because the second-place player needs at least two-thirds of the leading player's score going into Final Jeopardy! to be able to pull this off. (Click for more on the two-thirds rule.)
If the third player is close enough to worry about, as in the example above, you need to guard as much as possible against him. The following scenario from a recent show is a perfect illustration of this principle. Going into Final Jeopardy!, Melizza was in the lead with \$7,500. Second was Miles with \$7,300, and third was Judy with \$5,800.
Again, the leader's bet is easy to calculate, and Melizza did in fact wager the correct amount: \$7,101 (again, that gives Melizza twice Miles' score plus \$1 if she gets it right). Miles should bet \$4,301, while Judy should bet \$2,800.
Sports Nut
# Grandmaster Clash
One of the most amazing feats in chess history just happened, and no one noticed.
Sports Nut
# How Much Should You Loathe NFL Commissioner Roger Goodell?
Here are the facts.
Food
# How to Order Chinese Food
First, stop thinking of it as “Chinese food.”
# The Country Where Women Aren’t Allowed to Work Once They’re 36 Weeks’ Pregnant
The XX Factor
Sept. 18 2014 11:40 AM The Country Where Women Aren’t Allowed to Work Once They’re 36 Weeks’ Pregnant
Moneybox
Sept. 17 2014 5:10 PM The Most Awkward Scenario in Which a Man Can Hold a Door for a Woman
News & Politics
The World
Sept. 18 2014 1:34 PM Americans Fault Obama for Giving Them Exactly the Anti-ISIS Strategy They Want
Sept. 17 2014 1:36 PM Nate Silver Versus Princeton Professor: Who Has the Right Models?
Life
The Eye
Sept. 18 2014 12:47 PM How One of the Most Prolific Known Forgers in Modern History Faked Great Works of Art
Double X
The XX Factor
Sept. 18 2014 12:03 PM The NFL Opines on “the Role of the Female”
Slate Plus
Behind the Scenes
Sept. 18 2014 1:23 PM “It’s Not Everyday That You Can Beat the World Champion” An exclusive interview with chess grandmaster Fabiano Caruana.
Arts
Brow Beat
Sept. 18 2014 1:47 PM The Only 15 Netflix Hacks You’ll Ever Need
Technology
Future Tense
Sept. 18 2014 1:24 PM Can the Celebrities Whose Photos Were Stolen Really Sue Apple? It may be harder to prove “harm” than it seems.
Health & Science | 1,237 | 4,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2014-41 | longest | en | 0.972218 |
http://en.allexperts.com/q/Physics-1358/f/physics-46.htm | 1,490,408,034,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188773.10/warc/CC-MAIN-20170322212948-00471-ip-10-233-31-227.ec2.internal.warc.gz | 119,331,515 | 6,425 | You are here:
# Physics/physics
BallStacker wrote at 2014-03-22 00:06:50
It is real=practical Space with 4 spatial Dimensions and well known so called NormalSpace aka SpatialSpace is just one for many SubSpaces of HyperSpace like 'the' SubSpace aka PlanarSpace which is any of manifold of SubSpaces from NormalSpace. Compare with Sheet of Paper and Stack of Paper - there can be many different Sheets in same Room and dont have to be parallel. HyperSpace is therefore rather near as Spartial Space is inside. See Question on Paper: where is NormalSpace - asked from/at PlanarSpace.
BallStacker wrote at 2014-03-22 00:07:26
It is real=practical Space with 4 spatial Dimensions and well known so called NormalSpace aka SpatialSpace is just one for many SubSpaces of HyperSpace like 'the' SubSpace aka PlanarSpace which is any of manifold of SubSpaces from NormalSpace. Compare with Sheet of Paper and Stack of Paper - there can be many different Sheets in same Room and dont have to be parallel. HyperSpace is therefore rather near as Spartial Space is inside. See Question on Paper: where is NormalSpace - asked from/at PlanarSpace.
BallStacker wrote at 2014-03-22 00:08:15
It is real=practical Space with 4 spatial Dimensions and well known so called NormalSpace aka SpatialSpace is just one for many SubSpaces of HyperSpace like 'the' SubSpace aka PlanarSpace which is any of manifold of SubSpaces from NormalSpace. Compare with Sheet of Paper and Stack of Paper - there can be many different Sheets in same Room and dont have to be parallel. HyperSpace is therefore rather near as Spartial Space is inside. See Question on Paper: where is NormalSpace - asked from/at PlanarSpace.
BallStacker wrote at 2014-03-22 00:08:24
It is real=practical Space with 4 spatial Dimensions and well known so called NormalSpace aka SpatialSpace is just one for many SubSpaces of HyperSpace like 'the' SubSpace aka PlanarSpace which is any of manifold of SubSpaces from NormalSpace. Compare with Sheet of Paper and Stack of Paper - there can be many different Sheets in same Room and dont have to be parallel. HyperSpace is therefore rather near as Spartial Space is inside. See Question on Paper: where is NormalSpace - asked from/at PlanarSpace.
Physics
Volunteer
#### HAL838
##### Expertise
My expertise is in theoretical quantum physics and cosmology. No questions or homework problems on math application and engineering, please.
##### Experience
Intermediary science and general studies college degree with years of individual research and study.
Organizations | 601 | 2,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-13 | latest | en | 0.947008 |
http://www.ck12.org/book/CK-12-Algebra-I-Concepts/r2/section/4.8/ | 1,448,729,438,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398453576.62/warc/CC-MAIN-20151124205413-00295-ip-10-71-132-137.ec2.internal.warc.gz | 358,107,641 | 54,971 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts Go to the latest version.
# 4.8: Graphs Using Slope-Intercept Form
Difficulty Level: At Grade Created by: CK-12
0%
Progress
Practice Graphs Using Slope-Intercept Form
Progress
0%
What if you were given the equation of a line like y=2x7\begin{align*}y = -2x - 7\end{align*}? How could you determine its slope and y-intercept? After completing this Concept, you'll be able to identify the slope and y\begin{align*}y-\end{align*}intercept of linear equations like this one.
### Try This
To get a better understanding of what happens when you change the slope or the y\begin{align*}y-\end{align*}intercept of a linear equation, try playing with the Java applet at http://standards.nctm.org/document/eexamples/chap7/7.5/index.htm.
### Guidance
The total profit of a business is described by the equation y=15000x80000\begin{align*}y = 15000x - 80000\end{align*}, where x\begin{align*}x\end{align*} is the number of months the business has been running. How much profit is the business making per month, and what were its start-up costs? How much profit will it have made in a year?
Identify Slope and y\begin{align*}y-\end{align*}intercept
So far, we’ve been writing a lot of our equations in slope-intercept form—that is, we’ve been writing them in the form y=mx+b\begin{align*}y = mx + b\end{align*}, where m\begin{align*}m\end{align*} and b\begin{align*}b\end{align*} are both constants. It just so happens that m\begin{align*}m\end{align*} is the slope and the point (0,b)\begin{align*}(0, b)\end{align*} is the y\begin{align*}y-\end{align*}intercept of the graph of the equation, which gives us enough information to draw the graph quickly.
#### Example A
Identify the slope and y\begin{align*}y-\end{align*}intercept of the following equations.
a) y=3x+2\begin{align*}y = 3x + 2\end{align*}
b) y=0.5x3\begin{align*}y = 0.5x - 3\end{align*}
c) y=7x\begin{align*}y = -7x\end{align*}
d) y=4\begin{align*}y = -4\end{align*}
Solution
a) Comparing , we can see that m=3\begin{align*}m = 3\end{align*} and b=2\begin{align*}b = 2\end{align*}. So y=3x+2\begin{align*}y = 3x + 2\end{align*} has a slope of 3 and a y\begin{align*}y-\end{align*}intercept of (0, 2).
b) has a slope of 0.5 and a y\begin{align*}y-\end{align*}intercept of (0, -3).
Notice that the intercept is negative. The b\begin{align*}b-\end{align*}term includes the sign of the operator (plus or minus) in front of the number—for example, y=0.5x3\begin{align*}y = 0.5x - 3\end{align*} is identical to y=0.5x+(3)\begin{align*}y = 0.5x + (-3)\end{align*}, and that means that b\begin{align*}b\end{align*} is -3, not just 3.
c) At first glance, this equation doesn’t look like it’s in slope-intercept form. But we can rewrite it as y=7x+0\begin{align*}y = -7x + 0\end{align*}, and that means it has a slope of -7 and a y\begin{align*}y-\end{align*}intercept of (0, 0). Notice that the slope is negative and the line passes through the origin.
d) We can rewrite this one as y=0x4\begin{align*}y = 0x - 4\end{align*}, giving us a slope of 0 and a y\begin{align*}y-\end{align*}intercept of (0, -4). This is a horizontal line.
#### Example B
Identify the slope and y\begin{align*}y-\end{align*}intercept of the lines on the graph shown below.
The intercepts have been marked, as well as some convenient lattice points that the lines pass through.
Solution
a) The y\begin{align*}y-\end{align*}intercept is (0, 5). The line also passes through (2, 3), so the slope is ΔyΔx=22=1\begin{align*}\frac{\Delta y}{\Delta x}=\frac{-2}{2} = -1\end{align*}.
b) The y\begin{align*}y-\end{align*}intercept is (0, 2). The line also passes through (1, 5), so the slope is ΔyΔx=31=3\begin{align*}\frac{\Delta y}{\Delta x} = \frac {3}{1} = 3\end{align*}.
c) The y\begin{align*}y-\end{align*}intercept is (0, -1). The line also passes through (2, 3), so the slope is ΔyΔx=42=2\begin{align*}\frac{\Delta y}{\Delta x} = \frac {4}{2} = 2\end{align*}.
d) The y\begin{align*}y-\end{align*}intercept is (0, -3). The line also passes through (4, -4), so the slope is ΔyΔx=14=14\begin{align*}\frac{\Delta y}{\Delta x} = \frac {-1}{4} = - \frac{1}{4}\end{align*} or -0.25.
Graph an Equation in Slope-Intercept Form
Once we know the slope and intercept of a line, it’s easy to graph it. Just remember what slope means. Let's look back at this example from Lesson 4.1.
Ali is trying to work out a trick that his friend showed him. His friend started by asking him to think of a number, then double it, then add five to the result. Ali has written down a rule to describe the first part of the trick. He is using the letter x\begin{align*}x\end{align*} to stand for the number he thought of and the letter y\begin{align*}y\end{align*} to represent the final result of applying the rule. He wrote his rule in the form of an equation: y=2x+5\begin{align*}y = 2x + 5\end{align*}.
Help him visualize what is going on by graphing the function that this rule describes.
In that example, we constructed a table of values, and used that table to plot some points to create our graph.
We also saw another way to graph this equation. Just by looking at the equation, we could see that the y\begin{align*}y-\end{align*}intercept was (0, 5), so we could start by plotting that point. Then we could also see that the slope was 2, so we could find another point on the graph by going over 1 unit and up 2 units. The graph would then be the line between those two points.
Here’s another problem where we can use the same method.
#### Example C
Graph the following function: y=3x+5\begin{align*}y=-3x+5\end{align*}
Solution
To graph the function without making a table, follow these steps:
1. Identify the y\begin{align*}y-\end{align*}intercept: b=5\begin{align*}b = 5\end{align*}
2. Plot the intercept: (0, 5)
3. Identify the slope: m=3\begin{align*}m = -3\end{align*}. (This is equal to 31\begin{align*}\frac {-3}{1}\end{align*}, so the rise is -3 and the run is 1.)
4. Move over 1 unit and down 3 units to find another point on the line: (1, 2)
5. Draw the line through the points (0, 5) and (1, 2).
Notice that to graph this equation based on its slope, we had to find the rise and run—and it was easiest to do that when the slope was expressed as a fraction. That’s true in general: to graph a line with a particular slope, it’s easiest to first express the slope as a fraction in simplest form, and then read off the numerator and the denominator of the fraction to get the rise and run of the graph.
#### Example D
Find integer values for the rise and run of the following slopes, then graph lines with corresponding slopes.
a) m=3\begin{align*}m=3\end{align*}
b) m=2\begin{align*}m=-2\end{align*}
Solution
a)
b)
Changing the Slope or Intercept of a Line
The following graph shows a number of lines with different slopes, but all with the same y\begin{align*}y-\end{align*}intercept: (0, 3).
You can see that all the functions with positive slopes increase as we move from left to right, while all functions with negative slopes decrease as we move from left to right. Another thing to notice is that the greater the slope, the steeper the graph.
This graph shows a number of lines with the same slope, but different y\begin{align*}y-\end{align*}intercepts.
Notice that changing the intercept simply translates (shifts) the graph up or down. Take a point on the graph of y=2x\begin{align*}y = 2x\end{align*}, such as (1, 2). The corresponding point on y=2x+3\begin{align*}y = 2x + 3\end{align*} would be (1, 5). Adding 3 to the y\begin{align*}y-\end{align*}intercept means we also add 3 to every other y\begin{align*}y-\end{align*}value on the graph. Similarly, the corresponding point on the y=2x3\begin{align*}y = 2x - 3\end{align*} line would be (1, -1); we would subtract 3 from the y\begin{align*}y-\end{align*}value and from every other y\begin{align*}y-\end{align*}value.
Notice also that these lines all appear to be parallel. Are they truly parallel?
To answer that question, we’ll use a technique that you’ll learn more about in a later chapter. We’ll take 2 of the equations—say, y=2x\begin{align*}y = 2x\end{align*} and y=2x+3\begin{align*}y = 2x + 3\end{align*}—and solve for values of x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} that satisfy both equations. That will tell us at what point those two lines intersect, if any. (Remember that parallel lines, by definition, are lines that don’t intersect.)
So what values would satisfy both y=2x\begin{align*}y = 2x\end{align*} and y=2x+3\begin{align*}y = 2x + 3\end{align*}? Well, if both of those equations were true, then y\begin{align*}y\end{align*} would be equal to both 2x\begin{align*}2x\end{align*} and 2x+3\begin{align*}2x + 3\end{align*}, which means those two expressions would also be equal to each other. So we can get our answer by solving the equation 2x=2x+3\begin{align*}2x = 2x + 3\end{align*}.
But what happens when we try to solve that equation? If we subtract 2x\begin{align*}2x\end{align*} from both sides, we end up with 0=3\begin{align*}0 = 3\end{align*}. That can’t be true no matter what x\begin{align*}x\end{align*} equals. And that means that there just isn’t any value for x\begin{align*}x\end{align*} that will make both of the equations we started out with true. In other words, there isn’t any point where those two lines intersect. They are parallel, just as we thought.
And we’d find out the same thing no matter which two lines we’d chosen. In general, since changing the intercept of a line just results in shifting the graph up or down, the new line will always be parallel to the old line as long as the slope stays the same.
Any two lines with identical slopes are parallel.
Watch this video for help with the Examples above.
### Vocabulary
• A common form of a line (linear equation) is slope-intercept form: y=mx+b\begin{align*}y=mx+b\end{align*}, where m\begin{align*}m\end{align*} is the slope and the point (0,b)\begin{align*}(0, b)\end{align*} is the y\begin{align*}y-\end{align*}intercept
• Graphing a line in slope-intercept form is a matter of first plotting the y\begin{align*}y-\end{align*}intercept (0,b)\begin{align*}(0, b)\end{align*}, then finding a second point based on the slope, and using those two points to graph the line.
• Any two lines with identical slopes are parallel.
### Guided Practice
Find integer values for the rise and run of the following slopes, then graph lines with corresponding slopes.
a) m=0.75\begin{align*}m=0.75\end{align*}
b) m=0.375\begin{align*}m=-0.375\end{align*}
Solution:
a)
b)
### Practice
Identify the slope and y\begin{align*}y-\end{align*}intercept for the following equations.
1. y=2x+5\begin{align*}y=2x+5\end{align*}
2. y=0.2x+7\begin{align*}y=-0.2x+7\end{align*}
3. y=x\begin{align*}y=x\end{align*}
4. y=3.75\begin{align*}y=3.75\end{align*}
Identify the slope of the following lines.
Identify the slope and y\begin{align*}y-\end{align*}intercept for the following functions.
For 7-10, plot the following functions on a graph.
1. y=2x+5\begin{align*}y=2x+5\end{align*}
2. y=0.2x+7\begin{align*}y=-0.2x+7\end{align*}
3. y=x\begin{align*}y=x\end{align*}
4. y=3.75\begin{align*}y=3.75\end{align*}
1. Which two of the following lines are parallel?
1. y=2x+5\begin{align*}y=2x+5\end{align*}
2. y=0.2x+7\begin{align*}y=-0.2x+7\end{align*}
3. y=x\begin{align*}y=x\end{align*}
4. y=3.75\begin{align*}y=3.75\end{align*}
5. y=15x11\begin{align*}y= -\frac{1}{5}x-11\end{align*}
6. y=5x+5\begin{align*}y=-5x+5\end{align*}
7. y=3x+11\begin{align*}y=-3x+11\end{align*}
8. y=3x+3.5\begin{align*}y=3x+3.5\end{align*}
2. What is the y\begin{align*}y-\end{align*}intercept of the line passing through (1, -4) and (3, 2)?
3. What is the y\begin{align*}y-\end{align*}intercept of the line with slope -2 that passes through (3, 1)?
4. Line A\begin{align*}A\end{align*} passes through the points (2, 6) and (-4, 3). Line B\begin{align*}B\end{align*} passes through the point (3, 2.5), and is parallel to line A\begin{align*}A\end{align*}
1. Write an equation for line A\begin{align*}A\end{align*} in slope-intercept form.
2. Write an equation for line B\begin{align*}B\end{align*} in slope-intercept form.
5. Line C\begin{align*}C\end{align*} passes through the points (2, 5) and (1, 3.5). Line D\begin{align*}D\end{align*} is parallel to line C\begin{align*}C\end{align*}, and passes through the point (2, 6). Name another point on line D\begin{align*}D\end{align*}. (Hint: you can do this without graphing or finding an equation for either line.)
### Vocabulary Language: English
Cartesian Plane
Cartesian Plane
The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
linear equation
linear equation
A linear equation is an equation between two variables that produces a straight line when graphed.
Linear Function
Linear Function
A linear function is a relation between two variables that produces a straight line when graphed.
Slope-Intercept Form
Slope-Intercept Form
The slope-intercept form of a line is $y = mx + b,$ where $m$ is the slope and $b$ is the $y-$intercept.
Oct 01, 2012
Oct 08, 2015 | 4,402 | 13,363 | {"found_math": true, "script_math_tex": 111, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 4, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2015-48 | longest | en | 0.839544 |
http://www.andrews-lines.com/action-reaction-course-2/ | 1,542,173,099,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741628.8/warc/CC-MAIN-20181114041344-20181114063344-00342.warc.gz | 369,368,445 | 6,360 | # Action Reaction Course 2
THE FOUNDATION FOR
ECONOMIC STABILIZATION
Alan H. Andrews Trustee
5738 S.W. 53rd Terrace
South Miami FL 33155
You will find enclosed the first study of the Course concerning the ML (median line) Method. This enables you to know where the trend of anything that fluctuates at random is headed. What everyone wants to know is where the latest trend is headed, and where the next pivot (P) will be from which the reverse trend will start.
The probability of the next P being at the latest ML seems to be about 80%, and even without any additional rules that enable you to be constantly either long or short , the profit potential of this simple rule is tremendous for you.
Although Marechal never left us exactly how he was able to predict twenty years in advance what his copyrighted chart showed the Dow Jones Industrial Averages would do, you can draw in the MLs from each P bisecting the distance between the 2 latest alternate Ps, and see that nearly every time the new P occurred when prices met that latest ML. You’ll also see that on the right hand side of his chart prices were too strong to drop to the ML that started from the high in 1945, which always signals that a big rise is ahead unless the next trend fails to reach the new ML. This cancels out the prior signal and signals a big move contrary to the big move previously signaled. And as there was no contrary signal after prices failed to drop to reach the ML from the high in 1945, you could be confident of realizing a big gain from your long position taken as soon as prices crossed the parallel to the ML from the high of 1945. You draw this parallel from the third top that the ML was drawn half way below on the distance to Previous P.
You can now tell from the enclosed Glossary what the abbreviations mean, in the right hand column of each weekly letter. This enables you to understand the scientific reason for each new position taken based on simple geometry. When you change a position your new methods enable you to be one of the few persons who knows how to be constantly either long or short, in this way you make profits after each rise and fall that follows the rise. You may be whip-sawed a few times but if you get you order in before the market opens the next day, should prices move against the position you have just taken, your losses will be small and often show a small profit.
You will see all this after you’ve done some “paper trading” which you should start on right away showing on your chart where each position was taken. You should concentrate on the ML method applying that even if you have had experience with other methods. For we learn best by concentrating on one thing at a time. When you have a question mark where the question arose and send me a copy of your chart that should also list our profits from the two contracts you take each time you change position. When you write out your question leave a space where my answer can be written and mailed back to you.
After you see that your paper trading has made well over the 100% profit rate, it will indicate you are ready to learn the Action and Reaction Method to which my friend the late Roger Babson attributed his fortune of over \$50,000,000. Then after that let us know and you will be sent the rest of the Course Studies.
Sincerely.
Alan H. Andrews, Trustee FFES. | 720 | 3,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-47 | longest | en | 0.954687 |
https://clima.github.io/OceananigansDocumentation/stable/literated/two_dimensional_turbulence/ | 1,726,855,783,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00468.warc.gz | 146,172,290 | 7,893 | # Two dimensional turbulence example
In this example, we initialize a random velocity field and observe its turbulent decay in a two-dimensional domain. This example demonstrates:
• How to run a model with no tracers and no buoyancy model.
• How to use computed Fields to generate output.
## Install dependencies
First let's make sure we have all required packages installed.
using Pkg
pkg"add Oceananigans, CairoMakie"
## Model setup
We instantiate the model with an isotropic diffusivity. We use a grid with 128² points, a fifth-order advection scheme, third-order Runge-Kutta time-stepping, and a small isotropic viscosity. Note that we assign Flat to the z direction.
using Oceananigans
grid = RectilinearGrid(size=(128, 128), extent=(2π, 2π), topology=(Periodic, Periodic, Flat))
model = NonhydrostaticModel(; grid,
timestepper = :RungeKutta3,
closure = ScalarDiffusivity(ν=1e-5))
NonhydrostaticModel{CPU, RectilinearGrid}(time = 0 seconds, iteration = 0)
├── grid: 128×128×1 RectilinearGrid{Float64, Periodic, Periodic, Flat} on CPU with 3×3×0 halo
├── timestepper: RungeKutta3TimeStepper
├── advection scheme: Upwind Biased reconstruction order 5
├── tracers: ()
├── closure: ScalarDiffusivity{ExplicitTimeDiscretization}(ν=1.0e-5)
├── buoyancy: Nothing
└── coriolis: Nothing
## Random initial conditions
Our initial condition randomizes model.velocities.u and model.velocities.v. We ensure that both have zero mean for aesthetic reasons.
using Statistics
u, v, w = model.velocities
uᵢ = rand(size(u)...)
vᵢ = rand(size(v)...)
uᵢ .-= mean(uᵢ)
vᵢ .-= mean(vᵢ)
set!(model, u=uᵢ, v=vᵢ)
## Setting up a simulation
We set-up a simulation that stops at 50 time units, with an initial time-step of 0.1, and with adaptive time-stepping and progress printing.
simulation = Simulation(model, Δt=0.2, stop_time=50)
Simulation of NonhydrostaticModel{CPU, RectilinearGrid}(time = 0 seconds, iteration = 0)
├── Next time step: 200 ms
├── Elapsed wall time: 0 seconds
├── Wall time per iteration: NaN days
├── Stop time: 50 seconds
├── Stop iteration : Inf
├── Wall time limit: Inf
├── Callbacks: OrderedDict with 4 entries:
│ ├── stop_time_exceeded => Callback of stop_time_exceeded on IterationInterval(1)
│ ├── stop_iteration_exceeded => Callback of stop_iteration_exceeded on IterationInterval(1)
│ ├── wall_time_limit_exceeded => Callback of wall_time_limit_exceeded on IterationInterval(1)
│ └── nan_checker => Callback of NaNChecker for u on IterationInterval(100)
├── Output writers: OrderedDict with no entries
└── Diagnostics: OrderedDict with no entries
The TimeStepWizard helps ensure stable time-stepping with a Courant-Freidrichs-Lewy (CFL) number of 0.7.
wizard = TimeStepWizard(cfl=0.7, max_change=1.1, max_Δt=0.5)
simulation.callbacks[:wizard] = Callback(wizard, IterationInterval(10))
Callback of TimeStepWizard(cfl=0.7, max_Δt=0.5, min_Δt=0.0) on IterationInterval(10)
## Logging simulation progress
We set up a callback that logs the simulation iteration and time every 100 iterations.
using Printf
function progress_message(sim)
max_abs_u = maximum(abs, sim.model.velocities.u)
walltime = prettytime(sim.run_wall_time)
return @info @sprintf("Iteration: %04d, time: %1.3f, Δt: %.2e, max(|u|) = %.1e, wall time: %s\n",
iteration(sim), time(sim), sim.Δt, max_abs_u, walltime)
end
add_callback!(simulation, progress_message, IterationInterval(100))
## Output
We set up an output writer for the simulation that saves vorticity and speed every 20 iterations.
### Computing vorticity and speed
To make our equations prettier, we unpack u, v, and w from the NamedTuple model.velocities:
u, v, w = model.velocities
NamedTuple with 3 Fields on 128×128×1 RectilinearGrid{Float64, Periodic, Periodic, Flat} on CPU with 3×3×0 halo:
├── u: 128×128×1 Field{Face, Center, Center} on RectilinearGrid on CPU
├── v: 128×128×1 Field{Center, Face, Center} on RectilinearGrid on CPU
└── w: 128×128×1 Field{Center, Center, Face} on RectilinearGrid on CPU
Next we create two Fields that calculate (i) vorticity that measures the rate at which the fluid rotates and is defined as
$$$ω = ∂_x v - ∂_y u \, ,$$$
ω = ∂x(v) - ∂y(u)
BinaryOperation at (Face, Face, Center)
├── grid: 128×128×1 RectilinearGrid{Float64, Periodic, Periodic, Flat} on CPU with 3×3×0 halo
└── tree:
- at (Face, Face, Center)
├── ∂xᶠᶠᶜ at (Face, Face, Center) via identity
│ └── 128×128×1 Field{Center, Face, Center} on RectilinearGrid on CPU
└── ∂yᶠᶠᶜ at (Face, Face, Center) via identity
└── 128×128×1 Field{Face, Center, Center} on RectilinearGrid on CPU
We also calculate (ii) the speed of the flow,
$$$s = \sqrt{u^2 + v^2} \, .$$$
s = sqrt(u^2 + v^2)
UnaryOperation at (Face, Center, Center)
├── grid: 128×128×1 RectilinearGrid{Float64, Periodic, Periodic, Flat} on CPU with 3×3×0 halo
└── tree:
sqrt at (Face, Center, Center) via identity
└── + at (Face, Center, Center)
├── ^ at (Face, Center, Center)
│ ├── 128×128×1 Field{Face, Center, Center} on RectilinearGrid on CPU
│ └── 2
└── ^ at (Center, Face, Center)
├── 128×128×1 Field{Center, Face, Center} on RectilinearGrid on CPU
└── 2
We pass these operations to an output writer below to calculate and output them during the simulation.
filename = "two_dimensional_turbulence"
simulation.output_writers[:fields] = JLD2OutputWriter(model, (; ω, s),
schedule = TimeInterval(0.6),
filename = filename * ".jld2",
overwrite_existing = true)
JLD2OutputWriter scheduled on TimeInterval(600 ms):
├── filepath: ./two_dimensional_turbulence.jld2
├── 2 outputs: (ω, s)
├── array type: Array{Float64}
├── including: [:grid, :coriolis, :buoyancy, :closure]
├── file_splitting: NoFileSplitting
└── file size: 27.8 KiB
## Running the simulation
Pretty much just
run!(simulation)
[ Info: Initializing simulation...
[ Info: Iteration: 0000, time: 0.000, Δt: 1.00e-01, max(|u|) = 6.8e-01, wall time: 0 seconds
[ Info: ... simulation initialization complete (7.844 seconds)
[ Info: Executing initial time step...
[ Info: ... initial time step complete (4.352 seconds).
[ Info: Iteration: 0100, time: 6.413, Δt: 6.76e-02, max(|u|) = 3.0e-01, wall time: 13.653 seconds
[ Info: Iteration: 0200, time: 13.758, Δt: 7.51e-02, max(|u|) = 3.1e-01, wall time: 14.895 seconds
[ Info: Iteration: 0300, time: 20.625, Δt: 6.98e-02, max(|u|) = 2.9e-01, wall time: 16.388 seconds
[ Info: Iteration: 0400, time: 27.789, Δt: 9.50e-02, max(|u|) = 2.5e-01, wall time: 17.862 seconds
[ Info: Iteration: 0500, time: 36.253, Δt: 8.09e-02, max(|u|) = 2.6e-01, wall time: 19.177 seconds
[ Info: Iteration: 0600, time: 44.400, Δt: 8.21e-02, max(|u|) = 2.5e-01, wall time: 20.364 seconds
[ Info: Simulation is stopping after running for 21.447 seconds.
[ Info: Simulation time 50.000 seconds equals or exceeds stop time 50 seconds.
## Visualizing the results
ω_timeseries = FieldTimeSeries(filename * ".jld2", "ω")
s_timeseries = FieldTimeSeries(filename * ".jld2", "s")
times = ω_timeseries.times
0.0:0.6:49.8
Construct the $x, y, z$ grid for plotting purposes,
xω, yω, zω = nodes(ω_timeseries)
xs, ys, zs = nodes(s_timeseries)
and animate the vorticity and fluid speed.
using CairoMakie
set_theme!(Theme(fontsize = 24))
fig = Figure(size = (800, 500))
axis_kwargs = (xlabel = "x",
ylabel = "y",
limits = ((0, 2π), (0, 2π)),
aspect = AxisAspect(1))
ax_ω = Axis(fig[2, 1]; title = "Vorticity", axis_kwargs...)
ax_s = Axis(fig[2, 2]; title = "Speed", axis_kwargs...)
We use Makie's Observable to animate the data. To dive into how Observables work we refer to Makie.jl's Documentation.
n = Observable(1)
Observable(1)
Now let's plot the vorticity and speed.
ω = @lift ω_timeseries[$n] s = @lift s_timeseries[$n]
heatmap!(ax_ω, ω; colormap = :balance, colorrange = (-2, 2))
heatmap!(ax_s, s; colormap = :speed, colorrange = (0, 0.2))
title = @lift "t = " * string(round(times[\$n], digits=2))
Label(fig[1, 1:2], title, fontsize=24, tellwidth=false)
fig
Finally, we record a movie.
frames = 1:length(times)
@info "Making a neat animation of vorticity and speed..."
record(fig, filename * ".mp4", frames, framerate=24) do i
n[] = i
end
[ Info: Making a neat animation of vorticity and speed... | 2,620 | 8,150 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.704125 |
https://www.softmath.com/algebra-software-2/add-and-multiply-fractions.html | 1,721,041,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00004.warc.gz | 894,553,676 | 9,398 | Related topics:
printable worksheets for adding and subtracting positive/negative numbers | how to multiply divide add subtract radicals | algebraic expressions combine similar terms | exponents and factoring | junior high permutations and combinations | factoring trinomials with tic tac | limits calculator online | help adding and subtracting radical expressions solver | free year 8 maths tutor program | Algebra Programs | simplifying algebraic equations | algebra 2 with trigonometry book | step by step equation solver
Author Message
cwconnectal
Registered: 27.11.2001
From:
Posted: Tuesday 18th of Feb 08:21 I am in a real fix . Somebody assist me please. I am having a lot of troubles with graphing function, function domain and radical equations and especially with add and multiply fractions. I have to show some speedy change in my math. someone told me there are plenty of Applications available online which can assist you in algebra. I can shell out some money too for an effective and inexpensive software which helps me with my studies. Any link is greatly appreciated. Thanks.
Jahm Xjardx
Registered: 07.08.2005
From: Odense, Denmark, EU
Posted: Wednesday 19th of Feb 21:31 I understand your situation because I had the same issues when I went to high school. I was very weak in math, especially in add and multiply fractions and my grades were really poor. I started using Algebrator to help me solve problems as well as with my assignments and eventually I started getting A’s in math. This is a remarkably good product because it explains the problems in a step-by-step manner so we understand them well. I am absolutely sure that you will find it useful too.
DVH
Registered: 20.12.2001
From:
Posted: Thursday 20th of Feb 07:21 Algebrator is a incredible software. All I had to do with my problems with equivalent fractions, graphing circles and trinomials was to merely type in the problems; click the ‘solve’ and presto, the result just popped out step-by-step in an effortless manner. I have used this to problems in Intermediate algebra, Intermediate algebra and Algebra 1. I would confidently say that this is just the answer for you.
Ctonlyom
Registered: 23.05.2004
From:
Posted: Thursday 20th of Feb 20:13 Hello again. Thanks a lot for the beneficial advice. I usually never trust math software ; however, this piece of program seems worth trying. Can I get a link to it?
cmithy_dnl
Registered: 08.01.2002
From: Australia
Posted: Friday 21st of Feb 07:48 I am glad to hear that you are ready to change your attitude towards this subject. Please visit https://softmath.com/ordering-algebra.html to grab a copy of this software. | 612 | 2,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.945444 |
http://spankbush.com/comparing-fractions-on-a-number-line-worksheet/ | 1,558,842,544,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00549.warc.gz | 178,789,471 | 10,538 | # Comparing Fractions On A Number Line Worksheet
## Comparing Fractions On A Number Line Worksheet
Math worksheets decimals on a number line from Comparing Fractions On A Number Line Worksheet
, from: myscres.com
Ordering Fractions Worksheet Free Luxury Greater Than Less Than from Comparing Fractions On A Number Line Worksheet
, from: migidiobourifa.com
Relationships Between Fractions Decimals Ratios & Percentages from Comparing Fractions On A Number Line Worksheet
, from: utas.edu.au
Free Fractions A Number Line Worksheets Inspirational Fraction from Comparing Fractions On A Number Line Worksheet
, from: edinblogs.net
Fractions A Number Line Worksheet from Comparing Fractions On A Number Line Worksheet
, from: fronteirastral.com
Fractions A Number Line Worksheet from Comparing Fractions On A Number Line Worksheet
, from: fronteirastral.com
44 Best paring Fractions Worksheet 4th Grade from Comparing Fractions On A Number Line Worksheet
, from: swiftcantrellpark.org
Fractions A Number Line 3rd Grade Worksheets Luxury 3rd Grade from Comparing Fractions On A Number Line Worksheet
, from: lyricjamesbooks.net
fractions on number line worksheets math aids this fractions on number line worksheet will produce problems for the children to correctly mark the fractions on the given number lines the worksheet will produce five number lines per page each number line will contain one group from the fractions that are selected below fraction worksheets paring fractions to decimals with number line each worksheet has 10 problems determining which amount is larger using a number line create new sheet e atta time flash cards fractions on number line worksheets softschools fractions on number line coloring worksheet fraction addition worksheets improper fractions and mixed number worksheets paring fractions worksheets fractions on number line game fractions addition on number line game more topics handwriting spanish facts paring and ordering fractions super teacher worksheets paring and ordering fractions first find each fraction on the number line pare the fractions using the less than greater than or equal to symbols students can use the number line to help from this index page you can jump to worksheets on paring 4 digit numbers 5 digit numbers or 6 digit numbers also includes links to paring fractions number line worksheets printable paring fractions number line showing top 8 worksheets in the category paring fractions number line some of the worksheets displayed are fractions number line pare with circles name write a number sentence that paring and ordering fractions decimals and percents equivalent fractions and paring fractions are you my mixed fractions number line paring fractions number
Fractions A Number Line 3rd Grade Worksheets Luxury 3rd Grade from Comparing Fractions On A Number Line Worksheet
, from: lyricjamesbooks.net
Construction Estimating Spreadsheet with Annuity Worksheet 0d Tags from Comparing Fractions On A Number Line Worksheet
, from: logoterapia.info
Worksheets Ordering Fractions And Decimals Math Myscres Percents from Comparing Fractions On A Number Line Worksheet
Equivalent Fraction Worksheets 6th Grade Math from Comparing Fractions On A Number Line Worksheet
, from: thoughtco.com
Fraction line plot worksheets from Comparing Fractions On A Number Line Worksheet
, from: myscres.com
Fractions A Number Line Worksheet Pdf from Comparing Fractions On A Number Line Worksheet
, from: reviewwalls.co
Fraction Printables Math in the Classroom Pinterest from Comparing Fractions On A Number Line Worksheet
, from: pinterest.com
Math worksheets decimals on a number line from Comparing Fractions On A Number Line Worksheet
, from: myscres.com
Greater Than Less Than Worksheets Math Aids from Comparing Fractions On A Number Line Worksheet
, from: math-aids.com
Fractions A Number Line Worksheet fadeintofantasy from Comparing Fractions On A Number Line Worksheet | 793 | 3,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-22 | latest | en | 0.776794 |
http://www.gurufocus.com/term/lynchvalue/MWIV/Peter%2BLynch%2BFair%2BValue/MWI%2BVeterinary%2BSupply%252C%2BInc | 1,474,809,477,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660208.2/warc/CC-MAIN-20160924173740-00268-ip-10-143-35-109.ec2.internal.warc.gz | 498,487,074 | 26,614 | Switch to:
MWI Veterinary Supply Inc (NAS:MWIV)
Peter Lynch Fair Value
\$126.90 (As of Today)
Peter Lynch Fair Value applies to growing companies. The ideal range for the growth rate is between 10 - 20% a year. Peter Lynch thinks that the fair P/E value for a growth company equals its growth rate, that is PEG = 1. The earnings here is trailing twelve month (TTM) earnings. The growth rate we use is the average growth rate for earnings per share over the past 5 years. If 5-Year Earnings Growth Rate is greater than 25% a year, we use 25. If 5-Year Earnings Growth Rate is smaller than 5% a year, we do not calculate Peter Lynch Fair Value.
Here, as of today, MWI Veterinary Supply Inc's PEG is 1. MWI Veterinary Supply Inc's 5-Year EBITDA Growth Rate is 22.46. MWI Veterinary Supply Inc's Earnings Per Share without Non-Recurring Items (NRI) for the trailing twelve months (TTM) ended in Sep. 2014 was \$5.65. Therefore, the Peter Lynch Fair Value for today is \$126.90.
As of today, MWI Veterinary Supply Inc's share price is \$190.00. MWI Veterinary Supply Inc's Peter Lynch fair value is \$126.90. Therefore, MWI Veterinary Supply Inc's Price to Peter Lynch Fair Value Ratio for today is 1.50.
Note: Please don't confuse Peter Lynch Fair Value with the value reached in Peter Lynch Chart.
Definition
MWI Veterinary Supply Inc's Peter Lynch Fair Value for today is calculated as
Peter Lynch Fair Value = PEG * 5-Year EBITDA Growth Rate * Earnings Per Share (NRI) (TTM) = 1 * 22.46 * 5.65 = 126.90
MWI Veterinary Supply Inc's Earnings Per Share (NRI) for the trailing twelve months (TTM) ended in Sep. 2014 was 1.45 (Dec. 2013 ) + 1.32 (Mar. 2014 ) + 1.52 (Jun. 2014 ) + 1.36 (Sep. 2014 ) = \$5.65.
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
If 5-Year Earnings Growth Rate is greater than 25% a year, we use 25.
If 5-Year Earnings Growth Rate is smaller than 5% a year, we do not calculate Peter Lynch Fair Value.
Please note that we use the 5-year average growth rate of EBITDA per share as the growth rate. EBITDA growth is subject to less manipulations than net earnings per share. In the calculation, PEG=1 because Peter Lynch thinks that the fair P/E ratio of the growth stock is equal to its earnings growth rate.
Explanation
Peter Lynch Fair Value applies to growing companies. The ideal range for the growth rate is between 10 - 20% a year.
Peter Lynch thinks that the fair P/E value for a growth company equals its growth rate, that is PEG = 1. The earnings here is trailing twelve month (TTM) earnings. The growth rate we use is the average growth rate for earnings per share over the past 5 years.
Please don't confuse Peter Lynch Fair Value with the value reached in Peter Lynch Chart. In Peter Lynch chart, a fixed P/E ratio of 15 is used to draw the Earnings Line. Therefore the value reached has a P/E ratio of 15. But in Peter Lynch Fair Value calculation, P/E equals to the growth rate of EBITDA over the past 5 years, which is 22.46 instead of 15 in this case.
MWI Veterinary Supply Inc's Price to Peter Lynch Fair Value Ratio for today is calculated as
Price to Peter Lynch Fair Value = Share Price / Peter Lynch Fair Value = 190.00 / 126.90 = 1.50
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
MWI Veterinary Supply Inc Annual Data
Sep05 Sep06 Sep07 Sep08 Sep09 Sep10 Sep11 Sep12 Sep13 Sep14 lynchvalue N/A N/A N/A N/A N/A 43.85 83.55 105.75 123.75 123.34
MWI Veterinary Supply Inc Quarterly Data
Jun12 Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 lynchvalue 101.00 105.75 112.50 116.25 120.50 123.75 126.70 124.83 124.05 123.34
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,080 | 4,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2016-40 | longest | en | 0.901937 |
https://garagedoorrepairaurora.co/quality-garage-door-opener-chamberlain-centennial-colorado-call-now.html | 1,544,499,505,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823550.42/warc/CC-MAIN-20181211015030-20181211040530-00615.warc.gz | 614,023,801 | 4,470 | The low rating on the CS is due to the fact that inwas out of town and my wife called because the garage door wouldn't open. We had repairs to the only other exit, and the CS said because it could be opened manually, it was no emergency. She was stuck in the house for over Sixteen hours. No emergency crew came out. Steve, a tech who came out the first time, fixed the door. It is not his fault CS took their sweet time to help a 45 year customer. I commend Steve. I do NOT have anything good to say about CS.
Beware of improprer prior installations: Sometimes the existing door installation is not correct, and the old springs should not be used as a specification for replacements. For example, the old springs might have been replaced with incorrect sizes because the last repairman didn't have the right one on his truck. If your door has never worked quite right, something like this might be the cause. To correct this, you must use the weight of the door to specify the spring, either from a spring rate manual giving spring torque constants, or from the formulas below.
##### Dodging a falling door:: Reversing this equation gives us x=gt^2/2 as the fallen distance x for a given time t. How much time would you have to dodge a falling door if the spring were to suddenly break at the top of travel? Let us assume you are 5.5 feet tall, so the door will hit your head after falling 2 feet from its 7.5 foot fully-raised height. This 2-foot fall takes sqrt(2*2/32.2) = 0.35 seconds (350 milliseconds). The quickest human response time is about 200 milliseconds, so even if you are alert to the hazard, this leaves you only about 150 milliseconds to accelerate and move your noggin out of the way. If you are an Olympic gold medalist in the 100 meter dash, you can accelerate (horizontally) about 10 feet/second^2, and your 150 milliseconds of wide-eyed panic will move you all of 10*0.15^2/2 = 0.11 foot = 1.35 inch.
You might genuinely need some extra parts when you thought you simply needed a broken spring replaced, and a good serviceman will perform a simple inspection to identify such parts. Nor is it unreasonable for a business to charge separately for a service call versus repair work actually performed. But the best protection for you as a buyer, being somewhat at the mercy of whomever you decide to bring on site, is to understand what is being done, and ask intelligently for a clear explanation or demonstration of why extra parts are required.
Spring Issues - Unless it is due to an obstruction of some type, uneven movement of your garage door is likely caused by a problem with the spring mechanism. Roll-up doors generally have one center-mounted torsion spring. The tension on torsion springs is great. Therefore, it is necessary to call a professional for any repairs to this type of spring mechanism to avoid serious injury. Because replacement of these springs or cables in a garage door can take at least two or three hours of labor, costs can average \$300 or more.
Garage doors can cost \$200 for a single door and up to \$4,000 for two doors or more. Keep in mind that better materials will cost more. This price also includes purchase of new tracks, adhesives, connectors and fasteners. You could install the door yourself but the weight of some garage doors are extremely heavy, so make sure you have help if you are trying DIY garage door replacement. Contact a professional or do research online to find out the average weight of different types of garage doors.
Since 1989, Henderson Garage Doors has specialized in the sale, installation, service and repair of residential and commercial door systems throughout the Baytown area. A family-owned and operated company, we’re committed to offering the finest products and delivering best in class service. We’re proud of our reputation for prompt, professional, and reliable service and we’d like to welcome you to the Henderson Family!
Spring Issues - Unless it is due to an obstruction of some type, uneven movement of your garage door is likely caused by a problem with the spring mechanism. Roll-up doors generally have one center-mounted torsion spring. The tension on torsion springs is great. Therefore, it is necessary to call a professional for any repairs to this type of spring mechanism to avoid serious injury. Because replacement of these springs or cables in a garage door can take at least two or three hours of labor, costs can average \$300 or more.
Your ad had no cheap coupons or promises.. Just business. Service was more than prompt. No attempt to fix something that was not broke or did not try to sell me anything I did not need. He stated the price at the beginning and stuck to it. (I was so pleased, I did not bother to mention a senior discount.) Very, very pleased that I have found someone I can trust and rely on. GaryRichard | 1,044 | 4,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-51 | latest | en | 0.955855 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-11-cumulative-review-page-670/11c | 1,540,219,165,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515088.88/warc/CC-MAIN-20181022134402-20181022155902-00341.warc.gz | 936,499,868 | 12,996 | ## Intermediate Algebra (6th Edition)
$\sqrt[]{26}$
The expression $\sqrt[]{26}$ does not contain a factor that is a perfect square. Hence, it is already in simplified form. | 42 | 174 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-43 | latest | en | 0.911264 |
https://www.teacheron.com/store?s=statistics&l=intermediate | 1,716,974,504,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00355.warc.gz | 866,100,563 | 21,638 | Filters
Subject
Types
Intermediate Statistics resources
By nishitha kalathil
This is a powerpoint presentation on Basic Statistics. It covers the basic terms in statistics , its types and explanations.
Subjects:
Types:
Price:
3 USD
By Statistics and Probability
These documents contain selected solved exercises from the book "Introductory Statistics by Openstax College". Solution is provided from chapter 1 to chapter 13 and the details of the exercises is provided in...
Price:
20 USD
By Statistics and Probability
The following documents contain the graduate and postgraduate level problems related to Descriptive Statistics, Hypothesis Testing and Confidence Intervals, Regression Analysis and Inference in Regression and...
Price:
20 USD
By Statistics and Probability
This product cointains the solved worksheets including Descriptive Statistics, Probability, Estimation and Hypothesis Testing. The solutions are provided in detail with explanation of the steps so students will be able...
Types:
Price:
8 USD
By Statistics and Probability
This document contains selected solved exercises from the book "Statistical Methods For The Social Sciences (5th Edition) by Alan Agresti" from chapter 3 to chapter 7. All the steps are calculated and explained in...
Price:
12 USD
By M.Y Statistical Consulting Services
What is statistics and probabilities? Probability And Statistics are the two important concepts in Maths. Probability is all about chance. Whereas statistics is more about how we handle various data using different...
Subjects:
Level:
Types:
Price:
3.50 USD
By Statistics and Probability
These are 8 solved problems of Probability, Normal Distribution and Hypothesis testing topics. You can use these solutions to understand how to solve different problems relating to these topics. All the steps are...
Price:
5 USD
By Bright Onu
This guide helps you make a choice for statistical analysis to be performed on your research data. It focuses on nine key areas which include the following; Use Cases, Industry Practical Examples, Data Preprocessing, ...
Types:
Price:
10 USD
By Shinjini Ghosh
Contains details of different types of sampling techniques. Related theorems, proofs and examples. A note that can be helpful for college & university students as well as professionals.
Types:
Price:
12.09 USD | 463 | 2,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.896836 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.