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Бедственный index php elementary math. Решение задачи коммивояжера. SAT Math Test: основные факты

Catalog Information

Title

Elementary Linear Algebra.

(Credit Hours:Lecture Hours:Lab Hours)

Offered

Prerequisite

Minimal learning outcomes

Upon completion of this course, the successful student will be able to:

1. Use Gaussian elimination to do all of the following: solve a linear system with reduced row echelon form, solve a linear system with row echelon form and backward substitution, find the inverse of a given matrix, and find the determinant of a given matrix.
2. Demonstrate proficiency at matrix algebra. For matrix multiplication demonstrate understanding of the associative law, the reverse order law for inverses and transposes, and the failure of the commutative law and the cancellation law.
3. Use Cramer"s rule to solve a linear system.
4. Use cofactors to find the inverse of a given matrix and the determinant of a given matrix.
5. Determine whether a set with a given notion of addition and scalar multiplication is a vector space. Here, and in relevant numbers below, be familiar with both finite and infinite dimensional examples.
6. Determine whether a given subset of a vector space is a subspace.
7. Determine whether a given set of vectors is linearly independent, spans, or is a basis.
8. Determine the dimension of a given vector space or of a given subspace.
9. Find bases for the null space, row space, and column space of a given matrix, and determine its rank.
10. Demonstrate understanding of the Rank-Nullity Theorem and its applications.
11. Given a description of a linear transformation, find its matrix representation relative to given bases.
12. Demonstrate understanding of the relationship between similarity and change of basis.
13. Find the norm of a vector and the angle between two vectors in an inner product space.
14. Use the inner product to express a vector in an inner product space as a linear combination of an orthogonal set of vectors.
15. Find the orthogonal complement of a given subspace.
16. Demonstrate understanding of the relationship of the row space, column space, and nullspace of a matrix (and its transpose) via orthogonal complements.
17. Demonstrate understanding of the Cauchy-Schwartz inequality and its applications.
18. Determine whether a vector space with a (sesquilinear) form is an inner product space.
19. Use the Gram-Schmidt process to find an orthonormal basis of an inner product space. Be capable of doing this both in R n and in function spaces that are inner product spaces.
20. Use least squares to fit a line (y = ax + b ) to a table of data, plot the line and data points, and explain the meaning of least squares in terms of orthogonal projection.
21. Use the idea of least squares to find orthogonal projections onto subspaces and for polynomial curve fitting.
22. Find (real and complex) eigenvalues and eigenvectors of 2 × 2 or 3 × 3 matrices.
23. Determine whether a given matrix is diagonalizable. If so, find a matrix that diagonalizes it via similarity.
24. Demonstrate understanding of the relationship between eigenvalues of a square matrix and its determinant, its trace, and its invertibility/singularity.
25. Identify symmetric matrices and orthogonal matrices.
26. Find a matrix that orthogonally diagonalizes a given symmetric matrix.
27. Know and be able to apply the spectral theorem for symmetric matrices.
28. Know and be able to apply the Singular Value Decomposition.
29. Correctly define terms and give examples relating to the above concepts.
30. Prove basic theorems about the above concepts.
31. Prove or disprove statements relating to the above concepts.
32. Be adept at hand computation for row reduction, matrix inversion and similar problems; also, use MATLAB or a similar program for linear algebra problems.

Инструкция . Для получения решения транспортной задачи в онлайн режиме выберите размерность матрицы тарифов (количество поставщиков и количество магазинов).

Вместе с этим калькулятором также используют следующие:
Графический метод решения ЗЛП
Симплексный метод решения ЗЛП
Решение матричной игры
С помощью сервиса в онлайн режиме можно определить цену матричной игры (нижнюю и верхнюю границы), проверить наличие седловой точки, найти решение смешанной стратегии методами: минимакс, симплекс-метод, графический (геометрический) метод, методом Брауна.

Экстремум функции двух переменных
Задачи динамического программирования

Первым этапом решения транспортной задачи является определение ее типа (открытая или закрытая, или иначе сбалансированная или не сбалансированная). Приближенные методы (методы нахождения опорного плана ) позволяют на втором этапе решения за небольшое число шагов получить допустимое, но не всегда оптимальное, решение задачи. К данной группе методов относятся методы:

• вычеркивания (метод двойного предпочтения);
• северо-западного угла;
• минимального элемента;
• аппроксимации Фогеля.

Опорное решение транспортной задачи

Опорным решением транспортной задачи называется любое допустимое решение, для которого векторы условий, соответствующие положительным координатам, линейно независимы. Для проверки линейной независимости векторов условий, соответствующих координатам допустимого решения, используют циклы.
Циклом называется такая последовательность клеток таблицы транспортной задачи, в которой две и только соседние клетки расположены в одной строке или столбце, причем первая и последняя также находятся в одной строке или столбце. Система векторов условий транспортной задачи линейно независима тогда и только тогда, когда из соответствующих им клеток таблицы нельзя образовать ни одного цикла. Следовательно, допустимое решение транспортной задачи, i=1,2,...,m; j=1,2,...,n является опорным только в том случае, когда из занятых им клеток таблицы нельзя образовать ни одного цикла.

Приближенные методы решения транспортной задачи.
Метод вычеркивания (метод двойного предпочтения) . Если в строке или столбце таблицы одна занятая клетка, то она не может входить в какой-либо цикл, так как цикл имеет две и только две клетки в каждом столбце. Следовательно, можно вычеркнуть все строки таблицы, содержащие по одной занятой клетке, затем вычеркнуть все столбцы, содержащие по одной занятой клетке, далее вернуться к строкам и продолжить вычеркивание строк и столбцов. Если в результате вычеркивания все строки и столбцы будут вычеркнуты, значит, из занятых клеток таблицы нельзя выделить часть, образующую цикл, и система соответствующих векторов условий является линейно независимой, а решение опорным. Если же после вычеркиваний останется часть клеток, то эти клетки образуют цикл, система соответствующих векторов условий линейно зависима, а решение не является опорным.
Метод «северо-западного угла» состоит в последовательном переборе строк и столбцов транспортной таблицы, начиная с левого столбца и верхней строки, и выписывании максимально возможных отгрузок в соответствующие ячейки таблицы так, чтобы не были превышены заявленные в задаче возможности поставщика или потребности потребителя. На цены доставки в этом методе не обращают внимание, поскольку предполагается дальнейшая оптимизация отгрузок.
Метод «минимального элемента» . Отличаясь простотой данный метод все же эффективнее чем, к примеру, метод Северо-западного угла. Кроме того, метод минимального элемента понятен и логичен. Его суть в том, что в транспортной таблице сначала заполняются ячейки с наименьшими тарифами, а потом уже ячейки с большими тарифами. То есть мы выбираем перевозки с минимальной стоимостью доставки груза. Это очевидный и логичный ход. Правда он не всегда приводит к оптимальному плану.
Метод «аппроксимации Фогеля» . При методе аппроксимации Фогеля на каждой итерации по всем столбцам и по всем строкам находят разность между двумя записанными в них минимальными тарифами. Эти разности записывают в специально отведенных для этого строке и столбце в таблице условий задачи. Среди указанных разностей выбирают минимальную. В строке (или в столбце), которой данная разность соответствует, определяют минимальный тариф. Клетку, в которой он записан, заполняют на данной итерации.

Пример №1 . Матрица тарифов (здесь количество поставщиков равно 4 , количество магазинов равно 6):

	1	2	3	4	5	6	Запасы
1	3	20	8	13	4	100	80
2	4	4	18	14	3	0	60
3	10	4	18	8	6	0	30
4	7	19	17	10	1	100	60
Потребности	10	30	40	50	70	30

Решение . Предварительный этап решения транспортной задачи сводится к определению ее типа, открытой она является или закрытой. Проверим необходимое и достаточное условие разрешимости задачи.
∑a = 80 + 60 + 30 + 60 = 230
∑b = 10 + 30 + 40 + 50 + 70 + 30 = 230
Условие баланса соблюдается. Запасы равны потребностям. Итак, модель транспортной задачи является закрытой. Если бы модель получилась открытой, то потребовалось бы вводить дополнительных поставщиков или потребителей.
На втором этапе осуществляется поиск опорного плана методами, приведенными выше (наиболее распространенным является метод наименьшей стоимости).
Для демонстрации алгоритма приведем лишь несколько итераций.
Итерация №1. Минимальный элемент матрицы равен нулю. Для этого элемента запасы равны 60 , потребности 30 . Выбираем из них минимальное число 30 и вычитаем его (см. в таблице). При этом из таблицы вычеркиваем шестой столбец (потребности у него равны 0).

3	20	8	13	4	x	80
4	4	18	14	3	0	60 - 30 = 30
10	4	18	8	6	x	30
7	19	17	0	1	x	60
10	30	40	50	70	30 - 30 = 0	0

Итерация №2. Снова ищем минимум (0). Из пары (60;50) выбираем минимальное число 50. Вычеркиваем пятый столбец.

3	20	8	x	4	x	80
4	4	18	x	3	0	30
10	4	18	x	6	x	30
7	19	17	0	1	x	60 - 50 = 10
10	30	40	50 - 50 = 0	70	0	0

Итерация №3. Процесс продолжаем до тех пор, пока не выберем все потребности и запасы.
Итерация №N. Искомый элемент равен 8. Для этого элемента запасы равны потребностям (40).

3	x	8	x	4	x	40 - 40 = 0
x	x	x	x	3	0	0
x	4	x	x	x	x	0
x	x	x	0	1	x	0
0	0	40 - 40 = 0	0	0	0	0

	1	2	3	4	5	6	Запасы
1	3	20	8	13	4	100	80
2	4	4	18	14	3	0	60
3	10	4	18	8	6	0	30
4	7	19	17	0	1	100	60
Потребности	10	30	40	50	70	30

Подсчитаем число занятых клеток таблицы, их 8, а должно быть m + n - 1 = 9. Следовательно, опорный план является вырожденным. Строим новый план. Иногда приходится строить несколько опорных планов, прежде чем найти не вырожденный.

	1	2	3	4	5	6	Запасы
1	3	20	8	13	4	100	80
2	4	4	18	14	3	0	60
3	10	4	18	8	6	0	30
4	7	19	17	0	1	100	60
Потребности	10	30	40	50	70	30

В результате получен первый опорный план, который является допустимым, так как число занятых клеток таблицы равно 9 и соответствует формуле m + n - 1 = 6 + 4 - 1 = 9, т.е. опорный план является невырожденным .
Третий этап заключается в улучшении найденного опорного плана. Здесь используют метод потенциалов или распределительный метод . На этом этапе правильность решения можно контролировать через функцию стоимости F(x) . Если она уменьшается (при условии минимизации затрат), то ход решения верный.

Пример №2 . Используя метод минимального тарифа, представить первоначальный план для решения транспортной задачи. Проверить на оптимальность, используя метод потенциалов.

	30	50	70	10	30	10
40	2	4	6	1	1	2
80	3	4	5	9	9	6
60	4	3	2	7	8	7
20	5	1	3	5	7	9

Пример №3 . Четыре кондитерские фабрики могут производить три вида кондитерских изделий. Затраты на производство одного центнера (ц) кондитерских изделий каждой фабрикой, производственные мощности фабрик (ц в месяц) и суточные потребности в кондитерских изделиях (ц в месяц) указаны в таблице. Составить план производства кондитерских изделий, минимизирующий суммарные затраты на производство.

Примечание . Здесь предварительно можно транспонировать таблицу затрат, поскольку для классической постановки транспортной задачи сначала следуют мощности (производство), а потом потребители.

Пример №4 . На строительство объектов кирпич поступает с трех (I, II, III) заводов. Заводы имеют на складах соответственно 50, 100 и 50 тыс. шт. кирпича. Объекты требуют соответственно 50, 70, 40 и 40 тыс. шт. кирпича. Тарифы (ден. ед./тыс.шт.) приведены в таблице. Составьте план перевозок, минимизирующий суммарные транспортные расходы.

будет закрытой если:
А) a=40, b=45
Б) a=45, b=40
В) a=11, b=12
Условие закрытой транспортной задачи : ∑a = ∑b
Находим, ∑a = 35+20+b = 55+b; ∑b = 60+a
Получаем: 55+b = 60+a
Равенство будет соблюдаться только при a=40, b=45

Lectures on Elementary Mathematics (1898) is the earliest English translation of Joseph Louis Lagrange "s 1795 publication, Leçons élémentaires sur les mathematiques , containing a series of lectures delivered the same year at the Ecole Normale . The work was translated and edited by Thomas J. McCormack, and a second edition, from which the following quotes are taken, appeared in 1901.

Contents

Quotes [ edit ]

Lecture III. On Algebra, Particularly the Resolution of Equations of the Third and Fourth Degree [ edit ]

• Algebra is a science almost entirely due to the moderns... for we have one treatise from the Greeks, that of Diophantus ... the only one which we owe to the ancients in this branch of mathematics. ...I speak of the Greeks only, for the Romans have left nothing in the sciences, and to all appearances did nothing.

• His work contains the first elements of this science . He employed to express the unknown quantity a Greek letter which corresponds to our st and which has been replaced in the translations by N . To express the known quantities he employed numbers solely, for algebra was long destined to be restricted entirely to the solution of numerical problems.

• [H]e uses the known and the unknown quantities alike. And herein consists virtually the essence of algebra, which is to employ unknown quantities, to calculate with them as we do with known quantities, and to form from them one or several equations from which the value of the unknown quantities can be determined.

• Although the work of Diophantus contains indeterminate problems almost exclusively, the solution of which he seeks in rational numbers,- problems which have been designated after him Diophantine problems , -we nevertheless find in his work the solution of a number of determinate problems of the first degree, and even of such as involve several unknown quantities. In the latter case, however, the author invariably has recourse to... reducing the problem to a single unknown quantity, -which is not difficult.

• He gives, also, the solution of equations of the second degree , but is careful so to arrange them that they never assume the affected form containing the square and the first power of the unknown quantity. ...he always arrives at an equation in which he has only to extract a square root to reach the solution...

• Diophantus ... does not proceed beyond equations of the second degree, and we do not know if he or any of his successors... ever pushed... beyond this point.

• Diophantus was not known in Europe until the end of the sixteenth century, the first translation having been a wretched one by Xylander made in 1575. Bachet de Méziriac ... a tolerably good mathematician for his time, subsequently published (1621) a new translation... accompanied by lengthy commentaries, now superfluous. Bachet"s translation was afterwards reprinted with observations and notes by Fermat .

• Prior to the discovery and publication of Diophantus ... algebra had already found its way into Europe. Towards the end of the fifteenth century there appeared in Venice a work by... Lucas Paciolus on arithmetic and geometry in which the elementary rules of algebra were stated.

• [T]he Europeans, having received algebra from the Arabs, were in possession of it one hundred years before the work of Diophantus was known to them. They made, however, no progress beyond equations of the first and second degree.

• In the work of Paciolus ... the general resolution of equations of the second degree... was not given. We find in this work simply rules, expressed in bad Latin verses, for resolving each particular case according to the different combinations of the signs of the terms of equation, and even these rules applied only to the case where the roots were real and positive. Negative roots were still regarded as meaningless and superfluous.

• It was geometry really that suggested to us the use of negative quantities, and herein consists one of the greatest advantages that have resulted from the application of algebra to geometry, -a step which we owe to Descartes .

• In the subsequent period the resolution of equations of the third degree was investigated and the discovery for a particular case ultimately made by... Scipio Ferreus (1515). ...Tartaglia and Cardan subsequently perfected the solution of Ferreus and rendered it general for all equations of the third degree.

• At this period, Italy, which was the cradle of algebra in Europe, was still almost the sole cultivator of the science, and it was not until about the middle of the sixteenth century that treatises on algebra began to appear in France, Germany, and other countries.

• The works of Peletier and Buteo were the first which France produced in this science...

• Tartaglia expounded his solution in bad Italian verses in a work treating of divers questions and inventions printed in 1546, a work which enjoys the distinction of being one of the first to treat of modern fortifications by bastions .

• Cardan published his treatise Ars Magna , or Algebra ... Cardan was the first to perceive that equations had several roots and to distinguish them into positive and negative. But he is particularly known for having first remarked the so-called irreducible case in which the expression of the real roots appears in an imaginary form. Cardan convinced himself from several special cases in which the equation had rational divisors that the imaginary form did not prevent the roots from having a real value. But it remained to be proved that not only were the roots real in the irreducible case, but that it was impossible for all three together to be real except in that case. This proof was afterwards supplied by Vieta , and particularly by Albert Girard , from considerations touching the trisection of an angle .

• [T]he irreducible case of equations of the third degree ... presents a new form of algebraical expressions which have found extensive application in analysis... it is constantly giving rise to unprofitable inquiries with a view to reducing the imaginary form to a real form and... it thus presents in algebra a problem which may be placed upon the same footing with the famous problems of the duplication of the cube and the squaring of the circle in geometry.

• The mathematicians of the period under discussion were wont to propound to one another problems for solution. These... were... public challenges and served to excite and to maintain that fermentation which is necessary for the pursuit of science. The challenges... were continued down to the beginning of the eighteenth century Europe, and really did not cease until the rise of the Academies which fulfilled the same end... partly by the union of the knowledge of their various members, partly by the intercourse which they maintained... and... by the publication of their memoirs, which served to disseminate the new discoveries and observations...

• The Algebra of Bombelli contains not only the discovery of Ferrari but also divers other important remarks on equations of the second and third degree and particularly on the theory of radicals by means of which the author succeeded in several cases in extracting the imaginary cube roots of the two binomials of the formula of the third degree in the irreducible case, so finding a perfectly real result... the most direct proof possible of the reality of this species of expressions.

• The solution of equations of the third and fourth degree was quickly accomplished. But the successive efforts of mathematicians for over two centuries have not succeeded in surmounting the difficulties of the equation of the fifth degree.

• Yet these efforts are far from having been in vain. They have given rise to the many beautiful theorems... on the formation of equations, on the character and signs of the roots, on the transformation of a given equation into others of which the roots may be formed at pleasure from the roots of the given equation, and finally, to the beautiful considerations concerning the metaphysics of the resolution of equations from which the most direct method of arriving at their solution, when possible, has resulted.

• Vieta and Descartes ... Harriot ... and Hudde ... were the first after the Italians... to perfect the theory of equations, and since their time there is scarcely a mathematician of note that has not applied himself...

Lecture V. On the Employment of Curves in the Solution of Problems [ edit ]

• As long as algebra and geometry travelled separate paths their advance was slow and their applications limited. But when these two sciences joined company, they drew from each other fresh vitality and thenceforward marched on at a rapid pace towards perfection. It is to Descartes that we owe the application of algebra to geometry,-an application which has furnished the key to the greatest discoveries in all branches of mathematics.

• The method... for finding and demonstrating divers general properties of equations by considering the curves which represent them, is a species of application of geometry to algebra... [T]his method has extended applications, and is capable of readily solving problems whose direct solution would be extremely difficult or even impossible... [T]his subject... is not ordinarily found in elementary works on algebra.

• [A]n equation of any degree can be resolved by means of a curve, of which the abscissæ represent the unknown quantity of the equation, and the ordinates the values which the left-hand member assumes for every value of the unknown quantity. ...[T]his method can be applied generally to all equations, whatever their form, and... only requires them to be developed and arranged according to the different powers of the unknown quantity.
[ edit ]
• Lectures on Elementary Mathematics 2nd ed. (1901) @GoogleBooks

Lesia М. Ohnivchuk

Abstract

The article considers way to extend the functionality of LMS Moodle when creating e-learning courses for the mathematical sciences, in particular e-learning courses "Elementary Mathematics" by using flash technology and Java-applets. There are examples of the use of flash-applications and Java-applets in the course "Elementary Mathematics".

Keywords

LMS Moodle; e-learning courses; technology flash; Java-applet, GeoGebra

References

Brandão, L. O., "iGeom: a free software for dynamic geometry into the web", International Conference on Sciences and Mathematics Education, Rio de Janeiro, Brazil, 2002.

Brandão, L. O. and Eisnmann, A. L. K. “Work in Progress: iComb Project - a mathematical widget for teaching and learning combinatorics through exercises” Proceedings of the 39th ASEE/IEEE Frontiers in Education Conference, 2009, T4G_1–2

Kamiya, R. H and Brandão, L. O. “iVProg – a system for introductory programming through a Visual Model on the Internet. Proceedings of the XX Simpósio Brasileiro de Informática na Educação, 2009 (in Portuguese).

Moodle.org: open-source community-based tools for learning [Електронний ресурс]. – Режим доступу: http://www.moodle.org.

MoodleDocs [Електронний ресурс]. – Режим доступум: http://docs.moodle.org.

Інтерактивні технології навчання: теорія, практика, досвід: методичний посібник авт.-уклад.: О. Пометун, Л. Пироженко. – К. : АПН; 2004. – 136 с.

Dmitry Pupinin. Question Type: Flash [Электронный ресурс]. – Режим доступа: https://moodle.org/mod/data/view.php?d=13&rid=2493&filter=1 – 26.02.14.

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DOI: https://doi.org/10.33407/itlt.v48i4.1249

Copyright (c) 2015 Lesia М. Ohnivchuk

An elementary math curriculum for supplementary or home school should teach much more than the “how to” of simple arithmetic. A good math curriculum should have elementary math activities that build a solid foundation which is both deep and broad, conceptual and “how to”.

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Time4Learning’s Elementary Math Curriculum

Time4Learning’s math curriculum contains a wide range of elementary math activities, which cover more than just arithmetic, math facts, and operations. Our elementary math curriculum teaches these five math strands.*

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• Algebra – The ability to sort and order objects or numbers and recognizing and building on simple patterns are examples of ways children begin to experience algebra. This elementary math concept sets the groundwork for working with algebraic variables as a child’s math experience grows.
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Elementary math curriculums that cover just one or two of these five math strands are narrow and lead to a weak understanding of math. Help your child build a strong, broad math foundation.