Approximate calculations in mathematics curriculum

S> - area, if the true value is less than the resulting.

- Maximum relative error of the work.

Task 4: Compare the maximum relative error of the work and the amount of the limiting relative error factors. Frontal discussion. (5 minutes). The sum of the factors limiting relative error is less than the marginal product of the relative error.

Task 5: A formula for finding the maximum relative error of the work. Frontal discussion. (10 min.)

Formula can not be obtained. In this case, the work can be extended individually.

Task 6: Do you think how to write the formula for the n terms? Frontal discussion. (5 minutes). Formulation of the problem. The issue may be continuing.

Topic 8: "The approximate solution of quadratic equations."

(2 hours, 40 min.)

Purpose: - To give an idea of the approximate location of calculations;

- Meet with the ways of finding approximate solutions of a quadratic equation;

Task:

- Learning to use new ways to solve problems.

Mathematics elective creative

The plan:

1. Organizational aspects. 2 min.

2. The first stage. The place of approximate calculations. 10 min.

3. The second stage. Finding the approximate value of the square root. 20 min.

4. The fourth stage. Methods of approximate solutions of quadratic equations.

A) selection. 125 min.:

20 min.

C) the method of successive approximations. 45 min.

C) the method of bisection. 25 min.

D) A similar problem. 35 min.

6. Summing up. 3 min.

Substantive grounds:

Students the necessary knowledge required skills

Concepts:

1) accurate, approximate values;

2) rounding;

3) the root;

4) the quadratic equation: 1) find the values of the schedule;

2) convert quadratic equations;

3) round.

Works:

1. Front case;

2. The work of students at the board;

3. Independent work of students;

Means:

1. write on the board;

2. handout.

Plan (detailed).

Stages. Forms / tasks / resources. Time.

1. Organizational aspects. 2 min.

2. The first stage. The place of approximate calculations. Forms: joint debate. 10 min.

3. The second stage. Finding the approximate value of the square root. Objectives: Restoration of knowledge of approximate determination of value.

Forms: joint debate (students offering their version, then they discuss);

Means: write on the board. 20 min.

4. The third stage. Methods of approximate solutions of quadratic equations.

A) selection.

C) the method of successive approximations.

C) the method of bisection. 125 min.

Forms: self-study, a joint discussion. 20 min.

Objectives: to systematize the knowledge of finding the root of the method of successive approximations;

Forms: Lecture (a geometric interpretation of the method), joint discussion (insertion method of successive approximations to find the approximate root of the equation), independent work (finding the approximate root of the equation);

Means: write on the board, handout (graphs, theoretical cards). 45 min.

Objectives: to systematize the knowledge of finding the root of the method of bisection;

Forms: joint debate (being the value of the root), lecture (presentation method);

Means: handout (graphs, theoretical cards). 20 min.

D) A similar problem. Forms: joint debate. 40 min.

6. Summarizing Note: problem. 3 min.

Defining the form of comments

To answer the question: What is the cause of the approximate numbers? joint discussion (10 min.) (Students offer their version).

K: (to the proposed version supplemented the missing may be prompted to something new, never been identified in a) - c).

a) When measuring the lengths of the segments and areas of figures, weighing bodies and other measurements obtained numbers expressing these values. Due to errors in the measurement result is an approximate measured values.

b) can lead to errors inaccuracy of construction of mathematical models of the phenomenon. (Inaccurate formula).

c) the use of traditional methods for solving algebraic equations using the computer can not guarantee the accuracy of the results of calculations. (When using the simplest methods of computer calculations using eight bits, the overflow of which is an accumulation of error).

There is a problem:

any result of the calculation is given to the MC with a finite number of sign due to the limitations of the indicator bits, rounding errors accumulate.

Task 1: an irrational number. Is it possible to calculate the value of not taking advantage of the MC?

Find the minimum interval in which the value is the value entered. To facilitate the work is proposed to analyze the table. joint discussion (20 min.) Any real number can be made between two rational numbers and, while holding the upper and lower boundary, identify any accuracy, but, nevertheless, approximately.

Question: How do you think is the equation is square?

Task 2: Find the root of approximate to two decimal places.

x2 - 2x - 2 = 0.

To find the root of the proposed substitute values into the equation and analyze the results.

Task 3: Finding the root of the approximate method of successive approximations.

x2 - 2x - 2 = 0

x2 = 2 +2 / x

x = 2 + 2 / x

Task 4: Write an algorithm for finding roots. (Some of the students have to explain, or explain the teacher).

independent work (10 minutes), a joint discussion (10 min). Note that to find the root of the selection to choose the number, substitute it into the equation. If you get a value less than 0, then you need to take a number more, if possible value greater than 0, then take a number of smaller ones.

independent work (15 min.) On the example to find the value of the root. We take a value of x and substitute into the right side of the equation. Next, we substitute in the right side of the values that have turned to the left. Perform long until we find the root of the value up to hundredths, thousandths, etc.

joint discussion (15 min.) Handout 1: a) graphically or by trial are the first approximation of the root

x = x0.

x 0 = first approximation of the root

b) to the right side of the equation x = 2 + 2 / x substitute x0 then x1 = 2 + 2/h0 - the second approximation of the root.

c) substituted in the right-hand side of the equation x = 2 + 2 / x for x x1.

x2 = 2 + 2/h1 - third approximation of the root.

x3 = 2 + 2/h2

x4 = 2 + 2/h3 etc.

Or algorithm in a general form:

Let x = y (x) - the left side of the equation, 2 + 2 / x = j (x) - the right part of the equation.

a) graphically or by trial are the first approximation of the root x = x0.

x 0 = first approximation of the root.

b) to the right side of the equation x = j (x) x0 and then substitute x1 = j (x).

x1 - the second approximation of the root.

c) substituted in the right-hand side of the equation x =--j (x) x 1 for x.

x2 = j (x1) x2 - third approximation of the root.

etc.

Task 5: Use the template to construct the graphs of

y = x and y = 2 + 2 / x.

Statement of the problem: We found the value equation in two ways (the selection and the method of successive approximations). What do you think, what method is easier? Why? independent work (built graphs f-tions).

joint discussion, lecture (geom. Interpretation).

(15 min.) Handout: templates.

Geometric interpretation of the method:

1) construct the graph of y = x and y = 2 + 2 / x;

2) choose an approximate value of the root x0 (in our case x0 = 2)

3) draw the line x = x0.

4) direct the curves meet at two points, we choose the most appropriate point.

5) The estimated values x = x0 determine the value of y (x0) = 2 + 2/h0, x0 - a first approximation of the root.

6) through the point A0 [x0, y (0)] draws a direct parallel to the x-axis to the point of intersection with the B1 curve y = x;

7) is substituted into the equation y = x instead of y value y0 = x0, we solve the equation 2 + 2 / x = x0, and find the value of x 1 - the second approximation of the root.

8) that the value of y1 = x1, etc.

Ask students how they think, what is the method. After the parse method together.

K: If you find the root of the method of bisection selected segment containing the root, which in the process of shrinking.

Task 6: Find the value of the root of the equation x2 - 2x - 2 = 0 by bisection. (See Appendix 4).

Statement of the problem: We found the value of the root of the equation in three ways (by choosing the method of successive approximations, and the method of bisection). What do you think, what method is easier? Why? joint discussion (being the value of the root, 15 min), lecture (presentation of the method, 10 min.) The method of bisection.

Represent the equation x2 - 2x - 2 = 0 as x = 2 + 2 / x;

We construct the graph of y = x

Y = 2 + 2 / x;

a) The value of x will be the point of intersection of the graphs is the root of x2 - 2x - 2 = 0.

b) Choose a closed interval [a, b], contains the point of intersection.

c) the segment [a, b] is divided by the two-point z1 = (a + b) / 2.

d) If the z12-2z1 - 2 = 0 then z1 - the desired root. If z12-2z1 - 2 0, then the two intervals [a, z1] and [z1, b] choose the one for which the value of the function y = x 2 - 2 - 2 at the ends has different signs and denoted by [a1, b1]. Now if we take the point Z2 = (a1 + b1) / 2, then again or z22-2z2 - 2 = 0 or z22-2z2 - 2 ? 0, etc.

Task 7. Find the value of the root of the method of successive approximations: x2 - 7 = 0. To simplify the calculation, use the table. (See Annex 5).

Task 8: Using templates build graphs of y = x and y = 7 /, and check the resulting value.

Statement of the problem: Is any equation can be solved by successive approximations? Which equations method works?

Activity 9: Make up a quadratic equation, which can be solved by successive approximations.

Task 10: Come up with a quadratic equation, which can not be solved by successive approximations.

Activity 11: To present the resulting equations. joint discussion (10 min.)

Individual work (10 min.)

Individual work (10 min);

joint discussion (5 minutes). Constantly get only two values.

The value is not close to the point of intersection, only moves in a circle.

Can be divided into two groups. One group performs the task 9, the other - set 10.

6.Podvedenie results.

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