TO tasks with parameter This may include, for example, the search for solutions to linear and quadratic equations in general form, the study of the equation for the number of roots available depending on the value of the parameter.
Without giving detailed definitions, consider the following equations as examples:
y = kx, where x, y are variables, k is a parameter;
y = kx + b, where x, y are variables, k and b are parameters;
ax 2 + bx + c = 0, where x are variables, a, b and c are a parameter.
Solving an equation (inequality, system) with a parameter means, as a rule, solving an infinite set of equations (inequalities, systems).
Tasks with a parameter can be divided into two types:
A) the condition says: solve the equation (inequality, system) - this means, for all values of the parameter, find all solutions. If at least one case remains uninvestigated, such a solution cannot be considered satisfactory.
b) it is required to indicate the possible values of the parameter at which the equation (inequality, system) has certain properties. For example, it has one solution, has no solutions, has solutions belonging to the interval, etc. In such tasks, it is necessary to clearly indicate at what parameter value the required condition is satisfied.
The parameter, being an unknown fixed number, has a kind of special duality. First of all, it is necessary to take into account that the assumed popularity indicates that the parameter must be perceived as a number. Secondly, the freedom to manipulate the parameter is limited by its obscurity. For example, operations of dividing by an expression that contains a parameter or extracting the root of an even degree from such an expression require preliminary research. Therefore, care is required when handling the parameter.
For example, to compare two numbers -6a and 3a, you need to consider three cases:
1) -6a will be greater than 3a if a is a negative number;
2) -6a = 3a in the case when a = 0;
3) -6a will be less than 3a if a is a positive number 0.
The solution will be the answer.
Let the equation kx = b be given. This equation is a short form for an infinite number of equations with one variable.
When solving such equations there may be cases:
1. Let k be any real number not equal to zero and b be any number from R, then x = b/k.
2. Let k = 0 and b ≠ 0, the original equation will take the form 0 x = b. Obviously, this equation has no solutions.
3. Let k and b be numbers equal to zero, then we have the equality 0 x = 0. Its solution is any real number.
An algorithm for solving this type of equation:
1. Determine the “control” values of the parameter.
2. Solve the original equation for x for the parameter values that were determined in the first paragraph.
3. Solve the original equation for x for parameter values different from those chosen in the first paragraph.
4. You can write the answer in the following form:
1) for ... (parameter values), the equation has roots ...;
2) for ... (parameter values), there are no roots in the equation.
Example 1.
Solve the equation with the parameter |6 – x| = a.
Solution.
It is easy to see that a ≥ 0 here.
According to the rule of module 6 – x = ±a, we express x:
Answer: x = 6 ± a, where a ≥ 0.
Example 2.
Solve the equation a(x – 1) + 2(x – 1) = 0 with respect to the variable x.
Solution.
Let's open the brackets: aх – а + 2х – 2 = 0
Let's write the equation in standard form: x(a + 2) = a + 2.
If the expression a + 2 is not zero, i.e. if a ≠ -2, we have the solution x = (a + 2) / (a + 2), i.e. x = 1.
If a + 2 is equal to zero, i.e. a = -2, then we have the correct equality 0 x = 0, so x is any real number.
Answer: x = 1 for a ≠ -2 and x € R for a = -2.
Example 3.
Solve the equation x/a + 1 = a + x with respect to the variable x.
Solution.
If a = 0, then we transform the equation to the form a + x = a 2 + ax or (a – 1)x = -a(a – 1). The last equation for a = 1 has the form 0 x = 0, therefore x is any number.
If a ≠ 1, then the last equation will take the form x = -a.
This solution can be illustrated on the coordinate line (Fig. 1)
Answer: there are no solutions for a = 0; x – any number with a = 1; x = -a for a ≠ 0 and a ≠ 1.
Graphical method
Let's consider another way to solve equations with a parameter - graphically. This method is used quite often.
Example 4.
Depending on the parameter a, how many roots does the equation ||x| – 2| = a?
Solution.
To solve using the graphical method, we construct graphs of the functions y = ||x| – 2| and y = a (Fig. 2).
The drawing clearly shows possible cases of the location of the straight line y = a and the number of roots in each of them.
Answer: the equation will not have roots if a< 0; два корня будет в случае, если a >2 and a = 0; the equation will have three roots in the case of a = 2; four roots – at 0< a < 2.
Example 5.
At what a the equation 2|x| + |x – 1| = a has a single root?
Solution.
Let us depict the graphs of the functions y = 2|x| + |x – 1| and y = a. For y = 2|x| + |x – 1|, expanding the modules using the interval method, we obtain:
(-3x + 1, at x< 0,
y = (x + 1, for 0 ≤ x ≤ 1,
(3x – 1, for x > 1.
On Figure 3 It is clearly seen that the equation will have a single root only when a = 1.
Answer: a = 1.
Example 6.
Determine the number of solutions to the equation |x + 1| + |x + 2| = a depending on the parameter a?
Solution.
Graph of the function y = |x + 1| + |x + 2| will be a broken line. Its vertices will be located at points (-2; 1) and (-1; 1) (Figure 4).
Answer: if parameter a is less than one, then the equation will not have roots; if a = 1, then the solution to the equation is an infinite set of numbers from the interval [-2; -1]; if the values of parameter a are greater than one, then the equation will have two roots.
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Olga Otdelkina, 9th grade student
This topic is an integral part of the school algebra course. The purpose of this work is to study this topic in more depth, to identify the most rational solution that quickly leads to an answer. This essay will help other students understand the use of the graphical method for solving equations with parameters, learn about the origin and development of this method.
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Introduction2
Chapter 1. Equations with a parameter
History of the emergence of equations with parameter3
Vieta's theorem4
Basic concepts5
Chapter 2. Types of equations with parameters.
Linear equations6
Quadratic equations……………………………………………………………......7
Chapter 3. Methods for solving equations with a parameter
Analytical method….…………………………………………......8
Graphic method. History of origin….…………………………9
Solution algorithm by graphical method..…………….....…………….10
Solution of the equation with modulus………………...…………………………….11
Practical part…………………...……………………………………12
Conclusion……………………………………………………………………………….19
References……………………………………………………………20
Introduction.
I chose this topic because it is an integral part of the school algebra course. In preparing this work, I set the goal of a deeper study of this topic, identifying the most rational solution that quickly leads to an answer. My essay will help other students understand the use of the graphical method for solving equations with parameters, learn about the origin and development of this method.
In modern life, the study of many physical processes and geometric patterns often leads to solving problems with parameters.
For solving such equations, the graphical method is very effective when you need to determine how many roots the equation has depending on the parameter α.
Problems with parameters are of purely mathematical interest, contribute to the intellectual development of students, and serve as good material for practicing skills. They have diagnostic value, since they can be used to test knowledge of the main branches of mathematics, the level of mathematical and logical thinking, initial research skills and promising opportunities for successfully mastering a mathematics course in higher educational institutions.
My essay discusses frequently encountered types of equations, and I hope that the knowledge I gained in the process of work will help me when passing school exams, becauseequations with parametersare rightfully considered one of the most difficult problems in school mathematics. It is precisely these tasks that are included in the list of tasks in the Unified State Examination.
History of the emergence of equations with a parameter
Problems on equations with a parameter were already encountered in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined a general rule for solving quadratic equations reduced to a single canonical form:
αx 2 + bx = c, α>0
In the equation, the coefficients, except for the parameter, can also be negative.
Quadratic equations by al-Khwarizmi.
Al-Khorezmi's algebraic treatise gives a classification of linear and quadratic equations with parameter a. The author counts 6 types of equations, expressing them as follows:
1) “Squares are equal to roots,” i.e. αx 2 = bx.
2) “Squares are equal to numbers”, i.e. αx 2 = c.
3) “The roots are equal to the number,” i.e. αx = c.
4) “Squares and numbers are equal to roots,” i.e. αx 2 + c = bx.
5) “Squares and roots are equal to the number”, i.e. αx 2 + bx = c.
6) “Roots and numbers are equal to squares,” i.e. bx + c = αx 2 .
Formulas for solving quadratic equations according to al-Khwarizmi in Europe were first set forth in the “Book of Abacus,” written in 1202 by the Italian mathematician Leonardo Fibonacci.
The derivation of the formula for solving a quadratic equation with a parameter in general form is available from Vieta, but Vieta recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 12th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. Thanks to the works of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations took on its modern form.
Vieta's theorem
The theorem expressing the relationship between the parameters, coefficients of a quadratic equation and its roots, named after Vieta, was formulated by him for the first time in 1591. As follows: “If b + d multiplied by α minus α 2 , is equal to bc, then α is equal to b and equal to d.”
To understand Vieta, we should remember that α, like any vowel letter, meant the unknown (our x), while the vowels b, d are coefficients for the unknown. In the language of modern algebra, the above Vieta formulation means:
If there is
(α + b)x - x 2 = αb,
That is, x 2 - (α -b)x + αb =0,
then x 1 = α, x 2 = b.
By expressing the relationship between the roots and coefficients of equations by general formulas written using symbols, Vieta established uniformity in methods for solving equations. However, the symbolism of Viet is still far from its modern form. He did not recognize negative numbers and therefore, when solving equations, he considered only cases where all the roots were positive.
Basic Concepts
Parameter - an independent variable, the value of which is considered a fixed or arbitrary number, or a number belonging to the interval specified by the condition of the problem.
Equation with parameter— mathematicalthe equation, the appearance and solution of which depends on the values of one or more parameters.
Decide equation with parameter means for each valuefind the values of x that satisfy this equation, and also:
- 1. Investigate at what values of the parameters the equation has roots and how many there are for different values of the parameters.
- 2. Find all expressions for the roots and indicate for each of them those parameter values at which this expression actually determines the root of the equation.
Consider the equation α(x+k)= α +c, where α, c, k, x are variable quantities.
System acceptable values variables α, c, k, xis any system of variable values in which both the left and right sides of this equation take real values.
Let A be the set of all admissible values of α, K the set of all admissible values of k, X the set of all admissible values of x, C the set of all admissible values of c. If for each of the sets A, K, C, X we select and fix, respectively, one value α, k, c, and substitute them into the equation, then we obtain an equation for x, i.e. equation with one unknown.
The variables α, k, c, which are considered constant when solving an equation, are called parameters, and the equation itself is called an equation containing parameters.
The parameters are denoted by the first letters of the Latin alphabet: α, b, c, d, ..., k, l, m, n, and the unknowns are denoted by the letters x, y, z.
Two equations containing the same parameters are called equivalent if:
a) they make sense for the same parameter values;
b) every solution to the first equation is a solution to the second and vice versa.
Types of equations with parameters
Equations with parameters are: linear and square.
1) Linear equation. General form:
α x = b, where x is unknown;α, b - parameters.
For this equation, the special or control value of the parameter is the one at which the coefficient of the unknown becomes zero.
When solving a linear equation with a parameter, cases are considered when the parameter is equal to its special value and different from it.
A special value of the parameter α is the valueα = 0.
1.If, and ≠0, then for any pair of parametersα and b it has a unique solution x = .
2.If, and =0, then the equation takes the form:0 x = b . In this case the value b = 0 is a special parameter value b.
2.1. At b ≠ 0 the equation has no solutions.
2.2. At b =0 the equation will take the form:0 x =0.
The solution to this equation is any real number.
Quadratic equation with parameter.
General form:
α x 2 + bx + c = 0
where parameter α ≠0, b and c - arbitrary numbers
If α =1, then the equation is called a reduced quadratic equation.
The roots of a quadratic equation are found using the formulas
Expression D = b 2 - 4 α c is called a discriminant.
1. If D> 0, the equation has two different roots.
2. If D< 0 — уравнение не имеет корней.
3. If D = 0, the equation has two equal roots.
Methods for solving equations with a parameter:
- Analytical - a method of direct solution, repeating standard procedures for finding the answer in an equation without parameters.
- Graphic - depending on the conditions of the problem, the position of the graph of the corresponding quadratic function in the coordinate system is considered.
Analytical method
Solution algorithm:
- Before you begin solving a problem with parameters using the analytical method, you need to understand the situation for a specific numerical value of the parameter. For example, take the value of the parameter α =1 and answer the question: is the value of the parameter α =1 required for this task.
Example 1. Solve relatively X linear equation with parameter m:
According to the meaning of the problem (m-1)(x+3) = 0, that is, m= 1, x = -3.
Multiplying both sides of the equation by (m-1)(x+3), we get the equation
We get
Hence, at m= 2.25.
Now we need to check whether there are any values of m for which
the value of x found is -3.
solving this equation, we find that x is equal to -3 with m = -0.4.
Answer: with m=1, m =2.25.
Graphic method. History of origin
The study of common dependencies began in the 14th century. Medieval science was scholastic. With this nature, there was no room left for the study of quantitative dependencies; it was only about the qualities of objects and their connections with each other. But among the scholastics a school arose that argued that qualities can be more or less intense (the dress of a person who has fallen into a river is wetter than that of someone who has just been caught in the rain)
The French scientist Nikolai Oresme began to depict intensity with the lengths of segments. When he placed these segments perpendicular to a certain straight line, their ends formed a line, which he called the “line of intensity” or the “line of the upper edge” (graph of the corresponding functional dependence). Oresme even studied “planar” and “physical” qualities, i.e. functions , depending on two or three variables.
Oresme's important achievement was his attempt to classify the resulting graphs. He identified three types of qualities: Uniform (with constant intensity), uniform-uneven (with a constant rate of change in intensity) and uneven-uneven (all others), as well as the characteristic properties of the graphs of such qualities.
To create a mathematical apparatus for studying the graphs of functions, the concept of a variable was needed. This concept was introduced into science by the French philosopher and mathematician Rene Descartes (1596-1650). It was Descartes who came up with the ideas about the unity of algebra and geometry and the role of variables; Descartes introduced a fixed unit segment and began to consider the relationships of other segments to it.
Thus, graphs of functions over the entire period of their existence have gone through a number of fundamental transformations, which led them to the form to which we are accustomed. Each stage or stage in the development of graphs of functions is an integral part of the history of modern algebra and geometry.
The graphical method of determining the number of roots of an equation depending on the parameter included in it is more convenient than the analytical one.
Solving algorithm by graphical method
Graph of a function - a set of points at whichabscissaare valid argument values, A ordinates- corresponding valuesfunctions.
Algorithm for graphically solving equations with a parameter:
- Find the domain of definition of the equation.
- We express α as a function of x.
- In the coordinate system we build a graph of the functionα (x) for those values of x that are included in the domain of definition of this equation.
- Finding the intersection points of a lineα =с, with the graph of the function
α(x). If the line α =с crosses the graphα (x), then we determine the abscissas of the intersection points. To do this, it is enough to solve the equation c = α (x) relative to x.
- Write down the answer
Solving equations with modulus
When solving equations with a modulus containing a parameter graphically, it is necessary to construct graphs of functions and consider all possible cases for different values of the parameter.
For example, │x│= a,
Answer: if a < 0, то нет корней, a > 0, then x = a, x = - a, if a = 0, then x = 0.
Problem solving.
Problem 1. How many roots does the equation have?| | x | - 2 | = a depending on parameter a?
Solution. In the coordinate system (x; y) we will construct graphs of the functions y = | | x | - 2 | and y = a . Graph of the function y = | | x | - 2 | shown in the figure.
Graph of the function y =α a = 0).
From the graph it can be seen that:
If a = 0, then straight line y = a coincides with the Ox axis and has the graph of the function y = | | x | - 2 | two common points; this means that the original equation has two roots (in this case, the roots can be found: x 1,2
= + 2).
If 0<
a
< 2, то прямая y =
α
has with the graph of the function y = | | x | - 2 | four common points and, therefore, the original equation has four roots.
If a = 2, then the line y = 2 has three common points with the graph of the function. Then the original equation has three roots.
If a > 2, then straight line y = a will have two points with the graph of the original function, that is, this equation will have two roots.
Answer: if a < 0, то корней нет;
if a = 0, a > 2, then there are two roots;
if a = 2, then there are three roots;
if 0<
a
< 2, то четыре корня.
Problem 2. How many roots does the equation have?| x 2 - 2| x | - 3 | = a depending on parameter a?
Solution. In the coordinate system (x; y) we will construct graphs of the functions y = | x 2 - 2| x | - 3 | and y = a.
Graph of the function y = | x 2 - 2| x | - 3 | shown in the figure. Graph of the function y =α is a straight line parallel to Ox or coinciding with it (when a = 0).
From the graph you can see:
If a = 0, then straight line y = a coincides with the Ox axis and has the graph of the function y = | x2 - 2| x | - 3 | two common points, as well as the straight line y = a will have with the graph of the function y = | x 2
- 2| x | - 3 | two common points at a > 4. So, for a = 0 and a > 4 the original equation has two roots.
If 0<
a< 3, то прямая y =
a
has with the graph of the function y = | x 2
- 2| x | - 3 | four common points, as well as the straight line y= a will have four common points with the graph of the constructed function at a = 4. So, at 0<
a
< 3,
a
= 4 the original equation has four roots.
If a = 3, then straight line y = a intersects the graph of a function at five points; therefore, the equation has five roots.
If 3<
a< 4, прямая y =
α
intersects the graph of the constructed function at six points; This means that for these parameter values the original equation has six roots.
If a < 0, уравнение корней не имеет, так как прямая y =
α
does not intersect the graph of the function y = | x 2 - 2| x | - 3 |.
Answer: if a < 0, то корней нет;
if a = 0, a > 4, then there are two roots;
if 0<
a
< 3,
a
= 4, then there are four roots;
if a = 3, then five roots;
if 3<
a
< 4, то шесть корней.
Problem 3. How many roots does the equation have?
depending on parameter a?
Solution. Let us construct a graph of the function in the coordinate system (x; y)
but first let's present it in the form:
The lines x = 1, y = 1 are asymptotes of the graph of the function. Graph of the function y = | x | + a obtained from the graph of the function y = | x | displacement by a units along the Oy axis.
Function graphs intersect at one point at a > - 1; This means that equation (1) for these parameter values has one solution.
When a = - 1, a = - 2 graphs intersect at two points; This means that for these parameter values, equation (1) has two roots.
At - 2<
a< - 1,
a
< - 2 графики пересекаются в трех точках; значит, уравнение (1) при этих значениях параметра имеет три решения.
Answer: if a > - 1, then one solution;
if a = - 1, a = - 2, then there are two solutions;
if - 2<
a
< - 1,
a
< - 1, то три решения.
Comment. When solving the problem equation, special attention should be paid to the case when a = - 2, since the point (- 1; - 1) does not belong to the graph of the functionbut belongs to the graph of the function y = | x | + a.
Problem 4. How many roots does the equation have?
x + 2 = a | x - 1 |
depending on parameter a?
Solution. Note that x = 1 is not a root of this equation, since the equality 3 = a 0 cannot be true for any parameter value a . Let's divide both sides of the equation by | x - 1 |(| x - 1 |0), then the equation takes the formIn the coordinate system xOy we will plot the function
The graph of this function is shown in the figure. Graph of the function y = a is a straight line parallel to the Ox axis or coinciding with it (if a = 0).
Equations with parameters are rightfully considered one of the most difficult problems in school mathematics. It is precisely these tasks that, year after year, are included in the list of tasks of type B and C in the unified state exam of the Unified State Examination. However, among the large number of equations with parameters, there are those that can easily be solved graphically. Let's consider this method using the example of solving several problems.
Find the sum of integer values of the number a for which the equation |x 2 – 2x – 3| = a has four roots.
Solution.
To answer the question of the problem, let’s construct graphs of functions on one coordinate plane
y = |x 2 – 2x – 3| and y = a.
Graph of the first function y = |x 2 – 2x – 3| will be obtained from the graph of the parabola y = x 2 – 2x – 3 by symmetrically displaying with respect to the x-axis that part of the graph that is below the Ox-axis. The part of the graph located above the x-axis will remain unchanged.
Let's do this step by step. The graph of the function y = x 2 – 2x – 3 is a parabola, the branches of which are directed upward. To build its graph, we find the coordinates of the vertex. This can be done using the formula x 0 = -b/2a. Thus, x 0 = 2/2 = 1. To find the coordinate of the vertex of the parabola along the ordinate axis, we substitute the resulting value for x 0 into the equation of the function in question. We get that y 0 = 1 – 2 – 3 = -4. This means that the vertex of the parabola has coordinates (1; -4).
Next, you need to find the intersection points of the parabola branches with the coordinate axes. At the points of intersection of the branches of the parabola with the abscissa axis, the value of the function is zero. Therefore, we solve the quadratic equation x 2 – 2x – 3 = 0. Its roots will be the required points. By Vieta’s theorem we have x 1 = -1, x 2 = 3.
At the points of intersection of the parabola branches with the ordinate axis, the value of the argument is zero. Thus, the point y = -3 is the point of intersection of the branches of the parabola with the y-axis. The resulting graph is shown in Figure 1.
To obtain a graph of the function y = |x 2 – 2x – 3|, let us display the part of the graph located below the abscissa symmetrically relative to the x-axis. The resulting graph is shown in Figure 2.
The graph of the function y = a is a straight line parallel to the abscissa axis. It is depicted in Figure 3. Using the figure, we find that the graphs have four common points (and the equation has four roots) if a belongs to the interval (0; 4).
Integer values of number a from the resulting interval: 1; 2; 3. To answer the question of the problem, let’s find the sum of these numbers: 1 + 2 + 3 = 6.
Answer: 6.
Find the arithmetic mean of integer values of the number a for which the equation |x 2 – 4|x| – 1| = a has six roots.
Let's start by plotting the function y = |x 2 – 4|x| – 1|. To do this, we use the equality a 2 = |a| 2 and select the complete square in the submodular expression written on the right side of the function:
x 2 – 4|x| – 1 = |x| 2 – 4|x| - 1 = (|x| 2 – 4|x| + 4) – 1 – 4 = (|x |– 2) 2 – 5.
Then the original function will have the form y = |(|x| – 2) 2 – 5|.
To construct a graph of this function, we construct sequential graphs of functions:
1) y = (x – 2) 2 – 5 – parabola with vertex at point with coordinates (2; -5); (Fig. 1).
2) y = (|x| – 2) 2 – 5 – part of the parabola constructed in step 1, which is located to the right of the ordinate axis, is symmetrically displayed to the left of the Oy axis; (Fig. 2).
3) y = |(|x| – 2) 2 – 5| – the part of the graph constructed in point 2, which is located below the x-axis, is displayed symmetrically relative to the x-axis upward. (Fig. 3).
Let's look at the resulting drawings:
The graph of the function y = a is a straight line parallel to the abscissa axis.
Using the figure, we conclude that the graphs of functions have six common points (the equation has six roots) if a belongs to the interval (1; 5).
This can be seen in the following figure:
Let's find the arithmetic mean of the integer values of parameter a:
(2 + 3 + 4)/3 = 3.
Answer: 3.
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2. Determination of unknown distribution parameters.
Using a histogram, we can approximately plot the distribution density of a random variable. The appearance of this graph often allows us to make an assumption about the probability density distribution of a random variable. The expression of this distribution density usually includes some parameters that need to be determined from experimental data.
Let us dwell on the particular case when the distribution density depends on two parameters.
So let x 1 , x 2 , ..., x n- observed values of a continuous random variable, and let its probability distribution density depend on two unknown parameters A And B, i.e. looks like . One of the methods for finding unknown parameters A And B consists in the fact that they are chosen in such a way that the mathematical expectation and variance of the theoretical distribution coincide with the sample means and variance:
 |
(66) |
 |
(67) |
From the two obtained equations () the unknown parameters are found A And B. So, for example, if a random variable obeys the normal probability distribution law, then its probability distribution density
depends on two parameters a And . These parameters, as we know, are, respectively, the mathematical expectation and standard deviation of a random variable; therefore equalities () will be written like this:
| (68) |
Therefore, the probability distribution density has the form
Note 1. We have already solved this problem in . The measurement result is a random variable that obeys the normal distribution law with parameters a And . For approximate value a we chose the value , and for the approximate value - the value .
Note 2. With a large number of experiments, finding quantities and using formulas () is associated with cumbersome calculations. Therefore, they do this: each of the observed values of the quantity , falling into i th interval ] X i-1 , X i [ statistical series, is considered approximately equal to the middle c i this interval, i.e. c i =(X i-1 +X i)/2. Consider the first interval ] X 0 , X 1 [. It hit him m 1 observed values of the random variable, each of which we replace with a number from 1. Therefore, the sum of these values is approximately equal to m 1 s 1. Similarly, the sum of values falling into the second interval is approximately equal to m 2 with 2 etc. That's why
In a similar way we obtain the approximate equality
So, let's show that
 |
(71) |
