Approximation of experimental data is a method based on replacing experimentally obtained data with an analytical function that most closely passes or coincides at nodal points with the original values (data obtained during an experiment or experiment). Currently, there are two ways to define an analytical function:
By constructing an n-degree interpolation polynomial that passes directly through all points a given data array. In this case, the approximating function is presented in the form of: an interpolation polynomial in Lagrange form or an interpolation polynomial in Newton form.
By constructing an n-degree approximating polynomial that passes in the closest proximity to points from a given data array. Thus, the approximating function smoothes out all random noise (or errors) that may arise during the experiment: the measured values during the experiment depend on random factors that fluctuate according to their own random laws (measurement or instrument errors, inaccuracy or experimental errors). In this case, the approximating function is determined using the method least squares.
Least squares method(in the English-language literature Ordinary Least Squares, OLS) is a mathematical method based on determining the approximating function, which is constructed in the closest proximity to points from a given array of experimental data. The closeness of the original and approximating functions F(x) is determined by a numerical measure, namely: the sum of squared deviations of experimental data from the approximating curve F(x) should be the smallest.
Approximating curve constructed using the least squares method
The least squares method is used:
To solve overdetermined systems of equations when the number of equations exceeds the number of unknowns;
To find a solution in the case of ordinary (not overdetermined) nonlinear systems of equations;
To approximate point values with some approximating function.
The approximating function using the least squares method is determined from the condition of the minimum sum of squared deviations of the calculated approximating function from a given array of experimental data. This criterion of the least squares method is written as the following expression:
The values of the calculated approximating function at the nodal points,
A given array of experimental data at nodal points.
The quadratic criterion has a number of “good” properties, such as differentiability and providing a unique solution to the approximation problem with polynomial approximating functions.
Depending on the conditions of the problem, the approximating function is a polynomial of degree m
The degree of the approximating function does not depend on the number of nodal points, but its dimension must always be less than the dimension (number of points) of a given experimental data array.
∙ If the degree of the approximating function is m=1, then we approximate the tabular function with a straight line (linear regression).
∙ If the degree of the approximating function is m=2, then we approximate the table function with a quadratic parabola (quadratic approximation).
∙ If the degree of the approximating function is m=3, then we approximate the table function with a cubic parabola (cubic approximation).
In the general case, when it is necessary to construct an approximating polynomial of degree m for given table values, the condition for the minimum sum of squared deviations over all nodal points is rewritten in the following form:
- unknown coefficients of the approximating polynomial of degree m;
The number of table values specified.
A necessary condition for the existence of a minimum of a function is the equality to zero of its partial derivatives with respect to unknown variables . As a result, we obtain the following system of equations:
Let's transform the resulting linear system equations: open the brackets and move the free terms to the right side of the expression. As a result, the resulting system of linear algebraic expressions will be written in the following form:
This system linear algebraic expressions can be rewritten in matrix form:
As a result, a system of linear equations of dimension m+1 was obtained, which consists of m+1 unknowns. This system can be solved using any method for solving linear algebraic equations (for example, the Gaussian method). As a result of the solution, unknown parameters of the approximating function will be found that provide the minimum sum of squared deviations of the approximating function from the original data, i.e. best possible quadratic approximation. It should be remembered that if even one value of the source data changes, all coefficients will change their values, since they are completely determined by the source data.
Approximation of initial data by linear dependence
(linear regression)
As an example, let's consider the technique for determining the approximating function, which is specified in the form of a linear dependence. In accordance with the least squares method, the condition for the minimum of the sum of squared deviations is written in the following form:
Coordinates of table nodes;
Unknown coefficients of the approximating function, which is specified as a linear dependence.
A necessary condition for the existence of a minimum of a function is the equality to zero of its partial derivatives with respect to unknown variables. As a result, we obtain the following system of equations:
Let us transform the resulting linear system of equations.
We solve the resulting system of linear equations. The coefficients of the approximating function in analytical form are determined as follows (Cramer’s method):
These coefficients ensure the construction of a linear approximating function in accordance with the criterion of minimizing the sum of squares of the approximating function from the given tabular values (experimental data).
Algorithm for implementing the least squares method
1. Initial data:
An array of experimental data with the number of measurements N is specified
The degree of the approximating polynomial (m) is specified
2. Calculation algorithm:
2.1. The coefficients for constructing a system of equations with dimensions are determined
Coefficients of the system of equations (left side of the equation)
- index of the column number of the square matrix of the system of equations
Free terms of a system of linear equations (right side of the equation)
- index of the row number of the square matrix of the system of equations
2.2. Formation of a system of linear equations with dimension .
2.3. Solving a system of linear equations to determine the unknown coefficients of an approximating polynomial of degree m.
2.4. Determination of the sum of squared deviations of the approximating polynomial from the original values at all nodal points
The found value of the sum of squared deviations is the minimum possible.
Approximation using other functions
It should be noted that when approximating the original data in accordance with the least squares method, the logarithmic function, exponential function and power function are sometimes used as the approximating function.
Logarithmic approximation
Let's consider the case when the approximating function is given by a logarithmic function of the form:
Example.
Experimental data on the values of variables X And at are given in the table. 
As a result of their alignment, the function is obtained 
Using least squares method, approximate these data by a linear dependence y=ax+b(find parameters A And b). Find out which of the two lines better (in the sense of the least squares method) aligns the experimental data. Make a drawing.
The essence of the least squares method (LSM).
The task is to find the linear dependence coefficients at which the function of two variables A And b 
takes the smallest value. That is, given A And b the sum of squared deviations of the experimental data from the found straight line will be the smallest. This is the whole point of the least squares method.
Thus, solving the example comes down to finding the extremum of a function of two variables.
Deriving formulas for finding coefficients.
A system of two equations with two unknowns is compiled and solved. Finding partial derivatives of a function with respect to variables A And b, we equate these derivatives to zero. 
We solve the resulting system of equations using any method (for example by substitution method or ) and obtain formulas for finding coefficients using the least squares method (LSM). 
Given A And b function takes the smallest value. The proof of this fact is given.
That's the whole method of least squares. Formula for finding the parameter a contains the sums , , , and parameter n- amount of experimental data. We recommend calculating the values of these amounts separately. Coefficient b found after calculation a.
It's time to remember the original example.
Solution.
In our example n=5. We fill out the table for the convenience of calculating the amounts that are included in the formulas of the required coefficients. 
The values in the fourth row of the table are obtained by multiplying the values of the 2nd row by the values of the 3rd row for each number i.
The values in the fifth row of the table are obtained by squaring the values in the 2nd row for each number i.
The values in the last column of the table are the sums of the values across the rows.
We use the formulas of the least squares method to find the coefficients A And b. We substitute the corresponding values from the last column of the table into them: 
Hence, y = 0.165x+2.184- the desired approximating straight line.
It remains to find out which of the lines y = 0.165x+2.184 or better approximates the original data, that is, estimates using the least squares method.
Error estimation of the least squares method.
To do this, you need to calculate the sum of squared deviations of the original data from these lines 
And 
, a smaller value corresponds to a line that better approximates the original data in the sense of the least squares method. 
Since , then straight y = 0.165x+2.184 better approximates the original data.
Graphic illustration of the least squares (LS) method.
Everything is clearly visible on the graphs. The red line is the found straight line y = 0.165x+2.184, the blue line is , pink dots are the original data.
Why is this needed, why all these approximations?
I personally use it to solve problems of data smoothing, interpolation and extrapolation problems (in the original example they might be asked to find the value of an observed value y at x=3 or when x=6 using the least squares method). But we’ll talk more about this later in another section of the site.
Proof.
So that when found A And b function takes the smallest value, it is necessary that at this point the matrix of the quadratic form of the second order differential for the function was positive definite. Let's show it.
The second order differential has the form: 
That is
Therefore, the matrix of quadratic form has the form 
and the values of the elements do not depend on A And b.
Let us show that the matrix is positive definite. To do this, the angular minors must be positive.
First order angular minor 
. The inequality is strict because the points do not coincide. In what follows we will imply this.
Second order angular minor 
Let's prove that 
by the method of mathematical induction.
Conclusion: found values A And b correspond to the smallest value of the function , therefore, are the required parameters for the least squares method.
After leveling, we obtain a function of the following form: g (x) = x + 1 3 + 1 .
We can approximate this data using the linear relationship y = a x + b by calculating the corresponding parameters. To do this, we will need to apply the so-called least squares method. You will also need to make a drawing to check which line will best align the experimental data.
What exactly is OLS (least squares method)
The main thing we need to do is to find such coefficients of linear dependence at which the value of the function of two variables F (a, b) = ∑ i = 1 n (y i - (a x i + b)) 2 will be the smallest. In other words, for certain values of a and b, the sum of the squared deviations of the presented data from the resulting straight line will have a minimum value. This is the meaning of the least squares method. All we need to do to solve the example is to find the extremum of the function of two variables.
How to derive formulas for calculating coefficients
In order to derive formulas for calculating coefficients, you need to create and solve a system of equations with two variables. To do this, we calculate the partial derivatives of the expression F (a, b) = ∑ i = 1 n (y i - (a x i + b)) 2 with respect to a and b and equate them to 0.
δ F (a , b) δ a = 0 δ F (a , b) δ b = 0 ⇔ - 2 ∑ i = 1 n (y i - (a x i + b)) x i = 0 - 2 ∑ i = 1 n ( y i - (a x i + b)) = 0 ⇔ a ∑ i = 1 n x i 2 + b ∑ i = 1 n x i = ∑ i = 1 n x i y i a ∑ i = 1 n x i + ∑ i = 1 n b = ∑ i = 1 n y i ⇔ a ∑ i = 1 n x i 2 + b ∑ i = 1 n x i = ∑ i = 1 n x i y i a ∑ i = 1 n x i + n b = ∑ i = 1 n y i
To solve a system of equations, you can use any methods, for example, substitution or Cramer's method. As a result, we should have formulas that can be used to calculate coefficients using the least squares method.
n ∑ i = 1 n x i y i - ∑ i = 1 n x i ∑ i = 1 n y i n ∑ i = 1 n - ∑ i = 1 n x i 2 b = ∑ i = 1 n y i - a ∑ i = 1 n x i n
We have calculated the values of the variables at which the function
F (a , b) = ∑ i = 1 n (y i - (a x i + b)) 2 will take the minimum value. In the third paragraph we will prove why it is exactly like this.
This is the application of the least squares method in practice. Its formula, which is used to find the parameter a, includes ∑ i = 1 n x i, ∑ i = 1 n y i, ∑ i = 1 n x i y i, ∑ i = 1 n x i 2, as well as the parameter
n – it denotes the amount of experimental data. We advise you to calculate each amount separately. The value of the coefficient b is calculated immediately after a.
Let's go back to the original example.
Example 1
Here we have n equals five. To make it more convenient to calculate the required amounts included in the coefficient formulas, let’s fill out the table.
| i=1 | i=2 | i=3 | i=4 | i=5 | ∑ i = 1 5 | |
| x i | 0 | 1 | 2 | 4 | 5 | 12 |
| y i | 2 , 1 | 2 , 4 | 2 , 6 | 2 , 8 | 3 | 12 , 9 |
| x i y i | 0 | 2 , 4 | 5 , 2 | 11 , 2 | 15 | 33 , 8 |
| x i 2 | 0 | 1 | 4 | 16 | 25 | 46 |
Solution
The fourth row includes the data obtained by multiplying the values from the second row by the values of the third for each individual i. The fifth line contains the data from the second, squared. The last column shows the sums of the values of individual rows.
Let's use the least squares method to calculate the coefficients a and b we need. To do this, substitute the required values from the last column and calculate the amounts:
n ∑ i = 1 n x i y i - ∑ i = 1 n x i ∑ i = 1 n y i n ∑ i = 1 n - ∑ i = 1 n x i 2 b = ∑ i = 1 n y i - a ∑ i = 1 n x i n ⇒ a = 5 33, 8 - 12 12, 9 5 46 - 12 2 b = 12, 9 - a 12 5 ⇒ a ≈ 0, 165 b ≈ 2, 184
It turns out that the required approximating straight line will look like y = 0, 165 x + 2, 184. Now we need to determine which line will better approximate the data - g (x) = x + 1 3 + 1 or 0, 165 x + 2, 184. Let's estimate using the least squares method.
To calculate the error, we need to find the sum of squared deviations of the data from the straight lines σ 1 = ∑ i = 1 n (y i - (a x i + b i)) 2 and σ 2 = ∑ i = 1 n (y i - g (x i)) 2, the minimum value will correspond to a more suitable line.
σ 1 = ∑ i = 1 n (y i - (a x i + b i)) 2 = = ∑ i = 1 5 (y i - (0, 165 x i + 2, 184)) 2 ≈ 0, 019 σ 2 = ∑ i = 1 n (y i - g (x i)) 2 = = ∑ i = 1 5 (y i - (x i + 1 3 + 1)) 2 ≈ 0.096
Answer: since σ 1< σ 2 , то прямой, наилучшим образом аппроксимирующей исходные данные, будет
y = 0.165 x + 2.184.
The least squares method is clearly shown in the graphical illustration. The red line marks the straight line g (x) = x + 1 3 + 1, the blue line marks y = 0, 165 x + 2, 184. The original data is indicated by pink dots.
Let us explain why exactly approximations of this type are needed.
They can be used in tasks that require data smoothing, as well as in those where data must be interpolated or extrapolated. For example, in the problem discussed above, one could find the value of the observed quantity y at x = 3 or at x = 6. We have devoted a separate article to such examples.
Proof of the OLS method
In order for the function to take a minimum value when a and b are calculated, it is necessary that at a given point the matrix of the quadratic form of the differential of the function of the form F (a, b) = ∑ i = 1 n (y i - (a x i + b)) 2 is positive definite. Let's show you how it should look.
Example 2
We have a second order differential of the following form:
d 2 F (a ; b) = δ 2 F (a ; b) δ a 2 d 2 a + 2 δ 2 F (a ; b) δ a δ b d a d b + δ 2 F (a ; b) δ b 2 d 2 b
Solution
δ 2 F (a ; b) δ a 2 = δ δ F (a ; b) δ a δ a = = δ - 2 ∑ i = 1 n (y i - (a x i + b)) x i δ a = 2 ∑ i = 1 n (x i) 2 δ 2 F (a; b) δ a δ b = δ δ F (a; b) δ a δ b = = δ - 2 ∑ i = 1 n (y i - (a x i + b) ) x i δ b = 2 ∑ i = 1 n x i δ 2 F (a ; b) δ b 2 = δ δ F (a ; b) δ b δ b = δ - 2 ∑ i = 1 n (y i - (a x i + b)) δ b = 2 ∑ i = 1 n (1) = 2 n
In other words, we can write it like this: d 2 F (a ; b) = 2 ∑ i = 1 n (x i) 2 d 2 a + 2 2 ∑ x i i = 1 n d a d b + (2 n) d 2 b.
We obtained a matrix of the quadratic form M = 2 ∑ i = 1 n (x i) 2 2 ∑ i = 1 n x i 2 ∑ i = 1 n x i 2 n .
In this case, the values of individual elements will not change depending on a and b . Is this matrix positive definite? To answer this question, let's check whether its angular minors are positive.
We calculate the angular minor of the first order: 2 ∑ i = 1 n (x i) 2 > 0 . Since the points x i do not coincide, the inequality is strict. We will keep this in mind in further calculations.
We calculate the second order angular minor:
d e t (M) = 2 ∑ i = 1 n (x i) 2 2 ∑ i = 1 n x i 2 ∑ i = 1 n x i 2 n = 4 n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2
After this, we proceed to prove the inequality n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 > 0 using mathematical induction.
- Let's check whether this inequality is valid for an arbitrary n. Let's take 2 and calculate:
2 ∑ i = 1 2 (x i) 2 - ∑ i = 1 2 x i 2 = 2 x 1 2 + x 2 2 - x 1 + x 2 2 = = x 1 2 - 2 x 1 x 2 + x 2 2 = x 1 + x 2 2 > 0
We have obtained a correct equality (if the values x 1 and x 2 do not coincide).
- Let us make the assumption that this inequality will be true for n, i.e. n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 > 0 – true.
- Now we will prove the validity for n + 1, i.e. that (n + 1) ∑ i = 1 n + 1 (x i) 2 - ∑ i = 1 n + 1 x i 2 > 0, if n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 > 0 .
We calculate:
(n + 1) ∑ i = 1 n + 1 (x i) 2 - ∑ i = 1 n + 1 x i 2 = = (n + 1) ∑ i = 1 n (x i) 2 + x n + 1 2 - ∑ i = 1 n x i + x n + 1 2 = = n ∑ i = 1 n (x i) 2 + n x n + 1 2 + ∑ i = 1 n (x i) 2 + x n + 1 2 - - ∑ i = 1 n x i 2 + 2 x n + 1 ∑ i = 1 n x i + x n + 1 2 = = ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 + n x n + 1 2 - x n + 1 ∑ i = 1 n x i + ∑ i = 1 n (x i) 2 = = ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 + x n + 1 2 - 2 x n + 1 x 1 + x 1 2 + + x n + 1 2 - 2 x n + 1 x 2 + x 2 2 + . . . + x n + 1 2 - 2 x n + 1 x 1 + x n 2 = = n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 + + (x n + 1 - x 1) 2 + (x n + 1 - x 2) 2 + . . . + (x n - 1 - x n) 2 > 0
The expression enclosed in curly braces will be greater than 0 (based on what we assumed in step 2), and the remaining terms will be greater than 0, since they are all squares of numbers. We have proven the inequality.
Answer: the found a and b will correspond to the smallest value of the function F (a, b) = ∑ i = 1 n (y i - (a x i + b)) 2, which means that they are the required parameters of the least squares method (LSM).
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The essence of the least squares method is in finding the parameters of a trend model that best describes the tendency of development of any random phenomenon in time or space (a trend is a line that characterizes the tendency of this development). The task of the least squares method (LSM) comes down to finding not just some trend model, but to finding the best or optimal model. This model will be optimal if the sum of square deviations between the observed actual values and the corresponding calculated trend values is minimal (smallest):
where is the square deviation between the observed actual value
and the corresponding calculated trend value,
The actual (observed) value of the phenomenon being studied,
The calculated value of the trend model,
The number of observations of the phenomenon being studied.
MNC is used quite rarely on its own. As a rule, most often it is used only as a necessary technical technique in correlation studies. It should be remembered that the information basis of OLS can only be a reliable statistical series, and the number of observations should not be less than 4, otherwise the smoothing procedures of OLS may lose common sense.
The MNC toolkit boils down to the following procedures:
First procedure. It turns out whether there is any tendency at all to change the resultant attribute when the selected factor-argument changes, or in other words, is there a connection between “ at " And " X ».
Second procedure. It is determined which line (trajectory) can best describe or characterize this trend.
Third procedure.
Example. Let's say we have information about the average sunflower yield for the farm under study (Table 9.1).
Table 9.1
|
Observation number |
||||||||||
|
Productivity, c/ha |
Since the level of technology in sunflower production in our country has remained virtually unchanged over the past 10 years, it means that, apparently, fluctuations in yield during the analyzed period were very much dependent on fluctuations in weather and climatic conditions. Is this really true?
First OLS procedure. The hypothesis about the existence of a trend in sunflower yield changes depending on changes in weather and climatic conditions over the analyzed 10 years is tested.
In this example, for " y " it is advisable to take the sunflower yield, and for " x » – number of the observed year in the analyzed period. Testing the hypothesis about the existence of any relationship between " x " And " y "can be done in two ways: manually and using computer programs. Of course, with the availability of computer technology, this problem can be solved by itself. But in order to better understand the MNC tools, it is advisable to test the hypothesis about the existence of a relationship between “ x " And " y » manually, when only a pen and an ordinary calculator are at hand. In such cases, the hypothesis about the existence of a trend is best checked visually by the location of the graphical image of the analyzed series of dynamics - the correlation field:
The correlation field in our example is located around a slowly increasing line. This in itself indicates the existence of a certain trend in changes in sunflower yields. It is impossible to talk about the presence of any tendency only when the correlation field looks like a circle, a circle, a strictly vertical or strictly horizontal cloud, or consists of chaotically scattered points. In all other cases, the hypothesis about the existence of a relationship between “ x " And " y ", and continue research.
Second OLS procedure. It is determined which line (trajectory) can best describe or characterize the trend of changes in sunflower yield over the analyzed period.
If you have computer technology, the selection of the optimal trend occurs automatically. In “manual” processing, the selection of the optimal function is carried out, as a rule, visually - by the location of the correlation field. That is, based on the type of graph, the equation of the line that best fits the empirical trend (the actual trajectory) is selected.
As is known, in nature there is a huge variety of functional dependencies, so it is extremely difficult to visually analyze even a small part of them. Fortunately, in real economic practice, most relationships can be described quite accurately either by a parabola, or a hyperbola, or a straight line. In this regard, with the “manual” option of selecting the best function, you can limit yourself to only these three models.
|
Hyperbola: |
||
|
|
|
Second order parabola: 
:
It is easy to notice that in our example, the trend of changes in sunflower yield over the analyzed 10 years is best characterized by a straight line, so the regression equation will be the equation of a straight line.
Third procedure. The parameters of the regression equation characterizing this line are calculated, or in other words, an analytical formula is determined that describes best model trend.
Finding the values of the parameters of the regression equation, in our case the parameters and , is the core of the OLS. This process comes down to solving a system of normal equations.
(9.2)
This system of equations can be solved quite easily by the Gauss method. Let us recall that as a result of the solution, in our example, the values of the parameters and are found. Thus, the found regression equation will have the following form: 
If a certain physical quantity depends on another quantity, then this dependence can be studied by measuring y at different values of x. As a result of measurements, a number of values are obtained:
x 1, x 2, ..., x i, ..., x n;
y 1 , y 2 , ..., y i , ... , y n .
Based on the data of such an experiment, it is possible to construct a graph of the dependence y = ƒ(x). The resulting curve makes it possible to judge the form of the function ƒ(x). However, the constant coefficients that enter into this function remain unknown. They can be determined using the least squares method. Experimental points, as a rule, do not lie exactly on the curve. The least squares method requires that the sum of the squares of the deviations of the experimental points from the curve, i.e. 2 was the smallest.
In practice, this method is most often (and most simply) used in the case of a linear relationship, i.e. When
y = kx or y = a + bx.
Linear dependence is very widespread in physics. And even when the relationship is nonlinear, they usually try to construct a graph so as to get a straight line. For example, if it is assumed that the refractive index of glass n is related to the light wavelength λ by the relation n = a + b/λ 2, then the dependence of n on λ -2 is plotted on the graph.
Consider the dependency y = kx(a straight line passing through the origin). Let's compose the value φ the sum of the squares of the deviations of our points from the straight line
The value of φ is always positive and turns out to be smaller the closer our points are to the straight line. The least squares method states that the value for k should be chosen such that φ has a minimum
or
(19)
The calculation shows that the root-mean-square error in determining the value of k is equal to
, (20)
where n is the number of measurements.
Let us now consider a slightly more difficult case, when the points must satisfy the formula y = a + bx(a straight line not passing through the origin).
The task is to find the best values of a and b from the available set of values x i, y i.
Let us again compose a quadratic form φ, equal to the sum of the squared deviations of points x i, y i from the straight line
and find the values of a and b for which φ has a minimum
;
.
The joint solution of these equations gives
(21)
The root mean square errors of determination of a and b are equal
(23)
. (24)
When processing measurement results using this method, it is convenient to summarize all the data in a table in which all the amounts included in formulas (19)(24) are preliminarily calculated. The forms of these tables are given in the examples below.
Example 1. The basic equation of dynamics was studied rotational movementε = M/J (line passing through the origin). At different values of the moment M, the angular acceleration ε of a certain body was measured. It is required to determine the moment of inertia of this body. The results of measurements of the moment of force and angular acceleration are listed in the second and third columns table 5.
Table 5
| n | M, N m | ε, s -1 | M 2 | M ε | ε - kM | (ε - kM) 2 |
| 1 | 1.44 | 0.52 | 2.0736 | 0.7488 | 0.039432 | 0.001555 |
| 2 | 3.12 | 1.06 | 9.7344 | 3.3072 | 0.018768 | 0.000352 |
| 3 | 4.59 | 1.45 | 21.0681 | 6.6555 | -0.08181 | 0.006693 |
| 4 | 5.90 | 1.92 | 34.81 | 11.328 | -0.049 | 0.002401 |
| 5 | 7.45 | 2.56 | 55.5025 | 19.072 | 0.073725 | 0.005435 |
| ∑ | | | 123.1886 | 41.1115 | | 0.016436 |
Using formula (19) we determine:
.
To determine the root mean square error, we use formula (20)
0.005775kg-1 · m -2 .
According to formula (18) we have
S J = (2.996 0.005775)/0.3337 = 0.05185 kg m2.
Having set the reliability P = 0.95, using the table of Student coefficients for n = 5, we find t = 2.78 and determine the absolute error ΔJ = 2.78 0.05185 = 0.1441 ≈ 0.2 kg m2.
Let's write the results in the form:
J = (3.0 ± 0.2) kg m2;
Example 2. Let's calculate the temperature coefficient of metal resistance using the least squares method. Resistance depends linearly on temperature
R t = R 0 (1 + α t°) = R 0 + R 0 α t°.
The free term determines the resistance R 0 at a temperature of 0 ° C, and the slope coefficient is the product of the temperature coefficient α and the resistance R 0 .
The results of measurements and calculations are given in the table ( see table 6).
Table 6
| n | t°, s | r, Ohm | t-¯t | (t-¯t) 2 | (t-¯t)r | r - bt - a | (r - bt - a) 2 .10 -6 |
| 1 | 23 | 1.242 | -62.8333 | 3948.028 | -78.039 | 0.007673 | 58.8722 |
| 2 | 59 | 1.326 | -26.8333 | 720.0278 | -35.581 | -0.00353 | 12.4959 |
| 3 | 84 | 1.386 | -1.83333 | 3.361111 | -2.541 | -0.00965 | 93.1506 |
| 4 | 96 | 1.417 | 10.16667 | 103.3611 | 14.40617 | -0.01039 | 107.898 |
| 5 | 120 | 1.512 | 34.16667 | 1167.361 | 51.66 | 0.021141 | 446.932 |
| 6 | 133 | 1.520 | 47.16667 | 2224.694 | 71.69333 | -0.00524 | 27.4556 |
| ∑ | 515 | 8.403 | | 8166.833 | 21.5985 | | 746.804 |
| ∑/n | 85.83333 | 1.4005 | | | | | |
Using formulas (21), (22) we determine
R 0 = ¯ R- α R 0 ¯ t = 1.4005 - 0.002645 85.83333 = 1.1735 Ohm.
Let's find an error in the definition of α. Since , then according to formula (18) we have:
.
Using formulas (23), (24) we have
;
0.014126 Ohm.
Having set the reliability to P = 0.95, using the table of Student coefficients for n = 6, we find t = 2.57 and determine the absolute error Δα = 2.57 0.000132 = 0.000338 deg -1.
α = (23 ± 4) 10 -4 hail-1 at P = 0.95.
Example 3. It is required to determine the radius of curvature of the lens using Newton's rings. The radii of Newton's rings r m were measured and the numbers of these rings m were determined. The radii of Newton's rings are related to the radius of curvature of the lens R and the ring number by the equation
r 2 m = mλR - 2d 0 R,
where d 0 the thickness of the gap between the lens and the plane-parallel plate (or the deformation of the lens),
λ wavelength of incident light.
λ = (600 ± 6) nm;
r 2 m = y;
m = x;
λR = b;
-2d 0 R = a,
then the equation will take the form y = a + bx.
.The results of measurements and calculations are entered into table 7.
Table 7
| n | x = m | y = r 2, 10 -2 mm 2 | m -¯ m | (m -¯m) 2 | (m -¯ m)y | y - bx - a, 10 -4 | (y - bx - a) 2 , 10 -6 |
| 1 | 1 | 6.101 | -2.5 | 6.25 | -0.152525 | 12.01 | 1.44229 |
| 2 | 2 | 11.834 | -1.5 | 2.25 | -0.17751 | -9.6 | 0.930766 |
| 3 | 3 | 17.808 | -0.5 | 0.25 | -0.08904 | -7.2 | 0.519086 |
| 4 | 4 | 23.814 | 0.5 | 0.25 | 0.11907 | -1.6 | 0.0243955 |
| 5 | 5 | 29.812 | 1.5 | 2.25 | 0.44718 | 3.28 | 0.107646 |
| 6 | 6 | 35.760 | 2.5 | 6.25 | 0.894 | 3.12 | 0.0975819 |
| ∑ | 21 | 125.129 | | 17.5 | 1.041175 | | 3.12176 |
| ∑/n | 3.5 | 20.8548333 | | | | | |