After alignment, we get a function of the following form: g (x) = x + 1 3 + 1 .
We can approximate this data with a linear relationship y = a x + b by calculating the appropriate parameters. To do this, we will need to apply the so-called method least squares. You will also need to make a drawing to check which line will best align the experimental data.
Yandex.RTB R-A-339285-1
The main thing we need to do is to find such linear dependence coefficients 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 minimum value. This is the meaning of the least squares method. All we have to do to solve the example is to find the extremum of the function of two variables.
In order to derive formulas for calculating the coefficients, it is necessary to compose 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 (yi - (axi + b)) xi = 0 - 2 ∑ i = 1 n ( yi - (axi + b)) = 0 ⇔ a ∑ i = 1 nxi 2 + b ∑ i = 1 nxi = ∑ i = 1 nxiyia ∑ i = 1 nxi + ∑ i = 1 nb = ∑ i = 1 nyi ⇔ a ∑ i = 1 nxi 2 + b ∑ i = 1 nxi = ∑ i = 1 nxiyia ∑ i = 1 nxi + nb = ∑ i = 1 nyi
To solve a system of equations, you can use any methods, such as substitution or Cramer's method. As a result, we should get formulas that calculate the 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 for 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 so.
This is the application of the least squares method in practice. His 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 , and the parameter
n - it denotes the amount of experimental data. We advise you to calculate each amount separately. The coefficient value b is calculated immediately after a .
Let's go back to the original example.
Example 1
Here we have n equal to five. To make it more convenient to calculate the required amounts included in the coefficient formulas, we 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 contains 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 the individual rows.
Let's use the least squares method to calculate the coefficients a and b we need. For this we substitute desired values from the last column and calculate the sums:
n ∑ i = 1 nxiyi - ∑ i = 1 nxi ∑ i = 1 nyin ∑ i = 1 n - ∑ i = 1 nxi 2 b = ∑ i = 1 nyi - a ∑ i = 1 nxin ⇒ a = 5 33 , 8 - 12 12, 9 5 46 - 12 2 b = 12, 9 - a 12 5 ⇒ a ≈ 0, 165 b ≈ 2, 184
We got that the desired approximating straight line will look like y = 0 , 165 x + 2 , 184 . Now we need to determine which line will best approximate the data - g (x) = x + 1 3 + 1 or 0 , 165 x + 2 , 184 . Let's make an estimate using the least squares method.
To calculate the error, we need to find the sums of squared deviations of the data from the lines σ 1 = ∑ i = 1 n (yi - (axi + bi)) 2 and σ 2 = ∑ i = 1 n (yi - g (xi)) 2 , the minimum value will correspond to a more suitable line.
σ 1 = ∑ i = 1 n (yi - (axi + bi)) 2 = = ∑ i = 1 5 (yi - (0 , 165 xi + 2 , 184)) 2 ≈ 0 , 019 σ 2 = ∑ i = 1 n (yi - g (xi)) 2 = = ∑ i = 1 5 (yi - (xi + 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 graphic illustration. The red line marks the straight line g (x) = x + 1 3 + 1, the blue line marks y = 0, 165 x + 2, 184. Raw data are marked with pink dots.
Let us explain why exactly approximations of this type are needed.
They can be used in problems that require data smoothing, as well as in those where the data needs to 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.
For the function to take the minimum value for calculated a and b, 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 be 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 bdadb + δ 2 F (a ; b) δ b 2 d 2b
Solution
δ 2 F (a ; b) δ a 2 = δ δ F (a ; b) δ a δ a = = δ - 2 ∑ i = 1 n (yi - (axi + b)) xi δ a = 2 ∑ i = 1 n (xi) 2 δ 2 F (a ; b) δ a δ b = δ δ F (a ; b) δ a δ b = = δ - 2 ∑ i = 1 n (yi - (axi + b) ) xi δ b = 2 ∑ i = 1 nxi δ 2 F (a ; b) δ b 2 = δ δ F (a ; b) δ b δ b = δ - 2 ∑ i = 1 n (yi - (axi + b)) δ b = 2 ∑ i = 1 n (1) = 2 n
In other words, it can be written as follows: 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 have obtained a matrix of 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 individual elements will not change depending on a and b . Is this matrix positive definite? To answer this question, let's check if its angular minors are positive.
Calculate the first order angular minor: 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 that, we proceed to the proof of the inequality n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 > 0 using mathematical induction.
2 ∑ i = 1 2 (xi) 2 - ∑ i = 1 2 xi 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 made it true equality(if the values x 1 and x 2 do not match).
We calculate:
(n + 1) ∑ i = 1 n + 1 (xi) 2 - ∑ i = 1 n + 1 xi 2 = = (n + 1) ∑ i = 1 n (xi) 2 + xn + 1 2 - ∑ i = 1 nxi + xn + 1 2 = = n ∑ i = 1 n (xi) 2 + n xn + 1 2 + ∑ i = 1 n (xi) 2 + xn + 1 2 - - ∑ i = 1 nxi 2 + 2 xn + 1 ∑ i = 1 nxi + xn + 1 2 = = ∑ i = 1 n (xi) 2 - ∑ i = 1 nxi 2 + n xn + 1 2 - xn + 1 ∑ i = 1 nxi + ∑ i = 1 n (xi) 2 = = ∑ i = 1 n (xi) 2 - ∑ i = 1 nxi 2 + xn + 1 2 - 2 xn + 1 x 1 + x 1 2 + + xn + 1 2 - 2 xn + 1 x 2 + x 2 2 + . . . + xn + 1 2 - 2 xn + 1 x 1 + xn 2 = = n ∑ i = 1 n (xi) 2 - ∑ i = 1 nxi 2 + + (xn + 1 - x 1) 2 + (xn + 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 rest of the terms will be greater than 0 because 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 desired parameters of the least squares method (LSM).
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
The least squares method is a mathematical procedure for compiling linear equation, maximally corresponding to a set of ordered pairs, by finding values for a and b, the coefficients in the straight line equation. The goal of the least squares method is to minimize the total squared error between the y and ŷ values. If for each point we determine the error ŷ, the least squares method minimizes:
where n = number of ordered pairs around the line. most relevant to the data.
This concept is illustrated in Figure
Judging by the figure, the line that best fits the data, the regression line, minimizes the total squared error of the four points on the graph. I will show you how to determine this using the least squares method in the following example.
Imagine a young couple who recently live together and share a bathroom vanity table. The young man began to notice that half of his table was shrinking inexorably, losing ground to hair mousses and soy complexes. Over the past few months, the guy has been closely monitoring the rate at which the number of items on her part of the table is increasing. The table below shows the number of objects the girl has accumulated on the bathroom table over the past few months.
Since our goal is to find out if the number of items increases over time, "Month" will be the independent variable, and "Number of Items" will be the dependent variable.
Using the least squares method, we determine the equation that best fits the data by calculating the values of a, the segment on the y-axis, and b, the slope of the line:
a = y cf - bx cf
where x cf is the mean value of x, the independent variable, y cf is the mean value of y, the independent variable.
The table below summarizes the calculations required for these equations.
The effect curve for our bathtub example would be given by the following equation:
Since our equation has a positive slope of 0.976, the guy has proof that the number of items on the table increases over time with average speed 1 item per month. The graph shows the effect curve with ordered pairs.
The expected number of items for the next half year (month 16) will be calculated as follows:
ŷ = 5.13 + 0.976x = 5.13 + 0.976(16) ~ 20.7 = 21 items
So it's time for our hero to take some action.
As you may have guessed, Excel has a function to calculate a value from least squares method. This feature is called TREND. Its syntax is the following:
TREND (known Y values; known X values; new X values; const)
known values of Y - an array of dependent variables, in our case, the number of items on the table
known values of X - an array of independent variables, in our case it is a month
new X values – new X (month) values for which TREND function returns the expected value of dependent variables (number of items)
const - optional. boolean value, which indicates whether the constant b is required to be 0.
For example, the figure shows the TREND function used to determine the expected number of items on the bathroom table for the 16th month.
100 r first order bonus
Choose the type of work Thesis Course work Abstract Master's thesis Report on practice Article Report Review Test Monograph Problem solving Business plan Answers to questions creative work Essay Drawing Compositions Translation Presentations Typing Other Increasing the uniqueness of the text Candidate's thesis Laboratory work Help online
Ask for a price
The method of least squares is a mathematical (mathematical and statistical) technique that serves to align dynamic series, identify the form of a correlation between random variables, etc. It consists in the fact that the function that describes this phenomenon, is approximated by a simpler function. Moreover, the latter is selected in such a way that the standard deviation (see Variance) of the actual levels of the function at the observed points from the leveled ones is the smallest.
For example, according to available data ( xi,yi) (i = 1, 2, ..., n) such a curve is constructed y = a + bx, on which the minimum of the sum of squared deviations is reached
i.e., a function is minimized that depends on two parameters: a- segment on the y-axis and b- the slope of the straight line.
Equations giving necessary conditions for minimizing a function S(a,b), are called normal equations. As approximating functions, not only linear (alignment along a straight line), but also quadratic, parabolic, exponential, etc. are used. M.2, where the sum of squared distances ( y 1 – ȳ 1)2 + (y 2 – ȳ 2)2 .... is the smallest, and the resulting straight line best reflects the trend of the dynamic series of observations for some indicator over time.
For unbiased least squares estimators, it is necessary and sufficient that essential condition regression analysis: the mathematical expectation of a random error conditional on the factors must be equal to zero. This condition, in particular, is met if: 1.the expectation of random errors is zero, and 2.factors and random errors are independent random variables. The first condition can be considered to be always satisfied for models with a constant, since the constant takes on a non-zero mathematical expectation of errors. The second condition - the condition of exogenous factors - is fundamental. If this property is not satisfied, then we can assume that almost any estimates will be extremely unsatisfactory: they will not even be consistent (that is, even a very large amount of data does not allow obtaining qualitative estimates in this case).
The most common in the practice of statistical estimation of the parameters of regression equations is the method of least squares. This method is based on a number of assumptions about the nature of the data and the results of the model building. The main ones are a clear separation of the initial variables into dependent and independent ones, the uncorrelatedness of the factors included in the equations, the linearity of the relationship, the absence of autocorrelation of the residuals, their equality mathematical expectations zero and constant dispersion.
One of the main hypotheses of the LSM is the assumption that the dispersions of deviations ei are equal, i.e. their spread around the average (zero) value of the series should be a stable value. This property is called homoscedasticity. In practice, the variances of deviations are quite often not the same, that is, heteroscedasticity is observed. This may be a consequence different reasons. For example, there may be errors in the original data. Random inaccuracies in the source information, such as errors in the order of numbers, can have a significant impact on the results. Often a greater spread of deviations єi is observed at large values dependent variable(s). If the data contains a significant error, then, naturally, the deviation of the model value calculated from the erroneous data will also be large. In order to get rid of this error, we need to reduce the contribution of these data to the calculation results, set a lower weight for them than for all the rest. This idea is implemented in weighted least squares.
It has many applications, as it allows an approximate representation of a given function by other simpler ones. LSM can be extremely useful in processing observations, and it is actively used to estimate some quantities from the results of measurements of others containing random errors. In this article, you will learn how to implement least squares calculations in Excel.
Suppose there are two indicators X and Y. Moreover, Y depends on X. Since OLS is of interest to us from the point of view of regression analysis (in Excel, its methods are implemented using built-in functions), we should immediately proceed to consider a specific problem.
So let X be the sales area grocery store, measured in square meters, and Y is the annual turnover, defined in millions of rubles.
It is required to make a forecast of what turnover (Y) the store will have if it has one or another retail space. Obviously, the function Y = f (X) is increasing, since the hypermarket sells more goods than the stall.
Let's say we have a table built with data for n stores.
According to mathematical statistics, the results will be more or less correct if the data on at least 5-6 objects are examined. Also, "anomalous" results cannot be used. In particular, an elite small boutique can have a turnover many times greater than the turnover of large outlets of the “masmarket” class.
The table data can be displayed on the Cartesian plane as points M 1 (x 1, y 1), ... M n (x n, y n). Now the solution of the problem will be reduced to the selection of an approximating function y = f (x), which has a graph passing as close as possible to the points M 1, M 2, .. M n .
Of course, you can use the polynomial high degree, but this option is not only difficult to implement, but simply incorrect, since it will not reflect the main trend that needs to be detected. The most reasonable solution is to search for a straight line y = ax + b, which best approximates the experimental data, and more precisely, the coefficients - a and b.
For any approximation, the assessment of its accuracy is of particular importance. Denote by e i the difference (deviation) between the functional and experimental values for the point x i , i.e. e i = y i - f (x i).
Obviously, to assess the accuracy of the approximation, you can use the sum of the deviations, i.e., when choosing a straight line for an approximate representation of the dependence of X on Y, one should give preference to the one for which smallest value sums e i at all considered points. However, not everything is so simple, since along with positive deviations, negative ones will practically also be present.
You can solve the problem using the deviation modules or their squares. The latter method is the most widely used. It is used in many areas, including regression analysis (in Excel, its implementation is carried out using two built-in functions), and has long been proven to be effective.
In Excel, as you know, there is a built-in autosum function that allows you to calculate the values of all values located in the selected range. Thus, nothing will prevent us from calculating the value of the expression (e 1 2 + e 2 2 + e 3 2 + ... e n 2).
In mathematical notation, this looks like:
Since the decision was initially made to approximate using a straight line, we have:
Thus, the task of finding a straight line that best describes a specific relationship between X and Y amounts to calculating the minimum of a function of two variables:
This requires equating to zero partial derivatives with respect to new variables a and b, and solving a primitive system consisting of two equations with 2 unknowns of the form:
After simple transformations, including dividing by 2 and manipulating the sums, we get:
Solving it, for example, by Cramer's method, we obtain a stationary point with certain coefficients a * and b * . This is the minimum, i.e. to predict what turnover the store will have for a certain area, the straight line y = a * x + b * is suitable, which is a regression model for the example in question. Of course she won't let you find exact result, but it will help you get an idea of whether buying a store on credit for a particular area will pay off.
Excel has a function for calculating the value of the least squares. She has next view: "TREND" (known Y values; known X values; new X values; constant). Let's apply the formula for calculating the OLS in Excel to our table.
To do this, in the cell in which the result of the calculation by the least squares method in Excel should be displayed, enter the “=” sign and select the “TREND” function. In the window that opens, fill in the appropriate fields, highlighting:
In addition, there is a logical variable "Const" in the formula. If you enter 1 in the field corresponding to it, then this will mean that calculations should be carried out, assuming that b \u003d 0.
If you need to know the forecast for more than one x value, then after entering the formula, you should not press "Enter", but you need to type the combination "Shift" + "Control" + "Enter" ("Enter") on the keyboard.
Regression analysis can be accessible even to dummies. Excel formula to predict the value of an array of unknown variables - "TREND" - can be used even by those who have never heard of the least squares method. It is enough just to know some features of its work. In particular:
It is implemented using several functions. One of them is called "PREDICTION". It is similar to TREND, i.e. it gives the result of calculations using the least squares method. However, only for one X, for which the value of Y is unknown.
Now you know the Excel formulas for dummies that allow you to predict the value of the future value of an indicator according to a linear trend.
If some 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 series of values is 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 plot 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. The experimental points, as a rule, do not lie exactly on the curve. The method of least squares requires that the sum of the squared 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 dependency very common in physics. And even when the dependence is non-linear, they usually try to build a graph in such a way as to get a straight line. For example, if it is assumed that the refractive index of glass n is related to the wavelength λ of the light wave by the relation n = a + b/λ 2 , then the dependence of n on λ -2 is plotted on the graph.
Consider the dependence y=kx(straight line passing through the origin). Compose the value φ - the sum of the squared deviations of our points from the straight line
The value of φ is always positive and turns out to be the smaller, the closer our points lie to the straight line. The method of least squares states that for k one should choose such a value at which 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 somewhat 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 given set of values x i , y i best values a and b.
Let's compose again quadratic formφ equal to the sum of the squared deviations of the points x i , y i from the straight line
and find the values a and b for which φ has a minimum
;
.
The joint solution of these equations gives
(21)
The root-mean-square errors of determining a and b are equal
(23)
. (24)
When processing the measurement results by this method, it is convenient to summarize all the data in a table in which all the sums included in formulas (19)–(24) are preliminarily calculated. The forms of these tables are shown in the examples below.
Example 1 The basic equation of dynamics was studied rotary motionε = M/J (straight line passing through the origin). At various values of the moment M, it was measured angular accelerationε of some body. 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 tables 5.
| n | M, N m | ε, s-1 | M2 | 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 |
By formula (19) we determine:
.
To determine the root-mean-square error, we use formula (20)
0.005775kg-one · m -2 .
By formula (18) we have
SJ = (2.996 0.005775)/0.3337 = 0.05185 kg m 2.
Given the reliability P = 0.95, according to the table of Student's coefficients for n = 5, we find t = 2.78 and determine the absolute error ΔJ = 2.78 0.05185 = 0.1441 ≈ 0.2 kg m 2.
We write the results in the form:
J = (3.0 ± 0.2) kg m 2;
Example 2 We calculate the temperature coefficient of resistance of the metal using the least squares method. Resistance depends on temperature according to a linear law
R t \u003d R 0 (1 + α t °) \u003d R 0 + R 0 α t °.
The free term determines the resistance R 0 at a temperature of 0 ° C, and slope- work temperature coefficientα to the resistance R 0 .
The results of measurements and calculations are given in the table ( see 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 | | | | | |
By formulas (21), (22) we determine
R 0 = ¯ R- α R 0 ¯ t = 1.4005 - 0.002645 85.83333 = 1.1735 Ohm.
Let us 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.
Given the reliability P = 0.95, according to 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 from 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 is the thickness of the gap between the lens and the plane-parallel plate (or lens deformation),
λ is the wavelength of the 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 in table 7.
| n | x = m | y \u003d 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 | | | | | |
