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IMPROVING THE ACCURACY OF INTERPOLATION OF FUNCTIONS
IN LAGRANGE AND NEWTON'S METHOD
Mardonov Dilmurod
PhD of Samarkand State University named after Sharof Rashidov
Rashidov Akbar
PhD of Samarkand State University named after Sharof Rashidov
Xuramov Latif
PhD of Samarkand State University named after Sharof Rashidov
e-mail: latifxya@gmail.com,
https://doi.org/10.5281/zenodo.15584107
Annotation
. In this work in a limited range located from points harvest
was without a break
)
(
x
f
function for
𝐿
𝑛
(𝑥)
Lagrange and
)
(
x
N
n
In Newton
Interpolation Methods approximation has been made, and their interpolation
The errors were estimated . At the same time , an improved algorithm for
constructing Lagrange and Newton interpolation methods was developed. As a
result, it was shown which method is more effective in approximating functions.
Key words
: Lagrange interpolation, Newton interpolation error,
approximation.
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The urgency of the problem is that in planning, weather forecasting, and
determining land resources, it is necessary to use the values of the function at
several points. Based on these aspects, it is urgent to derive the mathematical
regularity of the problem on a computer using mathematical modeling and
computer simulation.
Most computational methods are based on replacing the functions
involved in the formulation of the problem with functions that are close to it in
some sense and simpler in structure. This article considers the simplest and
most widely used part of the problem of approximation of functions, the
problem of Lagrange and Newton interpolation of functions.
2
. Main part
In the Lagrange and Newton interpolation definition formula, the value of
the coefficient is expressed by the distance between the nodes of the function
and the nodes.
x
f
y
Given a function,
x
the value of is assigned
x
y
to any possible
value of .
y
It is not always easy to determine . For example, if ,
x
is a parameter,
x
y
it may be considered a solution to a complex problem, or
x
y
if the values of
have been determined as a result of expensive research. In this case, we can
construct a table of values of the function, but this is not possible for very large
values of the argument.
Such in cases usually interpolation formulas is applied .
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b
a
,
in the cut
n
x
x
x
,....,
,
1
0
of the argument
1
n
to different values suitable
visitor
x
f
y
function values
0
0
y
x
f
,
1
1
y
x
f
,…,
n
n
y
x
f
given Let it be .
n
x
x
x
,....,
,
1
0
given in knots
x
f
y
function with same price reception to do and
level
n
not exceeding
х
n
plural construction demand be done , that is
.
,...,
2
,
1
,
n
i
y
х
p
i
i
n
х
n
- interpolation often ,
n
x
x
x
,....,
,
1
0
interpolation knots , breaking issue
and interpolation issue because is conducted . Most of the time interpolation
formulas of the argument interval values for
x
f
y
function value in finding is
used .
In this
х
point
n
х
х
,
0
in the meantime while lying down the
question of interpolation because is being conducted .
Interpol or Zion of many general in appearance various expressions
available : Newton , Lagrange , Gauss , Sterling , Bessel and others N Yuton and
Lagrange formulas in calculations convenient computer also and in hand in
calculations accuracy control to do provides the rest other forms interpolation
knots location private in case appropriate . Optional located in knots in
interpolation Lagrange interpolation from the formula is used . N Yuton
interpolation formulas equal in the distance located knots for under
consideration of the interval head and at the end at points function value
calculation for convenient .
The matter to be placed
Faraz Let's do [a, b ] in the intersection.
n
at the point given Let it be .
𝑥
0
, 𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
This points interpolation knots because is called . Someone
f(x)
function
this in points value below equal let it be
n
n
1
1
0
0
y
)
f(x
,....
)
(
...
,.........
y
)
f(x
,
y
)
f(x
i
i
y
x
f
Known to class relevant was and interpolation at the nodes
f(x)
function
reception did values that is :
n
n
1
1
0
0
y
)
F(x
,....
)
(
...
,.........
y
)
F(x
,
y
)
F(x
i
i
y
x
F
values reception doer
F(x)
function construction and of error
determination demand Let it be done .
The matter solution method
Lagrange interpolation by the method using above cited issue for
F(x)
of
the multiverse harvest we will do
Every one interpolation knot
(𝑥
𝑖
, 𝑦
𝑖
)
for separately multi-level we will fix
it .
𝐹
𝑛
(𝑥) = 𝐿
𝑛
(𝑥) = 𝑦
0
∙ 𝐹
0𝑛
(𝑥) + 𝑦
1
∙ 𝐹
1𝑛
(𝑥) + ⋯ + 𝑦
𝑛
∙ 𝐹
𝑛𝑛
(𝑥) (1)
𝐹
𝑖𝑛
(𝑥)
– of every one is of degree n polynomial , then ( 1) is also n- degree
multi-level will be .
𝐹
𝑖𝑛
(𝑥)
– every one of them
𝐹
𝑖𝑛
(𝑥) = {
0, агар 𝑖 ≠ 𝑗 бўлганда
1, агар 𝑖 = 𝑗 бўлганда
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the conditions performant as we choose .
𝐹
𝑖𝑛
(𝑥)
- roots
𝑥
0
, 𝑥
1
, 𝑥
2
, … , 𝑥
𝑖−1
, 𝑥
𝑖+1
, … , 𝑥
𝑛
– n -th degree It will be a
polynomial .
Understandable multi-level
𝐹
𝑖𝑛
(𝑥) = 𝐴 ∙ (𝑥 − 𝑥
0
)(𝑥 − 𝑥
1
) ∙ … ∙ (𝑥 − 𝑥
𝑖−1
)(𝑥 − 𝑥
𝑖+1
) ∙ … ∙ (𝑥 − 𝑥
𝑛
)
In appearance this is it where A is some constant. Condition according to
𝑃
𝑖𝑛
(𝑥) = 1
will be if i=j , then the constant A is we determine :
𝐴 ∙ (𝑥
𝑗
− 𝑥
0
)(𝑥
𝑗
− 𝑥
1
) ∙ … ∙ (𝑥
𝑗
− 𝑥
𝑖−1
)(𝑥
𝑗
− 𝑥
𝑖+1
) ∙ … ∙ (𝑥
𝑗
− 𝑥
𝑛
) = 1 (2)
(2) A 's value harvest we will do
𝐴 =
1
(𝑥
𝑗
− 𝑥
0
)(𝑥
𝑗
− 𝑥
1
) ∙ … ∙ (𝑥
𝑗
− 𝑥
𝑖−1
)(𝑥
𝑗
− 𝑥
𝑖+1
) ∙ … ∙ (𝑥
𝑗
− 𝑥
𝑛
)
⟹
𝐹(𝑥) =
(𝑥 − 𝑥
0
)(𝑥 − 𝑥
1
) ∙ … ∙ (𝑥 − 𝑥
𝑖−1
)(𝑥 − 𝑥
𝑖+1
) ∙ … ∙ (𝑥 − 𝑥
𝑛
)
(𝑥
𝑗
− 𝑥
0
)(𝑥
𝑗
− 𝑥
1
) ∙ … ∙ (𝑥
𝑗
− 𝑥
𝑖−1
)(𝑥
𝑗
− 𝑥
𝑖+1
) ∙ … ∙ (𝑥
𝑗
− 𝑥
𝑛
)
in that case unequaled intervals for Lagrange interpolation most of them
as follows harvest we will do :
𝐿
𝑛
(𝑥) = 𝑦
0
∙ 𝑃
0𝑛
(𝑥) + 𝑦
1
∙ 𝑃
1𝑛
(𝑥) + ⋯ + 𝑦
𝑛
∙ 𝑃
𝑛𝑛
(𝑥) =
= 𝑦
0
∙
(𝑥 − 𝑥
1
) ∙ … ∙ (𝑥 − 𝑥
𝑛
)
(𝑥
0
− 𝑥
1
) ∙ … ∙ (𝑥
0
− 𝑥
𝑛
)
+ 𝑦
1
∙
(𝑥 − 𝑥
0
) ∙ (𝑥 − 𝑥
2
) ∙ … ∙ (𝑥 − 𝑥
𝑛
)
(𝑥
1
− 𝑥
0
) ∙ (𝑥
1
− 𝑥
2
) ∙ … ∙ (𝑥
0
− 𝑥
𝑛
)
+ ⋯ +
+𝑦
𝑛
∙
(𝑥 − 𝑥
0
) ∙ (𝑥 − 𝑥
1
) ∙ … ∙ (𝑥 − 𝑥
𝑛−1
)
(𝑥
𝑛
− 𝑥
0
) ∙ (𝑥
𝑛
− 𝑥
1
) ∙ … ∙ (𝑥
𝑛
− 𝑥
𝑛−1
)
=
= ∑ 𝑦
𝑖
∙
(𝑥 − 𝑥
0
)(𝑥 − 𝑥
1
) ∙ … ∙ (𝑥 − 𝑥
𝑖−1
)(𝑥 − 𝑥
𝑖+1
) ∙ … ∙ (𝑥 − 𝑥
𝑛
)
(𝑥
𝑗
− 𝑥
0
)(𝑥
𝑗
− 𝑥
1
) ∙ … ∙ (𝑥
𝑗
− 𝑥
𝑖−1
)(𝑥
𝑗
− 𝑥
𝑖+1
) ∙ … ∙ (𝑥
𝑗
− 𝑥
𝑛
)
𝑛
𝑖=0
;
Private in cases this formula following appearances takes :
n =1 when two to the point owner we will be
(𝑎, 𝑦
0
), (𝑏, 𝑦
1
)
, he in case
𝐿
1
(𝑥) = 𝑦
0
∙
(𝑥−𝑏)
(𝑎−𝑏)
+ 𝑦
1
∙
(𝑥−𝑎)
(𝑏−𝑎)
appearance will take .
When n =3 three to the point owner we will be
(𝑎, 𝑦
0
), (𝑏, 𝑦
1
), (𝑐, 𝑦
2
)
, then
𝐿
2
(𝑥) = 𝑦
0
∙
(𝑥 − 𝑏)(𝑥 − 𝑐)
(𝑎 − 𝑏)(𝑎 − 𝑐)
+ 𝑦
1
∙
(𝑥 − 𝑎)(𝑥 − 𝑐)
(𝑏 − 𝑎)(𝑏 − 𝑐)
+ 𝑦
2
∙
(𝑥 − 𝑎)(𝑥 − 𝑏)
(𝑐 − 𝑎)(𝑐 − 𝑏)
When n=4 four to the point owner we will be
(𝑎, 𝑦
0
), (𝑏, 𝑦
1
), (𝑐, 𝑦
2
)(𝑑, 𝑦
3
)
he/she in case
𝐿
2
(𝑥) = 𝑦
0
∙
(𝑥 − 𝑏)(𝑥 − 𝑐)
(𝑎 − 𝑏)(𝑎 − 𝑐)
+ 𝑦
1
∙
(𝑥 − 𝑎)(𝑥 − 𝑐)
(𝑏 − 𝑎)(𝑏 − 𝑐)
+ 𝑦
2
∙
(𝑥 − 𝑎)(𝑥 − 𝑏)
(𝑐 − 𝑎)(𝑐 − 𝑏)
+ 𝑦
3
∙
(𝑥 − 𝑎)(𝑥 − 𝑏)(𝑥 − 𝑐)
(𝑑 − 𝑎)(𝑑 − 𝑏)(𝑑 − 𝑐)
Experiment result
From points transient Lagrange interpolation plural to be formed and
remainder limit or error to be determined :
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1
2
x
x
y
function for
Lagrangian interpolation polynomial we
will
build
3
,
1
3
,
2
,
1
1
,
1
,
1
1
,
1
0
x
x
x
x
In knots
suitable function value we calculate (
Table
1
):
99
,
3
,
64
,
3
,
31
,
3
,
3
3
2
1
0
y
x
y
y
Table 1
Information table
x
y ( x )
0
1
3
1
1.1
3.31
2
1.2
3.64
3
1.3
3.99
Given in the next step in the cut
3
,
1
,
2
,
1
,
1
,
1
,
1
3
2
1
0
x
x
x
x
we
construct the Lagrange polynomial for the points.
858
2155
1800
500
)
3
,
1
1
(
)
2
,
1
1
(
)
1
,
1
1
(
)
3
,
1
(
)
2
,
1
(
)
1
,
1
(
3
)
(
2
3
0
x
x
x
x
x
x
x
l
(3)
8
,
2581
3
,
6719
25
,
579
1655
)
3
,
1
1
,
1
(
)
2
,
1
1
,
1
(
)
1
1
,
1
(
)
3
,
1
(
)
2
,
1
(
)
1
(
31
,
3
)
(
2
3
1
x
x
x
x
x
x
x
l
(4)
6
.
2602
6
.
6970
6188
1820
)
3
,
1
1
,
1
(
)
1
,
1
2
,
1
(
)
1
2
,
1
(
)
3
,
1
(
)
1
,
1
(
)
1
(
64
,
3
)
(
2
3
2
x
x
x
x
x
x
x
l
(5)
8
,
877
3
,
2407
5
,
2194
655
)
2
,
1
3
,
1
(
)
1
,
1
3
,
1
(
)
1
3
,
1
(
)
2
,
1
(
)
1
,
1
(
)
1
(
99
,
3
)
(
2
3
3
x
x
x
x
x
x
x
l
(6)
In the next step, we calculate the sum of (3), (4), (5), (6) and form the
expected polynomial.
1
8
,
877
3
,
2407
5
,
2194
655
6
.
2602
6
.
6970
6188
1820
8
,
2581
3
,
6719
25
,
579
1655
858
2155
1800
500
)
(
2
2
3
2
3
2
3
2
3
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
L
i
We
1
)
(
2
x
x
x
L
n
have a lot of crops
To determine the calculation error, we find the values of the function f(x)
corresponding to the intermediate value of x, and the values of the polynomial
L(x) derived from the initial function f(x) are taken, respectively, as the largest
absolute value of the difference (Table 2-3).
generate a polynomial for the above problem using Newton's interpolation
method
N(x)
.
3
3
31
,
3
0
1
0
y
y
y
(7)
33
,
0
31
,
3
64
,
3
1
2
1
y
y
y
(8)
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35
,
0
34
,
3
39
,
3
2
3
2
y
y
y
(9)
By subtracting the values of (8) and (7), we get (10), and by subtracting
the values of (9) and (8) we get (11):
02
,
0
31
,
0
33
,
0
0
1
2
0
y
y
y
(10)
02
,
0
33
,
0
35
,
0
1
2
2
1
y
y
y
(11)
(11) and (10 ) values Subtract (12) to get we will do :
0
02
,
0
02
,
0
2
0
2
1
3
0
y
y
y
(12)
701
,
2
289
,
0
01
,
0
3
2
1
)
2
,
1
(
)
1
,
1
(
)
1
(
0
2
1
)
1
,
1
(
)
1
(
02
,
0
1
)
1
(
31
,
0
3
!
3
)
2
,
1
(
)
1
,
1
(
)
1
(
!
2
)
1
,
1
(
)
1
(
!
1
)
1
(
3
)
(
2
3
0
2
0
0
x
x
x
x
x
x
x
x
x
x
x
y
x
x
y
x
y
x
N
n
;
4
,
1
6
;
3
,
1
5
;
29
,
1
4
;
2
,
1
3
;
15
,
1
2
;
1
,
1
1
;
1
0
x
x
x
x
x
x
x
(13) from the nodal points in the interval (13). harvest dirty many things
values table we draw (table 2):
Table 2 .
Information table
I
x i
y(x i )
L i (x i )
| y(x i )-
L i (x i )|
N i (x i )
| y(x i )- N
i (x i )|
0
1
3
3
0
3
0
1
1.1
3,310000 3.31
0000
0
3,031 0000 0.279000
2
1.15
3.489025 3.489025 0
3.047725 0 0.441300
3
1.2
3.64
0000
3.64
0000
0
3.0622 000 0.577800
4
1.29
3.954100 3.954100 0
3.0917725
0.862327
5
1.3
3.99
0000
3.99
0000
0
3.0936 000 0.896400
6
1.4
4.360000 4,360000 0
4.126600 0 0.233400
[1;1,4] interval according to maximum error :
In the Lagrangian method :
max (|𝑅
𝑛
(𝑥)|) = max (|
)
(
x
f
−
)
(
x
L
n
|) ≈ 0
;
Newtonian method :
𝒎𝒂𝒙|𝑅
𝑛
(𝑥)| = max (|
)
(
x
f
−
)
(
x
N
n
|) ≈ 0,8964
constitutes
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Using the theoretical concepts presented, the implementation of
approximations in the Lagrange and Newton interpolation methods, the
programming of the processes of finding their interpolation errors was
developed in the Python programming language environment, and the expected
results of the research work were obtained. The results show that the program
works correctly.
Lagrangian method algorithm
block diagram view
Newton 's method algorithm
block diagram view
Бошлаш
x(n),y(n)
j
=0,n
p=1
i=0,n
x,L
n
i≠j
L
тамоа
Бошлаш
x(n),y(n)
j
=0,k
z=1
i=0,k
x,N
k
i=j
N
тамоа
1-расм. Функциянинг Лагранж усулида
интерполяциялаш
натижаси
2
-расм. Функциянинг Ньютон усулида
интерполяциялаш натижаси
ACADEMIC RESEARCH IN MODERN SCIENCE
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conclusion
Polynomials were constructed from points (nodes) on the cross-section using
the Lagrange and Newton interpolation methods. An algorithm for increasing
the interpolation accuracy and a software product were developed, and the
results were obtained. The residual limit or error of the Lagrange interpolation
polynomial on the interval
𝐿(𝑥) ≈ 0
was the maximum. The residual limit or
error of the Newton interpolation polynomial
𝑁(𝑥) ≈ 0,8964
on the interval was the maximum. It was found that the Lagrange
interpolation method is more effective for functions constructed from points
located at an arbitrary distance from the cross-section, and the results were
obtained (Fig. 1-2).
Literature:
1. Akhatov A., Renavikar A., Rashidov A., Nazarov F. “Optimization of the number
of databases in the Big Data processing” Проблемы информатики, № 1(58)
2023, DOI: 10.24412/2073-0667-2023-1-33-47
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