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34
IMPROVING THE ACCURACY OF INTERPOLATION OF FUNCTIONS IN
LAGRANGE AND NEWTON'S METHOD
Gofurjonov M.R.
Tashkent University of Information
Technologies named after Muhammad al-Khwarizmi,
gofurjonov13@mail.ru
Xuramov L. Ya.
Samarkand State University named after Sharof Rashidov
Tel:+99(893) 546-21-41,e-mail: latifxya@gmail.com
Meliyeva Marifat
Samarkand State University named after Sharof Rashidov
https://doi.org/10.5281/zenodo.15709220
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.
Introduction.
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.
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.
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
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35
.
,...,
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, 𝑖 ≠ 𝑗
1, 𝑖 = 𝑗
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)
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(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:
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.
In the next step, we calculate the sum of (3), (4), (5), (6) and form the expected
polynomial.
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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)
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
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[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
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
тамом
Бошлаш
x(n),y(n)
j
=0,k
z=1
i=0,k
x,N
k
i=j
N
тамом
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1-расм. Функциянинг Лагранж
усулида интерполяциялаш натижаси
2-расм. Функциянинг Ньютон
усулида интерполяциялаш
натижаси
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 interpolationpolynomial
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).
References:
Используемая литература:
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2.
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