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4international scientific journal volume 1 issue 7 UIF-2022: 8.2 | ISSN: 2181-3337
EQUATION OF THE DERIVATIVE OF THE FUNCTION TO SOLVE EQUATIONS
Tillabayev B.Sh.
Fergana Politecnical Institute https://doi. org/10.5281/zenodo.7346610
Abstract. This article presents approximate calculations using the exact integral
Keywords: integral, equality, inequality.
УРАВНЕНИЕ ПРОИЗВОДНОЙ ФУНКЦИИ ДЛЯ РЕШЕНИЯ УРАВНЕНИЙ
Аннотация. В данной статье представлены приближенные расчеты с использованием точного интеграла
Ключевые слова: интеграл, равенство, неравенство.
INTRODUCTION
Let us be given a non-standard equation in the form of some kind of f (x) = 0 in section [a,b]. Let it be required to find the solution of this equation in Section [a,b] (if it
exists) S in accuracy.
MATERIALS AND METHODS
First of all, we check whether the function f (x) satisfies the conditions of Baltsano-Cauchy theorem 1 in the section [a, b] or not.
Theorem one of Balsano-Koshi. If the function f (x) is defined on the section [a,b] and is continuous and has different sign value at the extreme points of the interval, then there exists a point c e (a, b) such that f (c) = 0.
So, if the function f (x) fulfills the conditions of the theorem, then the section [a, b] contains the solution of the equation f (x) = 0.
RESULTS
Suppose that the section [a,b] contains the root of the equation f (x) = 0. That is, let x e[a,b] exist such that f (x) = 0 is equal. Also let x = a + Ax be. Then the equality f (a + Ax) = 0 holds. If we consider the equation f (x0 + Ax) « f (x0) + f (x0) • Ax,
f (a + Ax) « f (a) + f (a) • Ax « 0 equality, that is, Ax: this, if we consider the equality Ax = a — x,
f (a ) s
f( a)
is formed. From
x — a « —
f (a) /( a)
f (a)
x « a--_, „ ч equalities are formed.
f( a)
If the equality
f( a )
approximately with accuracy S.
f ( a )
< S is fulfilled, the solution of the equation will be found
international scientific journal volume 1 issue 7 UIF-2022: 8.2 | ISSN: 2181-3337
Suppose that this inequality does not hold, then we denote a —
fM
/( a )
by Xj and
consider the equation in the interval [x, b]. Denote the root of the equation as x = xl + Ax and
f (X)
calculate X ~ X,
f( Xl )
. If we denote it as x2 and x2 — xx | <S then Xj + Ax is the root.
Continuing in this way, after a finite step, the inequality XM — XM-11 <S is satisfied and xn is taken to be an approximate solution of the given equation around S.
Example 1. Prove that the equation f (x) = x3 — 3x — 6 has a unique solution in the
interval [ 2;3] and find it with precision 0.001. DISCUSSION
Solving. f (x) = x3 — 3x — 6 is a function defined on the interval [2;3], continuous and f (2) = —4 < 0, f (3) = 12 > 0. On the other hand Vx = ( 2;3) for f (x) = 3( x2 —1)> 0, that is, the function is increasing. Therefore, the equation
x3 — 3x — 6 = 0 has a unique solution in the interval [ 2;3]. Now let's say x = 2 + Ax and find Ax such that f (x) = f (2 + Ax) = 0. For sufficiently small Ax f ( 2 + Ax ) = f ( 2) + Af ( 2 ) « f ( 2) + f ( 2 ) • Ax.
Then we come to the equation f (2) + f (2) • Ax « 0 . By finding Ax
Z2) f( 2 )
from
this, we find the equation x = 2 + Ax « 2 —
Z(2)
f( 2)
for the solution. Finally, taking into account
that f (2) = —4, / (2) = (3x2 — 3)|x=2= 9, we find x « x = 2 ——4 = 2-4 . Using the same reasoning, we find the solution:
X = 2.4 — = 2.4 — . 2.360.
2 f' ( 2.4) 14,28
xq = 2.360 —
x, = 2.356 —
f (2.360) f(2.360)
f(2.356)
2.356.
2.356,
|x2 — xJ = 0.40 > 0.001;
x3 — X2 = 0.004 > 0.001;
x4 — x3 = 0.000 < 0.001.
f(2-356) CONCLUSION
Therefore, the number 2.356 is the solution of the equation with an accuracy of 0.001.
international scientific journal volume 1 issue 7 UIF-2022: 8.2 | ISSN: 2181-3337
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