> NewtonsMethod(exp(x)-10*x-2, x =3, view = [-2..5, DEFAULT], output =plot)

1-2 tajriba mashg'ulotlari. Algebraik va transendent tenglamalarni yechish usullari va algoritmlari


> NewtonsMethod(exp(x)-10*x-2, x =3, view = [-2..5, DEFAULT], output =plot)


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1-2 TAJRIBA ISHLARI

> NewtonsMethod(exp(x)-10*x-2, x =3, view = [-2..5, DEFAULT], output =plot);

ex-10x-2=0 tenglamani [a,b] oraliqda >0 aniqlikdagi yechimini urinmalar usuli bilan topishning hisoblash dasturini tuzamiz.


Masaladagi berilganlar asosida ko’rsatilgan usulda hisoblashning algoritmini 2.7-jadvalda beramiz :
2.7-jadval

Berilganlar

Belgilashlar

matn bo‘yicha

dastur bo‘yicha

Tenglama funksiyasi

f(x)=ex-10x-2

FNF(x)=exp(X)-10*X -2

Tenglama funksiyasining birinchi hosilasi

f '(x)=ex-10

FNF1(x)=exp(X)-10

Tenglama funksiyasining ikkinchi hosilasi

f '' (x)=ex

FNF2(x)=exp(X)

Ildiz yotgan kesma сhegarasi

a=-1, b=0

a=-1, b=0

Kesmani bo‘linish qadami

H=0.1

H=0.1

Ildiz yotgan kesma

(x1, x2)=(x1, x1+h)

(x1, x2)=(x1, x1+h)

Ildiz yotgan kesmani aniqlash sharti

f(x1)·f(x2)<0

fnf(x1)*fnf(x2)<0

Urinmalar usulini qo‘llash sharti

f(x) f ''(x)>0

fnf(x)*fnf2(x)>0

Urinmalar usulida hisoblash formulasi

x1= x1 – f(x1)/ f '(x1)



X=X-(A-X)*FNF(X)/FNF1(X)

Ildizga yaqinlashish sharti

x1 –x2<

ABS(x1-x2)<=E yoki FNF(X)<=E


4 ‘ ----------- 2.7- DASTUR -------------------
5 ‘Urinmalar usulida trantsendent tenglama
6 ‘--------- ildizini aniqlash--------------
10 REM SAVE”kas-1.bas”,a
20 DEF FNF(X)=EXP(X)-10*X-2
30 DEF FNF1(X)=EXP(X)-10
40 DEF FNF2(X)=EXP(X)
50 INPUT” ildiz chegarasi a,b=”; A,B
52 H=.1:E=.001
60 X1=A
70 X2=X1+H:X=X1:A=X2
80 IF X2>B THEN 180
90 IF FNF(X1)*FNF(X2)>0 THEN 170
100 IF FNF(X1)*FNF2(X1)>0 THEN 120
110 X=X2:A=X1
120 X=X-FNF(X)/FNF1(X)
130 IF FNF(X)>=E THEN 120
140 PRINT “(“;USING “##.###”;X1;
150 PRINT “,”;USING “##.###”;X2;
160 PRINT “) x= “;USING “##.######”;X
170 X1=X2:GOTO 70
180 END
Ok
RUN
? -2,5
(-0.200,-0.100) x= -0.110458
( 3.600, 3.700) x= 3.650891
Ok


(*-------- 2.7- DASTUR ---------*)
(* URINMALAR USULIDA *)
(*Trantsendent tenglamaning ildizi
yotgan oraliq va ildizni anuqlash *)
uses crt;
LABEL L1,L2,L3,L4;
function fnf(x:real):real;
begin fnf:=EXP(x)-10*x-2; end;
function fna(x:real):real;
begin fna:=EXP(x)-10; end;
function fnb(x:real):real;
begin fnb:=EXP(x); end;
var
a,b,h,EPS,x1,x2,x:real;
i:integer;
begin
clrscr;
writeln(‘ Ildiz yotgan Kesma (a,b)’);
writeln(‘ ildizni aniqlash qadami h ‘);
write(‘ a=’);readln(a);
write(‘ b=’);readln(b);
writeln(‘ URINMALAR usulida hisoblash ‘);
writeln(‘oraliq va ildiz’);
i:=1; EPS:=0.001; h:=0.1;
x1:=a;
L1: x2:=x1+h;
x:=x1; a:=x2;
if x2>b then goto L4;
if fnf(x1)*fnf(x2)>0 then goto L3;
if fnf(x1)*fnb(x1)>0 then goto L2;
x:=x2;a:=x1;
L2: x:=x-fnf(x)/fna(x);
if abs(fnf(x))>EPS then goto L2;
writeln;
writeln(‘ (‘,x1:6:4,’, ‘,x2:6:4,’)’,’ x=’,x:8:4);
i:=i+1;
L3: x1:=x2;
goto L1;
L4: readln;
end.
Ildiz yotgan Kesma (a,b) va ildizni aniqlash qadami h bo‘lganda
URINMALAR usulida hisoblash
a= -7
b= 7
oraliq va ildiz
(-0.2000, -0.1000) x= -0.1105
(3.6000, 3.7000) x= 3.6509


Mustaqil ishlar uchun topshiriqlar
Quyidagi jadvaldagi tenglamalar ildizini:
1. Ildizlarining qisqa atrofini
-analitik va grafik usulda aniqlang
-ildizni ajratish dasturini tuzing (BEYSIK tilida)
2.Aniqlangan oraliqda ildizni vatarlar, urinmalar va birgalashgan usulida hisoblang.


1.



1) 2x+5x-3=0
2) 3x4-4x3-12x2-5=0
3) 0.5x+1=(x-2)2
4) (x-3)Cosx=1,(-2x2)

2.



1) arctgx-1/(3x3)=0
2) 2x3-9x2-60x+1=0
3) [log2(-x)](x+2)=-1
4) Sin(x+/3)-0.5x=0

3.

1) 5x+3x=0
2) x4-x-1=0
3) 0.5x+x2= 2
4) (x-1)2Ln(x+1)=1

4.



1) 2ex=2+5x
2) 2x4-x2-10=0
3) xLog3(x+1)=1
4) Cos(x+0.5)=x3

5.

1) 3x-1-2-x=0
2) 3x4+8x3+6x2-10=0
3) (x-4)2log0.5(x-3) =-1
4) 5Sinx=x

6.



1) arctgx-1/(2x3)=0
2) x4-18x2+6=0
3) x22x=1
4) tgx=x+1, (-/2x/2)

7.



1) e-2x-2x+1=0
2) x4+4x3-8x2-17=0
3) 0.5x-1=(x+2)2
4) x2Cos2=-1

8.



1) 5x-6x-3=0
2) x4-x3-2x2+3x-3=0
3) 0.5x -2x2- 3=0
4) xLog(x+1)=1

9.



1) arctg(x-1)+2x=0
2) 3x4+ 4x3-12x2+1=0
3) (x-2)22x=1
4) x2-20Sinx=0

10.



1) 2arctgx-x+3=0
2) 3x4-8x3-18x2+3=0
3) 2Sin(x+/3)=0.5x2-1
4) 2Logx-x/2+1=0

11



1) 3x+2x-2=0
2) 2x4-8x3+8x2-1=0
3) [(x-2)2-1]2x =1
4) (x-2)Cosx=1

12.



1) 2arctgx-3x+2=0
2) 2x4+8x3+8x2 -1=0
3) Sin(x-0.5)-x+0.8=0
4) (x-1)Log2(x+2)=1

13.


1) 3x+2x-5=0
2) x4-4x3-8x2+1=0
3) 0.5x +x2- 3=0
4) (x-2)2Lg(x+1)=1

14.


1) 2ex+3x+3x+1=0
2) 3x4+4x3-12x2-5=0
3) Cos (x+0.3)=x2
4) xLog3(x+1)=2

15.



1) 3x-1-4-x=0
2) 2x3-9x2-60x+1=0
3) (x-3)2Log0.5(x-2)=-1
4) Sinx=x-1

16.

1) arctgx-1/(3x3)=0
2) x4-x–1=0
3) (x-1)22x=1
4) tg3x=x-1

17.



1) ex+x+1=0
2) 2x4-x2-1=0
3) 0.5x –3=(x+2)2
4) x2Cos2x=-1,(-2x2)

18.



1) 3x-2x+5=0
2) 3x4+8x3+6x2-10=0
3) 2x2-0.5x=0
4 xLg(x+1)=1

19.



1) arctg(x-1)+3x-2=0
2) x4-18x2+6=0
3) x2-20Sinx=0
4) (x-2)22x=1

20.

1) 2arctgx-x+3=0
2) x4+4x3-8x2 -17=0
3) 2Sin(x+/2)=x2-0.8
4) 2Lgx-x/2+1=0

21

1) 2x-3x-2=0.
2)x4-x3-2x2 +3x- 3=0;
3)(0.5)x+ 1=(x-2)2
4)(x-3)cosx = -1, -2’
22

1) arctgx+2x-1=0
2) 3x4+4x3-12x2 +1=0
3)(x+2)Log2(x)=1
4) Sin(x+1)=0.5x

23

1) 3x+2x-3=0.
2) 3x4-8x3-18x2 +2=0;
3) (0.5)x=4-x2
4) (x+2)2Lg(x+11)=1

24

1) 2ex-2x-3=0.
2) 3x4+4x3-12x2 -5=0;
3) xLog2(x+1)=1
4) cos(x+0.5)= x3

25

1) 3x+2+x=0.
2) 2x3-9x2 -60x+1=0;
3) (x-4)2Log0.5(x-3)=-1
4) 5Sinx=x-0.5

26

1) arctg(x-1)+2x-3=0
2) x4x -1=0;
3) (x-1)22x=1
4) tg3x=x-1, (-/2x/2)

27

1) 2ex-2x-3=0.
2) 2x4-x2 -10=0;
3) (0.5)x-3=-(x+1)2
4) x2cos2x = 1

28

1) 3x-2x-5=0.
2) 3x4+8x3+6x2 -10=0;
3) 2x2 -0.5x - 3=0
4) xLg(x+1)=1

29

1) arctg(x-1)+2x=0
2) x4-18x2+6=0
3) (x-2)22x=1
4) x2-10sinx = 0

30

1) 3x+5x-2=0.
2) 3x4+4x3-12x2 +1=0;
3) (x-2)2 =0.5x +1
4)(x+3)cosx = 1, -2

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