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

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1-2 TAJRIBA ISHLARI

=0.01 H=0.1: E=0.01
4 ‘ -------- 2.9- DASTUR -------------------------- 5 ‘—---Iteratsiya usulida trantsendent tenglama ---------- 6 ‘ ----------- ildizini aniqlash
40 X=X1 50 FOR I=1 TO N 60 Y=FNF(X) 70 IF ABS(Y-X) 80 X=Y 90 NEXT I
(*----------------- 2.9- DASTUR -----------------*) (* Iteratsiya usulida chiziksiz tenglamalarni *) (* Paskal tili dasturida hisoblash dasturi *) USES CRT;
2.4-Maple 7 dasturi > f:= x -> (exp(x)-2)/10; > x||0:=0; > for k from 1 to 16 do x||k: =evalf(f(x)||(k-1)): od
1. Vatarlar usuli 2.2-masala. e x -10x-2=0 tenglamaning =0.01 aniqlikdagi taqribiy ildizi topilsin. Yechish . f(x) =e
Berilganlar Belgilashlar matn bo‘yicha
(* -- 2.6 - Paskal tili dastur -----*) (* VATARLAR USULIDA *) (*Trantsendent tenglamaning ildizi yotgan oraliq va ildizni anuqlash *) USES CRT;
4 ‘ ------ 2.6- DASTUR --------------- 5 ‘—-Vatarlar usulida trantsendent tenglama

Berilganlar
Belgilashlar
matn bo‘yicha	dastur bo‘yicha
Tenglama funksiyasi	f(x)=e^x-10x-2	FNF(x)=EXP(X)-10*X -2
Iteratsiya usulini qo‘llash uchun tenglamani o‘zgartirilgan ko‘rinishi	x=(e^x-2)/10	X=(EXP(X)-2)/10
Tenglama funksiyasi ning ikkinchi hosilasi	f ’’ (x)=e^x-10	FNF2(x)=EXP(X)
Ildiz yotgan Kesma	a=-1, b=0	a=-1, b=0
Kesmani bo‘linish qadami va aniqlikda	H=0.1, =0.01	H=0.1: E=0.01
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 va vatarlar usulini qo‘llash sharti	f(x) f’’(x)>0 f(x) f’’(x)<0	fnf(x)fnf2(x)>0 fnf(x)fnf2(x)<0
Urinmalar va vatarlar usulida hisoblash formulasi	x₂= x₁ – f(x₁)/ f '(x₁), x₁= x₁ – (x₂ – x₁)f(x₁)/ (f(x₂)-f(x₁))	A=A-FNF(A)/ FNF1(A), X=X-(A-X)* FNF(X)/(FNF(A)-FNF(X))
Ildizga yaqinlashish sharti	x₁–x₂<	ABS(x₁-x₂)<=E yoki FNF(X)<=E

4 ‘ -------- 2.9- DASTUR --------------------------
5 ‘—---Iteratsiya usulida trantsendent tenglama ----------
6 ‘ ----------- ildizini aniqlash---------------------
10 PRINT “Boshlangich qiymatni, aniqlikni va iteratsiya sonin”
12 PRINT “kiriting X1 , E, N=:”
20 INPUT X1 , E, N
30 DEF FNF(X)=(EXP(X)-2)/10
40 X=X1
50 FOR I=1 TO N
60 Y=FNF(X)
70 IF ABS(Y-X)<=E THEN 120
80 X=Y
90 NEXT I
100 PRINT “ berilga N aniqlik etarli emas”
110 GOTO 10
120 PRINT” Xisob natijasi: “
130 PRINT” i=”; i,” da x=”;x ,”y=”;y
140 END
Berilganlar: x0=0 , =0.01 , N=100
Javob: I=3 , x= -0.1095 , y= -0.108965

(*----------------- 2.9- DASTUR -----------------*)
(* Iteratsiya usulida chiziksiz tenglamalarni *)
(* Paskal tili dasturida hisoblash dasturi *)
USES CRT;
LABEL L1,L2;
function fnf(x:real):real;
write(‘ х=’);readln(х);
begin fnf:= (EXP(x)-2)/10; end;
y:=fnf(x);
P:=x-y;
if abs(P)
write( i:2,’-yakinlashish’);
writeln(‘x(‘,i:2,’)=’,x:8:4);
i:=i+1;
x:=y;
goto L1;
L2: writeln(‘x(‘,i:2,’)=’,x:8:4);
readln;
end.
X=-1
1-yakinlashish x(1)= -1.0000
2-yakinlashish x(2)= -0.1632
3-yakinlashish x(3)= -0.1151

2.4-Maple 7 dasturi
> f:= x -> (exp(x)-2)/10;
> x||0:=0;
> for k from 1 to 16 do x||k: =evalf(f(x)||(k-1)): od:
> f(x)=x; solve(%, x);evalf(%,4);

MUSTAQIL ISHLAR UCHUN TOPSHIRIQLAR
Quyidagi tenglamalar:
1.Ildizlarning qisqa atrofini analitik yoki grafik usulda aniqlang;
2. Aniqlangan oraliqda ildizning taqribiy qiymatini 0.01 aniqlikda iteratsiya usuli bilan hisoblang;

1. 1) x-sinx=0.25
2) x³-3x²+9x-8=0

2 1) tg(0.58x+0.1)=x²
2) x³-6x-8=0

3 1) x-cos(0.378x)=0
2) x3-3x2+6x+3=0

4 1) tg(0.4x+0.4)=x²
2) x3-0.1x2+0.4x-1.5=0

5 1) lgx-7/ (2x+6)=0
2) x³+x-5=0

6 1) tg(0.5x+0.2)=x²
2) x³+x-5=0

7 1) 3x-cosx-1=0
2) x³+0.2x²+0.5x-1.2=0

8 1) x+lgx=0.5
2) x³+3x+1=0

9 1)tg(0.5x+0.1)=x²
2) x³+0.2x²=0.5x-2=0

10 1) x3+4sinx=0
2) x³-3x²+12x-9=0

11 1) ctg(1.05x)-x²=0
2) x³-0.2x²+0.3x-1.2=0

12 1) tg(0.4x-0.3)=x²
2) x³-3x²+6x-2=0

13 1)xlgx-1.2=0
2) x³-0.1x²+0.4x-1.5=0

14 1) 1.8x²-sin10x=0
2) x³+3x²+6x-1=0

15. 1) ctgx-x/4=0
2) x³-0.1x²+0.4x-1.2=0

16. 1) tg(0.3x+0.4)=0
2) x³+4x-6=0

17. 1) x²-20Sinx=0
2) x³+0.2x²+0.5x+0.8=0

18. 1) ctgx-x/3=0
2) x³-3x²+12x-12=0

19. 1) tg(0.47x+0.2)=x²
2) x³-0.2x²+0.3x+1.2=0

20. 1) x²+4sinx=0
2) x³-2x+4=0

21 1) ctgx-x/2=0
2) x³-0.2x²+0.5x-1.4=0

22 1) 2x-lgx-7=0
2) x³-3x²+6x-5=0

23 1) tg(0.44x+0.3)=x²
2) x³-0.1x²+0.4x+1.2=0

24 1) 3x-cosx-1=0
2) x³-0.2x²+0.5x-1=0

25 1) ctgx-x/10=0
2) x³+3x²+12x+3=0

26 1) x²+4Sinx=0
2) x³-0.1x²+0.4x+2=0

27 1) tg(0.36x+0.4)=x²
2) x³-0.2x²+0.4x-1.4=0

28 1) x+lgx=0.5
2) x³+0.4x²+0.6x-1.6=0

29 1) ctgx-x/5=0
2) x³+x-3=0

30 1) 2lgx-x/2+1=0
2) x³-0.2x²+0.5x+1.4=0

1. Vatarlar usuli
2.2-masala. e^x-10x-2=0 tenglamaning =0.01 aniqlikdagi taqribiy ildizi topilsin.
Yechish. f(x)=e^x-10x-2 funksiya [-1,0] oraliqda 2.4-teoremaning barcha shartlarini qanoatlantirishini tekshirib ko‘rish qiyinchilik tug‘dirmaydi. x[-1,0] da ikkinchi tartibli hosila f''(x) = e^x >0. undan tashqari, f(0)=-1, f(-1)=8.368 bo‘lganligi uchun, (2.5) shartga asosan f(0)f''(0)<0 bo‘lgani uchun {b_n} ketma-ketlik (2.7) jarayon bilan quriladi.
Berilganlar: a=-1, b=0, =0. 01
f(x)= e^x-10x-2, f (-1) =e^-1 -10(-1) -2=8.386, f (0) =e⁰-100-2= -1
(2.7) formulaga asosan:
b₀= 0
b₁= b₀ – (a- b₀) f(b₀)/ (f(a)-f(b₀))= -0.107
b₁–b₂> bo‘lganligi uchun b₂ yaqinlashishni hisoblaymiz. Buning uchun
b₁= -0.107, f (-0.107) =e^-0.107-10(-0.107)-2=-0.038, f(a)=f(-1)=8.386 larga asosan:
b₂= b₁ – (a- b₁) f(b₁)/ (f(a)-f(b₁)) = -0,111
 b₂- b₁=- 0.111+0.107=0.004<=0,01
Demak, 0.01 aniqlikdagi taqribiy yechim deb t b₂ =-0.11 ni olish mumkin.
e^x-10x-2=0 tenglamani [a,b] oraliqdagi yechimini vatarlar usuli bilan topishning Beysik tilidagi blok sxemaci va 2.6-dasturini tuzamiz.
Masaladagi berilganlar asosida ko’rsatilgan usulda hisoblashning algoritmini 2.6-jadvalda beramiz :
2.6-jadval

Berilganlar

Belgilashlar

matn bo‘yicha

dastur bo‘yicha

Tenglama funksiyasi

f(x)=e^x-10x-2

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

Tenglama funksiyasining birinchi hosilasi

f '(x)=e^x-10

FNF1(x)=exp(X)-10

Tenglama funksiyasining ikkinchi hosilasi

f '' (x)=e^x

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

Vatarlar usulini qo‘llash sharti

f(x) f ''(x)<0

fnf(x)*fnf2(x)<0

Vatarlar usulida hisoblash formulasi

x₁= x₁ – (x₂ – x₁)f(x₁)/ (f(x₂)-f(x₁))

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

Ildizga yaqinlashish sharti

x₁–x₂<

ABS(x₁-x₂)<=E yoki FNF(X)<=E

(* -- 2.6 - Paskal tili dastur -----*)
(* VATARLAR 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 qadavb h’);
write(‘ a=’);readln(a);
write(‘ b=’);readln(b);
write(‘ h=’);readln(h);
write(‘ildiz ytgan oraliq va ildiz ‘);
i:=1; EPS:=0.001;
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)*(a-x)/(fnf(a)-fnf(x));
if abs(fnf(x))>EPS then goto L2;
writeln;
write(‘ (‘,’x1=’,x1:6:3,’, ‘,’x2=’,x2:6:3,’)’,’ ‘,’x=’,x:8:4);
i:=i+1;
L3: x1:=x2;
goto L1;
L4: readln;
end.
Ildiz yotgan Kesma (a,b) va ildizni aniqlash qadavb h bo‘lganda
VATARLAR usulida hisoblash
a=-7
b=7
h=0.1
ildiz ytgan oraliq va ildiz
(x1= -0.200, x2= -0.100) x= -0.1104
(x1= 3.600, x2= 3.700) x= 3.6509

4 ‘ ------ 2.6- DASTUR ---------------
5 ‘—-Vatarlar usulida trantsendent tenglama ----

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