![204
Ch測ng 12 : Tnh gn 速坦ng 速孫o h袖m v袖 tch ph息n x存c 速nh
則1. 則孫o h袖m Romberg
則孫o h袖m theo ph測ng ph存p Romberg l袖 m辿t ph測ng ph存p ngo孫i suy 速 x存c 速nh 速孫o
h袖m v鱈i m辿t 速辿 chnh x存c cao . Ta xt khai trin Taylor c単a h袖m f(x) t孫i (x+h) v袖 (x-h) :
++霞霞+霞++=+ )x(f
!4
h
)x(f
!3
h
)x(f
2
h
)x(fh)x(f)hx(f )4(
432
(1)
+霞霞霞霞+霞= )x(f
!4
h
)x(f
!3
h
)x(f
2
h
)x(fh)x(f)hx(f )4(
432
(2)
Tr探 (1) cho (2) ta c達 :
++霞霞+=+ )x(f
!5
h2
)x(f
!3
h2
)x(fh2)hx(f)hx(f )5(
53
(3)
Nh vy r坦t ra :
霞霞霞
+
= )x(f
!5
h
)x(f
!3
h
h2
)hx(f)hx(f
)x(f )5(
42
(4)
hay ta c達 th vit l孫i :
[ ] +++++= 6
6
4
4
2
2 hahaha)hx(f)hx(f
h2
1
)x(f (5)
trong 速達 c存c h s竪 ai ph担 thu辿c f v袖 x .
Ta 速t :
)]hx(f)hx(f[
h2
1
)h( + =
(6)
Nh vy t探 (5) v袖 (6) ta c達 :
== 6
6
4
4
2
2 hahaha)x(f)h()1,1(D (7)
=
=
64
h
a
16
h
a
4
h
a)x(f
2
h
)1,2(D
6
6
4
4
2
2 (8)
v袖 t脱ng qu存t v鱈i hi = h/2i-1
ta c達 :
== 6
i6
4
i4
2
i2i hahaha)x(f)h()1,i(D (9)
Ta t孫o ra sai ph息n D(1,1) - 4D(2,1) v袖 c達 :
霞=
6
6
4
4 ha
16
15
ha
4
3
)x(f3
2
h
4)h( (10)
Chia hai v c単a (10) cho -3 ta nhn 速樽c :
+++=
= 6
6
4
4 ha
16
5
ha
4
1
)x(f
4
)1,1(D)1,2(D4
)2,2(D (11)
Trong khi D(1,1) v袖 D(2,1) sai kh存c f(x) ph担 thu辿c v袖o h2
th D(2,2) sai kh存c f(x) ph担
thu辿c v袖o h4
. B息y gi棚 ta l孫i chia 速束i b鱈c h v袖 nhn 速樽c :
D f x a h a h(2, ) ( ) ( / ) ( / ) ...2
1
4
2
5
16
24
4
6
6
= + + +
(12)
v袖 kh旦 s竪 h孫ng c達 h4
b損ng c存ch t孫o ra :
D D f x a h(2, ) ( , ) ( ) ( ) ...2 16 32 15
15
64 6
6
= + + +
(13)
Chia hai v c単a (13) cho -15 ta c達 :
D
D D
f x a h(3, )
(3, ) (2, )
( ) . ...3
16 2 2
15
1
64 6
6
= =
(14)](https://image.slidesharecdn.com/chuong12-150428171134-conversion-gate01/85/Chuong12-1-320.jpg)
![205
V鱈i ln tnh n袖y sai s竪 c単a 速孫o h袖m ch cn ph担 thu辿c v袖o h6
. L孫i tip t担c chia 速束i b鱈c h
v袖 tnh D(4,4) th sai s竪 ph担 thu辿c h8
. S測 速奪 tnh 速孫o h袖m theo ph測ng ph存p Romberg l袖 :
D(1,1)
D(2,1) D(2,2)
D(3,1) D(3,2) D(3,3)
D(4,1) D(4,2) D(4,3) D(4,4)
. . . . . . . . . . . .
trong 速達 m巽i gi存 tr sau l袖 gi存 tr ngo孫i suy c単a gi存 tr tr鱈c 速達 谷 h袖ng tr捉n .
V鱈i 2 j i n ta c達 :
D j
D j D jj
j(i, )
(i, ) (i , )
=
1
1
4 1 1 1
4 1
v袖 gi存 tr kh谷i 速u l袖 :
D h
h
f x h f x hi
i
i i
(i, ) ( ) [ ( ) ( )]1
1
2
= = +
v鱈i hi = h/2i-1
.
Ch坦ng ta ng探ng l孫i khi hiu gi歎a hai ln ngo孫i suy 速孫t 速辿 chnh x存c y捉u cu.
V d担 : Tm 速孫o h袖m c単a h袖m f(x) = x2
+ arctan(x) t孫i x = 2 v鱈i b鱈c tnh h = 0.5 . Tr chnh
x存c c単a 速孫o h袖m l袖 4.2
201843569.4)]75.1(f)25.2(f[
25.02
1
)1,2(D
207496266.4)]5.1(f)5.2(f[
5.02
1
)1,1(D
=
=
=
=
200458976.4)]875.1(f)125.2(f[
125.02
1
)1,3(D =
=
200492284.4
14
)2,2(D)2,3(D4
)3,3(D
200458976.4
14
)1,2(D)1,3(D4
)2,3(D
19995935.4
14
)1,1(D)1,2(D4
)2,2(D
21
2
1
1
1
1
==
==
==
Ch測ng trnh tnh 速孫o h袖m nh d鱈i 速息y . D誰ng ch測ng trnh tnh 速孫o h袖m c単a h袖m
cho trong function v鱈i b鱈c h = 0.25 t孫i xo = 0 ta nhn 速樽c gi存 tr 速孫o h袖m l袖 1.000000001.
Ch測ng trnh12-.1
//Daoham_Romberg;
#include <conio.h>
#include <stdio.h>
#include <math.h>
#define max 11
float h;
void main()
{
float d[max];
int j,k,n;
float x,p;
float y(float),dy(float);](https://image.slidesharecdn.com/chuong12-150428171134-conversion-gate01/85/Chuong12-2-320.jpg)
![206
clrscr();
printf("Cho diem can tim dao ham x = ");
scanf("%f",&x);
printf("Tinh dao ham theo phuong phap Rombergn");
printf("cua ham f(x) = th(x) tai x = %4.2fn",x);
n=10;
h=0.2;
d[0]=dy(x);
for (k=2;k<=n;k++)
{
h=h/2;
d[k]=dy(x);
p=1.0;
for (j=k-1;j>=1;j--)
{
p=4*p;
d[j]=(p*d[j+1]-d[j])/(p-1);
}
}
printf("y'= %10.5fn",d[1]);
getch();
}
float y(float x)
{
float a=(exp(x)-exp(-x))/(exp(x)+exp(-x));
return(a);
}
float dy(float x)
{
float b=(y(x+h)-y(x-h))/(2*h);
return(b);
}
則2. Kh存i nim v tch ph息n s竪
M担c 速ch c単a tnh tch ph息n x存c 速nh l袖 速存nh gi存 速nh l樽ng biu th淡c :
J f x
a
b
= ( )dx
trong 速達 f(x) l袖 h袖m li捉n t担c trong kho其ng [a,b] v袖 c達
th biu din b谷i 速棚ng cong y= f(x). Nh vy tch
ph息n x存c 速nh J l袖 din tch SABba , gi鱈i h孫n b谷i 速棚ng
cong f(x) , tr担c ho袖nh , c存c 速棚ng th村ng x = a v袖 x = b
. Nu ta chia 速o孫n [a,b] th袖nh n phn b谷i c存c 速im xi
th J l袖 g鱈i h孫n c単a t脱ng din tch c存c hnh ch歎 nht
f(xi).(xi+1 - xi) khi s竪 速im chia tin t鱈i , ngha l袖 :
a a
b
A
B
y
x](https://image.slidesharecdn.com/chuong12-150428171134-conversion-gate01/85/Chuong12-3-320.jpg)
![207
J f x x x
n
i
i
n
i i
=
=
+lim ( )( )
0
1
Nu c存c 速im chia xi c存ch 速u , th ( xi+1- xi ) =
h . Khi 速t f(xo) = fo , f(x1) = f1 ,... ta c達 t脱ng :
n i
i
n
S h f=
=
0
Khi n rt l鱈n , Sn tin t鱈i J . Tuy nhi捉n sai s竪 l袖m trn l孫i 速樽c tch lu端 . Do vy cn
ph其i tm ph測ng ph存p tnh chnh x存c h測n . Do 速達 ng棚i ta t khi d誰ng ph測ng ph存p hnh
ch歎 nht nh v探a n捉u .
則3. Ph測ng ph存p hnh thang
Trong ph測ng ph存p hnh thang , thay v chia din tch SABba th袖nh c存c hnh ch歎 nht ,
ta l孫i d誰ng hnh thang . V d担 nu chia th袖nh 3 速o孫n nh hnh v th :
S3 = t1 + t2 + t3
trong 速達 ti l袖 c存c din tch nguy捉n t竪 . M巽i din tch n袖y l袖 m辿t hnh thang :
ti = [f(xi) + f(xi-1)]/ (2h)
= h(fi - fi-1) / 2
Nh vy :
S3 = h[(fo+f1)+(f1+f2)+(f2+f3)] / 2
= h[fo+2f1+2f2+f3] / 2
M辿t c存ch t脱ng qu存t ch坦ng ta c達 :
)f2f2f2f(
n
ab
S n1n1on
++++=
hay :
}f2ff{
n
ab
S
n
1i
ion n ++
=
=
M辿t c存ch kh存c ta c達 th vit :
f x dx f x hf a kh f a k h
a
b
a kh
a k h
k
n
k
n
( ) ( )dx { ( ) / [ ( ) ] / }
( )
= + + + +
+
+ +
=
=
1
1
1
0
1
2 1 2
hay :
f x h f a f a h f a n h f b
a
b
( )dx { ( ) / ( ) [ ( ) ] ( ) / }= + + + + + + 2 1 2
Ch測ng trnh tnh tch ph息n theo ph測ng ph存p hnh thang nh sau :
Ch測ng trnh 12-2
//tinh tich phan bang phuong phap hinh_thang;
#include <conio.h>
#include <stdio.h>
#include <math.h>
float f(float x)
{
float a=exp(-x)*sin(x);
return(a);
};](https://image.slidesharecdn.com/chuong12-150428171134-conversion-gate01/85/Chuong12-4-320.jpg)
![208
void main()
{
int i,n;
float a,b,x,y,h,s,tp;
clrscr();
printf("Tinh tich phan theo phuong phap hinh thangn");
printf("Cho can duoi a = ");
scanf("%f",&a);
printf("Cho can tren b = ");
scanf("%f",&b);
printf("Cho so buoc n = ");
scanf("%d",&n);
h=(b-a)/n;
x=a;
s=(f(a)+f(b))/2;
for (i=1;i<=n;i++)
{
x=x+h;
s=s+f(x);
}
tp=s*h;
printf("Gia tri cua tich phan la : %10.6fn",tp);
getch();
}
D誰ng ch測ng trnh n袖y tnh tch ph息n c単a h袖m cho trong function trong kho其ng [0 ,
1] v鱈i 20 速im chia ta c達 J = 0.261084.
則4. C束ng th淡c Simpson
Kh存c v鱈i ph測ng ph存p hnh thang , ta chia 速o孫n [a,b] th袖nh 2n phn 速u nhau b谷i
c存c 速im chia xi :
a = xo < x1 < x2 < ....< x2n = b
xi = a+ih ; h = (b - a)/ 2n v鱈i i =0 , . . , 2n
Do yi = f(xi) n捉n ta c達 :
+++=
x
x
fdx...
x
x
fdx
b
a
x
x
fdxdx)x(f
n2
2n2
4
2
2
0
則 tnh tch ph息n n袖y ta thay h袖m f(x) 谷 v ph其i b損ng 速a th淡c n辿i suy Newton tin
bc 2 :
y
t2
)1t(t
yty)x(P 0
2
002
++=
v袖 v鱈i tch ph息n th淡 nht ta c達 :
dx)x(Pdx)x(f
x
x
x
x
2
0
2
0
2 =
則脱i bin x = x0+th th dx = hdt , v鱈i x0 th t =0 v袖 v鱈i x2 th t = 2 n捉n :](https://image.slidesharecdn.com/chuong12-150428171134-conversion-gate01/85/Chuong12-5-320.jpg)
![209
|]y)
2
t
3
t(
2
1
y
2
tty[h
dt)y
2
)1t(1
yty(hdx)x(P
2t
0t0
2
23
0
2
0
0
2
0
2
0
02
x
x
2
0
=
=
++=
++=
]yy4y[
3
h
]y)
2
4
3
8(
2
1
y2y2[h
210
0
2
00
++=
++=
則竪i v鱈i c存c tch ph息n sau ta c嘆ng c達 kt qu其 t測ng t湛 :
]yy4y[
3
h
dx)x(f 2i21i2i2
x
x
2i2
i2
++
++=
+
C辿ng c存c tch ph息n tr捉n ta c達 :
]y)yyy(2)yyy(4y[
3
h
dx)x(f n22n2421n231o
b
a
+++++++++=
Ch測ng trnh d誰ng thut to存n Simpson nh sau :
Ch測ng trnh 12-3
//Phuong phap Simpson;
#include <conio.h>
#include <stdio.h>
#include <math.h>
float y(float x)
{
float a=4/(1+x*x);
return(a);
}
void main()
{
int i,n;
float a,b,e,x,h,x2,y2,x4,y4,tp;
clrscr();
printf("Tinh tich phan theo phuong phap Simpsonn");
printf("Cho can duoi a = ");
scanf("%f",&a);
printf("Cho can tren b = ");
scanf("%f",&b);
printf("Cho so diem tinh n = ");
scanf("%d",&n);
h=(b-a)/n;
x2=a+h;
x4=a+h/2;
y4=y(x4);
y2=y(x2);
for (i=1;i<=n-2;i++)
{](https://image.slidesharecdn.com/chuong12-150428171134-conversion-gate01/85/Chuong12-6-320.jpg)
![210
x2+=h;
x4+=h;
y4+=y(x4);
y2+=y(x2);
}
y2=2*y2;
y4=4*(y4+y(x4+h));
tp=h*(y4+y2+y(a)+y(b))/6;
printf("Gia tri cua tich phan la : %10.8fn",tp);
getch();
}
D誰ng ch測ng trnh n袖y tnh tch ph息n c単a h袖m trong function trong 速o孫n [0,1] v鱈i 20
kho其ng chia cho ta kt qu其 J = 3.14159265.](https://image.slidesharecdn.com/chuong12-150428171134-conversion-gate01/85/Chuong12-7-320.jpg)
Chuong12
- 1. 204 Ch測ng 12 : Tnh gn 速坦ng 速孫o h袖m v袖 tch ph息n x存c 速nh 則1. 則孫o h袖m Romberg 則孫o h袖m theo ph測ng ph存p Romberg l袖 m辿t ph測ng ph存p ngo孫i suy 速 x存c 速nh 速孫o h袖m v鱈i m辿t 速辿 chnh x存c cao . Ta xt khai trin Taylor c単a h袖m f(x) t孫i (x+h) v袖 (x-h) : ++霞霞+霞++=+ )x(f !4 h )x(f !3 h )x(f 2 h )x(fh)x(f)hx(f )4( 432 (1) +霞霞霞霞+霞= )x(f !4 h )x(f !3 h )x(f 2 h )x(fh)x(f)hx(f )4( 432 (2) Tr探 (1) cho (2) ta c達 : ++霞霞+=+ )x(f !5 h2 )x(f !3 h2 )x(fh2)hx(f)hx(f )5( 53 (3) Nh vy r坦t ra : 霞霞霞 + = )x(f !5 h )x(f !3 h h2 )hx(f)hx(f )x(f )5( 42 (4) hay ta c達 th vit l孫i : [ ] +++++= 6 6 4 4 2 2 hahaha)hx(f)hx(f h2 1 )x(f (5) trong 速達 c存c h s竪 ai ph担 thu辿c f v袖 x . Ta 速t : )]hx(f)hx(f[ h2 1 )h( + = (6) Nh vy t探 (5) v袖 (6) ta c達 : == 6 6 4 4 2 2 hahaha)x(f)h()1,1(D (7) = = 64 h a 16 h a 4 h a)x(f 2 h )1,2(D 6 6 4 4 2 2 (8) v袖 t脱ng qu存t v鱈i hi = h/2i-1 ta c達 : == 6 i6 4 i4 2 i2i hahaha)x(f)h()1,i(D (9) Ta t孫o ra sai ph息n D(1,1) - 4D(2,1) v袖 c達 : 霞= 6 6 4 4 ha 16 15 ha 4 3 )x(f3 2 h 4)h( (10) Chia hai v c単a (10) cho -3 ta nhn 速樽c : +++= = 6 6 4 4 ha 16 5 ha 4 1 )x(f 4 )1,1(D)1,2(D4 )2,2(D (11) Trong khi D(1,1) v袖 D(2,1) sai kh存c f(x) ph担 thu辿c v袖o h2 th D(2,2) sai kh存c f(x) ph担 thu辿c v袖o h4 . B息y gi棚 ta l孫i chia 速束i b鱈c h v袖 nhn 速樽c : D f x a h a h(2, ) ( ) ( / ) ( / ) ...2 1 4 2 5 16 24 4 6 6 = + + + (12) v袖 kh旦 s竪 h孫ng c達 h4 b損ng c存ch t孫o ra : D D f x a h(2, ) ( , ) ( ) ( ) ...2 16 32 15 15 64 6 6 = + + + (13) Chia hai v c単a (13) cho -15 ta c達 : D D D f x a h(3, ) (3, ) (2, ) ( ) . ...3 16 2 2 15 1 64 6 6 = = (14)
- 2. 205 V鱈i ln tnh n袖y sai s竪 c単a 速孫o h袖m ch cn ph担 thu辿c v袖o h6 . L孫i tip t担c chia 速束i b鱈c h v袖 tnh D(4,4) th sai s竪 ph担 thu辿c h8 . S測 速奪 tnh 速孫o h袖m theo ph測ng ph存p Romberg l袖 : D(1,1) D(2,1) D(2,2) D(3,1) D(3,2) D(3,3) D(4,1) D(4,2) D(4,3) D(4,4) . . . . . . . . . . . . trong 速達 m巽i gi存 tr sau l袖 gi存 tr ngo孫i suy c単a gi存 tr tr鱈c 速達 谷 h袖ng tr捉n . V鱈i 2 j i n ta c達 : D j D j D jj j(i, ) (i, ) (i , ) = 1 1 4 1 1 1 4 1 v袖 gi存 tr kh谷i 速u l袖 : D h h f x h f x hi i i i (i, ) ( ) [ ( ) ( )]1 1 2 = = + v鱈i hi = h/2i-1 . Ch坦ng ta ng探ng l孫i khi hiu gi歎a hai ln ngo孫i suy 速孫t 速辿 chnh x存c y捉u cu. V d担 : Tm 速孫o h袖m c単a h袖m f(x) = x2 + arctan(x) t孫i x = 2 v鱈i b鱈c tnh h = 0.5 . Tr chnh x存c c単a 速孫o h袖m l袖 4.2 201843569.4)]75.1(f)25.2(f[ 25.02 1 )1,2(D 207496266.4)]5.1(f)5.2(f[ 5.02 1 )1,1(D = = = = 200458976.4)]875.1(f)125.2(f[ 125.02 1 )1,3(D = = 200492284.4 14 )2,2(D)2,3(D4 )3,3(D 200458976.4 14 )1,2(D)1,3(D4 )2,3(D 19995935.4 14 )1,1(D)1,2(D4 )2,2(D 21 2 1 1 1 1 == == == Ch測ng trnh tnh 速孫o h袖m nh d鱈i 速息y . D誰ng ch測ng trnh tnh 速孫o h袖m c単a h袖m cho trong function v鱈i b鱈c h = 0.25 t孫i xo = 0 ta nhn 速樽c gi存 tr 速孫o h袖m l袖 1.000000001. Ch測ng trnh12-.1 //Daoham_Romberg; #include <conio.h> #include <stdio.h> #include <math.h> #define max 11 float h; void main() { float d[max]; int j,k,n; float x,p; float y(float),dy(float);
- 3. 206 clrscr(); printf("Cho diem can tim dao ham x = "); scanf("%f",&x); printf("Tinh dao ham theo phuong phap Rombergn"); printf("cua ham f(x) = th(x) tai x = %4.2fn",x); n=10; h=0.2; d[0]=dy(x); for (k=2;k<=n;k++) { h=h/2; d[k]=dy(x); p=1.0; for (j=k-1;j>=1;j--) { p=4*p; d[j]=(p*d[j+1]-d[j])/(p-1); } } printf("y'= %10.5fn",d[1]); getch(); } float y(float x) { float a=(exp(x)-exp(-x))/(exp(x)+exp(-x)); return(a); } float dy(float x) { float b=(y(x+h)-y(x-h))/(2*h); return(b); } 則2. Kh存i nim v tch ph息n s竪 M担c 速ch c単a tnh tch ph息n x存c 速nh l袖 速存nh gi存 速nh l樽ng biu th淡c : J f x a b = ( )dx trong 速達 f(x) l袖 h袖m li捉n t担c trong kho其ng [a,b] v袖 c達 th biu din b谷i 速棚ng cong y= f(x). Nh vy tch ph息n x存c 速nh J l袖 din tch SABba , gi鱈i h孫n b谷i 速棚ng cong f(x) , tr担c ho袖nh , c存c 速棚ng th村ng x = a v袖 x = b . Nu ta chia 速o孫n [a,b] th袖nh n phn b谷i c存c 速im xi th J l袖 g鱈i h孫n c単a t脱ng din tch c存c hnh ch歎 nht f(xi).(xi+1 - xi) khi s竪 速im chia tin t鱈i , ngha l袖 : a a b A B y x
- 4. 207 J f x x x n i i n i i = = +lim ( )( ) 0 1 Nu c存c 速im chia xi c存ch 速u , th ( xi+1- xi ) = h . Khi 速t f(xo) = fo , f(x1) = f1 ,... ta c達 t脱ng : n i i n S h f= = 0 Khi n rt l鱈n , Sn tin t鱈i J . Tuy nhi捉n sai s竪 l袖m trn l孫i 速樽c tch lu端 . Do vy cn ph其i tm ph測ng ph存p tnh chnh x存c h測n . Do 速達 ng棚i ta t khi d誰ng ph測ng ph存p hnh ch歎 nht nh v探a n捉u . 則3. Ph測ng ph存p hnh thang Trong ph測ng ph存p hnh thang , thay v chia din tch SABba th袖nh c存c hnh ch歎 nht , ta l孫i d誰ng hnh thang . V d担 nu chia th袖nh 3 速o孫n nh hnh v th : S3 = t1 + t2 + t3 trong 速達 ti l袖 c存c din tch nguy捉n t竪 . M巽i din tch n袖y l袖 m辿t hnh thang : ti = [f(xi) + f(xi-1)]/ (2h) = h(fi - fi-1) / 2 Nh vy : S3 = h[(fo+f1)+(f1+f2)+(f2+f3)] / 2 = h[fo+2f1+2f2+f3] / 2 M辿t c存ch t脱ng qu存t ch坦ng ta c達 : )f2f2f2f( n ab S n1n1on ++++= hay : }f2ff{ n ab S n 1i ion n ++ = = M辿t c存ch kh存c ta c達 th vit : f x dx f x hf a kh f a k h a b a kh a k h k n k n ( ) ( )dx { ( ) / [ ( ) ] / } ( ) = + + + + + + + = = 1 1 1 0 1 2 1 2 hay : f x h f a f a h f a n h f b a b ( )dx { ( ) / ( ) [ ( ) ] ( ) / }= + + + + + + 2 1 2 Ch測ng trnh tnh tch ph息n theo ph測ng ph存p hnh thang nh sau : Ch測ng trnh 12-2 //tinh tich phan bang phuong phap hinh_thang; #include <conio.h> #include <stdio.h> #include <math.h> float f(float x) { float a=exp(-x)*sin(x); return(a); };
- 5. 208 void main() { int i,n; float a,b,x,y,h,s,tp; clrscr(); printf("Tinh tich phan theo phuong phap hinh thangn"); printf("Cho can duoi a = "); scanf("%f",&a); printf("Cho can tren b = "); scanf("%f",&b); printf("Cho so buoc n = "); scanf("%d",&n); h=(b-a)/n; x=a; s=(f(a)+f(b))/2; for (i=1;i<=n;i++) { x=x+h; s=s+f(x); } tp=s*h; printf("Gia tri cua tich phan la : %10.6fn",tp); getch(); } D誰ng ch測ng trnh n袖y tnh tch ph息n c単a h袖m cho trong function trong kho其ng [0 , 1] v鱈i 20 速im chia ta c達 J = 0.261084. 則4. C束ng th淡c Simpson Kh存c v鱈i ph測ng ph存p hnh thang , ta chia 速o孫n [a,b] th袖nh 2n phn 速u nhau b谷i c存c 速im chia xi : a = xo < x1 < x2 < ....< x2n = b xi = a+ih ; h = (b - a)/ 2n v鱈i i =0 , . . , 2n Do yi = f(xi) n捉n ta c達 : +++= x x fdx... x x fdx b a x x fdxdx)x(f n2 2n2 4 2 2 0 則 tnh tch ph息n n袖y ta thay h袖m f(x) 谷 v ph其i b損ng 速a th淡c n辿i suy Newton tin bc 2 : y t2 )1t(t yty)x(P 0 2 002 ++= v袖 v鱈i tch ph息n th淡 nht ta c達 : dx)x(Pdx)x(f x x x x 2 0 2 0 2 = 則脱i bin x = x0+th th dx = hdt , v鱈i x0 th t =0 v袖 v鱈i x2 th t = 2 n捉n :
- 6. 209 |]y) 2 t 3 t( 2 1 y 2 tty[h dt)y 2 )1t(1 yty(hdx)x(P 2t 0t0 2 23 0 2 0 0 2 0 2 0 02 x x 2 0 = = ++= ++= ]yy4y[ 3 h ]y) 2 4 3 8( 2 1 y2y2[h 210 0 2 00 ++= ++= 則竪i v鱈i c存c tch ph息n sau ta c嘆ng c達 kt qu其 t測ng t湛 : ]yy4y[ 3 h dx)x(f 2i21i2i2 x x 2i2 i2 ++ ++= + C辿ng c存c tch ph息n tr捉n ta c達 : ]y)yyy(2)yyy(4y[ 3 h dx)x(f n22n2421n231o b a +++++++++= Ch測ng trnh d誰ng thut to存n Simpson nh sau : Ch測ng trnh 12-3 //Phuong phap Simpson; #include <conio.h> #include <stdio.h> #include <math.h> float y(float x) { float a=4/(1+x*x); return(a); } void main() { int i,n; float a,b,e,x,h,x2,y2,x4,y4,tp; clrscr(); printf("Tinh tich phan theo phuong phap Simpsonn"); printf("Cho can duoi a = "); scanf("%f",&a); printf("Cho can tren b = "); scanf("%f",&b); printf("Cho so diem tinh n = "); scanf("%d",&n); h=(b-a)/n; x2=a+h; x4=a+h/2; y4=y(x4); y2=y(x2); for (i=1;i<=n-2;i++) {
- 7. 210 x2+=h; x4+=h; y4+=y(x4); y2+=y(x2); } y2=2*y2; y4=4*(y4+y(x4+h)); tp=h*(y4+y2+y(a)+y(b))/6; printf("Gia tri cua tich phan la : %10.8fn",tp); getch(); } D誰ng ch測ng trnh n袖y tnh tch ph息n c単a h袖m trong function trong 速o孫n [0,1] v鱈i 20 kho其ng chia cho ta kt qu其 J = 3.14159265.