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用數(shù)值積分吧 ' Z/ f5 X3 p( O; n- Q
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clear all5 q3 z0 c x$ a% ~$ T5 [ i
format long9 f+ B, R$ o; x2 N9 ?3 E! K
a=0;4 f; K) h) S7 Z1 F6 X3 Z4 B, O
b=1;
' B: B" Q. [0 u% ~, e' Uepsilon=10^(-6);
& C! B% D4 E" O2 Z' wsyms x;
2 J/ g8 j0 I+ x9 n* Ofun=log(x^2 + 1)/(x + 1);: L; q1 z, K7 ?
Hfun=@ Remberg;
6 W) R6 [; `: u3 X" SIvalue= feval(Hfun,fun,a,b,epsilon);- S! F" r4 _% \! l1 @
5 M T) O! `; K+ y% B: k- D%Remberg.m
G: e+ p5 N& v0 t% L%a,b為積分限,,epsilon為精度,,s為返回積分值,fun為被積函數(shù)
2 {& U6 h7 R3 s# y%R(n,m)表示計(jì)算值,,(n-1)為變步長指標(biāo),,(m-1)為加速次數(shù): G" B8 \3 s1 A2 I4 ]0 n
function s=Remberg(fun,a,b,epsilon)
! p/ |. R/ A# V1 w0 Y) C+ _& Gsyms x ;
+ U$ z/ W* i, [1 v0 q& Rfvalue=zeros(1,1000);2 a3 J- J4 ^# m$ y$ f
R=zeros(100,100);
# I$ ?- }) u; S: \fvaluea=double(subs(fun,x,a));: [. u4 y/ {# k
fvalueb=double(subs(fun,x,b));& c( b" _/ s8 @0 S& g/ L: p# x. e
R(1,1)=(b-a)/2*(fvaluea+fvalueb); %梯形公式3 u+ ?7 v! x9 U' Q, v! p# b4 a
km=1;, k. C' s7 f) F3 o5 b, q7 U( k! ]
for k1=1:100; %設(shè)置一個(gè)比較大的循環(huán)量: E# A) L# Y8 E, W6 b
h=(b-a)/(2^(k1));
0 t- B1 V9 @$ h) D+ _0 j% T R(k1+1,1)=1/2*R(k1,1);
3 U4 W- X! U6 N- p" }' t2 j for k2=1:2^(k1-1);6 C/ f- a/ _( e0 Q0 r
fvalue(2*k2)=double(subs(fun,x,a+(2*k2-1)*h));4 O; E; A3 Y! D
R(k1+1,1)=h*fvalue(2*k2)+R(k1+1,1); %變步長值1 d; b& @) V* b( C
end
) G0 ] s) Q2 u; [( X for k3=1:km; %加速計(jì)算' ^# s7 l. ]# W, q1 x2 A3 ^# s2 X
R(k1+1,k3+1)=1/(4^(k3)-1)*(4^(k3)*R(k1+1,k3)-R(k1,k3));
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if abs(R(k1+1,km+1)-R(k1+1,km))<epsilon %控制精度7 t' H5 Z: @: `1 E7 S
s=R(k1+1,km+1);
$ V+ ~# W, A' Q break;
3 p, s4 D/ |# } else8 B- x0 ^5 I8 n
km=km+1;( h" z' Y4 p# e2 h( I- k" s
end
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end
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