數(shù)值分析的MATLAB程序設(shè)計(jì)

1、數(shù)值分析作業(yè)題目 利用MATLAB編寫數(shù)值算法程序班級/學(xué)號04140101/20100414028學(xué)生姓名 張陽 指導(dǎo)教師 劉勇進(jìn)編寫lagrange插值法L( x) = Wi=0(x x )(x x )(x x )(x x )(x x )01ii+1L yi(x) = (x x )(x x )(x x)(x x )(x x )n01in1n計(jì)算拉氏基函數(shù)，再求和。開始輸入插值點(diǎn) 向量x和拉氏 基函數(shù)序號kii =1 : nii = k圖1拉氏基函數(shù)求解流程圖Lagrange_base_function.m 文件%Copyright(c) 2012 ShenYang Areospace Un

2、iversity%All rights reserved%file name:Lagrange_base_function.m%summary%:%current edition:%programmer:Yung Zhung%accomplish time:2012-04-15%function name:Lagrange_base_function.m%function describe%function in detailed% function Y = Lagrange_base_function(x, k) n = length(x);y = ones(n, 2);Y = 1;for

3、ii = 1 : n if ii = k continue;endy(ii, 2) = -x(ii);Y = conv(Y, y(ii, :);endend開臺輸入插值點(diǎn)向量x和相應(yīng)函數(shù)值二一.一.檢查輸入出錯ii = 1 : nY計(jì)算第ii個(gè)拉氏基函數(shù)|由拉氏基函數(shù)計(jì)算代入點(diǎn)|值和x(ii)值N兩值相除乘以y(ii)再累結(jié)束圖2計(jì)算lagrange插值函數(shù)及差值值Lagrange.m 文件%Copyright(c) 2012 ShenYang Areospace University%All rights reserved%file name:Lagrange.m%summary%curre

4、nt edition:Yung Zhung:2012-04-15%programmer%accomplish time%function name:Lagrange.m%function describe : %function in detailed : % function Y = Lagrange(x, y, point) if min(size(x) 1 | min(size(y) 1error the muxt be two vectors!); end n = length(x); if n = length(y)error the dimention of the two vec

5、tors must aggree!); end Y = 0;for ii = 1 : ny_1 = Lagrange_base_function(x, ii);= polyadd(Y, y(ii) * y_1 / polyval(y_1, x(ii); endif nargin = 2 disp(Y);else= polyval(Y, point); end編寫復(fù)合二重simpson求積公式圖3二重復(fù)合simpson公式流程圖 轉(zhuǎn)化為一重simpson求積公式。%Copyright(c) 2012 ShenYang Areospace University %All rights reserv

6、ed%file name:simpson_con.m%summary% %current edition:Yung Zhung:2012-06-25%programmer%accomplish time%function name:simpson_con.m%function describe :%function in detailed :% function val = simpson_con(f, a, b, c, d, m, n) if b aflag_1 = a;a = b;b = flag_1;flag_1 = -1;elseflag_1 = 1;end if d cflag_2

7、= d;d = c;c = flag_2;flag_2 = -1;elseflag_2 = 1;end TOC o 1-5 h z h_1 =(b- a)/m;h_2 =(d- c)/n;x_1 =a :h_1/2 :b;x_2 =c :h_2/2 :d;x_1, x_2 = meshgrid(x_1, x_2);y = f(x_1, x_2);1, :) + 41) + 4 火y_1 = (y(1, :) + y(2 火 n + 1, :) + 2 火 sum(y(3 : 2 : 2 火 n * sum(y(2 : 2 : 2 火 n, :) / 6 火 h_2;val = (y_1(1)

8、+ y_1(2 火 m + 1) + 2 火 sum(y_1(3 : 2 : 2 火 m sum(y_1(2 : 2 : 2 火 m) / 6 火 h_1 * flag_1 * flag_2;end三、編寫Gauss消去法開始輸入被積函數(shù)，積分區(qū)間和分段數(shù)一 檢查輸入出錯N利用第ii+1到第n行由第 ii列到第n列與第ii行第 ii列到第n列將第ii列化 為0圖4求解Gauss消去法Gauss_01.m 文件%Copyright(c) 2012 ShenYang Areospace University %All rights reserved%file name:Gauss_01.m%summary%:%current edition:%programmer:Yung Zhung%accomplish time:2012-06-14%function name:Gauss_01.m%function describe :%function in detailed :%function A = Gauss_01(A)n = size(A, 1);if n = size(A, 2)error (error!);endfor ii = 1 : n - 1%for jj = ii + 1 : n% A(jj,ii : n) = A(jj, ii : n) -