Matlab数值分析.docVIP

第一章 尋找f(x) = 0 的解 在本章中包含下列 Matlab 的m-file bisection.m bisection1.m deltasqr.m fixpt.m newton.m 2-1 newtonnew.m newtion1.m (for system of nonlinear equations F(X)=0 ) basin3.m secant.m falsep.m müller.m bairstow.m laguerre.m 將須要的m-file之檔案夾加入搜尋路徑中 path(d:\numerical, path) 註: 如果你有安裝Matlab Notebook 要執行下列input cells (綠色指令敘述) 之前必須先執行上面的cell – [path (…) ] 藍色的內容是Matlab [output cells] 本章同時利用Extended Symbolic ToolBox內的指令或函數 1. bisection .m 顯示bisection.m的內容 type bisection.m function bisection( s, a, b, n, tor) %Bisection Method % epsilon = 10^-12 ; syms x ; fun= sym(s); fa = subs(fun,x,a); fb = subs(fun,x,b); if fa * fb =0 fprintf(a= %10.6f f(a)= %0.7g\n,a, fa) fprintf(b= %10.6f f(a)= %0.7g\n,b, fb) fprintf(There is some trouble at endpoints\n) return; end %start bisection fprintf( n c f(c) error\n) err = b-a; for j = 1: n err = err/2 ; c = a + err; fc = subs(fun,x,c); fprintf(%3d %10.6f %0.7g %0.7g\n,j,c,fc,err) if abs(err) tor | abs(fc) epsilon fprintf(\nconvergent or f(x) is almost 0\n) return; end if fa * fc 0 b=c; fb = fc; else a=c; fa = fc; end err =b-a ; end %for 例題 1: To find the approximation root of x3 - 2 sin(x) on [0.5, 2] 首先把f(x)的圖形畫出來 t = 0.5:0.1:2 ; y = t.^3 - 2*sin(t) ; %注意是t.^3 plot(t,y,[0.5 2],[0 0],r-) 準備輸入必要的參數, 如函數f(x), 區間端點a, b, 計算的次數n 與誤差範圍tor a = 0.5 ; b = 2; n = 20; tor = 10^(-4) ; s = x^3 - 2*sin(x) ; bisection(s, a, b, n, tor) n c f(c) error 0 1.250000 0 0.75 1 0.875000 -0.8651651 0.375 2 1.062500 -0.547687 0.1875 3 1.156250 -0.2847914