应用牛顿法求方程cos(x)cosh(x)-1=0的头五个非零的正根

1.应用牛顿法求方程cos(x)cosh(x)-1=0的头五个非零的正根。 先求出方程近似根 function R = approot (f,X,epsilon) Y=f(X); yrange = max(Y)-min(Y); epsilon2 = yrange*epsilon; n=length(X); m=0; X(n+1)=X(n); Y(n+1)=Y(n); for k=2:n if Y(k-1)*Y(k) <= 0, m=m+1; R(m)=(X(k-1)+X(k))/2; end s=(Y(k)-Y(k-1))*(Y(k+1)-Y(k)); if (abs(Y(k)) < epsilon2) & (s <= 0) m=m+1; R(m)=X(k); end end syms x; X=0:.001:20; fun(x)=cos(x)*cosh(x) - 1; ans = 0.0005 3.9270 4.7305 7.0690 7.8535 10.2100 10.9955 14.1375 17.2785 function [p0,err,k,y]=newton(f,df,p0,delta,epsilon,max1) for k=1:max1 p1=p0-feval(f,p0)/feval(df,p0); err=abs(p1-p0); relerr=2*err/(abs(p1)+delta); p0=p1; y=f(p0); if (err<delta)|(relerr<delta)|(abs(y)<epsilon) break end end 代入第一个近似根： newton(fun,dfun,0.0005,0.00001,0.00001,5) ans =3.7477e-04 故3.7477e-04为方程的第一个非零正根 代入第二个近似根： dfun（3.9270）=0，不能用牛顿法 代入第三个近似根： newton(fun,dfun,4.7305,0.0001,0.0001,3) ans =4.7300 故4.7300为方程的第二个非零正根 代入第四个近似根： dfun（7.0690）=0，不能用牛顿法 代入第五个近似根： newton(fun,dfun,7.8535,0.0001,0.0001,3) ans = 7.8532 故7.8532为方程的第三个非零正根 代入第六个近似根： dfun（10.2100）=0，不能用牛顿法 代入第七个近似根： newton(fun,dfun,10.9955,0.0001,0.0001,3) ans =10.9956 故10.9956为方程的第四个非零正根 代入第八个近似根： newton(fun,dfun,14.1375,0.0001,0.0001,3) ans =14.1372 故14.1372为方程的第五个非零正根 2.用二分法求方程在区间[0,1]内的根，要求。 function [c,err,yc]=bisect(f,a,b,delta) ya=f(a); yb=f(b); if ya*yb > 0,return,end max1=1+round((log(b-a)-log(delta))/log(2)); for k=1:max1 c=(a+b)/2; yc=f(c); if yc==0 a=c; b=c; elseif yb*yc>0 b=c; yb=yc; else a=c; ya=yc; end if b-a < delta, break,end end c=(a+b)/2; err=abs(b-a); yc=f(c); syms x; fun(x)=2*exp(-x)-sin(x); >> bisect(fun,0.0,1.0,0.000005) ans = 1.9073e-06