常州大学数值分析第五章习题解答.doc

化专012 闵建中 121029144.解：(1) : T=0.5*(3+1)=2 S=2 C=2 err1 = 0 err2=0(2) : T=1*(1-3)=-2 S=-10/3 C=-10/3 err 1= 4/3 err2=0(3) : T=0 S=0 C=0 err1=0 err2=0(4) : T=e/2 S=1/6*(e+2exp(0.5) C= 1 err1=-1.7183 err2= 0.00267.解： 复合梯形公式T2n的matlab实现：function I = trapezoid(fun,a,b,n)n = 2*n;h = (b-a)/(n-1);x = a : h :b;f = feval(fun , x);I = h * (0.5*f(1)+sum(f(2:n-1)+ 0.5*f(n);分别用复合梯形公式和复合辛普森公式计算积分，并与精确值比较，这里积分的精确值由matlab int函数求出，不手工计算。function trapezoid_and_sinpsomclc;format longsyms xIexact = int(x*exp(x2),x,0,1);a=0;b=1;for n=2:1:4t = trapezoid(f,a,b,n)s = simpson(f,a,b,n)err1 = vpa(Iexact - t,5)err2 = vpa(Iexact - s,5)endfunction y=f(x)y = x*exp(x2);return从而得出第一小题的结果：n = 2t = 1.000576811286697s = 0.860997139578795err1 = -0.14144 err2 = -0.0018562n = 3t = 0.923798756293777s = 0.859533825596209err1 = -0.064658err2 = -0.00039291n = 4t = 0.895892057505771s = 0.859268455239111err1 = -0.036751err2 = -0.00012754将f(x)简单修改就可以得到第2，3小题的结果8.解：function test_romberg()a=1;b=3;n=2;maxit=10;tol=0.5e-4;R=romberg(f,a,b,n,tol,maxit);R = R(1,4)Rexact = log(3)err = R-Rexactreturn function y=f(x)y=1/x;return10.解：高斯-勒让德法求解积分的matlab实现：function I = guasslegendre(fun,a,b,n,tol)% guasslegendre: guass-legendre method to calculation integral value%Input : fun is the integrand input as a sring fun% a and b are upper and lower limits of integration% n is the initial number of subintervals% tol is the tolerance%Output : I is Integral result % calculate quadrature nodesyms xp=sym2poly(diff(x2-1)(n+1),n+1)/(2n*factorial(n);tk=roots(p); % calculate quadrature coefficientAk=zeros(n+1,1);for i=1:n+1 xkt=tk; xkt(i)=; pn=poly(xkt); fp=(x)polyval(pn,x)/polyval(pn,tk(i); Ak(i)=quadl(fp,-1,1,tol);end% integral variable substitution, transformat a, b to 1, 1xk=(b-a)/2*tk+(b+a)/2;% calculate the value of the integral function after variable substitutionfx=fun(xk)*(b-a)/2;%calculation integral valueI=sum(Ak.*fx);对第一小题，精确值由matlab int函数求出，不手工计算；fun = inline(x.5+3.*x.2+2);a = 0;b = 1;tol = 1e-5;for n =1:2; n I = guasslegendre(fun,a,b,n,tol) Iexact = int(x5+3*x2+2,x,0,1) err = Iexact-Iend计算结果如下：n = 1I = 3.152777777777778Iexact = 19/6 = 3.166666666666667err = 1/72n = 2I = 3.166666666666666Iexact = 19/6 = 3.166666666666667 err = 0由此可见高斯-勒让德方法可以在较少的求积节点数下得到较为精确的数值积分解！第2,3小题同理可得。11.解：(1):fun = inline(x.*exp(x.2);a = 0;b = 0.5;n = 1;tol = 1e-4; I1 = guasslegendre(fun,a,b,n,tol);a = 0.5;b = 1;n = 1;tol = 1e-4; I2 = guasslegendre(fun,a,b,n,tol);I = I1+I2I