常州大學(xué)數(shù)值分析第五章習(xí)題解答(共4頁)

1、精選優(yōu)質(zhì)文檔-傾情為你奉上化專012 閔建中 4.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： 復(fù)合梯形公式T2n的matlab實現(xiàn)：function I = trapezoid(fun,a,b,n)n = 2*n;h = (b-a)/(n-1);x = a

2、 : h :b;f = feval(fun , x);I = h * (0.5*f(1)+sum(f(2:n-1)+ 0.5*f(n);分別用復(fù)合梯形公式和復(fù)合辛普森公式計算積分，并與精確值比較，這里積分的精確值由matlab int函數(shù)求出，不手工計算。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 =

3、 vpa(Iexact - s,5)endfunction y=f(x)y = x*exp(x2);return從而得出第一小題的結(jié)果：n = 2t = 1.6697s = 0.8795err1 = -0.14144 err2 = -0.n = 3t = 0.3777s = 0.6209err1 = -0.err2 = -0.n = 4t = 0.5771s = 0.9111err1 = -0.err2 = -0.將f(x)簡單修改就可以得到第2，3小題的結(jié)果8.解：function test_romberg()a=1;b=3;n=2;maxit=10;tol=0.5e-4;R=romberg

4、(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-勒讓德法求解積分的matlab實現(xiàn)：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 ar

5、e 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; xk

6、t(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)

7、;對第一小題，精確值由matlab int函數(shù)求出，不手工計算；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計算結(jié)果如下：n = 1I = 3.7778Iexact = 19/6 = 3.6667err = 1/72n = 2I = 3.6666Iexact = 19/6 = 3.6667 err = 0由此可見高斯-勒讓德方法可以在較少的求積節(jié)點數(shù)下得到較為精確的數(shù)值積分解！第2,3小題同理可得。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