南京邮电大学数学实验A 作业4 符号计算 答案 | 《MATLAB数学实验》第三版 第七章 课后习题答案

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1(课本习题2)

做因式分解$f(x)=x^4-5x^3+5x^2+5x-6$.

代码：

syms x;
f = x^4-5*x^3+5*x^2+5*x-6;
factor(f)

输出：

$(\begin{array}{cccc}x-1& x-2& x-3& x+1\end{array})$

2(课本习题3)

求矩阵$A=\left(\begin{array}{cc}1& 2\\ 2& a\end{array}\right)$的逆和特征值.

代码：

syms a;
A = [1 2;2 a];
iA = inv(A)
e = eig(A)

答：

该矩阵的逆(iA)为$(\begin{array}{cc}\frac{a}{a-4}& -\frac{2}{a-4}\\ \frac{2}{a-4}& \frac{1}{a-4}\end{array})$，
特征值(e)为$(\begin{array}{r}\frac{a}{2}-\frac{\sqrt{a^2-2a+17}}{2}+\frac{1}{2}\\ \frac{a}{2}+\frac{\sqrt{a^2-2a+17}}{2}+\frac{1}{2}\end{array})$。

3(课本习题4)

计算极限

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left(3^x+9^x\right)}^{\frac{1}{x}},\underset{y\to 0^{+}}{\lim }\underset{x\to 0^{+}}{\lim }\frac{\ln (2x+e^{-y})}{\sqrt{x^3+y^2}}， \\ \underset{x\to \infty }{\lim }\frac{\ln (1+\frac{1}{x})}{arccotx},\underset{x\to 0}{\lim }\frac{1-\sqrt{1-x^2}}{e^x-\cos x} \end{gathered}$$

代码：

syms x y;
ans1 = limit((3^x+9^x)^(1/x),x,inf)
temp1 = limit(log(2*x+exp(-y))/(sqrt(x^3+y^2)-1),x,0);
ans2 = limit(s1,y,0)
ans3 = limit(log(1+1/x)/acot(x),x,inf)
ans4 = limit((1-sqrt(1-x^2))/(exp(x)-cos(x)),x,inf)

输出：

ans1 = 9

ans2 = 0

ans3 = 1

ans4 = 0

4(课本习题5)

计算

$$\sum _{k=1}^n{k}^2,\sum _{k=1}^{\infty }\frac{1}{k^2},\sum _{n=0}^{\infty }\frac{1}{(2n+1)(2x+1{)}^{2n+1}}.$$

代码：

syms k n x;
s1=symsum(k^2,k,1,n);
s2=symsum(k^(-2),k,1,inf);
s3=symsum(1/(2*n+1)/(2*x+1)^(2*n+1),n,0,inf);
s1=simplify(s1)
s2=simplify(s2)
s3=simplify(s3)

输出：

s1 = $\frac{n(2n+1)(n+1)}{6}$

s2 = $\frac{{\pi }^2}{6}$

s3 = $atanh(\frac{1}{2x+1})\text{ if }1<\Vert 2x+1\Vert$

5(课本习题6)

求${\frac{{\partial }^3}{\partial x^2\partial y}\sin \left(x^{2}yz\right)\Vert }_{x=1,y=1,z=3}$.

代码：

syms x y z;
s=sin(x^2*y*z);
s=diff(s,x,2);
s=diff(s,y,1);
s=subs(s,{x,y,z},{1,1,3})

输出：

s = $log(x+\sqrt{x^2+1})$

6(课本习题7)

(Taylor展开)求下列函数在$x=0$的Taylor幂级数展开式(n=8):

$$e^x,\ln (1+x),\sin x,$$

$$\ln (x+\sqrt{1+x^2}),\frac{1}{x^2-3x+2}.$$

输出：

syms x
f1 = exp(x);
f2 = log(1+x);
f3 = sin(x);
f4 = log(x+sqrt(1+x^2));
f5 = 1/(x^2-3*x+2);
f = [f1 f2 f3 f4 f5];
for i=1:length(f)
    taylor(f(i), x, 'Order', 9, 'ExpansionPoint', 0)
end

输出：

ans =
$\frac{x^8}{40320}+\frac{x^7}{5040}+\frac{x^6}{720}+\frac{x^5}{120}+\frac{x^4}{24}+\frac{x^3}{6}+\frac{x^2}{2}+x+1$

ans =
$-\frac{x^8}{8}+\frac{x^7}{7}-\frac{x^6}{6}+\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}-\frac{x^2}{2}+x$

ans = $-\frac{x^7}{5040}+\frac{x^5}{120}-\frac{x^3}{6}+x$

ans =
$-\frac{5x^7}{112}+\frac{3x^5}{40}-\frac{x^3}{6}+x$

ans =
$\frac{511x^8}{512}+\frac{255x^7}{256}+\frac{127x^6}{128}+\frac{63x^5}{64}+\frac{31x^4}{32}+\frac{15x^3}{16}+\frac{7x^2}{8}+\frac{3x}{4}+\frac{1}{2}$

7(课本习题9)

计算下列不定积分并用diff验证:

$$\int e^{2x}(\tan x+1{)}^{2}dx,\int \frac{e^{2y}}{e^y+2}dy,\int \frac{x^2}{\sqrt{a^2-x^2}}dx,$$

$$\int e^{x^{-2}}dx,\int \frac{dx}{x(\sqrt{\ln x+a}+\sqrt{\ln x+b})}(a\ne b).$$

代码：

function intf(f,symbol)
    fi = int(f,symbol)
    s = simplify(diff(fi));
    if (f / s == 1)
        fprintf('经验证，运算结果正确。')
    end
end


% 7.1
syms x;
f1 = exp(2*x)*(tan(x)+1)^2;
intf(f1,x);

% 7.2
syms y;
f2 = exp(2*y)/(exp(y)+2);
intf(f2,y)


% 7.3
syms x a;
f3 = x^2/sqrt(a^2-x^2);
intf(f3,x)

% 7.4
syms x;
f4 = exp(x^(-2));
intf(f4,x)

% 7.5
syms x;
syms a b;
assume(a ~= b);
f5 = 1/x/(sqrt(log(x)+a)+sqrt(log(x)+b));
intf(f5,x)

输出：

fi = $e^{2x}\tan (x)$
经验证，运算结果正确。

fi = $e^y-2\log (e^y+2)$

fi =
$\frac{a^{2}atan(\frac{x}{\sqrt{a^2-x^2}})}{2}-\frac{x\sqrt{a^2-x^2}}{2}$

经验证，运算结果正确。

fi = $\frac{x\text{expint}(\frac{3}{2},-\frac{1}{x^2})}{2}$

fi =
$\sqrt{a+\log (x)}\left(\frac{2a}{3a-3b}+\frac{2\log (x)}{3a-3b}\right)-\sqrt{b+\log (x)}\left(\frac{2b}{3a-3b}+\frac{2\log (x)}{3a-3b}\right)$

8(课本习题10)

计算积分$I(x)={\int }_{-x}^x(x-y{)}^3\sin (x+2y)dy$.

代码：

syms x y;
f=(x-y)^3*sin(x+2*y);
Ix=simplify(int(f,y,-x,x))

输出：

Ix =
$4x^{3}cos(x)-3x^{2}sin(x)+\frac{3{cos(x)}^{2}sin(x)}{2}-\frac{3xcos(x)}{2}$

9(课本习题12)

用solve和vpasolve求解：

(1) $x^2+x+1$;

(2) $3x^5-4x^3+2x-1$;

(3)$5x^{23}-6x^7+8x^6-5x^2$;

(4) $\{\begin{array}{l}a=0.7\sin a+0.2\cos b\\ b=0.7\cos a-0.2\sin b\end{array}$

代码：

syms x;
f1 = x^2+x+1;
f2 = 3*x^5-4*x^3+2*x-1;
f3 = 5*x^23-6*x^7+8*x^6-5*x^2;
syms a b;
f4_1 = a-0.7*sin(a)-0.2*cos(b);
f4_2 = b-0.7*cos(a)+0.2*sin(b);

ans1 = solve(f1)
ans1_vpa = vpasolve(f1)
ans2 = solve(f2)
ans2_vpa = vpasolve(f2)
ans3 = solve(f3)
ans3_vpa = vpasolve(f3)
ans4=solve(f4_1,f4_2);
a = ans4.a, b = ans4.b
ans4_vpa = vpasolve([f4_1,f4_2],[a,b]);
a = ans4_vpa.a, b = ans4_vpa.b

输出：

ans1 =

- (3^(1/2)*1i)/2 - 1/2

(3^(1/2)*1i)/2 - 1/2

ans1_vpa =

- 0.5 - 0.86602540378443864676372317075294i

- 0.5 + 0.86602540378443864676372317075294i

ans2 =

1

root(z^4 + z^3 - z^2/3 - z/3 + 1/3, z, 1)

root(z^4 + z^3 - z^2/3 - z/3 + 1/3, z, 2)

root(z^4 + z^3 - z^2/3 - z/3 + 1/3, z, 3)

root(z^4 + z^3 - z^2/3 - z/3 + 1/3, z, 4)

ans2_vpa =

1.0

- 0.94789546187456058989982247394741 +
0.38447007122004299382156325898354i

…

b =

0.50791971903684924497183722688768

10(课本习题13)

用dsolve求解：

(1) $y^{{}^{\prime }}=x+y,y(0)=1$

(2) $-x^{{}^{\prime }}=2x+3y,y^{{}^{\prime }}=2x+y,x(0)=-2.7,y(0)=2.8$

(3) $y^{{}^{\prime \prime }}-0.01y^{{}^{\prime }2}+2y=\sin t,y(0)=1,y^{{}^{\prime }}(0)=0$

(4) $2x^{{}^{\prime \prime }}(t)-5x^{{}^{\prime }}(t)-3x(t)=90e^{2t},x(0)=2,x^{{}^{\prime }}(0)=1$

(5) $x^{{}^{\prime \prime }}=-\frac{2x^{{}^{\prime }}}{t}+\frac{2x}{t^2}+\frac{10\cos \ln t}{t^2},x(1)=1,x(3)=3.$

代码：

% 10.1
syms y(x)
Dy = diff(y);
eqn = Dy == x + y;
cond = y(0) == 1;
S = dsolve(eqn, cond)

% 10.2
syms x(t) y(t)
eqn1 = diff(x, t) == -2*x - 3*y;
eqn2 = diff(y, t) == 2*x + y;
eqns = [eqn1, eqn2];
cond1 = x(0) == -2.7;
cond2 = y(0) == 2.8;
conds = [cond1, cond2];
S = dsolve(eqns, conds);
Sx = S.x, Sy = S.y

% 10.3 (解不出)
syms y(t)
eqn = diff(y, t, t) - 0.01*(diff(y, t))^2 + 2*y == sin(t);
cond1 = y(0) == 1;
Dy = diff(y);
cond2 = Dy(0) == 0;
conds = [cond1, cond2];
S = dsolve(eqn, conds)

% 10.4
syms x(t)
eqn = 2*diff(x, t, t) - 5*diff(x, t) - 3*x == 90*exp(2*t);
cond1 = x(0) == 2;
Dx = diff(x);
cond2 = Dx(0) == 1;
conds = [cond1, cond2];
S = dsolve(eqn, conds)

% 10.5
syms x(t)
eqn = diff(x, t, t) == -(2*diff(x, t))/t + 2*x/t^2 + (10*cos(log(t)))/t^2;
cond1 = x(1) == 1;
cond2 = x(3) == 3;
conds = [cond1, cond2];
S = dsolve(eqn, conds)

输出：

S = $2e^x-x-1$

Sx =

$\begin{array}{r}\frac{4\sqrt{15}\left(\frac{3e^{-\frac{t}{2}}{\sigma }_1}{4}-\frac{\sqrt{15}{e}^{-\frac{t}{2}}{\sigma }_2}{4}\right)}{25}-\frac{21e^{-\frac{t}{2}}{\sigma }_2}{10}-\frac{7\sqrt{15}{e}^{-\frac{t}{2}}{\sigma }_1}{10}\\ \\ where\\ \\ {\sigma }_1=\sin \left(\frac{\sqrt{15}t}{2}\right)\\ \\ {\sigma }_2=\cos \left(\frac{\sqrt{15}t}{2}\right)\end{array}$

Sy =
$\frac{14e^{-\frac{t}{2}}\cos (\frac{\sqrt{15}t}{2})}{5}-\frac{4\sqrt{15}{e}^{-\frac{t}{2}}\sin (\frac{\sqrt{15}t}{2})}{25}$

警告: Unable to find symbolic solution.

S = [ empty sym ]

S =
$\frac{2e^{-\frac{t}{2}}\left(47e^{\frac{7t}{2}}-63e^{\frac{5t}{2}}+23\right)}{7}$

S = $\begin{array}{r}\frac{\frac{t^3\left(\frac{27\sqrt{10}{\sigma }_1}{26}+\frac{69}{26}\right)}{3}-\sqrt{10}{t}^2\cos (atan(\frac{1}{3})+\log (t))}{t^2}-\frac{\frac{9\sqrt{10}{\sigma }_1}{26}-\frac{81}{26}}{t^2}\\ \\ where\\ \\ {\sigma }_1=\cos (atan(\frac{1}{3})+\log (3))\end{array}$

11

计算导数：$y=\frac{1+\sin x}{1-\cos x}$, $y=\left[\begin{array}{cc}\arcsin x& \arccos x\\ \arctan x& arccotx\end{array}\right]$.

代码：

% 对于函数 y = (1 + sin(x))/(1 - cos(x))
syms x;
y1 = (1 + sin(x))/(1 - cos(x));
y_prime_1 = diff(y1, x)

% 对于向量函数 y = [arcsin(x), arccos(x), arctan(x), arccot(x)]
y2 = [asin(x), acos(x); atan(x), acot(x)];
y_prime_2 = diff(y2, x)

输出：

y_prime_1 = $-\frac{\cos (x)}{\cos (x)-1}-\frac{\sin (x)\left(\sin (x)+1\right)}{{\left(\cos (x)-1\right)}^2}$

y_prime_2 = $(\begin{array}{cc}\frac{1}{\sqrt{1-x^2}}& -\frac{1}{\sqrt{1-x^2}}\\ \frac{1}{x^2+1}& -\frac{1}{x^2+1}\end{array})$

12

计算下列定积分：

$${\int }_0^{\frac{\pi }{4}}\frac{x}{1+\cos 2x}dx,{\int }_0^{1}x{\left(1-x^4\right)}^{\frac{3}{2}}dx$$

代码：

syms x;
f1 = x / (1 + cos(2*x));
int(f1, 0, pi/4)

f2 = x * (1 - x^4)^(3/2);
int(f2, 0, 1)

输出：

ans = $\frac{\pi }{8}-\frac{\log (2)}{4}$

ans = $\frac{3\pi }{32}$

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