matlab语言与应用 03 微积分类

现代科学运算-matlab语言与应用

$\qquad \qquad$ 东北大学 http://www.icourses.cn/home/ (薛定宇)
《高等应用数学问题的MATLAB求解（第四版）》
代码可在matlab r2016b 运行。

03 微积分问题的计算机求解

03.01 极限计算

单变量函数的极限
极限的定义：$L = \lim \limits_{x \to x_0} f(x)$
matlab函数: L = limit(fun, x, $x_0$)
左右极限: $L_1 = \lim \limits_{x \to x_0^-} f(x) \qquad L_2 = \lim \limits_{x \to x_0^+} f(x)$
matlab函数: L = limit(fun, x, $x_0$, ‘left’ or ‘right’)

例3-1 求极限 $\lim \limits_{x \to 0} \dfrac{\sin x}{x}$
三步:声明符号变量、描述函数、直接求解

% matlab 求极限
syms x; f = sin(x)/x; limit(f, x, 0)

例3-2 求极限 $\lim \limits_{x \to \infty} x(1 + \dfrac{a}{x})^x \sin \dfrac{b}{x}$

syms x a b; % 变量之间用空格，不能用,
f = x*(1+a/x)^x*sin(b/x);
limit(f, x, inf)

例3-3 求单边极限 $\lim \limits_{x \to 0^+} \dfrac{e^{x^3} - 1}{1 - \cos{\sqrt{x - \sin x}}}$

syms x;
limit((exp(x^3)-1)/(1-cos(sqrt(x-sin(x)))), x, 0, 'right')

极限及拓展知识

x = -0.1: 0.001: 0.1;
y = (exp(x .^3) - 1) ./ (1-cos(sqrt(x - sin(x))));
plot(x, y, '-', [0], [12], 'o')

拓展:复数的余弦 – Euler公式
$cos{j\alpha} = (e^{\alpha} + e^{-\alpha})/2$

例3-4 正切函数的单边极限
函数 tan t
求关于 $\pi/2$ 点处的左右极限

syms t;
f = tan(t);
L1 = limit(f, t, pi/2, 'left'),
L2 = limit(f, t, pi/2, 'right')
% 间断点

例3-5 序列的极限
给定序列: $\lim \limits _{n \to infty} \dfrac{\sqrt[3]{n^2} \sin{n!}}{n + 1}$

syms n;
f = n^(2/3)*sin(factorial(n))/(n+1);
F = limit(f, n, inf)

例3-6 求序列函数极限
$\lim \limits_{n \to \infty} n \arctan(\dfrac{1}{n(x^2+1) + x}) \tan^n(\dfrac{\pi}{4} + \dfrac{x}{2n})$

syms x, n;
f = n*atan(1/(n*(x^2+1)+x))*tan(pi/4+x/2/n)^n;
limit(f, n, inf)

03.02 区间极限与多变量极限

例3-7 求极限 $\lim \limits_{x \to \infty} x^n, \lim \limits_{n \to \infty} x^n$

syms x n real;
f = x^n;
L1 = limit(f, x, inf), L2 = limit(f, n, inf)

MuPAD的底层limit函数
底层limit函数可以求解区间极限问题
L = feval(symengine, ‘limit’, f, ‘x=infinity’, ‘Intervals’)

例3-8 a,b > 0,求 $\lim \limits_{x \to \infty} a \sin{(8x^2)} + b\cos{(2x-2)}$

syms a b positive, syms x;
f = a*sin(8*x^2)+b*cos(2*x-2);
L=feval(symengine, 'limit', f, 'x=infinity', 'Intervals')

分段函数的解析描述

% 分段函数的解析描述
function f=piecewise(varargin), str=[];
try
    for i=1:2:length(varargin),
        str = [str, '[', varargin{i}, ',', varargin[i+1], '],'];
    end
catch
    error('Input arguments should be given in pairs.'),
end
f = feval(symengine, 'piecewise', str(1:end-1));
% 调用格式
% f = piecewise(var1, var2, ...)

例3-9 分段函数绘图
分段函数 $y = \left \{ \begin{array}{l}1.1 \mathbb{sign}(x), |x| > 1.1 \\ x, \qquad \qquad |x| \leq 1.1 \end{array} \right.$

f = piecewise('abs(x)>1.1', ...
    '1.1*sign(x)', 'abs(x) <= 1.1', 'x');
syms x;
x0 = -3: 0.01: 3;
f1 = subs(f, x, x0); 
plot(x0, f1)

多变量函数的极限
函数f(x, y)的累极限
$L_1 = \lim \limits_{x \to x_0} \Big[ \lim \limits_{y \to y_0} f(x, y) \Big], L_2 = \lim \limits_{y \to y_0} \Big[ \lim \limits_{x \to x_0} f(x, y) \Big]$
累极限求解
$L_1$ = limit(limit(f, y, $y_0$), x, $x_0$)
$L_2$ = limit(limit(f, x, $x_0$), y, $y_0$)
积分顺序不同，结果可能不同

例3-10 二元函数的累极限
$\lim \limits_{y \to \infty} \Big[ \lim \limits_{x \to {1/\sqrt{y}}} e^{-1/(y^2+x^2)}\dfrac{\sin^2 x}{x^2}(1 + \dfrac{1}{y^2})^{x+a^2y^2} \Big]$

syms x a;
syms y positive;
f = exp(-1/(y^2+x^2))*sin(x)^2/x^2 ...
    *(1+1/y^2)^(x+a^2*y^2);
L = limit(limit(f, x, 1/sqrt(y)), y, inf)

多元函数重极限
数学表达式 $L = \lim \limits_{x \to x_0 \\ y \to y_0} f(x, y)$
不易求解
理论上，沿所有方向均得出相同极限才可
不能用累极限方法求解
累极限存在单不相等，没有重极限
有时累极限存在且相等，但无重极限

例3-11 计算重极限 $\lim \limits_{x \to \infty \\ y \to \infty}(\dfrac{xy}{x^2+y^2})^{x^2}$
试图按4个不同方向和速度求

syms x y;
f = (x*y/(x^2+y^2))^(x^2);
L1 = limit(limit(f, x, inf), y, inf),
L2 = limit(limit(f, y, inf), x, inf)
L3 = limit(limit(f, x, y^2), y, inf),
L4 = limit(limit(f, y, x^2), x, inf)

例3-12 判断重极限是否存在 $\lim \limits_{x \to 0 \\ y \to 0} \dfrac{xy}{x^2+y^2}$
沿y = rx 方向趋近

syms r x y;
f = x*y/(x^2+y^2);
L = limit(subs(f, y, r*x), x, 0)
% L = r/(r^2 + 1),r不同，极限不同，重极限不存在

03.03 单变量函数求导

函数的导数和高阶导数
函数求导的三个步骤 $\dfrac{d f(x)}{dx} = \lim \limits_{h \to 0} \dfrac{f(t+h) - f(t)}{h}$
声明变量、输入函数表达式、调用diff()函数直接求导
函数语法$\dfrac{d f(x)}{dx}$: y = diff(fun, x)
或 $\dfrac{d^n f(x)}{d x^n}$: y = diff(fun, x, n)
自变量为唯一符号变量，可以省去x

例3-13 一阶导数及其曲线
给定函数$f(x) = \dfrac{\sin x}{x^2 + 4x + 3}$, 求$\dfrac{d^4 f(x)}{d x^4}$

% 求1阶导数和4阶导数
syms x;
f = sin(x)/(x^2 + 4*x + 3);
f1 = diff(f)
f4 = diff(f, x, 4)
ezplot(f, [0, 5]),
hold on; ezplot(f1, [0, 5]),
hold on; ezplot(f4, [0, 5])
hold off;
% 化简
ff1 = collect(simplify(f4), sin(x))
ff4 = collect(simplify(f4), cos(x))
% 函数diff()的高效率
tic, diff(f, x, 50); toc

例3-14 复合泛函求导
已知函数$F(t) = t^2 \sin t f(t)$
推导其3阶导函数公式
当$f(t) = e^{-t}$时,求3阶导数
难点:如何定义f(t) – syms f(t)

syms t f(t);
G = simplify(diff(t^2*sin(t)*f, t, 3))
% f(t) = e^{-t}
simplify(subs(G, f, exp(-t))),
% 验证
simplify(diff(t^2*sin(t)*exp(-t), 3) -ans)

例3-15 矩阵函数求导
矩阵函数$H(x) = \begin{bmatrix}4\sin 5x & e^{-4x^2} \\ 3x^2 + 4x + 1 & \sqrt{4x^2 + 2} \end{bmatrix}$
直接求解，对每个矩阵元素直接求导

syms x;
H=[4*sin(5*x), exp(-4*x^2);
  3*x^2+4*x+1, sqrt(4*x^2+2)];
H13 = diff(H, x, 3)

参数方程的导数
参数方程: $y = f(t), x = g(t)$
由递推公式求: $\dfrac{d^n y}{d x^n}$
$\dfrac {dy}{dx} = \dfrac{f^{\prime}(t)}{g^{\prime}(t)}$
$\dfrac{d^2 y}{d x^2} = \dfrac{d}{dt} \Big(\dfrac{f^{\prime}(t)}{g^{\prime}(t)} \Big) \dfrac{1}{g^{\prime}(t)} = \dfrac{d}{dt} \Big(\dfrac{dy}{dx} \Big) \dfrac{1}{g^{\prime}(t)}$
$\vdots$
$\dfrac{d^n y}{dx^n} = \dfrac{d}{dt} \Big(\dfrac{d^{n-1} y}{dx^{n-1}} \Big) \dfrac{1}{g^{\prime}(t)}$

参数方程求导的递归实现

% 参数方程求导的递归实现函数
function result = paradiff(y, x, t, n)
  if mod(n, 1) ~= 0
    error('n should positive integer, please correct'),
  else
    if n == 1
      result = diff(y, t)/diff(x, t);
    else
      result = diff(paradiff(y, x, t, n-1), t)/diff(x, t);
    end;
  end;
end
% y1 = paradiff(y, x, t, n)

例3-16 参数方程求导
已知参数方程 $y = \dfrac{\sin t}{(t+1)^3}, x = \dfrac{\cos t}{(t+1)^3}$, 求$\dfrac{d^3 y}{d x^3}$。

syms t;
y = sin(t)/(t+1)^3;
x = cos(t)/(t+1)^3;
f = paradiff(y, x, t, 3);
[n, d] = numden(f); % 提取分子、分母
F = simplify(n)/simplify(d) %分子、分母分别化简，再相除

03.04 偏导数计算

多元函数偏导数
双变量函数$f(x, y)$
求高阶偏导数: $\dfrac{\partial^{m+n}}{\partial x^m \partial y^n}f(x, y)$
matlab语法:
f = diff(diff(fun, x, m), y, n)
或 f = diff(diff(fun, y, n), x, m)

例3-17 一阶偏导数
二元函数偏导数图形表示
$z = f(x, y) = (x^2 - 2x)e^{-x^2 - y^2 - xy}$
求两个偏导数:$\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}$

syms x y;
z = (x^2 - 2*x)*exp(-x^2-y^2-x*y);
zx = simplify(diff(z, x, 1)),
zy = diff(z, y, 1)

梯度函数图形表示
绘制三维曲面

[x0, y0] = meshgrid(-3: 0.2: 2, -2:  .2: 2);
z0 = double(subs(z, {x, y}, {x0, y0}));
surf(x0, y0, z0), zlim([-0.7 1.5])

引力线(负梯度)

contour(x0, y0, z0, 30), hold on;
zx0 = subs(zx, {x, y}, {x0, y0});
zy0 = subs(zy, {x, y}, {x0, y0});
quiver(x0, y0, -zx0, -zy0)

例3-18 三元函数求偏导
三元函数 $f(x, y, z) = \sin(x^2y) e^{-x^2y-z^2}$
求偏导数：$\dfrac{\partial ^4 f(x, y, z)}{\partial x^2 \partial y \partial z}$

syms x y z;
f = sin(x^2*y)*exp(-x^2*y-z^2);
df = diff(diff(diff(f, x, 2), y), z);
df = simplify(df)

隐函数偏导数
隐函数: $f(x_1, x_2, \cdots, x_n) = 0$
一阶偏导数: $\dfrac{\partial x_i}{\partial x_j} = -\dfrac{\dfrac{\partial }{\partial x_j} f(x_1, x_2, \cdots, x_n)}{\dfrac{\partial }{\partial x_i}f(x_1, x_2, \cdots, x_n)}$
matlab代码: F = - diff(f, $x_j$) / diff(f, $x_i$)

二元隐函数的高阶偏导数
二元函数 f(x, y)
一阶偏导数: $\partial y / \partial x = F_1(x, y)$
二阶偏导数:
$F_2(x, y) = \dfrac{\partial ^2 y}{\partial x^2} = \dfrac{\partial F_1(x, y)}{\partial x} + \dfrac{\partial F_1(x, y)}{\partial y} F_1(x, y)$
高阶偏导数:
$F_n(x, y) = \dfrac{\partial ^n y}{\partial x^n} = \dfrac{\partial F_{n-1}(x, y)}{\partial x} + \dfrac{\partial F_{n-1}(x,y)}{\partial y}F_1(x, y)$

高阶偏导数求解
递推公式: $F_n(x, y) = \dfrac{\partial^n y}{\partial x^n} = \dfrac{\partial F_{n-1}(x, y)}{\partial x} + \dfrac{\partial F_{n-1}(x, y)}{\partial y} F_1(x, y)$
matlab求解: $f_1$ = impldiff(f, x, y, n)

% 求高阶偏导数
function dy = impldiff(f, x, y, n)
  if mod(n, 1) ~= 0
    error('n should positive integer, please correct')
  else
    F1 = -simplify(diff(f, x)/diff(f, y));
    dy = F1;
    for i=2:n
      dy = simplify(diff(dy, x) + diff(dy, y)*F1);
    end;
  end;
end

例3-19 隐函数求导
二元隐函数 $z = f(x, y) = (x^2 - 2x) e^{-x^2-y^2-xy}$

% 一阶偏导数
syms x y;
f = (x^2-2*x)*exp(-x^2-y^2-x*y);
F1 = impldiff(f, x, y, 1)
% 二阶偏导数
F2 = impldiff(f, x, y, 2)
% 三阶偏导数
F3 = impldiff(f, x, y, 3)
[n, d] = numden(F3), simplify(n)

例3-20 隐函数求导与简化
隐函数: $x^2 + xy + y^2 = 3$
各阶导数:

f = x^2 + x*y + y^2 - 3;
F1 = impldiff(f, x, y, 1)
f2 = impldiff(f, x, y, 2);
F2 = subs(f2, x^2+x*y+y^2, 3)
f3 = impldiff(f, x, y, 3);
F3 = subs(f3, x^2+x*y+y^2, 3)
f4 = impldiff(f, x, y, 4);
F4 = subs(f4, x^2+x*y+y^2, 3)

多元函数的Jacobi矩阵
多元函数
$\left \{ \begin{array}{l}y_1 = f_1(x_1, x_2, \cdots, x_n) \\ y_2 = f_2(x_1, x_2, \cdots, x_n) \\ \vdots \quad \vdots \\ y_m = f_m(x_1, x_2, \cdots, x_n) \end{array} \right.$
Jacobi矩阵
$J = \begin{bmatrix} \partial y_1 / \partial x_1 & \partial y_1 / \partial x_2 & \cdots & \partial y_1 / \partial x_n \\ \partial y_2 / \partial x_1 & \partial y_2 / \partial x_2 & \cdots & \partial y_2 / \partial x_n \\ \vdots & \vdots & \ddots & \vdots \\ \partial y_m / \partial x_1 & \partial y_m / \partial x_2 & \cdots & \partial y_m / \partial x_m \end{bmatrix}$
matlab求解: J = jacobian(y, x)

例3-21 Jacobi矩阵
直角坐标和极坐标变化公式
$x = r \sin \theta \cos \phi, y = r \sin \theta \sin \phi, z = r \cos \theta$
求 Jacobi矩阵

syms r theta phi;
x = r*sin(theta)*cos(phi);
y = r*sin(theta)*sin(phi);
z = r*cos(theta);
J = jacobian([x;y;z], [r theta phi])

Hessian偏导数矩阵
给定n元函数 $f(x_1, x_2, \cdots, x_n)$
Hessian矩阵
$H = \begin{bmatrix}\partial^2 f /\partial x_1^2 & \partial^2 f / \partial x_1 \partial x_2 & \cdots & \partial^2 f / \partial x_1 \partial x_n \\ \partial^2 f / \partial x_2 \partial x_1 & \partial^2 f / \partial x_2^2 & \cdots & \partial^2 f / \partial x_2 \partial xn \\ \vdots & \vdots & \ddots & \vdots \\ \partial^2 f / \partial x_n \partial x_1 & \partial^2 f / \partial x_n \partial x_2 & \cdots & \partial^2 f / \partial x_n^2 \end{bmatrix}$
matlab：H = hessian(f, x)
早期matlab版本: H = jacobian(jacobian(f, x), x)

例3-22 Hessian矩阵
试求二元函数的Hessian矩阵
$z = f(x, y) = (x^2-2x)e^{-x^2-y^2-xy}$

syms x y;
f = (x^2-2*x)*exp(-x^2-y^2-x*y);
H = simplify(hessian(f, [x, y]))

% 另一种方法
X = [x, y];
H1 = simplify(jacobian(jacobian(f, X), X))

03.05 积分运算

* 求不定积分*
不定积分: $\int f(x) dx, \int \cdots \int f(x) dx^n$
matlab函数:
积分: F = int(fun, x)
多重积分，嵌套调用;更多重应该用循环
F = int(… int(fun, x))
积分得出的是原函数，结果再加C

例3-23 先求导再积分还原
函数: $f(x) = \dfrac{\sin x}{x^2 + 4x + 3}$
求其一阶导数，再积分

syms x; 
y = sin(x) / (x^2 + 4*x + 3);
y1 = diff(y);
y0 = int(y1)

% 求其4阶导数，再积分，检验结果
y4 = diff(y, 4);
y0 = int(int(int(int(y4))));
simplify(y0)

例3-26 积分公式证明
试证明: $\int x^3 \cos^2 ax dx = \dfrac{x^4}{8} + (\dfrac{x^3}{4a} - \dfrac{3x}{8a^3})\sin 2ax + (\dfrac{3x^2}{8a^2} - \dfrac{3}{16a^4}) \cos 2ax + C$

% 对等号左侧进行化简
syms a x;
f = simplify(int(x^3*cos(a*x)^2, x));
% 输入右侧，比较并化简
f1 = x^4/8 + (x^3/(4*a) - 3*x/(8*a^3))*sin(2*a*x) + (3*x^2/(8*a^2) - 3/(16*a^4))*cos(2*a*x);
simplify(f-f1)

例3-27 不可积函数的不定积分计算
考虑如下两个不可积问题
$f(x) = e^{-x^2/2}, g(x) = x\sin(ax^4)e^{x^2/2}$
matlab求解:

syms x;
I = int(exp(-x^2/2))

结果: $I = \dfrac{erf(\sqrt{2}x)}{\sqrt{2 \pi}}$
特殊函数: $erf(x) = \dfrac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt$

含参数的不可积函数
求解$\int g(x) dx$, 其中 $g(x) = x \sin(ax^4) e^{x^2/2}$

syms a x;
int(x*sin(a*x^4)*exp(x^2/2))
% 并不是所有的积分都能被计算出，因为原始函数不一定存在
% ans = int(x*sin(a*x^4)*exp(x^2/2), x)

定积分与无穷积分计算
定积分与无穷积分
$\int_a^b f(x) dx \qquad \int_a^{\infty} f(x) dx$
语句格式:
I = int(fun, x, a, b)
I = int(fun, x, a, inf)
有时需要Newton-Leibniz公式 - 原函数 F(x)
$\int_a^b f(x) dx = F(b) - F(a)$

例3-28 定积分求解
函数 $f(x) = e^{-x^2/2}$
求a = 0, b = 1.5 或 $\infty$ 时的定积分值

syms x;
I1 = int(exp(-x^2/2), x, 0, 1.5)
vpa(I1, 70),
I2 = int(exp(-x^2/2), x, 0, inf)

例3-30 反常积分计算
数学问题 $\int_1^{2e} \dfrac{1}{x\sqrt{1 - \ln ^2 x}} dx$

syms x;
f = 1/(x*sqrt(1 - log(x)^2));
I = int(f, x, 1, 2*exp(sym(1)))

多重积分问题
多重积分
$I = \int _{x_m}^{x_M} \int_{y_m(x)}^{y_M(x)} \int_{z_m(x, y)}^{z_M(x,y)} f(x, y, z) dzdydx$
利用int()函数直接求解
注意：某些问题需要根据实际情况先选择积分顺序，可积的部分作为内积分，然后再处理外积分，否则会的不出解析解

例3-32 三元函数积分
已知三元函数
$F(x, y, z) = -4ze^{-x^2y-z^2}[\cos x^2y - 10yx^2 \cos x^2y + 4x^4y^2 \sin x^2y + 4x^4y^2 \cos x^2 y - \sin x^2y]$
试求 $\int \cdots \int F(x, y, z) dx^2 dy dz$

syms x y z;
f0 = -4*z*exp(-x^2*y-z^2)*(cos(x^2*y) - 10*cos(x^2*y)*y*x^2 ...
    +4*sin(x^2*y)*x^4*y^2 + 4*cos(x^2*y)*x^4*y^2 - sin(x^2*y));
f1 = int(f0, z);
f1 = int(f1, y);
f1 = int(f1, x);
f1 = simplify(int(f1, x))
% f1 = sin(x^2*y)*exp(- y*x^2 - z^2)

例3-33 多重积分
求解积分问题 $\int_0^2 \int_0^{\pi} \int_0^{\pi} 4xze^{-x^2y-z^2} dzdydx$

syms x y z;
I = int(int(int(4*x*z*exp(-x^2*y-z^2), z, 0, pi), y, 0, pi), x, 0, 2)
% I=-(2*exp(-pi^2) - 2)*(eulergamma/2 + log(2) + log(pi)/2 - ei(-4*pi)/2)

注意:
eulergamma 为euler常数$\gamma$
$Ei(n, z) = \int_1^{\infty} e^{-zt}t^{-n} dt$

03-06 Fourier级数逼近

Fourier级数展开
给定周期函数 $f(x), x in [-L, L], T = 2L$
满足: $f(x) = f(x + kT)$
一般 $(a, b)$ 区间，变量替换 $L = \dfrac{b-a}{2}, x_1 = x + L + a$
Fourier级数展开
$f(x) = \dfrac{a_0}{2} + \sum \limits_{n=1}^{\infty} (a_n \cos \dfrac{n \pi}{L} x + b_n \sin \dfrac{n \pi}{L} x)$
系数: $\left \{ \begin{array}{l} a_n = \dfrac{1}{L} \int_{-L}^{L} f(x) \cos \dfrac{n\pi x}{L} dx, n = 0, 1, 2, \cdots \\ b_n = \dfrac{1}{L} \int_{-L}^{L} f(x) \sin \dfrac{n \pi x}{L} dx, n = 1, 2, 3, \cdots \end{array} \right.$

Fourier级数展开函数编写
求解 [F, A, B] = fseries(f, x, p, a, b)

% 求Fourier级数展开的matlab代码
function [F, A, B] = fseries(f, x, varargin)
  [p, a, b] = default_vals({6, -pi, pi}, varargin{:});
  L = (b-a)/2;
  if a+b
    f = subs(f, x, x+L+a);
  end;
  A = int(f, x, -L, L)/L;
  B = [];
  F = A/2;
  for n = 1:p
    an = int(f*cos(n*pi*x/L), x, -L, L)/L;
    A = [A, an];
    bn = int(f*sin(n*pi*x/L), x, -L, L)/L;
    B = [B, bn];
    F = F + an*cos(n*pi*x/L) + bn*sin(n*pi*x/L);
  end;
  if a+b
    F = subs(F, x, x-L-a);
  end;
end

function varargout = default_vals(vals, varargin)
  if nargout ~= length(vals)
    error('number of arguments mismatch');
  else
    nn = length(varargin) + 1;
    varargout = varargin;
    for i = nn:nargout
      varargout{i} = vals{i};
    end;
  end;
end

例3-33 函数的Fourier近似
函数 $y = x(x-\pi)(x - 2\pi), x \in (0, 2\pi)$
求它的Fourier级数展开

syms x;
f = x*(x-pi)*(x - 2*pi);
[F, A, B] = fseries(f, x, 12, 0, 2*pi);
F
%比较结果
ezplot(f, [0, 2*pi]),
hold on,
ezplot(F, [0, 2*pi]), hold off;

展开结果和效果
更大的区间 $x \in (-\pi, 3\pi)$

ezplot(f, [-pi, 3*pi]),
hold on,
ezplot(F, [-pi, 3*pi]),
hold off;

例3-34 分段函数的近似
给定函数 $y = \left \{ \begin{array}{l} 1 \quad x \geq 0 \\ -1 \quad x < 0 \end{array} \right. \quad x \in (-\pi, \pi)$
原函数可以表示成 $f(x) = \dfrac{|x|}{x}$

syms x;
f = abs(x)/x;
xx = [-pi: pi/200: pi];
xx = xx(xx ~= 0);
xx = sort([xx, -eps, eps]);
yy = subs(f, x, xx);
plot(xx, yy), 
hold on;
for n = 1: 20
  [f1, a, b] = fseries(f, x, n);
  y1 = subs(f1, x, xx);
  plot(xx, y1)
end;

展开效果与函数
前14项的Fourier级数展开

[f1, a, b] = fseries(f, x, 14); f1

数学形式:
$f(x) \approx 4\dfrac{sin x}{\pi} + \dfrac{4 \sin 3x}{3\pi} + \dfrac{4 \sin 5x}{5 \pi} + \dfrac{4 \sin 7x}{7 \pi} + \dfrac{4 \sin 9x}{9 \pi} + \dfrac{4 \sin 11x}{11 \pi} + \dfrac{4 \sin 13x}{13 \pi}$
一般形式：$f(x) = \dfrac{4}{\pi} \sum \limits_{k=1}^{\infty} \dfrac{\sin(2k-1)x}{2k-1}$

周期函数假设
在区间$[-2\pi, 2\pi]$进行拟合效果比较

xx = [-2*pi: pi/200:2*pi];
xx = xx(xx ~= 0);
xx = sort([xx, -eps, eps]);
yy = subs(f, x, xx);
plot(xx, yy),
hold on;
y1 = subs(f1, x, xx);
plot(xx, y1),
hold off;

开始假设: $f(t) = f(kT + t)$

03.07 Taylor级数逼近

单变量函数的Taylor幂级数展开
数学表示
在 x = 0 点附近的Taylor 幂级数
$f(x) = a_1 + a_2x + a_3x^2 + \cdots + a_kx^{k-1} + o(x^k)$
其中 $a_i = \dfrac{1}{i!} \lim \limits_{x \to 0} \dfrac{d^{i-1}}{d x^{i-1}}f(x), i = 1, 2, 3, \cdots$
matlab格式： F = taylor(fun, x, ‘Order’, k)
早期版本: F = taylor(fun, x, k)

关于 x = a 的 Taylor 展开
关于 x = a 点的 Taylor 展开
$f(x) = b_1 + b_2(x-a) + b_3(x-a)^2 + \cdots + b_k(x-a)^{k-1} + o([(x-a)^k]$
其中 $b_i = \dfrac{1}{i!} \lim \limits_{x \to a} \dfrac{d^{i=1}}{d x^{i-1}}f(x), i = 1, 2, 3, \cdots$
matlab格式: F = taylor(fun, x, a, ‘Order’, k)
早期版本: F = taylor(fun, x, a, k)

例3-35 函数的Taylor展开
函数 $f(x) = \dfrac{\sin x}{x^2 + 4x + 3}$
在 x = 0, x = 2 和 x = a 处的 Taylor 展开.

% 在 x = 0 的 Taylor 展开
syms x;
f = sin(x)/(x^2 + 4*x + 3);
y = taylor(f, x, 'Order', 9)
% 在x = 2 的 Taylor 展开
y = taylor(f, x, 2, 'Order', 9)
% 在x = a 的 Taylor 展开
syms a;
y = taylor(f, x, a, 'Order', 9)

近似效果

% 在 x = 0 的 Taylor 展开
syms x;
f = sin(x)/(x^2 + 4*x + 3);
y = taylor(f, x, 'Order', 9)
% 在区间[-1, 1]内绘图
ezplot(f, [-1, 1]),
hold on;
h = ezplot(y, [-1, 1]);
set(h, 'Color', 'r'),
hold off;
% 更小的区间
figure;
ezplot(f, [-0.6, 0.6]),
hold on;
h = ezplot(y, [-0.6, 0.6]);
set(h, 'Color', 'r');
hold off;

例3-36 正弦函数的幂级数逼近
原函数 $y(t) = \sin(t)$

syms x;
y = sin(x);
ezplot(y),
hold on;
for n = [6: 2: 16]
  p = taylor(y, x, 'Order', n),
  ezplot(p);
end;
hold off;

多变量函数的Taylor级数展开
多元函数 $f(x_1, x_2, \cdots, x_n)$
Taylor幂级数展开
$f(x) = f(a) + [(x_1 - a_1) \dfrac{\partial}{\partial x_1} + \cdots + (x_n - a_n)\dfrac{\partial}{\partial x_n}] f(x) |_{x = a} +$
$\qquad \dfrac{1}{2!}[(x_1-a_1)\dfrac{\partial}{\partial x_1} + \cdots + (x_n-a_n)\dfrac{\partial}{\partial x_n}]^2 f(x)|_{x = a} + \cdots +$
$\qquad \dfrac{1}{k!}[(x_1-a_1)\dfrac{\partial}{\partial x_1} + \cdots + (x_n-a_n)\dfrac{\partial}{\partial x_n}]^k f(x)|_{x=a} + \cdots$
F = taylor(f, [$x_1, x_2, \cdots, x_n$], [$a_1, x_2, \cdots, a_n$], ‘Order’, k)

例3-37 二元函数Taylor展开
函数 $z = f(x, y) = (x^2 - 2x)e^{-x^2-y^2-xy}$

% 在原点[0, 0]展开Taylor级数
syms x y;
f = (x^2-2*x)*exp(-x^2-y^2-x*y);
F = taylor(f, [x, y], [0, 0], 'Order', 9);
F1 = collect(F, x)
% 关于在(1, a)点展开
syms a;
F = taylor(f, [x,y], [1, a], 'Order', 9);
F2 = simplify(F)

03.08 级数求和与序列求积

级数求和
已知通项的有穷或无穷级数的和
数学表示: $S = \sum \limits_{k=k_0}^{k_n} f_k$
matlab语句: S = symsum($f_k, k, k_0, k_n$)

例子3-38 由古老的故事引出级数求和问题
计算 $S = 2^0 + 2^1 2^2 + \cdots + 2^62 + 2^63 = \sum \limits_{i=0}^{63} 2^i$

format long;
sum(2 .^ [0:63])
% 使用symsum()
syms k;
symsum(2^k, 0, 63)
% 更多项的扩展
sum(sym(2) .^ [0:200])

例3-39 级数求和
求无穷级数的和
$S = \dfrac{1}{1 \times 4} + \dfrac{1}{4 \times 7} + \cdots + \dfrac{1}{(3n-2)(3n+1)} + \cdots$

syms n;
s = symsum(1/((3*n-2)*(3*n+1)), n, 1, inf)
% 使用数值方法
m = 1:10000000;
s1 = sum(1 ./((3*m-2) .* (3*m+1)));
format long;
s1

例3-40 级数函数的求和
试求解含有变量x的无穷级数
$J = 2 \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)(2x+1)^{2n+1}}$
符号运算方法
早期版本不能给出级数的收敛区域

syms x n;
s1 = symsum(2/((2*n+1)*(2*x+1)^(2*n+1)), n, 0, inf);
simplify(s1)
% piecewise(0 < x | x < -1, 2*atanh(1/(2*x + 1)))

例3-41 综合问题
试求级数与极限综合问题
$\lim \limits_{n \to \infty} [(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \cdots + \dfrac{1}{n}) - \ln n]$

syms m n;
res = limit(symsum(1/m, m, 1, n) - log(n), n, inf)
vpa(res)
% 注意：该问题不能先求解无穷级数的和，在减去 ln n，都是无穷大
% ans = eulergamma

例3-42 级数的极限
$S = \lim \limits_{n \to \infty} [(1+\dfrac{1}{n^2}\sin \dfrac{\pi}{n^2} + \cdots + (1+\dfrac{n-1}{n^2}) \sin \dfrac{(n-1) \pi}{n^2}]$
通项提取: $(1 + \dfrac{k}{n^2})\sin \dfrac{k \pi}{n^2}, k = 1, 2, \cdots, n-1$

syms n k;
S = simplify(limit(symsum((1+k/n^2)*sin(k*pi/n^2), k, 1, n-1), n, inf))

序列求积问题
序列求积运算：$P = \prod \limits_{n=a}^{b} f(n)$
matlab语句:
新符号运算工具箱: P = symprod($f_n$, n, a, b)
早期版本: P = maple(‘product’, fun, ‘n=a..b’)

例3-43 序列求积
计算序列乘积：$P_n = \prod \limits_{k=1}^n (1 + \dfrac{1}{k^3})$

% 序列求积
syms k n;
P1 = symprod(1 + 1/k^3, k, 1, n);
P1 = simplify(P1)
P2 = symprod(1 + 1/k^3, 1, inf);
P2 = simplify(P2)

例3-44 序列求积
$S = 1 - \dfrac{1}{2} + \dfrac{1 \times 3}{2 \times 4} - \dfrac{1 \times 3 \times 5}{2 \times 4 \times 6} + \dfrac{1 \times 3 \times 5 \times 7}{2 \times 4 \times 6 \times 8} - \dfrac{1 \times 3 \times 5 \times 7 \times 9}{2 \times 4 \times 6 \times 8 \times 10} + \cdots$
通项: $(-1)^n \prod \limits_{k=1}^{n} \dfrac{2k-1}{2k}, k = 1, 2, \cdots, \infty$

syms k n;
S = 1 + symsum((-1)^n*symprod((2*k-1)/(2*k), k, 1, n), n, 1, inf)
% 2^(1/2)/2

03.09 曲线积分与曲面积分

第一类曲线积分
第一类曲线积分：$I_1 = \int_l f(x, y, z) ds$
曲线l满足参数方程 x = x(t), y = y(t), z = z(t)
$ds = \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2 + (\dfrac{dz}{dt})^2} dt$
转换成定积分问题
$I = \int_{t_m}^{t_M} f[x(t), y(t), z(t)] \sqrt{x_t^2 + y_t^2 + z_t^2}dt$

曲线积分函数

% 通用曲线积分函数
function I = path_integral(F, vars, t, a, b)
  if length(F) == 1
    I = int(F*sqrt(sum(diff(vars, t) .^2)), t, a, b);
  else
    F = F(:) .';
    vars = vars(:);
    I = int(F*diff(vars, t), t, a, b);
  end;
end

调用格式:
I = path_integral(f, [x, y], t, $t_m, t_M$)
I = path_integral(f, [x, y, z], t, $t_m, t_M$)

例3-49 曲线积分求解
计算 $\int _l \dfrac{z^2}{x^2 + y^2} ds, l$ 是如下定义的螺线
$\left \{ \begin{array}{l}x = a \cos t, \\ y = a \sin t, \\ z = at, \end{array} \right. 0 \leq t \leq 2\pi, a > 0$
公式: $I = \int_{t_m}^{t_M} f[x(t), y(t), z(t)] \sqrt{x_t^2 + y_t^2 + z_t^2} dt$

% 计算曲线积分
syms t;
syms a positive;
x = a*cos(t);
y = a* sin(t);
z = a*t;
f = z^2/(x^2+y^2);
I = path_integral(f, [x, y, z], t, 0, 2*pi)
% (8*2^(1/2)*a*pi^3)/3

第二类曲线积分
第二类曲线积分 $I_2 = \int_l \vec f(x, y, z) \cdot d \vec s$
其中 $\vec f(x, y, z) = [P(x, y, z), Q(x, y, z), R(x, y, z)]$
并且: $d \vec s = [\dfrac{dx}{dt}, \dfrac{dy}{dt}, \dfrac{dz}{dt}] ^Tdt$
上式化为: $\int_a^b [P(x, y, z), Q(x, y, z), R(x, y, z)] [\dfrac{dx}{dt}, \dfrac{dy}{dt}, \dfrac{dz}{dt}] ^T dt$
matlab语句：
I = path_integral([P, Q], [x, y], t, a, b)
I = path_integral([P, Q, R], [x, y, z], t, a, b)
I = path_integral(F, v, t, a, b)

例3-51 第二类曲线积分
曲线积分：$\int_l \dfrac{x+y}{x^2 + y^2} dx - \dfrac{x-y}{x^2+y^2} dy$
$l 为正向圆周x^2 + y^2 = a^2$
正向圆周的参数函数描述
$x = a \cos t, y = a \sin t, (0 \leq t \leq 2\pi)$

syms t;
syms a positive;
x = a * cos(t);
y = a * sin(t);
F = [(x+y)/(x^2+y^2), -(x-y)/(x^2+y^2)];
I = path_integral(F, [x, y], t, 2*pi, 0)

曲面积分

% 曲面积分函数
function I=surf_integral(f,xx,uu,um,vm)
if length(f)==1 % scalar integrand
   if length(xx)==1 % surface described by an explicit function
      I=int(int(f*sqrt(1+diff(xx,uu(1))^2+diff(xx,uu(2))^2),...
            uu(2),um(1),um(2)),uu(1),vm(1),vm(2));
   else   % surface described by an implicit function
      xx=[xx(:).' 1]; x=xx(1); y=xx(2); z=xx(3); u=uu(1); v=uu(2);
      E=diff(x,u)^2+diff(y,u)^2+diff(z,u)^2;
      F=diff(x,u)*diff(x,v)+diff(y,u)*diff(y,v)+diff(z,u)*diff(z,v);
      G=diff(x,v)^2+diff(y,v)^2+diff(z,v)^2;
      I=int(int(f*sqrt(E*G-F^2),u,um(1),um(2)),v,vm(1),vm(2));
   end
else % vector integrand
   if length(xx)==1 % surface described by an explicit function
      syms x y z; ua=sqrt(1+diff(xx,x)^2+diff(xx,y)^2);
      cA=-diff(xx,x)/ua; cB=-diff(xx,y)/ua; cC=1/ua;
      I=surf_integral(f(:).'*[cA; cB; cC],xx,uu,um,vm);
   else, x=xx(1); y=xx(2); z=xx(3); u=uu(1); v=uu(2);
      A=diff(y,u)*diff(z,v)-diff(z,u)*diff(y,v);
      B=diff(z,u)*diff(x,v)-diff(x,u)*diff(z,v);
      C=diff(x,u)*diff(y,v)-diff(y,u)*diff(x,v); F=A*f(1)+B*f(2)+C*f(3);
      I=int(int(F,uu(1),um(1),um(2)),uu(2),vm(1),vm(2));
    end;
end

第一类曲面积分
I = surf_integral($f, z, [x, y], [y_m, y_M], [x_m, x_M]$)
I = surf_integral($f, [x, y, z], [u, v], [u_m, u_M], [v_m, v_M]$)
第二类曲面积分
I = surf_integral($[P, Q, R], z, [u, v], [u_m, u_M], [v_m, v_M]$)
I = surf_integral($[P, Q, R], [x, y, z], [u, v], [u_m, u_M], [v_m, v_M]$)

例3-54 曲面积分
曲面积分 $\iint (x^2y + zy^2) ds$
积分曲面
$x = u \cos v, y = u \sin v, z = v, 0 \leq u \leq a, 0 \leq v \leq 2 \pi$

syms u v;
syms a positive;
x = u * cos(v);
y = u * sin(v);
z = v;
f = x^2*y + z*y^2;
I = surf_integral(f, [x, y, z], [u, v], [0, a], [0, 2*pi])

03.10 数值微分

一阶数值微分算法
导数定义: $y^{\prime}(t) = \lim \limits_{\Delta t \to 0} \dfrac{y(t+\Delta t) - y(t)}{\Delta t}$
前向差分公式: $y_i^{\prime} \approx \dfrac{\Delta y_i}{\Delta t} = \dfrac{y_{i+1} - y_i}{\Delta t}$
后向差分公式: $y_i^{\prime} \approx \dfrac{\Delta y_i}{\Delta t} = \dfrac{y_i - y_{i-1}}{\Delta t}$
算法精度: $o(\Delta t)$

中心差分算法高精度公式
$y_i^{\prime} \approx \dfrac{-y_{i+2} + 8y_{i+1} - 8y_{i-1} + y_{i-2}}{12\Delta t}$
$y_i^{\prime \prime} \approx \dfrac{-y_{i+2} + 16y_{i+1} - 30y_{i} + 16y_{i-1} - y_{i-2}}{12 \Delta t^2}$
$y_i^{\prime \prime \prime} \approx \dfrac{-y_{i+3} + 8y_{i+2} - 13 y_{i+1} + 13y_{i-1} - 8y_{i-2} + y_{i-3}}{8\Delta t^3}$
$y_i^{(4)} \approx \dfrac{-y_{i+3} + 12y_{i+2} - 39y_{i+1} + 56y_{i} - 39y_{i-1} + 12y_{i-2} - y_{i-3}}{6\Delta t^4}$
算法精度$o(\Delta t^4)$

中心差分算法

% 中心差分算法
function [dy,dx]=diff_ctr(y,Dt,n)
    y1=[y 0 0 0 0 0 0];
    y2=[0 y 0 0 0 0 0];
    y3=[0 0 y 0 0 0 0];
    y4=[0 0 0 y 0 0 0];
    y5=[0 0 0 0 y 0 0];
    y6=[0 0 0 0 0 y 0];
    y7=[0 0 0 0 0 0 y]; %{\rm\,尽量采用向量化运算，准备若干平移的向量}
    switch n            %{\rm\,按给定的阶次由开关结构求数值微分}
       case 1, dy=(-y1+8*y2-8*y4+y5)/12/Dt;
       case 2, dy=(-y1+16*y2-30*y3+16*y4-y5)/12/Dt^2;
       case 3, dy=(-y1+8*y2-13*y3+13*y5-8*y6+y7)/8/Dt^3;
       case 4, dy=(-y1+12*y2-39*y3+56*y4-39*y5+12*y6-y7)/6/Dt^4;
    end
    dy=dy(5+2*(n>2):end-4-2*(n>2)); dx=([2:length(dy)+1]+(n>2))*Dt;
end

函数调用: $[d_y, d_x]$ = diff_ctr($y, \Delta t, n$)

例3-56 数值微分逼近
对函数 $f(x) = \dfrac{\sin x}{x^2 + 4x + 3}$

h = 0.05;
x = 0: h: pi;
syms x1;
y = sin(x1)/(x1^2+4*x1+3);
yy1 = diff(y);
f1 = subs(yy1, x1, x);
yy2 = diff(yy1);
f2 = subs(yy2, x1, x);
yy3 = diff(yy2); 
f3 = subs(yy3, x1, x);
yy4 = diff(yy3);
f4 = subs(yy4, x1, x);

与解析解的精确比较

% 比较不同阶导数
y = sin(x1)/(x1^2+4*x1+3);
y = subs(y, x1, x);
[y1, dx1] = diff_ctr(y, h, 1);
subplot(221), plot(x, f1, dx1, y1, ':');
[y2, dx2] = diff_ctr(y, h, 2);
subplot(222), plot(x, f2, dx2, y2, ':');
[y3, dx3] = diff_ctr(y, h, 3);
subplot(223), plot(x, f3, dx3, y3, ':');
[y4, dx4] = diff_ctr(y, h, 4);
subplot(224), plot(x, f4, dx4, y4, ':');
% 分析误差
norm(double((y4-f4(4:60))./f4(4:60)))

二元函数梯度计算
多变量函数$f(x_1, x_2, \cdots, x_n)$
梯度是其对各个自变量偏导数构成的向量
$v = [\dfrac{\partial f}{\partial x_1}, \dfrac{\partial f}{\partial x_2}, \cdots, \dfrac{\partial f}{\partial x_n}]$
函数gradient()的调用格式 $[f_x, f_y] = \mathbb{gradient}(z)$
计算梯度$f_x = f_x/\Delta x, f_y = f_y / \Delta y$
$其中 \Delta x 和 \Delta y 分别是x和y生成网格的步距$

例3-57 梯度数值计算
已知$z = f(x, y) = (x^2 - 2x) e^{-x^2-y^2-xy}$
生成数据
用数值方法求梯度并分析误差

syms x y;
f = (x^2-2*x)*exp(-x^2-y^2-x*y);
[x1, y1] = meshgrid(-3: .2: 3, -2: .2: 2);
z1 = double(subs(f, {x, y}, {x1, y1}));
[zx1, zy1] = gradient(z1);
zx1 = zx1/0.2;
zy1 = zy1/0.2;
contour(x1, y1, z1, 30);
hold on;
quiver(x1, y1, -zx1, -zy1),
hold off;

03.11 单变量函数的数值积分

由给定数据进行梯形求积
梯形近似方法的基本思想
$\int_a^b f(x) dx = \sum \limits_{i=1}^{N} \int_{x_i}^{x_{i+1}} f(x) dx = \sum \limits _{i=1}^{N} \Delta f_i$
matlab调用格式: sum((2*y(1:end-1, :) + diff(y)) .* diff(x)) /2
或 S = trapz(x, y)

例3-59 定步长求积分
用定步长方法求解积分 $\int_0^{\frac{3\pi}{2}} \cos 15x dx$
并比较不同步长下的结果
首先绘图

% 在求解区域内被积函数有很强的震荡
x = [0:0.01:3*pi/2, 3*pi/2];
y = cos(15*x);
plot(x, y)

不同步距积分结果

syms x, A = int(cos(15*x), 0, 3*pi/2);
h0 = [0.1, 0.01, 0.001, 0.0001, 0.00001, 0.000001];
v = [];
for h=h0
  x = [0:h:3*pi/2, 3*pi/2];
  y = cos(15*x);
  I = trapz(x, y);
  v = [v; h, I, 1/15-I];
end;
v

数值积分求解函数
调用格式
新版本数值积分
I = integral(f, a, b, property pairs)
早期版本
求定积分
$[y, k] = quad(fun, a, b)$
$[y, k] = quad(fun, a, b, \varepsilon)$
其它数值定积分函数：quadl(), quadgk(), quadv()

例3-60 被积函数描述
计算积分: erf(x) = $\dfrac{2}{\sqrt{\pi}}\int_0^{1.5} e^{-t^2} dt$
方法1：一般函数方法(点运算)

function y = c3ffun(x)
  y = 2/sqrt(pi)*exp(-x.^2);

方法2：inline函数方法(不建议使用)

f = inline('2/sqrt(pi)*exp(-x.^2)', 'x');

方法3：匿名函数(matlab 7.0+)

f = @(x)2/sqrt(pi)*exp(-x.^2);

数值求解

y = integral(f, 0, 1.5)

解析解:运用符号工具箱

syms x;
y0 = vpa(int(2/sqrt(pi)*exp(-x^2), 0, 1.5), 60)

默认选项下数值解函数 integral()即可保证高精度的数值解

f = @(x)2/sqrt(pi)*exp(-x.^2);
y = integral(f, 0, 1.5)
syms x;
y0 = vpa(int(2/sqrt(pi)*exp(-x^2), 0, 1.5), 60)

例3-61 分段函数积分
给定如下分段函数
$f(x) = \left \{ \begin{array}{l}e^{x^2}, \qquad \qquad \quad 0 \leq x \leq 2 \\ \dfrac{80}{4 - \sin(16 \pi x)}, 2 < x \leq 4 \end{array} \right.$
计算积分值: $I = \int_0 ^4 f(x) dx$
定积分示意图:

x = [0:0.01:2, 2+eps: 0.01: 4, 4];
y = exp(x.^2).*(x <= 2) + ...
  80./(4-sin(16*pi*x)).*(x>2);
y(end) = 0;
x = [eps, x];
y = [0,y]
fill(x, y, 'g')

求解与验证

% 求解语句
f = @(x)exp(x.^2).*(x<=2) + 80 ./ ...
  (4-sin(16*pi*x)).*(x>2);
I = integral(f, 0, 4)
% 提高精度(检验)
I2 = integral(f, 0, 4, 'RelTol', 1e-20)
%解析解
syms x;
f = piecewise(x<=2, exp(x^2), ...
  x>2, 80/(4-sin(16*pi*x)));
I3 = vpa(int(f, x, 0, 4))

例3-62 与梯形法比较
重新计算积分 $\int_0^{3\pi/2} \cos(15x) dx$

f = @(x) cos(15*x);
tic,
S = integral(f, 0, 3*pi/2, 'RelTol', 1e-20),
toc

结论:和梯形法相比, 速度今读都明显提高
理论值: 1/15

例3-64 大范围积分
积分问题: $\int_0^{100} cos(15x) dx$
早期版本的几个积分函数不适用

% 数值解
f = @(x)cos(15*x);
I1 = integral(f, 0, 100, 'RelTol', 1e-20)
% 解析解
syms x;
I = int(cos(15*x), x, 0, 100),
vpa(I)

广义数值积分问题求解
Integral()函数可以直接用于广义积分运算
函数调用格式与前面介绍的完全一致
I = integral(fun, a, b)
早期版本, 采用Gauss-Kronrod算法
[y, k] = quadgk(fun, a, b, $\varepsilon$)

例3-66 广义积分的数值计算
数值计算：$\int_0^{\infty} e^{-x^2} dx$
数值解与解析解
解析解不可求，可以引出特殊函数erf()
数值解可以直接得出数值

f = @(x)exp(-x.^2);
I = integral(f, 0, inf, 'RelTol', 1e-20)
syms x;
I1 = int(exp(-x^2), 0, inf),
vpa(I1)

例3-67 含参数函数积分
积分函数: 绘制I(a)曲线
$I(a) = \int_0^{\infty} e^{a x^2} \sin(a^2 x) dx$
新版本支持向量参数a
早期版本需要对a循环

% 含参数函数积分
a = 0: 0.1: 4;
f = @(x) exp(-a*x.^2).*sin(a.^2*x);
I = integral(f, 0, inf, 'RelTol', 1e-20, 'ArrayValued', true);
plot(a, I)

03.12 双重数值积分

双重积分问题的数值解
数学标准型 $I = \int_{x_m}^{x_M} \int_{y_m(x)}^{y_M(x)} f(x, y) dydx$
注意积分顺序,先y后x
I = integral2(f, $x_m, x_M, y_m, y_M$, par pairs)
$y_m与y_M还可以为函数句柄$

例3-68 双重积分
试求出双重定积分
$J = \int_{-1}^{1}\int_{-2}^{2} e^{-x^2/2} \sin(x^2+y) dxdy$
积分区域为矩形区域, 对应好边界即可

f = @(x, y) exp(-x.^2/2).*sin(x.^2+y);
J = integral2(f, -2, 2, -1, 1, 'RelTol', 1e-20)

积分曲面绘制
仿照积分曲线绘制的函数，
[x, y, F] = intfunc2(f, $x_m, x_M, y_m, y_M$, n, m)

% 积分曲面
function [yv, xv, F] = intfunc2(f, xm, xM, varargin)
  [ym, yM, n, m] = default_vals({xm,xM,50,50},varargin{:}); 
  xv = linspace(xm, xM, n);
  yv = linspace(ym, yM, m);
  d = yv(2) - yv(1);
  [x y] = meshgrid(xv, yv);
  F = zeros(n,m);
  for i = 2:n
    for j = 2:m
      F(i,j) = integral2(f, xv(1), xv(i), yv(1), yv(j), 'RelTol', 1e-20);
  end;
end

例3-69 二元函数积分曲面
绘制积分函数曲面
$J = \int_{-1}^{1}\int_{-2}^{2} e^{-x^2/2} \sin(x^2+y) dxdy$
积分区域为矩形区域, 对应好边界即可

f = @(x, y) exp(-x.^2/2).*sin(x.^2+y);
[x, y, z] = intfunc2(f, -2, 2, -1, 1);
surf(x, y, z);
I = z(end, end)

例3-70 双重积分计算
是求出双重定积分
$J = \int_{-\frac{1}{2}}^{1} \int_{-\sqrt{1-x^2/2}}^{\sqrt{1-x^2/2}} e^{-\frac{x^2}{2}}\sin(x^2+y) dydx$
先y后x，与标准型一致, 可以直接求解

fh = @(x) sqrt(1-x.^2/2);
fl = @(x) -sqrt(1-x.^2/2);
f = @(x, y) exp(-x.^2/2).*sin(x.^2+y);
y = integral2(f, -1/2, 1, fl, fh)

先x后y的数值积分计算
与标准型积分顺序相反的二元积分
$I = \int_{y_m}^{y_M} \int_{x_m(y)}^{x_M(y)} f(x, y) dx dy$
被积函数描述时替换次序, 令 $\hat x = y, \hat y = x$
则 $I = \int_{\hat x_m}^{\hat x_M} \int_{\hat y_m(\hat x)}^{\hat y_M(\hat x)} f(\hat y, \hat x) d \hat y d\hat x$
被积函数: f = @(y, x)

例3-71 解析不可积
先x后y积分
$J = \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} e^{-x^2/2} \sinh(x^2 + y) dxdy$
解析不可求解，只能求数值解
交换x，y次序

tic,
f = @(y, x)exp(-x.^2/2).*sinh(x.^2+y);
fh = @(y) sqrt(1-y.^2);
fl = @(y)-sqrt(1-y.^2);
I = integral2(f, -1, 1, fl, fh, 'RelTol', 1e-20),
toc;

例3-72 非矩形区域积分区域的积分曲面
数学模型 $J = \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} e^{x^2/2}sin(x^2+y) dxdy$
绘制积分曲面
先计算矩形区域，再删去区域外函数值(NaN)

f = @(x, y)exp(-x.^2/2).*sin(x.^2+y);
fh = @(x)sqrt(1-x.^2/2);
fl = @(x)-sqrt(1-x.^2/2);
[x, y, z] = intfunc2(f, -1/2, 1, -1.2, 1.2);
for i = 1:length(x)
  x0 = x(i);
  xx = sort([fl(x0), fh(x0)]);
  ii = find(y>xx(2) | y < xx(1));
  z(ii, i) = NaN;
end;
surf(x, y, z);

03.13 三重与多重数值积分

三重定积分数值求解
长方体区域的三重积分标准型
$I = \int_{x_m}^{x_M} \int_{y_m(x)}^{y_M(x)} \int_{z_m(x,y)}^{z_M(x,y)} f(x, y, z) dzdydx$
函数调用:
$I = \mathbb{integral3}(f, x_m, x_M, y_m, y_M, z_m, z_M, pars)$
早期版本:
$I = \mathbb{triplequad}(f, x_m, x_M, y_m, y_M, z_m, z_M, \epsilon, @quadl)$

例3-73 三重积分计算
用数值方法求三重定积分问题
$\int_0^2 \int_0^{\pi} \int_0^{\pi} 4xze^{-x^2y-z^2} dzdydx$
长方体区域

f = @(x, y, z)4*x.*z.*exp(-x.*x.*y - z.*z);
I = integral3(f, 0, 2, 0, pi, 0, pi, 'RelTol', 1e-20)

多重积分数值求解
NIT工具箱(数值积分工具箱)还可以解决多重超长方体边界定积分问题
$I = \int_{x_{1m}}^{x_{1M}} \int_{x_{2m}}^{x_{2M}} \cdots \int_{x_{pm}}^{x_{pM}} f(x_1, x_2, \cdots, x_p) dx_p \cdots dx_2 dx_1$
调用格式：
$I = \mathbb{quadndg}(fun, [x_{1m}, x_{2m}, \cdots, x_{pm}], [x_{1M}, x_{2M}, \cdots, x_{pM}], \epsilon)$

例3-74 重新计算三重积分
前面的长方体区域三重积分问题
$\int_0^2 \int_0^{\pi} \int_0^{\pi} 4xze^{-x^2y-z^2} dzdydx$
变量替换：$x_1 = x, x_2 = y, x_3 = z$
被积函数：$f(x) = 4x_1 x_3 e^{-x_1^2x_2-x_3^2}$
重新求解(更快)：

f = @(x) 4*x(1)*x(3)*exp(-x(1)^2*x(2)-x(3)^2);
tic,
I = quadndg(f, [0 0 0], [2, pi, pi]),
toc

例3-75 五重积分的数值计算
5 重定积分问题
$I = \int_0^5 \int_0^4 \int_0^1 \int_0^2 \int_0^3 \sqrt[3]{v} \sqrt{w} x^2 y^3 z dz dy dx dw dv$
变量替换: $x_1 = v, x_2 = w, x_3 = x, x_4 = y, x_5 = z$
被积函数: $f(x) = \sqrt[3]{x_1} \sqrt{x_2} x_3^2 x_4^3 x_5$

% 五重积分计算
f = @(x)(x(1))^(1/3)*sqrt(x(2))*x(3)^2*x(4)^3*x(5);
I = quadndg(f, [0 0 0 0 0], [5, 4, 1, 2, 3])

五重积分解析解
本例存在解析解

syms x y z w v;
F = v^(1/3)*sqrt(w)*x^2*y^3*z;
I = int(int(int(int(int(F,z,0,3), y, 0, 2), x, 0, 1), w,0, 4), v, 0, 5)

注意:该工具箱单重积分函数duadg()调用格式和quad()一致,其效率也高于duadl()，故在进行数值求积分时建议使用此工具箱。

例3-76 解析不可积的五重积分计算
解析不可积多重积分
$I = \int_0^5 \int_0^4 \int_0^1 \int_0^2 \int_0^3 (e^{-\sqrt[3]{v}} \sin {\sqrt{w}} + e^{-x^2y^3z}) dzdydxdwdv$
变量替换：$x_1 = v, x_2 = w, x_e = x, x_4 = y, x_5 = z$
被积函数: $f(x) = e^{-\sqrt[3]{x_1}} \sin{\sqrt{x_2}} + e^{-x_e^2x_4^3x_5}$

f = @(x) exp(-(x(1))^(1/3))*sin(sqrt(x(2)))+exp(-x(3)^2*x(4)^3*x(5));
tic,
I = quadndg(f, [0 0 0 0 0], [5, 4, 1, 2, 3])
toc