数值分析思考题（钟尔杰版）参考解答

第 1 型曲线积分两种离散化公式 左矩形公式: ∫ L f ( x , y ) d x ≈ ∑ k = 0 N − 1 f ( x k , y k ) Δ s k \int_L f(x, y) d x \approx \sum_{k=0}^{N-1} f\left(x_k, y_k\right) \Delta s_k ∫L​f(x,y)dx≈∑k=0N−1​f(xk​,yk​)Δsk​ 右矩形公式: ∫ L f ( x , y ) d x ≈ ∑ k = 1 N f ( x k , y k ) Δ s k − 1 \int_L f(x, y) d x \approx \sum_{k=1}^N f\left(x_k, y_k\right) \Delta s_{k-1} ∫L​f(x,y)dx≈∑k=1N​f(xk​,yk​)Δsk−1​ 区别: 左矩形的微元的体积要小于右矩形的微元的体积

两点线性揷值公式: f ( x ) ≈ b − x b − a f ( a ) + x − a b − a f ( b ) f(x) \approx \frac{b-x}{b-a} f(a)+\frac{x-a}{b-a} f(b) f(x)≈b−ab−x​f(a)+b−ax−a​f(b) 简单梯形公式: ∫ a b f ( x ) d x ≈ b − a 2 [ f ( a ) + f ( b ) ] \int_a^b f(x) d x \approx \frac{b-a}{2}[f(a)+f(b)] ∫ab​f(x)dx≈2b−a​[f(a)+f(b)] ∫ a b f ( x ) d x = [ ∫ a b b − x b − a d x ] f ( a ) + [ ∫ a b x − a b − a d x ] f ( b ) A 0 = ∫ a b b − x b − a d x = 1 2 ( b − a ) A 1 = ∫ a b x − a b − a d x = 1 2 ( b − a ) ⇒ ∫ a b f ( x ) d x ≈ b − a 2 [ f ( a ) + f ( b ) ] \begin{aligned} &\int_a^b f(x) d x=\left[\int_a^b \frac{b-x}{b-a} d x\right] f(a)+\left[\int_a^b \frac{x-a}{b-a} d x\right] f(b) \\ &A_0=\int_a^b \frac{b-x}{b-a} d x=\frac{1}{2}(b-a) \quad A_1=\int_a^b \frac{x-a}{b-a} d x=\frac{1}{2}(b-a) \Rightarrow \\ &\int_a^b f(x) d x \approx \frac{b-a}{2}[f(a)+f(b)] \end{aligned} ​∫ab​f(x)dx=[∫ab​b−ab−x​dx]f(a)+[∫ab​b−ax−a​dx]f(b)A0​=∫ab​b−ab−x​dx=21​(b−a)A1​=∫ab​b−ax−a​dx=21​(b−a)⇒∫ab​f(x)dx≈2b−a​[f(a)+f(b)]​

在 [ a , b ] [\mathrm{a}, \mathrm{b}] [a,b] 内揷入点: a ≤ x 0 < x 1 < x 2 < x n ≤ b a \leq x_00, 考虑函数 f ( x ) f(x) f(x) 的 Tay1or 展式 f ( a + h ) = f ( a ) + h f ′ ( a ) + h 2 2 ! f ′ ′ ( a ) + h 3 3 ! f ( 3 ) ( a ) + h 4 4 ! f ( 4 ) ( a ) + ⋯ f ( a − h ) = f ( a ) − h f ′ ( a ) + h 2 2 ! f ′ ′ ( a ) − h 3 3 ! f ( 3 ) ( a ) + h 4 4 ! f ( 4 ) ( a ) + ⋯ \begin{aligned} &f(a+h)=f(a)+h f^{\prime}(a)+\frac{h^2}{2 !} f^{\prime \prime}(a)+\frac{h^3}{3 !} f^{(3)}(a)+\frac{h^4}{4 !} f^{(4)}(a)+\cdots \\ &f(a-h)=f(a)-h f^{\prime}(a)+\frac{h^2}{2 !} f^{\prime \prime}(a)-\frac{h^3}{3 !} f^{(3)}(a)+\frac{h^4}{4 !} f^{(4)}(a)+\cdots \end{aligned} ​f(a+h)=f(a)+hf′(a)+2!h2​f′′(a)+3!h3​f(3)(a)+4!h4​f(4)(a)+⋯f(a−h)=f(a)−hf′(a)+2!h2​f′′(a)−3!h3​f(3)(a)+4!h4​f(4)(a)+⋯​ 截取等式右端前两项, 得数值微分公式 f ′ ( a ) ≈ f ( a + h ) − f ( a ) h f ′ ( a ) ≈ f ( a ) − f ( a − h ) h f^{\prime}(a) \approx \frac{f(a+h)-f(a)}{h} \quad f^{\prime}(a) \approx \frac{f(a)-f(a-h)}{h} f′(a)≈hf(a+h)−f(a)​f′(a)≈hf(a)−f(a−h)​ 这两式分别是 f ( x ) f(x) f(x) 在 x = a x=a x=a 处的右导数和左导数近似值, 被称为向前差商和 向后差商公式。如果将两式取算术平均, 则有一阶导数计算的中心差商公式 f ′ ( a ) ≈ f ( a + h ) − f ( a − h ) 2 h f^{\prime}(a) \approx \frac{f(a+h)-f(a-h)}{2 h} f′(a)≈2hf(a+h)−f(a−h)​

由泰勤展式, 利用二阶中心差商得 f ′ ( x k ) ≈ f ( x k + 1 ) − f ( x k − 1 ) 2 h − 1 6 [ f ′ ( x k + 1 ) − 2 f ′ ( x k ) + f ′ ( x k − 1 ) ] + o ( h 2 )  令  m k = f ′ ( x k ) ( k = 0 , 1 , 2 , … , n )  则有  m k − 1 + 4 m k + m k + 1 = 3 h [ f ( x k + 1 ) − f ( x k − 1 ) ] + o ( h 4 ) \begin{aligned} &f^{\prime}\left(x_k\right) \approx \frac{f\left(x_{k+1}\right)-f\left(x_{k-1}\right)}{2 h}-\frac{1}{6}\left[f^{\prime}\left(x_{k+1}\right)-2 f^{\prime}\left(x_k\right)+f^{\prime}\left(x_{k-1}\right)\right]+o\left(h^2\right) \\ &\text { 令 } m_k=f^{\prime}\left(x_k\right)(k=0,1,2, \ldots, n) \text { 则有 } \\ &m_{k-1}+4 m_k+m_{k+1}=\frac{3}{h}\left[f\left(x_{k+1}\right)-f\left(x_{k-1}\right)\right]+o\left(h^4\right) \end{aligned} ​f′(xk​)≈2hf(xk+1​)−f(xk−1​)​−61​[f′(xk+1​)−2f′(xk​)+f′(xk−1​)]+o(h2) 令 mk​=f′(xk​)(k=0,1,2,…,n) 则有 mk−1​+4mk​+mk+1​=h3​[f(xk+1​)−f(xk−1​)]+o(h4)​ 在上式中, 一阶导数值隐含在线性方程组中, 当给定 m 0 , m 1 m_0, m_1 m0​,m1​ 时方程组有唯 一解。上述方法为数值求导的隐式方法。

对于插值型求导公式: 误差余项: f ′ ( x ) − P n ′ ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ω n + 1 ′ ( x ) + ω n + 1 ( x ) ( n + 1 ) ! d d x f ( n + 1 ) ( ξ ) f^{\prime}(x)-P_n^{\prime}(x)=\frac{f^{(n+1)}(\xi)}{(n+1) !} \omega_{n+1}^{\prime}(x)+\frac{\omega_{n+1}(x)}{(n+1) !} \frac{d}{d x} f^{(n+1)}(\xi) f′(x)−Pn′​(x)=(n+1)!f(n+1)(ξ)​ωn+1′​(x)+(n+1)!ωn+1​(x)​dxd​f(n+1)(ξ) 式中: ω n + 1 ( x ) = ∏ i = 0 n ( x − x i ) \omega_{n+1}(x)=\prod_{i=0}^n\left(x-x_i\right) ωn+1​(x)=∏i=0n​(x−xi​)