1 知识点
1.1 要求
为简化计算, 通常用多项式近似表达复杂函数:
设函数 f ( x ) f(x) f(x) 在含有 x 0 x_0 x0 的开区间内具有 ( n + 1 ) (n+1) (n+1) 阶导数, 试找出一个关于 ( x − x 0 ) (x-x_0) (x−x0) 的 n n n 次多项式 p n ( x ) p_n(x) pn(x) 近似表达 f ( x ) f(x) f(x);
要求 p n ( x ) p_n(x) pn(x) 与 f ( x ) f(x) f(x) 之差是比 ( x − x 0 ) n (x-x_0)^n (x−x0)n 高阶的无穷小, 并给出误差 ∣ f ( x ) − p n ( x ) ∣ |f(x)-p_n(x)| ∣f(x)−pn(x)∣ 的具体表达式.
1.2 泰勒多项式
设 p n ( x ) p_n(x) pn(x) 的形式为:
p n ( x ) = a 0 + a 1 ( x − x 0 ) + a 2 ( x − x 0 ) 2 + ⋯ + a n ( x − x 0 ) n ( 1 ) p_n(x)=a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots + a_n(x-x_0)^n \qquad (1) pn(x)=a0+a1(x−x0)+a2(x−x0)2+⋯+an(x−x0)n(1)
假设 p n ( x ) p_n(x) pn(x) 与 f ( x ) f(x) f(x) 在 x 0 x_0 x0 处的函数值相等, 且同阶导数在 x 0 x_0 x0 处的值也相等, 得:
f ( x 0 ) = p n ( x 0 ) f ′ ( x 0 ) = p n ′ ( x 0 ) f ′ ′ ( x 0 ) = p n ′ ′ ( x 0 ) ⋯ f ( n ) ( x 0 ) = p n ( n ) ( x 0 ) } ( 2 ) \left.\begin{aligned}f(x_0)=p_n(x_0)\\f'(x_0)=p'_n(x_0)\\f''(x_0)=p''_n(x_0)\\ \cdots\\f^{(n)}(x_0)=p_n^{(n)}(x_0)\end{aligned}\right\}\qquad (2) f(x0)=pn(x0)f′(x0)=pn′(x0)f′′(x0)=pn′′(x0)⋯f(n)(x0)=pn(n)(x0)⎭ ⎬ ⎫(2)
由 ( 1 ) (1) (1) 式和 ( 2 ) (2) (2) 式得:
f ( x 0 ) = p n ( x 0 ) = a 0 + 0 + ⋯ + 0 = a 0 f ′ ( x 0 ) = p n ′ ( x 0 ) = 0 + a 1 ⋅ 1 + 0 + ⋯ + 0 = a 1 f ′ ′ ( x 0 ) = p n ′ ′ ( x 0 ) = 0 + 0 + a 2 ⋅ 2 ! + 0 + ⋯ + 0 = a 2 ⋅ 2 ! ⋯ f ( n ) ( x 0 ) = p ( n ) ( x 0 ) = 0 + ⋯ + 0 + a n ⋅ n ! = a n ⋅ n ! } ( 3 ) \left.\begin{aligned}f(x_0)=p_n(x_0)=a_0+0+\cdots +0=a_0 \\ f'(x_0)=p'_n(x_0)=0+a_1\cdot 1+0+\cdots +0=a_1 \\ f''(x_0)=p''_n(x_0)=0+0+a_2\cdot 2! +0+\cdots +0 = a_2\cdot2!\\ \cdots \\ f^{(n)}(x_0)=p^{(n)}(x_0)=0+\cdots +0+a_n\cdot n!= a_n\cdot n!\end{aligned}\right\}\qquad (3) f(x0)=pn(x0)=a0+0+⋯+0=a0f′(x0)=pn′(x0)=0+a1⋅1+0+⋯+0=a1f′′(x0)=pn′′(x0)=0+0+a2⋅2!+0+⋯+0=a2⋅2!⋯f(n)(x0)=p(n)(x0)=0+⋯+0+an⋅n!=an⋅n!⎭ ⎬ ⎫(3)
将 ( 3 ) (3) (3) 式变换, 得:
a 0 = f ( x 0 ) a 1 = f ′ ( x 0 ) a 2 = f ′ ′ ( x 0 ) 2 ! ⋯ a n = f ( n ) ( x 0 ) n ! } ( 4 ) \left.\begin{aligned}a_0=f(x_0) \\ a_1=f'(x_0) \\ a_2=\frac{f''(x_0)}{2!} \\ \cdots \\ a_n=\frac{f^{(n)}(x_0)}{n!}\end{aligned}\right\}\qquad (4) a0=f(x0)a1=f′(x0)a2=2!f′′(x0)⋯an=n!f(n)(x0)⎭ ⎬ ⎫(4)
将 ( 4 ) (4) (4) 式代入 ( 1 ) (1) (1) 式, 得:
p n ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n ( 5 ) p_n(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\qquad (5) pn(x)=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!f(n)(x0)(x−x0)n(5)
( 5 ) (5) (5) 式称为函数 f ( x ) f(x) f(x) 按 ( x − x 0 ) (x-x_0) (x−x0) 的幂展开的 n n n 次泰勒多项式, 是函数 f ( x ) f(x) f(x) 的近似表达.
1.3 带有拉格朗日型余项的泰勒公式
[泰勒(Taylor)中值定理] 如果函数 f ( x ) f(x) f(x) 在含有 x 0 x_0 x0 的某个开区间 ( a , b ) (a,b) (a,b) 内具有直到 ( n + 1 ) (n+1) (n+1) 阶的导数，则对任一 x ∈ ( a , b ) x\in(a,b) x∈(a,b)，有：
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n + R n ( x ) ( 6 ) f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x) \qquad (6) f(x)=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!f(n)(x0)(x−x0)n+Rn(x)(6)
其中,
R n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) n + 1 ( 7 ) R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}\qquad (7) Rn(x)=(n+1)!f(n+1)(ξ)(x−x0)n+1(7)
公式 ( 6 ) (6) (6) 称为函数 f ( x ) f(x) f(x) 按 ( x − x 0 ) (x-x_0) (x−x0) 的幂展开的带有拉格朗日型余项的 n n n 阶泰勒公式.
公式 ( 7 ) (7) (7) 称为拉格朗日型余项，其中的 ξ \xi ξ 介于 x 0 x_0 x0 与 x x x 之间.
当 n = 0 n=0 n=0 时，泰勒公式变成拉格朗日中值公式 f ( x ) = f ( x 0 ) + f ′ ( ξ ) ( x − x 0 ) f(x)=f(x_0)+f'(\xi)(x-x_0) f(x)=f(x0)+f′(ξ)(x−x0) （ ξ \xi ξ 介于 x 0 x_0 x0 与 x x x 之间）.
如果 x 0 = 0 x_0=0 x0=0，则 ξ ∈ ( 0 , x ) \xi\in(0,x) ξ∈(0,x)，可以令 ξ = θ x ( 0 < θ < 1 ) \xi=\theta x(0<\theta <1) ξ=θx(0<θ<1) .
1.4 泰勒公式的误差
以多项式 p n ( x ) p_n(x) pn(x) 近似表达函数 f ( x ) f(x) f(x) 时, 其误差为 ∣ R n ( x ) ∣ |R_n(x)| ∣Rn(x)∣ .
如果对于某个固定的 n n n, 当 x ∈ ( a , b ) x\in (a,b) x∈(a,b) 时, ∣ f ( n + 1 ) ( x ) ∣ ≤ M |f^{(n+1)}(x)|\leq M ∣f(n+1)(x)∣≤M, 则有:
∣ R n ( x ) ∣ = ∣ f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) n + 1 ∣ ≤ M ( n + 1 ) ! ∣ x − x 0 ∣ n + 1 ( 8 ) \begin{vmatrix}R_n(x)\end{vmatrix}=\begin{vmatrix}\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}\end{vmatrix}\leq \frac{M}{(n+1)!}\begin{vmatrix}x-x_0\end{vmatrix}^{n+1}\qquad (8) Rn(x) = (n+1)!f(n+1)(ξ)(x−x0)n+1 ≤(n+1)!M x−x0 n+1(8)
及
lim x → x 0 R n ( x ) ( x − x 0 ) n = 0 ( 9 ) \lim_{x\rightarrow x_0}\frac{R_n(x)}{(x-x_0)^n}=0\qquad (9) x→x0lim(x−x0)nRn(x)=0(9)
可见, 当 x → x 0 x\rightarrow x_0 x→x0 时, 误差 ∣ R n ( x ) ∣ |R_n(x)| ∣Rn(x)∣ 是比 ( x − x 0 ) n (x-x_0)^n (x−x0)n 高阶的无穷小, 即:
R n ( x ) = o [ ( x − x 0 ) n ] ( 10 ) R_n(x)=o[(x-x_0)^n]\qquad (10) Rn(x)=o[(x−x0)n](10)
公式 ( 10 ) (10) (10) 称为佩亚诺(Peano)型余项.
1.5 带有佩亚诺型余项的泰勒公式
在不需要余项的精确表达时, n n n 阶泰勒公式也可写成：
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n + o [ ( x − x 0 ) n ] ( 11 ) f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+o[(x-x_0)^n]\qquad (11) f(x)=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!f(n)(x0)(x−x0)n+o[(x−x0)n](11)
公式 ( 11 ) (11) (11) 称为 f ( x ) f(x) f(x) 按 ( x − x 0 ) (x-x_0) (x−x0) 的幂展开的带有佩亚诺型余项的 n n n 阶泰勒公式.
1.6 带有拉格朗日型余项的麦克劳林公式
公式 ( 6 ) (6) (6) 中，如果取 x 0 = 0 x_0=0 x0=0，则 ξ \xi ξ 在 0 0 0 与 x x x 之间. 因此可以令 ξ = θ x ( 0 < θ < 1 ) \xi=\theta x(0<\theta <1) ξ=θx(0<θ<1) ，从而泰勒公式变为带有拉格朗日型余项的麦克劳林公式：
f ( x ) = f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 ! x 2 + ⋯ + f ( n ) ( 0 ) n ! x n + f ( n + 1 ) ( θ x ) ( n + 1 ) ! x n + 1 ( 0 < θ < 1 ) ( 12 ) . f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots +\frac{f^{(n)}(0)}{n!}x^n+\frac{f^{(n+1)}(\theta x)}{(n+1)!}x^{n+1}(0<\theta<1)\qquad (12). f(x)=f(0)+f′(0)x+2!f′′(0)x2+⋯+n!f(n)(0)xn+(n+1)!f(n+1)(θx)xn+1(0<θ<1)(12).
1.7 带有佩亚诺型余项的麦克劳林公式
如果取 x 0 = 0 x_0=0 x0=0，则公式 ( 11 ) (11) (11) 变换为带有佩亚诺型余项的麦克劳林公式：
f ( x ) = f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 ! x 2 + ⋯ + f ( n ) ( 0 ) n ! x n + o ( x n ) ( 13 ) f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots +\frac{f^{(n)}(0)}{n!}x^n+o(x^n) \qquad (13) f(x)=f(0)+f′(0)x+2!f′′(0)x2+⋯+n!f(n)(0)xn+o(xn)(13)
2 练习题
[题1] 按 ( x − 4 ) (x-4) (x−4) 的幂展开多项式 f ( x ) = x 4 − 5 x 3 + x 2 − 3 x + 4 f(x)=x^4-5x^3+x^2-3x+4 f(x)=x4−5x3+x2−3x+4.
[解]
根据公式 ( 5 ) (5) (5) 得:
p n ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n p_n(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n pn(x)=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!f(n)(x0)(x−x0)n
此题 x 0 = 4 x_0=4 x0=4:
原式 = f ( 4 ) + f ′ ( 4 ) ( x − 4 ) + f ′ ′ ( 4 ) 2 ! ( x − 4 ) 2 + f ′ ′ ′ ( 4 ) 3 ! ( x − 4 ) 3 + f ( 4 ) 4 ! ( x − 4 ) 4 + 0 + ⋯ + 0 =f(4)+f'(4)(x-4)+\frac{f''(4)}{2!}(x-4)^2+\frac{f'''(4)}{3!}(x-4)^3+\frac{f^{(4)}}{4!}(x-4)^4+0+\cdots +0 =f(4)+f′(4)(x−4)+2!f′′(4)(x−4)2+3!f′′′(4)(x−4)3+4!f(4)(x−4)4+0+⋯+0
= − 56 + 21 ( x − 4 ) + 37 ( x − 4 ) 2 + 11 ( x − 4 ) 3 + ( x − 4 ) 4 =-56+21(x-4)+37(x-4)^2+11(x-4)^3+(x-4)^4 =−56+21(x−4)+37(x−4)2+11(x−4)3+(x−4)4
[题2] 应用麦克劳林公式，按 x x x 的幂展开函数 f ( x ) = ( x 2 − 3 x + 1 ) 3 f(x)=(x^2-3x+1)^3 f(x)=(x2−3x+1)3.
[解]
根据麦克劳林公式得:
p n ( x ) p_n(x) pn(x)
= f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 ! x + ⋯ + f ( 6 ) ( 0 ) 6 ! x + 0 + ⋯ + 0 =f(0)+f'(0)x+\frac{f''(0)}{2!}x+\cdots +\frac{f^{(6)}(0)}{6!}x+0+\cdots +0 =f(0)+f′(0)x+2!f′′(0)x+⋯+6!f(6)(0)x+0+⋯+0
= 1 − 9 x + 30 x 2 − 45 x 3 + 30 x 4 − 9 x 5 + x 6 =1-9x+30x^2-45x^3+30x^4-9x^5+x^6 =1−9x+30x2−45x3+30x4−9x5+x6
[题3] 求函数 f ( x ) = x f(x)=\sqrt{x} f(x)=x 按 ( x − 4 ) (x-4) (x−4) 的幂展开的带有拉格朗日型余项的3阶泰勒公式.
[解]
根据公式 ( 6 ) (6) (6), 且 x 0 = 4 x_0=4 x0=4 得:
f ( x ) f(x) f(x)
= x =\sqrt{x} =x
= f ( 4 ) + f ′ ( 4 ) ( x − 4 ) + f ′ ′ ( 4 ) 2 ! ( x − 4 ) 2 + f ′ ′ ′ ( 4 ) 3 ! ( x − 4 ) 3 + f ( 4 ) ( ξ ) 4 ! ( x − 4 ) 4 =f(4)+f'(4)(x-4)+\frac{f''(4)}{2!}(x-4)^2+\frac{f'''(4)}{3!}(x-4)^3+\frac{f^{(4)}(\xi)}{4!}(x-4)^4 =f(4)+f′(4)(x−4)+2!f′′(4)(x−4)2+3!f′′′(4)(x−4)3+4!f(4)(ξ)(x−4)4
= 2 + x − 4 4 − ( x − 4 ) 2 64 + ( x − 4 ) 3 512 − 5 ξ − 7 2 ( x − 4 ) 4 128 =2+\frac{x-4}{4}-\frac{(x-4)^2}{64}+\frac{(x-4)^3}{512}-\frac{5\xi^{-\frac{7}{2}}(x-4)^4}{128} =2+4x−4−64(x−4)2+512(x−4)3−1285ξ−27(x−4)4（ ξ \xi ξ 介于 4 4 4 与 x x x 之间）
[题4] 求函数 f ( x ) = l n x f(x)=lnx f(x)=lnx 按 ( x − 2 ) (x-2) (x−2) 的幂展开的带有佩亚诺型余项的 n n n 阶泰勒公式.
[解]
根据公式 ( 11 ) (11) (11) ，且 x 0 = 0 x_0=0 x0=0 得:
f ( x ) f(x) f(x)
= l n x =lnx =lnx
= f ( 2 ) + f ′ ( 2 ) ( x − 2 ) + f ′ ′ ( 2 ) 2 ! ( x − 2 ) 2 + ⋯ + f ( n ) ( 2 ) n ! ( x − 2 ) n + o [ ( x − 2 ) n ] =f(2)+f'(2)(x-2)+\frac{f''(2)}{2!}(x-2)^2+\cdots + \frac{f^{(n)}(2)}{n!}(x-2)^n+o[(x-2)^n] =f(2)+f′(2)(x−2)+2!f′′(2)(x−2)2+⋯+n!f(n)(2)(x−2)n+o[(x−2)n]
= l n 2 + x − 2 2 − ( x − 2 ) 2 2 3 + ⋯ + ( − 1 ) n − 1 ( x − 2 ) n n ⋅ 2 n + o [ ( x − 2 ) n ] =ln2+\frac{x-2}{2}-\frac{(x-2)^2}{2^3}+\cdots + \frac{(-1)^{n-1}(x-2)^n}{n\cdot 2^n}+o[(x-2)^n] =ln2+2x−2−23(x−2)2+⋯+n⋅2n(−1)n−1(x−2)n+o[(x−2)n]
[题5] 求函数 f ( x ) = 1 x f(x)=\frac{1}{x} f(x)=x1 按 ( x + 1 ) (x+1) (x+1) 的幂展开的带有拉格朗日型余项的 n n n 阶泰勒公式.
[解]
根据公式 ( 6 ) (6) (6) ,且 x 0 = − 1 x_0=-1 x0=−1 得:
f ( x ) f(x) f(x)
= 1 x =\frac{1}{x} =x1
= f ( − 1 ) + f ′ ( − 1 ) ( x + 1 ) + f ′ ′ ( − 1 ) 2 ! ( x + 1 ) 2 + ⋯ + f ( n ) ( − 1 ) n ! ( x + 1 ) n + f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x + 1 ) n + 1 =f(-1)+f'(-1)(x+1)+\frac{f''(-1)}{2!}(x+1)^2+\cdots +\frac{f^{(n)}(-1)}{n!}(x+1)^n+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x+1)^{n+1} =f(−1)+f′(−1)(x+1)+2!f′′(−1)(x+1)2+⋯+n!f(n)(−1)(x+1)n+(n+1)!f(n+1)(ξ)(x+1)n+1
= − 1 − ( x + 1 ) − ( x + 1 ) 2 + ⋯ + ( − 1 ) ( x + 1 ) n + ( − 1 ) n + 1 ( x + 1 ) n + 1 ξ n + 2 =-1-(x+1)-(x+1)^2+\cdots +(-1)(x+1)^n+(-1)^{n+1}\frac{(x+1)^{n+1}}{\xi^{n+2}} =−1−(x+1)−(x+1)2+⋯+(−1)(x+1)n+(−1)n+1ξn+2(x+1)n+1( ξ \xi ξ 介于 − 1 -1 −1 与 x x x 之间).
[题6] 求函数 f ( x ) = t a n x f(x)=tanx f(x)=tanx 的带有佩亚诺型余项的 3 3 3 阶麦克劳林公式.
[解]
根据公式 ( 13 ) (13) (13) 得:
f ( x ) f(x) f(x)
= t a n x =tanx =tanx
= f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 x 2 + f ′ ′ ′ ( 0 ) 6 x 3 + o ( x 3 ) =f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+o(x^3) =f(0)+f′(0)x+2f′′(0)x2+6f′′′(0)x3+o(x3)
= 0 + x s e c 2 0 + 2 s e c 2 0 t a n 0 2 x 2 + 4 s e c 2 x t a n 2 0 + 2 s e c 4 0 6 x 3 + o ( x 3 ) =0+xsec^20+\frac{2sec^20tan0}{2}x^2+\frac{4sec^2xtan^20+2sec^40}{6}x^3+o(x^3) =0+xsec20+22sec20tan0x2+64sec2xtan20+2sec40x3+o(x3)
= x + 1 3 x 3 + o ( x 3 ) =x+\frac{1}{3}x^3+o(x^3) =x+31x3+o(x3)
[题7] 求函数 f ( x ) = x e x f(x)=xe^x f(x)=xex 的带有佩亚诺型余项的 n n n 阶麦克劳林公式.
[解]
根据公式 ( 13 ) (13) (13) 得:
f ( x ) f(x) f(x)
= x e x =xe^x =xex
= f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 ! x 2 + ⋯ + f ( n ) ( 0 ) n ! x n + o ( x n ) =f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots + \frac{f^{(n)}(0)}{n!}x^n+o(x^n) =f(0)+f′(0)x+2!f′′(0)x2+⋯+n!f(n)(0)xn+o(xn)
= 0 + x + 2 x 2 2 ! + 3 x 3 3 ! + ⋯ + n n ! x n + o ( x n ) =0+x+\frac{2x^2}{2!}+\frac{3x^3}{3!}+\cdots +\frac{n}{n!}x^n+o(x^n) =0+x+2!2x2+3!3x3+⋯+n!nxn+o(xn)
= x + x 2 + x 3 2 + ⋯ + x n ( n − 1 ) ! + o ( x n ) =x+x^2+\frac{x^3}{2}+\cdots +\frac{x^n}{(n-1)!}+o(x^n) =x+x2+2x3+⋯+(n−1)!xn+o(xn)
[题8] 验证当 0 < x ≤ 1 2 0<x\leq \frac{1}{2} 0<x≤21 时，按公式 e x ≈ 1 + x + x 2 2 + x 3 6 e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6} ex≈1+x+2x2+6x3 计算 e x e^x ex 的近似值时，所产生的误差小于 0.01 0.01 0.01，并求 e \sqrt{e} e 的近似值，使误差小于 0.01 0.01 0.01.
[解]
根据泰勒中值定理得：
e x e^x ex 与 1 + x + x 2 2 + x 3 6 1+x+\frac{x^2}{2}+\frac{x^3}{6} 1+x+2x2+6x3 之间得误差为 ∣ e ξ 24 x 4 ∣ \begin{vmatrix}\frac{e^{\xi}}{24}x^4\end{vmatrix} 24eξx4 ( ξ \xi ξ 介于 0 0 0 与 x x x 之间)，即拉格朗日型余项.
∵ 0 < x ≤ 1 2 \because 0<x\leq \frac{1}{2} ∵0<x≤21
∴ 0 < ξ ≤ 1 2 \therefore 0<\xi \leq \frac{1}{2} ∴0<ξ≤21
∴ ∣ e ξ 24 x 4 ∣ ≤ e 1 2 24 ⋅ 2 4 < 0.01 \therefore \begin{vmatrix}\frac{e^\xi}{24}x^4\end{vmatrix}\leq \frac{e^\frac{1}{2}}{24\cdot 2^4}<0.01 ∴ 24eξx4 ≤24⋅24e21<0.01
∴ e = e 1 2 ≈ 1 + 1 2 + 1 2 ⋅ ( 1 2 ) 3 + 1 6 ⋅ ( 1 2 ) 3 = 1 + 1 2 + 1 8 + 1 48 = 79 48 ≈ 1.646 \therefore \sqrt{e}=e^{\frac{1}{2}}\approx 1+\frac{1}{2}+\frac{1}{2}\cdot (\frac{1}{2})^3+\frac{1}{6}\cdot (\frac{1}{2})^3=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{48}=\frac{79}{48}\approx 1.646 ∴e =e21≈1+21+21⋅(21)3+61⋅(21)3=1+21+81+481=4879≈1.646
[题9] 应用 3 3 3 阶泰勒公式求下列各数的近似值，并估计误差：
(1) 30 3 \sqrt[3]{30} 330
(2) s i n 1 8 ∘ sin18^\circ sin18∘
[解(1)]
令 f ( x ) = 3 1 + x 3 f(x)=3\sqrt[3]{1+x} f(x)=331+x ，则 f ( 1 9 ) = 30 3 f(\frac{1}{9})=\sqrt[3]{30} f(91)=330
f ( x ) = 3 1 + x 3 = f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 x 2 + f ′ ′ ′ ( 0 ) 6 x 3 + f ( 4 ) ( ξ ) 24 x 4 f(x)=3\sqrt[3]{1+x}=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+\frac{f^{(4)}(\xi)}{24}x^4 f(x)=331+x =f(0)+f′(0)x+2f′′(0)x2+6f′′′(0)x3+24f(4)(ξ)x4( ξ \xi ξ 介于 0 0 0 与 1 9 \frac{1}{9} 91 之间)
∴ 30 3 = f ( 1 9 ) ≈ f ( 0 ) + f ′ ( 0 ) ⋅ 1 9 + f ′ ′ ( 0 ) 2 ⋅ 1 81 + f ′ ′ ′ ( 0 ) 6 ⋅ 1 729 \therefore \sqrt[3]{30}=f(\frac{1}{9})\approx f(0)+f'(0)\cdot \frac{1}{9}+\frac{f''(0)}{2}\cdot \frac{1}{81}+\frac{f'''(0)}{6}\cdot \frac{1}{729} ∴330 =f(91)≈f(0)+f′(0)⋅91+2f′′(0)⋅811+6f′′′(0)⋅7291
= 3 + 1 9 − 2 3 ⋅ 1 2 ⋅ 1 9 2 + 10 9 ⋅ 1 6 ⋅ 1 9 3 ≈ 3.1092 =3+\frac{1}{9}-\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{1}{9^2}+\frac{10}{9}\cdot \frac{1}{6}\cdot \frac{1}{9^3}\approx 3.1092 =3+91−32⋅21⋅921+910⋅61⋅931≈3.1092
其误差为 ∣ f ( 4 ) ( ξ ) 24 x 4 ∣ = ∣ 80 27 ⋅ 4 ! ⋅ 9 4 ( 1 + ξ ) − 11 3 ∣ < 80 27 ⋅ 4 ! ⋅ 9 4 ≈ 1.8817 × 1 0 − 5 \begin{vmatrix}\frac{f^{(4)}(\xi)}{24}x^4\end{vmatrix}=\begin{vmatrix}\frac{80}{27\cdot 4!\cdot 9^4}(1+\xi)^{-\frac{11}{3}}\end{vmatrix}<\frac{80}{27\cdot 4!\cdot 9^4}\approx1.8817\times10^{-5} 24f(4)(ξ)x4 = 27⋅4!⋅9480(1+ξ)−311 <27⋅4!⋅9480≈1.8817×10−5
[解(2)]
s i n 1 8 ∘ = s i n π 10 sin18^\circ=sin\frac{\pi}{10} sin18∘=sin10π
令 f ( x ) = s i n x f(x)=sinx f(x)=sinx
利用带拉格朗日型余项的泰勒公式进行计算：
f ( x ) = s i n x = f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 x 2 + f ′ ′ ′ ( 0 ) 6 x 3 + f ( 4 ) ( ξ ) 24 x 4 f(x)=sinx=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+\frac{f^{(4)}(\xi)}{24}x^4 f(x)=sinx=f(0)+f′(0)x+2f′′(0)x2+6f′′′(0)x3+24f(4)(ξ)x4（ ξ \xi ξ 介于 0 0 0 与 x x x 之间）
∴ s i n 1 8 ∘ = f ( π 10 ) = sin π 10 ≈ 0 + 1 ⋅ π 10 − 0 − 1 6 ⋅ ( π 10 ) 3 ≈ 0.3090 \therefore sin18^\circ=f(\frac{\pi}{10})=\sin\frac{\pi}{10}\approx 0+1\cdot \frac{\pi}{10}-0-\frac{1}{6}\cdot (\frac{\pi}{10})^3\approx 0.3090 ∴sin18∘=f(10π)=sin10π≈0+1⋅10π−0−61⋅(10π)3≈0.3090
利用拉格朗日型余项表示上述计算的误差：
∣ R n ( x ) ∣ = ∣ f ( 4 ) ( ξ ) 24 x 4 ∣ = ∣ s i n ξ 24 ( π 10 ) 4 ∣ < s i n π 10 24 ( π 10 ) 4 ≈ 0.000125 \begin{vmatrix}R_n(x)\end{vmatrix}=\begin{vmatrix}\frac{f^{(4)}(\xi)}{24}x^4\end{vmatrix}=\begin{vmatrix}\frac{sin\xi}{24}(\frac{\pi}{10})^4\end{vmatrix}<\frac{sin\frac{\pi}{10}}{24}(\frac{\pi}{10})^4\approx 0.000125 Rn(x) = 24f(4)(ξ)x4 = 24sinξ(10π)4 <24sin10π(10π)4≈0.000125
[题10] 利用泰勒公式极限 lim x → + ∞ ( x 3 + 3 x 2 3 − x 4 − 2 x 3 4 ) \lim_{x\rightarrow +\infty}(\sqrt[3]{x^3+3x^2}-\sqrt[4]{x^4-2x^3}) limx→+∞(3x3+3x2 −4x4−2x3 ).
[解]
令 t = 1 x t=\frac{1}{x} t=x1
原式 = lim t → 0 1 + 3 t 3 − 1 − 2 t 4 t =\lim_{t\rightarrow 0}\frac{\sqrt[3]{1+3t}-\sqrt[4]{1-2t}}{t} =limt→0t31+3t −41−2t
利用带佩亚诺型余项的麦克劳林公式得：
1 + 3 t 3 = f ( 0 ) + f ′ ( 0 ) t + o ( t ) = 1 + t + o ( t ) \sqrt[3]{1+3t}=f(0)+f'(0)t+o(t)=1+t+o(t) 31+3t =f(0)+f′(0)t+o(t)=1+t+o(t)
1 − 2 t 4 = 1 − 1 2 t + o ( t ) \sqrt[4]{1-2t}=1-\frac{1}{2}t+o(t) 41−2t =1−21t+o(t)
∴ \therefore ∴ 原式 = lim t → 0 1 + t + o ( t ) − 1 + 1 2 t − o ( t ) t = lim t → 0 3 2 t + o ( t ) t = 3 2 =\lim_{t\rightarrow 0}\frac{1+t+o(t)-1+\frac{1}{2}t-o(t)}{t}=\lim_{t\rightarrow 0}\frac{\frac{3}{2}t+o(t)}{t}=\frac{3}{2} =limt→0t1+t+o(t)−1+21t−o(t)=limt→0t23t+o(t)=23
[学习资料]
1.《高等数学（第六版）》 ，同济大学数学系 编
感谢您的点赞、收藏和关注，更欢迎您的批评、指正和指导！