2013年考研数二第15题解析：等价无穷小

解析

方法一

$$\cos x \sim 1-\frac{1}{2} x^{2}.$$

$$\cos 2x \sim 1- \frac{1}{2} (2x)^{2} \sim 1-2 x^{2};$$

$$\cos 3x \sim 1-\frac{1}{2}(3x)^{2} \sim 1-\frac{9}{2}x^{2}.$$

$$1-\cos x \cdot \cos 2x \cdot \cos 3x =$$

$$1-(1-\frac{1}{2}x^{2})\cdot (1-2x^{2}) \cdot (1-\frac{9}{2}x^{2}) =$$

$$1-(1-2x^{2}-\frac{1}{2}x^{2} + x^{4})(1-\frac{9}{2}x^{2}).$$

$$1-(1-2x^{2}-\frac{1}{2}x^{2} + x^{4})(1-\frac{9}{2}x^{2}) =$$

$$1-(1-2x^{2}-\frac{1}{2}x^{2})(1-\frac{9}{2}x^{2})$$

$$1-(1-\frac{5}{2}x^{2})(1-\frac{9}{2}x^{2}) =$$

$$\frac{5+9}{2}x^{2} – \frac{5\cdot 9}{4}x^{4}.$$

$$\frac{5+9}{2}x^{2} – \frac{5\cdot 9}{4}x^{4} =$$

$$\frac{14}{2}x^{2} = 7x^{2}.$$

$$\frac{7x^{2}}{ax^{n}} = 1.$$

$$n=2, a=7.$$

方法二

$$\cos x \cos 2x \cos 3x =$$

$$(\cos 2x \cos x) \cos 3x=$$

$$\frac{1}{2}(\cos 3x + \cos x) \cos 3x =$$

$$\frac{1}{2} \cos 3x \cdot \cos 3x + \frac{1}{2} \cos 3x \cos x =$$

$$\frac{1}{2} \cdot \frac{1}{2} (\cos 6x + 0) + \frac{1}{2} \cdot \frac{1}{2} (\cos 4x + \cos 2x) =$$

$$\frac{1}{4} \cos 6x + \frac{1}{4} \cos 4x + \frac{1}{4} \cos 2x.$$

$$\frac{1-\cos x \cdot \cos 2x \cdot \cos 3x}{ax^{n}} =$$

$$\frac{1- \frac{1}{4} \cos 6x – \frac{1}{4} \cos 4x – \frac{1}{4} \cos 2x}{ax^{n}} \Rightarrow 洛必达 \Rightarrow$$

$$\frac{\frac{6}{4} \sin 6x + \sin 4x + \frac{1}{2} \sin 2x}{a \cdot n x^{n-1}} \Rightarrow 等价无穷小 \Rightarrow$$

$$\frac{9x + 4x + x}{a \cdot nx^{n-1}} =$$

$$\frac{14x}{a \cdot nx^{n-1}} = 1$$

$$n=2,a=7.$$

方法三

$$1-\cos x \cos 2x \cos 3x =$$

$$1-\cos x \cos 2x + \cos x \cos 2x – \cos x \cos 2x \cos 3x =$$

$$(1-\cos x) + (\cos x – \cos x \cos 2x) +$$

$$(\cos x \cos 2x – \cos x \cos 2x \cos 3x) =$$

$$(1-\cos x) + \cos x(1- \cos 2x) +$$

$$\cos x \cos 2x (1-\cos 3x).$$

$$\cos x = 1;$$

$$\cos x \cos 2x = 1 \cdot 1 = 1.$$

$$(1-\cos x) + \cos x(1- \cos 2x) +$$

$$\cos x \cos 2x (1-\cos 3x)=$$

$$(1-\cos x) + (1- \cos 2x) + (1-\cos 3x) =$$

$$\frac{1}{2}x^{2} + \frac{1}{2}(2x)^{2} + \frac{1}{2}(3x)^{2} =$$

$$\frac{1+4+9}{2}x^{2} = 7x^{2}.$$

$$\frac{7x^{2}}{ax^{n}} = 1.$$

$$n=2, a=7.$$