利用x=0处二阶泰勒级数（麦克劳林级数）与拉格朗日余项确定tan0.7的区间

Let’s walk through how to find the interval where tan(0.7) must lie using the second-order Maclaurin series (Taylor series centered at x=0) and the Lagrange remainder term.

First, we calculate the necessary derivatives of ( f(x) = \tan x ) and their values at x=0:

• ( f(x) = \tan x ), so ( f(0) = 0 )
• ( f'(x) = 1 + \tan^2 x = \sec^2 x ), so ( f'(0) = 1 )
• ( f''(x) = 2\tan x (1 + \tan^2 x) = 2\sec^2 x \tan x ), so ( f''(0) = 0 )

The second-order Maclaurin polynomial ( P_2(x) ) simplifies to:
[ P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 = 0 + 1 \cdot x + 0 = x ]

Evaluating this at x=0.7 gives ( P_2(0.7) = 0.7 ).

Next, we use the Lagrange remainder term to quantify the error between ( \tan(0.7) ) and ( P_2(0.7) ). The Lagrange remainder ( R_2(x) ) for the second-order series is:
[ R_2(x) = \frac{f'''(c)}{3!}x^3 ]
where ( c ) is some value between 0 and x (so 0 < c < 0.7 here).

First, compute the third derivative of ( f(x) ):
[ f'''(x) = 2(1 + \tan^2 x)(1 + 3\tan^2 x) ]

Bounding ( f'''(c) ) #

Since ( \tan x ) is strictly increasing on ( (0, \pi/2) ) (and 0.7 radians is less than ( \pi/4 \approx 0.785 ) radians, so well within this interval), ( \tan(c) ) increases as c goes from 0 to 0.7.

• At c=0: ( f'''(0) = 2(1 + 0)(1 + 0) = 2 )
• For c < 0.7, ( \tan(c) < \tan(0.7) ). We don’t know ( \tan(0.7) ), but we know ( \tan(\pi/4) = 1 ), and since 0.7 < π/4, ( \tan(c) < 1 ). Using this upper bound:
  [ f'''(c) < 2(1 + 1^2)(1 + 3 \cdot 1^2) = 2(2)(4) = 16 ]

We can also note ( f'''(x) ) is strictly increasing on ( (0, \pi/2) ) (all terms in its expression are positive and increasing with x), so ( 2 < f'''(c) < 16 ) for 0 < c < 0.7.

Calculating the Remainder Bounds #

Plug these bounds into the remainder formula for x=0.7:

• Lower bound for ( R_2(0.7) ): ( \frac{2}{6} \cdot (0.7)^3 = \frac{1}{3} \cdot 0.343 \approx 0.1143 )
• Upper bound for ( R_2(0.7) ): ( \frac{16}{6} \cdot 0.343 \approx 2.6667 \cdot 0.343 \approx 0.9147 )

Final Interval for tan(0.7) #

Since ( \tan(0.7) = P_2(0.7) + R_2(0.7) ), substitute the bounds:
[ 0.7 + 0.1143 < \tan(0.7) < 0.7 + 0.9147 ]
[ 0.8143 < \tan(0.7) < 1.6147 ]

If you want a tighter upper bound, you can use the next term in the Maclaurin series (which is positive for x>0) to estimate ( \tan(0.7) < 0.8367 ), leading to a tighter remainder upper bound of ~0.6025 and an interval of ( 0.8143 < \tan(0.7) < 1.3025 ). But the wider interval above is still valid and easier to derive without extra approximations.

内容的提问来源于stack exchange，提问作者A. Lindberg