Q. 28

Question

In Exercises 21–30 in Section 8.2 you were asked to find the fourth Maclaurin polynomial $P_{4} (x)$ for the specified function. In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_{4} (x)$ .

. $\tan x$

Step-by-Step Solution

Verified

Answer

The remainder is $\frac{2 s e c^{6} c + 11 s e c^{4} c \tan^{2} c + 2 s e c^{2} c \tan^{4} c}{15} x^{5}$

1Step 1 : Given Information

Given equation : $\tan x$

Theory used : For $n > 0, if | f^{(n + 1)} (c) | \leq 1$ for every value of $x$ , then using the Lagrange's form for the remainder, we have

$R_{n} (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} x^{n + 1}$

So, the Lagrange's form for the remainder, $R_{4} (x) is$

$R_{4} (x) = \frac{f^{(5)} (c)}{5!} x^{5}$

2Step 2 : Calculating Lagrange’s form for the corresponding remainder

Calculating the derivatives :

1) $f^{(1)} (\tan x) = s e c^{2} x$

2) $f^{(2)} (\tan x) = 2 s e c^{2} x \tan x$

3) $f^{(3)} (\tan x) = 2 (s e c^{4} x + 2 s e c^{2} x \tan x)$

4) $f^{(4)} (\tan x) = 8 (2 s e c^{6} x + 11 s e c^{4} x \tan^{2} x + 2 s e c^{2} x \tan^{4} x)$

So,

$R_{4} (x) = \frac{8 (2 s e c^{6} c + 11 s e c^{4} c \tan^{2} c + 2 s e c^{2} c \tan^{4} c)}{5!} x^{5} = \frac{2 s e c^{6} c + 11 s e c^{4} c \tan^{2} c + 2 s e c^{2} c \tan^{4} c}{15} x^{5}$

Q. 27

Q. 29

Other exercises in this chapter

Q. 26

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, R4(x)ln(1+x)

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Q. 27

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, R4(x)tan-1x

View solution

Q. 29

In Exercises 21–30 in Section 8.2 you were asked to find the fourth Maclaurin polynomial P4(x) for the specified function. In Exercises 23–32 w

View solution

Q. 30

In Exercises 21–30 in Section 8.2 you were asked to find the fourth Maclaurin polynomial P4(x) for the specified function. In Exercises 23–32 w

View solution