Answered: f(x) = tan 2.a = 0

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter3: The Derivative

Section3.4: Definition Of The Derivative

Problem 36E

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Question

Taylor series and interval of convergence
a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.
b. Write the power series using summation notation.
c. Determine the interval of convergence of the series.

Expert Solution

Step 1

The given function is $f (x) = \tan^{- 1} \frac{x}{2}$ , a=0.

(a) To find: First four terms of Taylor's series/Maclaurin series centered at a.

(b) To Write: Power series using summation notation.

Step 2

Taylor's series for a function f is given by,

$f (x) = f (0) + \frac{1}{1!} f' (0) (x - 0) + \frac{1}{2!} f'' (0) {(x - 0)}^{2} + \frac{1}{3!} f''' (0) {(x - 0)}^{3} + . . .$ (i)

$f (x) = \tan^{- 1} \frac{x}{2}$ .

$a_{0} = f (0) = 0$

$f' (x) = \frac{1}{2} \frac{1}{1 + \frac{x^{2}}{4}} = \frac{2}{4 + x^{2}}$ ,

$f' (0) = \frac{4}{4} = 1$

$a_{1} = f' (0) = 1$

$f'' (x) = \frac{- 2 (2 x)}{{(4 + x^{2})}^{2}} = \frac{- 4 x}{{(4 + x^{2})}^{2}}$

$f'' (0) = 0$ ,

$a_{2} = \frac{1}{2!} f'' (0) = 0$

$f''' (x) = \frac{{(4 + x^{2})}^{2} \times (- 4) - (- 4 x) 2 (4 + x^{2}) \times 2 x}{{(4 + x^{2})}^{4}} = \frac{- 4 (4 + x^{2}) + 16 x}{{(4 + x^{2})}^{3}}$

$f''' (0) = \frac{- 16}{64} = \frac{- 1}{4}$

$a_{3} = \frac{1}{3!} f''' (0) = \frac{- 1}{4 \times 3!} = \frac{- 1}{24}$ ,

$f^{(1 v)} (x) = \frac{{(4 + x^{2})}^{3} ((- 8 x) + 16) - (- 4 (4 + x^{2}) + 16 x) 3 {(4 + x^{2})}^{2} \times 2 x}{{(4 + x^{2})}^{6}} = - \frac{- 48 x (x^{2} - 4)}{{(4 + x^{2})}^{4}}$

$f^{(i v)} (0) = 0$

$a_{4} = \frac{1}{4!} f^{(i v)} (0) = 0$ ,

$f^{(v)} (x) = \frac{48 (5 x^{4} - 40 x^{2} + 16)}{{(x^{2} + 4)}^{5}}$

$f^{v} (0) = \frac{48 \times 16}{4^{5}} = \frac{3 \times 4^{2} \times 4^{2}}{4^{5}} = \frac{3}{4}$ .

$a_{5} = \frac{1}{5!} f^{(v)} (0) = \frac{1}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{3}{4} = \frac{1}{160}$ .

$f^{v i} (x) = \frac{- 480 x (3 x^{4} - 40 x^{2} + 48)}{{(x^{2} + 4)}^{6}}$

$f^{(v i)} (0) = 0$ .

$a_{6} = \frac{1}{6!} f^{(v i)} (0) = 0$ .

$f^{(v i i)} (x) = \frac{1440 (7 x^{6} - 140 x^{4} + 336 x^{2} - 64)}{{(x^{2} + 4)}^{7}}$

$f^{(v i i)} (0) = \frac{1440 \times (- 64)}{4^{7}} = - \frac{3^{2} \times 4^{2} \times 10 \times 4^{3}}{4^{7}} = - \frac{90}{4^{2}}$

$a_{7} = \frac{1}{7!} f^{(v i i)} (0) = \frac{- 90}{4^{2}} \times \frac{1}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{- 3 \times 3 \times 2 \times 5}{4^{2} \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{- 1}{4 \times 2 \times 7 \times 4^{2}} = - \frac{1}{896}$ .

Thus, From (i),

$f (x) = \frac{x}{2} - \frac{x^{3}}{24} + \frac{x^{5}}{160} - \frac{x^{7}}{896}$ +...

these are the first four terms of Taylor's expansion of the function $f (x) = \tan^{- 1} \frac{x}{2}$ .

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