Q. 50 In Exercises 41鈥�50, find Macla... [FREE SOLUTION]

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Chapter 8: Q. 50 (page 701)

In Exercises 41鈥�50, find Maclaurin series for the given pairs of functions, using these steps:
(a) Use substitution and/or multiplication and the appropriate Maclaurin series to find the Maclaurin series for the given function f .
(b) Use Theorem 8.12 and your answer from part (a) to find the Maclaurin series for the antiderivative $F = 鈭� f$ that satisfies the specified initial condition.
$(a) f (x) = x^{4} \tan^{- 1} (3 x^{3}) (b) F (0) = 0$

Short Answer

Expert verified

Part (a) :

$x^{4} \tan^{- 1} (3 x^{3}) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} x^{6 k + 7}$

Part (b) :

$F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} (\frac{x^{6 k + 8}}{6 k + 8})$

Step by step solution

Part (a) Step 1. Given information

Let us consider the given function $f (x) = x^{4} \tan^{- 1} (3 x^{3})$

Part (a) Step 2. Maclaurin series for given function

$The Maclaurin series for g (x) = \tan^{- 1} x is : \tan^{- 1} x = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} x^{2 k + 1} So, the Maclaurin series for \tan^{- 1} (3 x^{3}) can be found by substituting x by 3 x^{3} \tan^{- 1} (3 x^{3}) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3 x^{3})}^{2 k + 1} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} x^{6 k + 3} Now, the Maclaurin series for x^{4} \tan^{- 1} (3 x^{3}) is : x^{4} \tan^{- 1} (3 x^{3}) = x^{4} {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} x^{6 k + 3} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} x^{6 k + 7}$

Part (b) Step 1. Given information

Let us consider the given function $F = 鈭� f$

Part (b) Step 2. Maclaurin series for the antiderivative

$F (x) = 鈭� ({鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} x^{6 k + 7}) dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} 鈭� x^{6 k + 7} dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} (\frac{x^{6 k + 7 + 1}}{6 k + 7 + 1}) + C = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} (\frac{x^{6 k + 8}}{6 k + 8}) + C Since the initial condition is F (0) = 0; This implies that, C = 0 Therefore, F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} (\frac{x^{6 k + 8}}{6 k + 8})$

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Short Answer

Step by step solution

Part (a) Step 1. Given information

Part (a) Step 2. Maclaurin series for given function

Part (b) Step 1. Given information

Part (b) Step 2. Maclaurin series for the antiderivative

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