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

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Chapter 8: Q. 41 (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
$(a) f (x) = \sin {(x)}^{3} (b) F (0) = 2$

Short Answer

Expert verified

Part (a) $\sin {(x)}^{3} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{6 k + 3}$

Part (b) $F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} (\frac{x^{6 k + 4}}{6 k + 4}) + 2$

Step by step solution

Part (a) Step 1. Given information

Let us consider the given function $f (x) = \sin {(x)}^{3}$

Part (a) Step 2. Use substitution and/or multiplication and the appropriate Maclaurin series to find the Maclaurin series for the given function f .

The maclaurin series for $g (x) = \sin x$ is :

$\sin x = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{2 k + 1}$

So,the maclaurin series for $f (x) = \sin (x^{3})$ can be founded by substituting $x$ by $x^{3}$

Thus,

role="math" localid="1650651702293" $\sin (x^{3}) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(x^{3})}^{2 k + 1} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{6 k + 3}$

Part (b) Step 1. Given information

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

Part (b) Step 2. Use Theorem 8.12 and your answer from part (a) to find the Maclaurin series for the antiderivative F=∫f that satisfies the specified initial condition

Put the value of function $f (x)$

role="math" localid="1650651633970" $F (x) = 鈭� {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{6 k + 3} d x = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} 鈭� x^{6 k + 3} d x = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} (\frac{x^{6 k + 3 + 1}}{6 k + 3 + 1}) + C = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} (\frac{x^{6 k + 4}}{6 k + 4}) + C$

Since,the initial condition is $F (0) = 2$

This implies that:

$C = 2$

Therefore,

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

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Short Answer

Step by step solution

Part (a) Step 1. Given information

Part (a) Step 2. Use substitution and/or multiplication and the appropriate Maclaurin series to find the Maclaurin series for the given function f .

Part (b) Step 1. Given information

Part (b) Step 2. Use Theorem 8.12 and your answer from part (a) to find the Maclaurin series for the antiderivative F=∫f that satisfies the specified initial condition

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