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Chapter 1: Q27p (page 32)

$e^{s i n x}$

Short Answer

Expert verified

$e^{\sin (x)} = 1 + x + \frac{x^{2}}{2} - \frac{x^{4}}{8} + 鈥�$

Step by step solution

01

Given information.

The Maclaurin expansion of the function $e^{\sin x}$ is to be evaluated.

02

Definition of Maclaurin series.

The definition of the Maclaurin series is defined.

$s i n (x) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + 鈥�$

$e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + 鈥�$

03

Part-A Step 3: Begin by stating the Maclaurin series.

State the Maclaurin series.

$\sin (x) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + 鈥�$

$e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + 鈥�$

04

Part-B Step 3: Use the first Maclaurin series to evaluate.

Substitute $x$ with $[x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + 鈥�]$ in the second series and evaluate.

$e^{\sin (x)} = 1 + [x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + 鈥�] + \frac{{[x - \frac{x^{3}}{6} + \frac{x^{3}}{120} + 鈥�]}^{2}}{2} + \frac{{[x - \frac{x^{3}}{6} + \frac{x^{0}}{120} + 鈥�]}^{3}}{6} + \frac{{[x - \frac{x^{3}}{6} + \frac{x^{3}}{120} + 鈥�]}^{4}}{24} + 鈥�$

$e^{\sin (x)} = 1 + \frac{155520 x}{155520} + \frac{77760 x^{2}}{155520} - \frac{19440 x^{4}}{155520} + 鈥�$

$e^{\sin (x)} = 1 + x + \frac{x^{2}}{2} - \frac{x^{4}}{8} + 鈥�$

Thus, the final answer is $e^{\sin (x)} = 1 + x + \frac{x^{2}}{2} - \frac{x^{4}}{8} + 鈥�$

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Most popular questions from this chapter

$In \cos x$ Hints:Method1:Write $\cos x = 1 + (\cos x - 1) = 1 + u$ ;use the series you know for $In (1 + u)$ ;replace u by the Maclaurin series for $(\cos x - 1)$

Method2: $In \cos x =$ $- {鈭�}_{0}^{x} \tan udu$ Use the series of Example 2 in method B.

Find a two-term approximation for each of the following integrals and an error bound for the given t interval ${鈭�}_{0}^{t} \sqrt{X} e^{- x} dx, 0 < t < 0.01$

$\cosh x = \frac{e^{x} + e^{- x}}{2}$

In the following problems, find the limit of the given sequence as

$n 鈫� 鈭�$

$f (x) = e^{x}, a = 3 .$

See all solutions

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