Chapter 1: Q27p (page 32)

$e^{s i n x}$

Short Answer

Expert verified

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

Step by step solution

Given information.

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

Definition of Maclaurin series.

The definition of the Maclaurin series is defined.

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

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

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

State the Maclaurin series.

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

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

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

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

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

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

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

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

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$e^{s i n x}$

Short Answer

Step by step solution

Given information.

Definition of Maclaurin series.

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

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

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Short Answer

Step by step solution

Given information.

Definition of Maclaurin series.

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

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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$e^{s i n x}$