Chapter 1: Q16MP (page 45)

Find the Maclaurin series $\frac{1}{\sqrt{1 + s i n (x)}}$ .

Short Answer

Expert verified

The answer is $\frac{1}{\sqrt{1 + \sin (x)}} = 1 - \frac{x}{2} + \frac{3 x^{2}}{8} - \frac{11 x^{3}}{48} + . . .$

Step by step solution

Given information.

The Maclaurin expansion of the function $\frac{1}{\sqrt{1 + \sin (x)}}$ is to be evaluated.

Definition of the Maclaurin series

The definition of Maclaurin series is defined as:

$\frac{1}{\sqrt{1 + \sin (x)}} = 1 - \frac{1}{2} x + \frac{3}{8} x^{2} - \frac{5}{16} x^{3} + . . . \sin (x) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + . . .$

Begin by stating the Maclaurin series.

State the Maclaurin series as:

$\frac{1}{\sqrt{1 + \sin (x)}} = 1 - \frac{1}{2} x + \frac{3}{8} x^{2} - \frac{5}{16} x^{3} + . . . \sin (x) = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + . . .$

Combine the two Maclaurin series and evaluate.

Combine the second Maclaurin series in the first and evaluate.

$\frac{1}{\sqrt{1 + \sin (x)}} = 1 - \frac{1}{2} [x - \frac{x^{3}}{6} + \frac{x^{5}}{120} + . . .] + \frac{3}{8} {[x - \frac{x^{3}}{6} + . . .]}^{2} - \frac{5}{16} {[x + . . .]}^{3} + . . . \frac{1}{\sqrt{1 + \sin (x)}} = 1 + \frac{- 240 x + 180 x^{2} - 110 x^{3}}{480} \frac{1}{\sqrt{1 + \sin (x)}} = 1 - \frac{x}{2} + \frac{3 x^{2}}{8} + \frac{11 x^{3}}{48} + . . .$

Thus, the final answer is $\frac{1}{\sqrt{1 + \sin (x)}} = 1 - \frac{x}{2} + \frac{3 x^{2}}{8} + \frac{11 x^{3}}{48} + . . .$

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Find the Maclaurin series $\frac{1}{\sqrt{1 + s i n (x)}}$ .

Short Answer

Step by step solution

Given information.

Definition of the Maclaurin series

Begin by stating the Maclaurin series.

Combine the two Maclaurin series and evaluate.

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Find the Maclaurin series 11+sin(x).

Short Answer

Step by step solution

Given information.

Definition of the Maclaurin series

Begin by stating the Maclaurin series.

Combine the two Maclaurin series and evaluate.

One App. One Place for Learning.

Most popular questions from this chapter

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Find the Maclaurin series $\frac{1}{\sqrt{1 + s i n (x)}}$ .