Chapter 8: Q. 60 (page 693)

Find the Maclaurin series for the functions in Exercises 51–60
by substituting into a known Maclaurin series. Also, give the
interval of convergence for the series.
$\sin^{2} x$ (Hint: use the identity $\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ )

Short Answer

Expert verified

The answer is $\sin^{2} x = \sum_{k = 1}^{\infty} {(- 1)}^{k + 1} \frac{2^{2 k - 1} x^{2 k}}{(2 k)!}$

Step by step solution

Step 1. Given Information

We consider the function $f (x) = \sin^{2} x$

Simplification

The formula for $\sin^{2} x$ is as follows $\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$

The Maclaurin series for the function $c o s x$ is $\cos x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k)!} x^{2 k}$

Substituting $2 x$ with $x$ in the above equation we get $\cos 2 x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k)!} {(2 x)}^{2 k}$

Calculation

We now evaluate the value of $\frac{1 - \cos 2 x}{2}$ using the above series

Thus, $\begin{array}{rcl} \frac{1 - \cos 2 x}{2} & = & \frac{1 - \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k)!} {(2 k)}^{2 k}}{2} \\ = & \frac{1}{2} - \frac{1}{2} \sum_{k = 0}^{\infty} {(- 1)}^{k} 2 \end{array}$ ^2kx^2k(2k)! =12-12-12∑k=1∞(-1)k22kx2k(2k)!

The interval of convergence of the Maclaurin series of the given function is $ℝ$ , as the interval of convergence for the Maclaurin series of $\cos x$ is $ℝ$

Therefore the required Maclaurin series for the given function is $\sin^{2} x = \sum_{k = 1}^{\infty} {(- 1)}^{k + 1} 2$ ^2k-1x^2k(2k)!

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Find the Maclaurin series for the functions in Exercises 51–60
by substituting into a known Maclaurin series. Also, give the
interval of convergence for the series.
$\sin^{2} x$ (Hint: use the identity $\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ )

Short Answer

Step by step solution

Step 1. Given Information

Simplification

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

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Find the Maclaurin series for the functions in Exercises 51–60by substituting into a known Maclaurin series. Also, give theinterval of convergence for the series.sin2x(Hint: use the identity sin2x=12(1-cos2x))

Short Answer

Step by step solution

Step 1. Given Information

Simplification

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Calculus

Geometry

Pure Maths

Applied Mathematics

Decision Maths

Study anywhere. Anytime. Across all devices.

Find the Maclaurin series for the functions in Exercises 51–60
by substituting into a known Maclaurin series. Also, give the
interval of convergence for the series.
$\sin^{2} x$ (Hint: use the identity $\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ )