Math / Calculus

Question2. ( $21 / 2$ points) The first 3 terms of the Maclaurin series of the function $f(x)=\sin (x) \cos (x)$ are: A. $x-\frac{2 x^{3}}{3}+\frac{2 x^{4}}{15}$ B. $1-\frac{2 x^{2}}{3}+\frac{2 x^{4}}{15}$ C. $x-\frac{2 x^{3}}{3}+\frac{2 x^{5}}{15}$ D. $1-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}$

Studdy Solution

STEP 1

Assumptions
1. The function given is $f(x) = \sin(x) \cos(x)$ .
2. We need to find the first three terms of the Maclaurin series for this function.
3. The Maclaurin series is a Taylor series expansion of a function about $x = 0$ .
4. The Maclaurin series for a function $f(x)$ is given by $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$ .
5. We will calculate the derivatives of $f(x)$ and evaluate them at $x = 0$ .

STEP 2

First, express $f(x) = \sin(x) \cos(x)$ using the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$ .
$f(x) = \frac{1}{2} \sin(2x)$

STEP 3

Find the first derivative of $f(x)$ .
$f'(x) = \frac{1}{2} \cdot 2 \cos(2x) = \cos(2x)$

STEP 4

Find the second derivative of $f(x)$ .
$f''(x) = -2 \sin(2x)$

STEP 5

Find the third derivative of $f(x)$ .
$f'''(x) = -4 \cos(2x)$

STEP 6

Evaluate $f(x)$ , $f'(x)$ , $f''(x)$ , and $f'''(x)$ at $x = 0$ .
1. $f(0) = \frac{1}{2} \sin(0) = 0$
2. $f'(0) = \cos(0) = 1$
3. $f''(0) = -2 \sin(0) = 0$
4. $f'''(0) = -4 \cos(0) = -4$

STEP 7

Substitute these values into the Maclaurin series formula.
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

STEP 8

Calculate the first three terms of the Maclaurin series.
1. $f(0) = 0$
2. $f'(0)x = 1 \cdot x = x$
3. $\frac{f''(0)}{2!}x^2 = \frac{0}{2}x^2 = 0$
4. $\frac{f'''(0)}{3!}x^3 = \frac{-4}{6}x^3 = -\frac{2}{3}x^3$

STEP 9

Combine the terms to get the first three non-zero terms of the Maclaurin series.
$f(x) = x - \frac{2}{3}x^3$

STEP 10

Compare the calculated series with the given options.
The series $x - \frac{2}{3}x^3$ matches the first two terms of option A: $x - \frac{2 x^{3}}{3} + \frac{2 x^{4}}{15}$ .
Therefore, the correct answer is option A.

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Question2. ( $21 / 2$ points) The first 3 terms of the Maclaurin series of the function $f(x)=\sin (x) \cos (x)$ are: A. $x-\frac{2 x^{3}}{3}+\frac{2 x^{4}}{15}$ B. $1-\frac{2 x^{2}}{3}+\frac{2 x^{4}}{15}$ C. $x-\frac{2 x^{3}}{3}+\frac{2 x^{5}}{15}$ D. $1-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}$

Studdy Solution

STEP 1

STEP 2

First, express $f(x) = \sin(x) \cos(x)$ using the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$ .
$f(x) = \frac{1}{2} \sin(2x)$

STEP 3

Find the first derivative of $f(x)$ .
$f'(x) = \frac{1}{2} \cdot 2 \cos(2x) = \cos(2x)$

STEP 4

Find the second derivative of $f(x)$ .
$f''(x) = -2 \sin(2x)$

STEP 5

Find the third derivative of $f(x)$ .
$f'''(x) = -4 \cos(2x)$

STEP 6

Evaluate $f(x)$ , $f'(x)$ , $f''(x)$ , and $f'''(x)$ at $x = 0$ .
1. $f(0) = \frac{1}{2} \sin(0) = 0$
2. $f'(0) = \cos(0) = 1$
3. $f''(0) = -2 \sin(0) = 0$
4. $f'''(0) = -4 \cos(0) = -4$

STEP 7

Substitute these values into the Maclaurin series formula.
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

STEP 8

Calculate the first three terms of the Maclaurin series.
1. $f(0) = 0$
2. $f'(0)x = 1 \cdot x = x$
3. $\frac{f''(0)}{2!}x^2 = \frac{0}{2}x^2 = 0$
4. $\frac{f'''(0)}{3!}x^3 = \frac{-4}{6}x^3 = -\frac{2}{3}x^3$

STEP 9

Combine the terms to get the first three non-zero terms of the Maclaurin series.
$f(x) = x - \frac{2}{3}x^3$

STEP 10

Compare the calculated series with the given options.
The series $x - \frac{2}{3}x^3$ matches the first two terms of option A: $x - \frac{2 x^{3}}{3} + \frac{2 x^{4}}{15}$ .
Therefore, the correct answer is option A.

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Studdy Solution

STEP 1

STEP 2

First, express f(x)=sin⁡(x)cos⁡(x)f(x) = \sin(x) \cos(x)f(x)=sin(x)cos(x) using the trigonometric identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x)sin(2x)=2sin(x)cos(x).f(x)=12sin⁡(2x) f(x) = \frac{1}{2} \sin(2x) f(x)=21​sin(2x)

STEP 3

Find the first derivative of f(x)f(x)f(x).f′(x)=12⋅2cos⁡(2x)=cos⁡(2x) f'(x) = \frac{1}{2} \cdot 2 \cos(2x) = \cos(2x) f′(x)=21​⋅2cos(2x)=cos(2x)

STEP 4

Find the second derivative of f(x)f(x)f(x).f′′(x)=−2sin⁡(2x) f''(x) = -2 \sin(2x) f′′(x)=−2sin(2x)

STEP 5

Find the third derivative of f(x)f(x)f(x).f′′′(x)=−4cos⁡(2x) f'''(x) = -4 \cos(2x) f′′′(x)=−4cos(2x)

STEP 6

STEP 7

Substitute these values into the Maclaurin series formula.f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+⋯ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots f(x)=f(0)+f′(0)x+2!f′′(0)​x2+3!f′′′(0)​x3+⋯

STEP 8

STEP 9

Combine the terms to get the first three non-zero terms of the Maclaurin series.f(x)=x−23x3 f(x) = x - \frac{2}{3}x^3 f(x)=x−32​x3

STEP 10

Compare the calculated series with the given options.The series x−23x3x - \frac{2}{3}x^3x−32​x3 matches the first two terms of option A: x−2x33+2x415x - \frac{2 x^{3}}{3} + \frac{2 x^{4}}{15}x−32x3​+152x4​.Therefore, the correct answer is option A.

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First, express $f(x) = \sin(x) \cos(x)$ using the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$ .
$f(x) = \frac{1}{2} \sin(2x)$

Find the first derivative of $f(x)$ .
$f'(x) = \frac{1}{2} \cdot 2 \cos(2x) = \cos(2x)$

Find the second derivative of $f(x)$ .
$f''(x) = -2 \sin(2x)$

Find the third derivative of $f(x)$ .
$f'''(x) = -4 \cos(2x)$

Substitute these values into the Maclaurin series formula.
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

Combine the terms to get the first three non-zero terms of the Maclaurin series.
$f(x) = x - \frac{2}{3}x^3$

Compare the calculated series with the given options.
The series $x - \frac{2}{3}x^3$ matches the first two terms of option A: $x - \frac{2 x^{3}}{3} + \frac{2 x^{4}}{15}$ .
Therefore, the correct answer is option A.