Math / Calculus

Question(1 point)
Use the third-order Taylor polynonial for $e^{x} \sin (3 x)$ at $x=0$ to approximate $e^{\frac{1}{7}} \sin (3 / 7)$ by a rational number. $e^{\frac{1}{7}} \sin (3 / 7) \approx$ $\square$

Studdy Solution

STEP 1

1. We are asked to approximate $e^{\frac{1}{7}} \sin \left(\frac{3}{7}\right)$ .
2. We will use the third-order Taylor polynomial for $e^x \sin(3x)$ centered at $x = 0$ .
3. The Taylor polynomial will be used to approximate the function value at $x = \frac{1}{7}$ .

STEP 2

1. Find the Taylor series expansion for $e^x$ up to the third order.
2. Find the Taylor series expansion for $\sin(3x)$ up to the third order.
3. Multiply the two series to get the Taylor polynomial for $e^x \sin(3x)$ .
4. Evaluate the polynomial at $x = \frac{1}{7}$ .

STEP 3

Find the Taylor series expansion for $e^x$ up to the third order:
$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

STEP 4

Find the Taylor series expansion for $\sin(3x)$ up to the third order:
$\sin(3x) = 3x - \frac{(3x)^3}{6} + \cdots$ $\sin(3x) = 3x - \frac{27x^3}{6}$ $\sin(3x) = 3x - \frac{27x^3}{6}$

STEP 5

Multiply the two series to get the Taylor polynomial for $e^x \sin(3x)$ :
$(1 + x + \frac{x^2}{2} + \frac{x^3}{6})(3x - \frac{27x^3}{6})$
We only need terms up to $x^3$ .
First, calculate the product term by term:
- Constant term: $0$ - Linear term: $3x$ - Quadratic term: $3x^2 + \frac{3x^3}{2}$ - Cubic term: $\frac{3x^3}{6} - \frac{27x^3}{6} = -\frac{24x^3}{6} = -4x^3$
Combine these:
$e^x \sin(3x) \approx 3x + 3x^2 - 4x^3$

STEP 6

Evaluate the polynomial at $x = \frac{1}{7}$ :
$3\left(\frac{1}{7}\right) + 3\left(\frac{1}{7}\right)^2 - 4\left(\frac{1}{7}\right)^3$
Calculate each term:
- $3\left(\frac{1}{7}\right) = \frac{3}{7}$ - $3\left(\frac{1}{7}\right)^2 = 3 \times \frac{1}{49} = \frac{3}{49}$ - $4\left(\frac{1}{7}\right)^3 = 4 \times \frac{1}{343} = \frac{4}{343}$
Combine these:
$\frac{3}{7} + \frac{3}{49} - \frac{4}{343}$
Convert to a common denominator and simplify:
$\frac{147}{343} + \frac{21}{343} - \frac{4}{343} = \frac{164}{343}$
The approximation of $e^{\frac{1}{7}} \sin \left(\frac{3}{7}\right)$ is:
$\boxed{\frac{164}{343}}$

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