Question

The coefficient of \( x^4 \) in the power series expansion of \( e^{\sin x} \) about \( x = 0 \) is ............. (correct up to three decimal places).

Hint:

For series expansions, remember to substitute known series expansions and combine terms to find the desired coefficient.

Answer 1

Correct Answer: -0.12 - -0.13

Using Series Expansions

We can use the known Maclaurin series for $e^u$ and $\sin x$.

$e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \frac{u^4}{4!} + \dots$

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

Let $u = \sin x$. We need terms up to $x^4$.

$$e^{\sin x} \approx 1 + (\sin x) + \frac{(\sin x)^2}{2} + \frac{(\sin x)^3}{6} + \frac{(\sin x)^4}{24}$$

Substitute the series for $\sin x$:

$$e^{\sin x} \approx 1 + \left(x - \frac{x^3}{6}\right) + \frac{1}{2}\left(x - \frac{x^3}{6}\right)^2 + \frac{1}{6}\left(x - \frac{x^3}{6}\right)^3 + \frac{1}{24}\left(x - \frac{x^3}{6}\right)^4 + \dots$$

We only collect the coefficients of the $x^4$ terms from each part:

From $\sin x$: No $x^4$ term.

From $\frac{1}{2}(\sin x)^2 = \frac{1}{2}\left(x - \frac{x^3}{6}\right)^2$:

$$\frac{1}{2} \left[ x^2 - 2(x)\left(\frac{x^3}{6}\right) + \dots \right] = \frac{1}{2} \left[ x^2 - \frac{x^4}{3} + \dots \right]$$

$$x^4 \text{ term}: \frac{1}{2} \left( -\frac{1}{3} x^4 \right) = -\frac{1}{6} x^4$$

From $\frac{1}{6}(\sin x)^3 = \frac{1}{6}\left(x - \frac{x^3}{6}\right)^3$:

The lowest power from $(\sin x)^3$ is $x^3$. The next is $x^5$. No $x^4$ term.

From $\frac{1}{24}(\sin x)^4 = \frac{1}{24}\left(x - \frac{x^3}{6}\right)^4$:

The lowest power from $(\sin x)^4$ is $x^4$.

$$\frac{1}{24} (x^4 + \dots)$$

$$x^4 \text{ term}: \frac{1}{24} x^4$$

Total coefficient of $x^4$:

$$\text{Coefficient} = \left(-\frac{1}{6}\right) + \left(\frac{1}{24}\right) = \frac{-4}{24} + \frac{1}{24} = \frac{-3}{24} = -\frac{1}{8}$$

Final Answer

The coefficient of $x^4$ is $-\frac{1}{8}$.

In decimal form:

$$-\frac{1}{8} = -0.125$$