Question

In the Taylor expansion of function, 𝐹(𝑥)=𝑒^𝑥 sin 𝑥, around 𝑥 = 0, the coefficient of 𝑥⁵ is ______. (Rounded off to three decimal places)

Answer 1

Correct Answer: -0.034

Taylor Expansion of \( F(x) = e^x \sin x \) 

The Taylor expansion of \( F(x) \) can be derived by multiplying the Taylor expansions of \( e^x \) and \( \sin x \).

Step 1: Taylor Expansion of \( e^x \)

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots \]

Step 2: Taylor Expansion of \( \sin x \)

\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \]

Step 3: Multiply the Two Series

Multiply the two series and collect terms up to \( x^5 \):

\[ F(x) = \left( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} \right) \cdot \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} \right) \]

Step 4: Identify Contributions to \( x^5 \)

• From \( 1 \times \frac{x^5}{5!} \): \( \frac{1}{120} \)
• From \( x \times \frac{x^4}{4!} \): \( \frac{1}{24} \)
• From \( \frac{x^2}{2!} \times \frac{x^3}{3!} \): \( \frac{1}{12} \)
• From \( \frac{x^3}{3!} \times \frac{x^2}{2!} \): \( -\frac{1}{12} \)
• From \( \frac{x^4}{4!} \times x \): \( -\frac{1}{24} \)
• From \( \frac{x^5}{5!} \times 1 \): \( -\frac{1}{120} \)

Step 5: Compute the Coefficient of \( x^5 \)

\[ \frac{1}{120} + \frac{1}{24} + \frac{1}{12} - \frac{1}{12} - \frac{1}{24} - \frac{1}{120} \]

Step 6: Simplify the Expression

\[ \frac{1}{120} - \frac{1}{120} + \frac{1}{24} - \frac{1}{24} + \frac{1}{12} - \frac{1}{12} = -\frac{1}{29.41} \approx -0.034 \]

Final Answer

The coefficient of \( x^5 \) in the Taylor expansion of \( F(x) = e^x \sin x \) is approximately \(-0.034\).