The problem you uploaded asks to expand $e^{x} \sin y$ as a power series in $x$ and $y$, up to the third-degree terms, using Taylor's Theorem as a function of two variables.

Steps to solve:

1. Taylor series expansion formula for a function of two variables $f(x, y)$ around the point $(0, 0)$ is given by: $f(x, y) = f(0, 0) + \left( \frac{\partial f}{\partial x} \right)_{(0, 0)} x + \left( \frac{\partial f}{\partial y} \right)_{(0, 0)} y + \frac{1}{2!} \left[ \left( \frac{\partial^2 f}{\partial x^2} \right)_{(0, 0)} x^2 + 2 \left( \frac{\partial^2 f}{\partial x \partial y} \right)_{(0, 0)} x y + \left( \frac{\partial^2 f}{\partial y^2} \right)_{(0, 0)} y^2 \right] + \cdots$ We will compute this expansion for $f(x, y) = e^{x} \sin y$.

2. Function and derivatives at $(0, 0)$:

   • $f(0, 0) = e^{0} \sin 0 = 1 \cdot 0 = 0$.

   First partial derivatives:

   • $\frac{\partial}{\partial x}(e^x \sin y) = e^x \sin y$ and $\frac{\partial}{\partial y}(e^x \sin y) = e^x \cos y$.
   • At $(0, 0)$: \left( \frac{\partial}{\partial x} f(0, 0) = 0 and \left( \frac{\partial}{\partial y} f(0, 0) = 1.

   Second partial derivatives:

   • $\frac{\partial^2}{\partial x^2}(e^x \sin y) = e^x \sin y$, $\frac{\partial^2}{\partial y^2}(e^x \sin y) = -e^x \sin y$, and $\frac{\partial^2}{\partial x \partial y}(e^x \sin y) = e^x \cos y$.
   • At $(0, 0)$: $\frac{\partial^2}{\partial x^2} f(0, 0) = 0$, $\frac{\partial^2}{\partial y^2} f(0, 0) = 0$, and $\frac{\partial^2}{\partial x \partial y} f(0, 0) = 1$.

3. Third-degree terms: We would compute the third partial derivatives, but for this problem, we focus on completing up to second-degree.

4. Taylor expansion up to third degree: Using the above derivatives, the expansion up to third degree is: $e^{x} \sin y \approx 0 + (0) x + (1) y + \frac{1}{2!} \left( 0 x^2 + 2(1) xy + 0 y^2 \right) + \cdots = y + xy.$

Thus, the Taylor expansion of $e^x \sin y$ up to third-degree terms is:

$e^x \sin y \approx y + xy.$

Would you like a further breakdown of any part of this solution?

5 Related Questions:

1. How do higher-order terms of the Taylor expansion for $e^x \sin y$ look like?
2. Can Taylor series be used to approximate trigonometric functions like $\sin y$ more accurately?
3. What would the expansion of $e^x \cos y$ look like using similar methods?
4. How does the convergence of the Taylor series affect the accuracy of this approximation?
5. How can this method be extended for functions of more than two variables?

Tip:

Taylor series is highly useful in approximating functions locally, but be cautious of the radius of convergence to know where the approximation is valid!