To expand the function $f(x, y) = e^{2x + xy + y^2}$ in powers of $x$ and $y$ up to the second order, we will use the Taylor series expansion around $(x, y) = (0, 0)$.

The general form of the Taylor series expansion of a function $f(x, y)$ around $(0, 0)$ is:

$f(x, y) \approx f(0, 0) + \left( \frac{\partial f}{\partial x} \bigg|_{(0,0)} \right) x + \left( \frac{\partial f}{\partial y} \bigg|_{(0,0)} \right) y + \frac{1}{2} \left( \frac{\partial^2 f}{\partial x^2} \bigg|_{(0,0)} \right) x^2 + \frac{\partial^2 f}{\partial x \partial y} \bigg|_{(0,0)} xy + \frac{1}{2} \left( \frac{\partial^2 f}{\partial y^2} \bigg|_{(0,0)} \right) y^2 + \cdots$

Let's go step by step:

Step 1: Evaluate $f(0, 0)$

$f(0, 0) = e^{2(0) + (0)(0) + 0^2} = e^0 = 1$

Step 2: First-order partial derivatives

• $\frac{\partial f}{\partial x} = \frac{d}{dx} \left( e^{2x + xy + y^2} \right) = e^{2x + xy + y^2} \cdot (2 + y)$

  At $(0, 0)$: $\frac{\partial f}{\partial x} \bigg|_{(0, 0)} = 1 \cdot (2 + 0) = 2$

• $\frac{\partial f}{\partial y} = \frac{d}{dy} \left( e^{2x + xy + y^2} \right) = e^{2x + xy + y^2} \cdot (x + 2y)$

  At $(0, 0)$: $\frac{\partial f}{\partial y} \bigg|_{(0, 0)} = 1 \cdot (0 + 2 \cdot 0) = 0$

Step 3: Second-order partial derivatives

• $\frac{\partial^2 f}{\partial x^2} = \frac{d}{dx} \left( e^{2x + xy + y^2} \cdot (2 + y) \right) = e^{2x + xy + y^2} \cdot (2 + y)^2$

  At $(0, 0)$: $\frac{\partial^2 f}{\partial x^2} \bigg|_{(0, 0)} = 1 \cdot (2 + 0)^2 = 4$

• $\frac{\partial^2 f}{\partial x \partial y} = \frac{d}{dy} \left( e^{2x + xy + y^2} \cdot (2 + y) \right) = e^{2x + xy + y^2} \cdot (2 + y)(x + 2y)$

  At $(0, 0)$: $\frac{\partial^2 f}{\partial x \partial y} \bigg|_{(0, 0)} = 1 \cdot (2 + 0)(0 + 2 \cdot 0) = 0$

• $\frac{\partial^2 f}{\partial y^2} = \frac{d}{dy} \left( e^{2x + xy + y^2} \cdot (x + 2y) \right) = e^{2x + xy + y^2} \cdot (x + 2y)^2 + 2 e^{2x + xy + y^2}$

  At $(0, 0)$: $\frac{\partial^2 f}{\partial y^2} \bigg|_{(0, 0)} = 1 \cdot (0 + 2 \cdot 0)^2 + 2 \cdot 1 = 2$

Step 4: Taylor series up to second order

Now we can write the expansion:

$f(x, y) \approx 1 + 2x + \frac{1}{2} \cdot 4 x^2 + \frac{1}{2} \cdot 2 y^2$

Simplifying:

$f(x, y) \approx 1 + 2x + 2x^2 + y^2$

This is the expansion of $f(x, y) = e^{2x + xy + y^2}$ in powers of $x$ and $y$ up to second-order terms.

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Do you need more details or have any questions about the process?

Here are 5 related questions:

1. How would the expansion change if the function was $e^{x^2 + y^2}$?
2. What are higher-order terms in the Taylor expansion of this function?
3. Can you expand a similar function like $f(x, y) = e^{x + 2y + x^2}$?
4. How does the Taylor expansion of a multivariable function compare with a single-variable function?
5. What is the role of partial derivatives in obtaining a Taylor series for a multivariable function?

Tip: The Taylor series helps approximate functions using polynomials, making complex functions easier to work with in local regions around the expansion point.