Expand f(x, y) = e
(2x+xy+y
2
)
in powers of x and y upto second order term.

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.

---

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.

Expand f(x, y) = e^(2x + xy + y^2) in powers of x and y up to second order term.

This page explores the Taylor series expansion of the function f(x, y) = e^(2x + xy + y^2) in powers of x and y up to the second order. Learn how to compute first and second-order partial derivatives and use them to construct the expansion, covering core concepts in multivariable calculus and Taylor series.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation