Math Problem Statement

Using Taylor’s series expand e x log (1 + y) up to the second order terms about (0, 0).

Solution

To expand exlog(1+y)e^x \log(1 + y) up to the second-order terms around (0,0)(0, 0) using a Taylor series, we follow these steps:

1. Function Definition

Let f(x,y)=exlog(1+y)f(x, y) = e^x \log(1 + y).

We want to expand this function around (x,y)=(0,0)(x, y) = (0, 0), including terms up to second order.

2. Taylor Expansion Formula

The Taylor expansion of a function of two variables f(x,y)f(x, y) around (0,0)(0, 0) is given by:

f(x,y)f(0,0)+fx(0,0)x+fy(0,0)y+12(2fx2(0,0)x2+22fxy(0,0)xy+2fy2(0,0)y2)f(x, y) \approx f(0, 0) + \frac{\partial f}{\partial x}(0, 0) x + \frac{\partial f}{\partial y}(0, 0) y + \frac{1}{2} \left( \frac{\partial^2 f}{\partial x^2}(0, 0) x^2 + 2 \frac{\partial^2 f}{\partial x \partial y}(0, 0) xy + \frac{\partial^2 f}{\partial y^2}(0, 0) y^2 \right)

3. Compute Function and Derivatives at (0,0)(0, 0)

f(0,0)f(0, 0):

f(0,0)=e0log(1+0)=10=0f(0, 0) = e^0 \log(1 + 0) = 1 \cdot 0 = 0

First-order partial derivatives:

  • fx=exlog(1+y)\frac{\partial f}{\partial x} = e^x \log(1 + y), so at (0,0)(0, 0): fx(0,0)=e0log(1+0)=10=0\frac{\partial f}{\partial x}(0, 0) = e^0 \log(1 + 0) = 1 \cdot 0 = 0

  • fy=ex1+y\frac{\partial f}{\partial y} = \frac{e^x}{1 + y}, so at (0,0)(0, 0): fy(0,0)=e01+0=1\frac{\partial f}{\partial y}(0, 0) = \frac{e^0}{1 + 0} = 1

Second-order partial derivatives:

  • 2fx2=exlog(1+y)\frac{\partial^2 f}{\partial x^2} = e^x \log(1 + y), so at (0,0)(0, 0): 2fx2(0,0)=e0log(1+0)=10=0\frac{\partial^2 f}{\partial x^2}(0, 0) = e^0 \log(1 + 0) = 1 \cdot 0 = 0

  • 2fy2=ex(1+y)2\frac{\partial^2 f}{\partial y^2} = -\frac{e^x}{(1 + y)^2}, so at (0,0)(0, 0): 2fy2(0,0)=e0(1+0)2=1\frac{\partial^2 f}{\partial y^2}(0, 0) = -\frac{e^0}{(1 + 0)^2} = -1

  • 2fxy=ex1+y\frac{\partial^2 f}{\partial x \partial y} = \frac{e^x}{1 + y}, so at (0,0)(0, 0): 2fxy(0,0)=e01+0=1\frac{\partial^2 f}{\partial x \partial y}(0, 0) = \frac{e^0}{1 + 0} = 1

4. Combine the Terms

Now, using the Taylor expansion formula, we plug in the values we've computed:

f(x,y)0+0x+1y+12(0x2+21xy+(1)y2)f(x, y) \approx 0 + 0 \cdot x + 1 \cdot y + \frac{1}{2} \left( 0 \cdot x^2 + 2 \cdot 1 \cdot xy + (-1) \cdot y^2 \right)

Simplifying:

f(x,y)y+xy12y2f(x, y) \approx y + xy - \frac{1}{2} y^2

Final Answer:

The second-order Taylor expansion of exlog(1+y)e^x \log(1 + y) around (0,0)(0, 0) is:

f(x,y)y+xy12y2f(x, y) \approx y + xy - \frac{1}{2} y^2

Would you like to explore the details further or ask any questions about the steps?

Here are 5 questions to expand on this topic:

  1. How do we generalize Taylor series for higher dimensions?
  2. What is the significance of second-order terms in a Taylor expansion?
  3. Can you expand exlog(1+y)e^x \log(1 + y) to third order?
  4. How does the mixed partial derivative 2fxy\frac{\partial^2 f}{\partial x \partial y} contribute to the expansion?
  5. Why is the expansion centered around (0,0)(0, 0) instead of another point?

Tip: When expanding around a point, ensure you calculate all partial derivatives correctly up to the desired order.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Taylor Series Expansion
Partial Derivatives
Multivariable Calculus

Formulas

f(x, y) ≈ f(0, 0) + (∂f/∂x)(0, 0) x + (∂f/∂y)(0, 0) y + ½ [(∂²f/∂x²)(0, 0) x² + 2 (∂²f/∂x∂y)(0, 0) xy + (∂²f/∂y²)(0, 0) y²]

Theorems

Taylor's Theorem

Suitable Grade Level

University Level (Calculus II / Multivariable Calculus)