Math Problem Statement

Expand to 3rd degree (e^x)*log(1+y)

Solution

To expand exlog(1+y)e^x \cdot \log(1 + y) up to the 3rd degree in terms of xx and yy, we'll use the series expansions of exe^x and log(1+y)\log(1 + y).

Step 1: Taylor series expansion of exe^x

The Taylor series of exe^x around x=0x = 0 is: ex=1+x+x22!+x33!+O(x4)e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \mathcal{O}(x^4) So, up to the 3rd degree: ex=1+x+x22+x36e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}

Step 2: Taylor series expansion of log(1+y)\log(1 + y)

The Taylor series of log(1+y)\log(1 + y) around y=0y = 0 is: log(1+y)=yy22+y33+O(y4)\log(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} + \mathcal{O}(y^4) So, up to the 3rd degree: log(1+y)=yy22+y33\log(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3}

Step 3: Multiply the two expansions

Now, we multiply exe^x and log(1+y)\log(1 + y) up to the 3rd degree: exlog(1+y)=(1+x+x22+x36)(yy22+y33)e^x \cdot \log(1 + y) = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) \cdot \left( y - \frac{y^2}{2} + \frac{y^3}{3} \right) We'll expand this multiplication, keeping terms up to the 3rd degree.

1. Multiply 11 by log(1+y)\log(1 + y):

1(yy22+y33)=yy22+y331 \cdot \left( y - \frac{y^2}{2} + \frac{y^3}{3} \right) = y - \frac{y^2}{2} + \frac{y^3}{3}

2. Multiply xx by log(1+y)\log(1 + y):

x(yy22+y33)=xyxy22+xy33x \cdot \left( y - \frac{y^2}{2} + \frac{y^3}{3} \right) = xy - \frac{xy^2}{2} + \frac{xy^3}{3}

3. Multiply x22\frac{x^2}{2} by log(1+y)\log(1 + y):

x22(yy22+y33)=x2y2x2y24+x2y36\frac{x^2}{2} \cdot \left( y - \frac{y^2}{2} + \frac{y^3}{3} \right) = \frac{x^2 y}{2} - \frac{x^2 y^2}{4} + \frac{x^2 y^3}{6}

4. Multiply x36\frac{x^3}{6} by log(1+y)\log(1 + y):

x36(yy22+y33)=x3y6x3y212+x3y318\frac{x^3}{6} \cdot \left( y - \frac{y^2}{2} + \frac{y^3}{3} \right) = \frac{x^3 y}{6} - \frac{x^3 y^2}{12} + \frac{x^3 y^3}{18}

Step 4: Collect all terms up to the 3rd degree

Now, combining all terms up to the 3rd degree: exlog(1+y)=yy22+y33+xyxy22+xy33+x2y2x2y24+x2y36+x3y6x3y212+x3y318e^x \cdot \log(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} + xy - \frac{xy^2}{2} + \frac{xy^3}{3} + \frac{x^2 y}{2} - \frac{x^2 y^2}{4} + \frac{x^2 y^3}{6} + \frac{x^3 y}{6} - \frac{x^3 y^2}{12} + \frac{x^3 y^3}{18}

This is the expansion of exlog(1+y)e^x \cdot \log(1 + y) up to the 3rd degree.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

  1. How does the radius of convergence affect the series expansion?
  2. What happens when expanding to higher degrees (e.g., 4th or 5th degree)?
  3. Can you explain the error term in Taylor expansions?
  4. How do you derive the expansion for more complex functions?
  5. How does a multivariable Taylor series expansion work?

Tip: To improve accuracy, always keep track of the highest degree terms needed when multiplying series expansions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Series Expansion
Taylor Series
Logarithmic Functions
Exponential Functions

Formulas

Taylor series for e^x: e^x = 1 + x + x^2/2! + x^3/3! + ...
Taylor series for log(1+y): log(1+y) = y - y^2/2 + y^3/3 + ...

Theorems

Taylor Series Expansion

Suitable Grade Level

Grades 11-12, Undergraduate Calculus