Question

The third-degree Taylor polynomial about x = 0 of In(1 - x) is

Answer 1

Answer:

$\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

• Functions
• Function Notation

Calculus

Derivatives

Derivative Notation

Derivative Rule [Quotient Rule]:                                                                                $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]:                                                                                    $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

MacLaurin/Taylor Polynomials

• Approximating Transcendental and Elementary functions
• MacLaurin Polynomial:                                                                                     $\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^{(n)}(0)}{n!}x^n$
• Taylor Polynomial:                                                                                            $\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

Step 1: Define

Identify

f(x) = ln(1 - x)

Center: x = 0

n = 3

Step 2: Differentiate

1. [Function] 1st Derivative:                                                                                  $\displaystyle f'(x) = \frac{1}{x - 1}$
2. [Function] 2nd Derivative:                                                                                $\displaystyle f''(x) = \frac{-1}{(x - 1)^2}$
3. [Function] 3rd Derivative:                                                                                 $\displaystyle f'''(x) = \frac{2}{(x - 1)^3}$

Step 3: Evaluate Functions

1. Substitute in center x [Function]:                                                                     $\displaystyle f(0) = ln(1 - 0)$
2. Simplify:                                                                                                             $\displaystyle f(0) = 0$
3. Substitute in center x [1st Derivative]:                                                             $\displaystyle f'(0) = \frac{1}{0 - 1}$
4. Simplify:                                                                                                             $\displaystyle f'(0) = -1$
5. Substitute in center x [2nd Derivative]:                                                           $\displaystyle f''(0) = \frac{-1}{(0 - 1)^2}$
6. Simplify:                                                                                                             $\displaystyle f''(0) = -1$
7. Substitute in center x [3rd Derivative]:                                                            $\displaystyle f'''(0) = \frac{2}{(0 - 1)^3}$
8. Simplify:                                                                                                             $\displaystyle f'''(0) = -2$

Step 4: Write Taylor Polynomial

1. Substitute in derivative function values [MacLaurin Polynomial]:                 $\displaystyle P_3(x) = \frac{0}{0!} + \frac{-1}{1!}x + \frac{-1}{2!}x^2 + \frac{-2}{3!}x^3$
2. Simplify:                                                                                                             $\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

Topic: AP Calculus BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations

Book: College Calculus 10e