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Chapter 10: Problem 2

Find the third Taylor polynomial at \(x=0\) of each function. $$ f(x)=\ln (1-x) $$

Short Answer

Expert verified
The third Taylor polynomial is \( P_3(x) = -x - \frac{1}{2}x^2 - \frac{1}{3}x^3 \).

Step by step solution

01

Understanding the Taylor Series

The Taylor series expansion of a function \( f(x) \) at the point \( a \) is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] In our case, we need the third Taylor polynomial of the function \( f(x) = \ln(1-x) \) at \( x = 0 \). Therefore, we will consider the first four derivatives evaluated at 0 and the polynomial terms up to the third degree.
02

Compute Initial Function Value

Evaluate \( f(x) \) at \( x = 0 \): \[ f(0) = \ln(1-0) = \ln(1) = 0 \] This is our zeroth term of the Taylor polynomial.
03

First Derivative and Evaluation

Compute the first derivative \( f'(x) \): \[ f'(x) = \frac{-1}{1-x} \] Evaluating at \( x = 0 \): \[ f'(0) = \frac{-1}{1-0} = -1 \] Thus, the first-degree term is \( -x \).
04

Second Derivative and Evaluation

Compute the second derivative \( f''(x) \): \[ f''(x) = \frac{-1}{(1-x)^2} \] Evaluating at \( x = 0 \): \[ f''(0) = \frac{-1}{(1-0)^2} = -1 \] The second-degree term is \( \frac{-1}{2!}x^2 = \frac{-1}{2}x^2 \).
05

Third Derivative and Evaluation

Compute the third derivative \( f'''(x) \): \[ f'''(x) = \frac{-2}{(1-x)^3} \] Evaluating at \( x = 0 \): \[ f'''(0) = \frac{-2}{(1-0)^3} = -2 \] The third-degree term is \( \frac{-2}{3!}x^3 = \frac{-1}{3}x^3 \).
06

Formulate the Third Taylor Polynomial

Combine all calculated terms: \[ P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \] Substituting in the values: \[ P_3(x) = 0 - x - \frac{1}{2}x^2 - \frac{1}{3}x^3 \] Thus, the third Taylor polynomial is: \[ P_3(x) = -x - \frac{1}{2}x^2 - \frac{1}{3}x^3 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
When working with Taylor polynomials, the concept of derivatives plays a crucial role. Derivatives allow us to understand how a function changes at any given point. Essentially, a derivative represents the slope of the tangent line to the function at a particular point.
  • First Derivative: Denoted as \( f'(x) \), tells us how the function changes with respect to \( x \). For our function, \( f(x) = \ln(1-x) \), the first derivative \( f'(x) \) is \( \frac{-1}{1-x} \). Evaluating this at \( x = 0 \), we find \( f'(0) = -1 \).
  • Second Derivative: Gives us information about the concavity of the function. For the second derivative, \( f''(x) = \frac{-1}{(1-x)^2} \), we again evaluate at \( x = 0 \) to get \( f''(0) = -1 \).
  • Third Derivative: This higher-order derivative helps in determining the rate of change of the rate of change (curvature). For \( f(x) = \ln(1-x) \), the third derivative \( f'''(x) = \frac{-2}{(1-x)^3} \), evaluated at \( x = 0 \), results in \( f'''(0) = -2 \).
These derivatives are key ingredients in constructing the Taylor polynomial.
Taylor Series
The Taylor series is a powerful tool in mathematics used to represent functions as infinite sums of their derivatives evaluated at a point. A Taylor series for a function \( f(x) \) centered at \( a \) is written as:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]
In this series, each term involves:
  • The function or its derivatives evaluated at the point \( a \).
  • The order of the derivative determines the power of \( (x-a) \) in the term.
  • A factorial in the denominator to balance the expansion.
For our exercise, we are looking at the Taylor series expansion at \( x = 0 \), known as the Maclaurin series. We calculate derivatives up to the third order because we need a third-degree polynomial. The role of the Taylor series is to approximate functions with polynomials which are often easier to compute with.
Polynomial Expansion
Polynomial expansion involves expressing a function as a sum of polynomial terms. In the context of Taylor polynomials, we use derivatives to create these terms. The idea is to find a polynomial that approximates the function near a particular point, offering easier computation or estimation.
For the function \( f(x) = \ln(1-x) \), we want a third-degree Taylor polynomial at \( x = 0 \). This means:
  • The zero-order term is \( f(0) = 0 \).
  • The first-order term is derived from the first derivative \( f'(0)x = -x \).
  • The second-order term comes from \( \frac{f''(0)}{2!}x^2 = \frac{-1}{2}x^2 \).
  • The third-order term is \( \frac{f'''(0)}{3!}x^3 = \frac{-1}{3}x^3 \).
These terms combined give the polynomial: \[ P_3(x) = -x - \frac{1}{2}x^2 - \frac{1}{3}x^3 \]
Polynomial expansion simplifies complex functions into an approximate polynomial form for easier handling and interpretation.

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Most popular questions from this chapter

Find the fifth Taylor polynomial for \(\sin 2 x\) by taking the fifth Taylor polynomial for \(\sin x\) (page 661 ) and replacing \(x\) by \(2 x\).

For a power series that converges to a function \(f\), $$ f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\cdots $$ show that the coefficients must be \(a_{0}=f(0)\) \(a_{1}=f^{\prime}(0), a_{2}=\frac{f^{\prime \prime}(0)}{2 !}, \ldots\) Since these are exactly the Taylor series coefficients, this shows that a power series that converges to a function is the Taylor series for that function. [Hint: For \(a_{0}\), just evaluate the given equation at \(0 .\) For \(a_{1}, a_{2}, \ldots,\) differentiate one or more times and then evaluate at \(x=0\). We are assuming the fact that a power series can be differentiated term by term inside its radius of convergence.]

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