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Chapter 11: Problem 25

Find Maclaurin's formula with remainder for the given \(f(x)\) and \(n\). $$ f(x)=1 /(x-1)^{2}, \quad n=5 $$

Short Answer

Expert verified
The Maclaurin polynomial is \( P_5(x) = 1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 \) with remainder \( R_5(x) = \frac{5040(c-1)^{-8}}{6!} x^6 \).

Step by step solution

01

Understand the Problem

We need to find the Maclaurin series expansion for the function \( f(x) = \frac{1}{(x-1)^2} \) up to \( n = 5 \) with the remainder term. Recall that Maclaurin's series is the Taylor series around \( a = 0 \).
02

Calculate the Derivatives

Compute the first few derivatives of \( f(x) = \frac{1}{(x-1)^2} \).1. \( f(x) = (x-1)^{-2} \)2. \( f'(x) = -2(x-1)^{-3} \)3. \( f''(x) = 6(x-1)^{-4} \)4. \( f'''(x) = -24(x-1)^{-5} \)5. \( f^{(4)}(x) = 120(x-1)^{-6} \)6. \( f^{(5)}(x) = -720(x-1)^{-7} \)
03

Evaluate Derivatives at a=0

Compute the derivative values at \( x = 0 \), since we are working with the Maclaurin series.- \( f(0) = 1 \)- \( f'(0) = -2 \)- \( f''(0) = 6 \)- \( f'''(0) = -24 \)- \( f^{(4)}(0) = 120 \)- \( f^{(5)}(0) = -720 \)
04

Construct the Maclaurin Polynomial

Use the derivatives calculated at \( x = 0 \) to construct the polynomial:\[P_5(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5\]Replacing the values:\[P_5(x) = 1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5\]
05

Determine the Remainder

For Maclaurin series, the remainder term \( R_n(x) \) is given by:\[R_5(x) = \frac{f^{(6)}(c)}{6!} x^6\]where \( c \) is some value between 0 and \( x \).Compute \( f^{(6)}(x) = 5040(x-1)^{-8} \) and evaluate at some \( c \) to approximate or identify the behavior of the remainder explicitly.

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Key Concepts

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

Taylor series
A Taylor series is a way to represent a function as an infinite sum of terms calculated from the derivatives of the function at a specific point. This mathematical technique is particularly useful in calculus for approximating complex functions with a simpler polynomial function.
  • The Taylor series centered at a point, usually denoted as 'a', requires knowing the function's derivatives at that point.
  • A special case of the Taylor series centered at zero is known as the Maclaurin series.
  • The general formula for a Taylor series is:\[T(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots\]
This allows for a series expansion that can approximate the original function around the point 'a'. When creating a Maclaurin series for a function, we set 'a' to be zero, thus simplifying the Taylor series formula.
calculus
Calculus, a branch of mathematics, deals with the study of change. It's fundamentally composed of two parts: differentiation and integration.
  • Differentiation: The process of finding derivatives, which give us the rate at which a quantity changes.
  • Integration: The reverse of differentiation; it allows us to find the area under curves.
In the context of Taylor and Maclaurin series, calculus is employed in calculating successive derivatives of a function. These derivatives are then used in creating polynomial approximations to the function itself. Thus, understanding calculus is crucial for effectively using these series.
derivatives
Derivatives are a cornerstone in calculus, serving as a measure of how a function changes as its input changes. In simpler terms, a derivative provides the rate of change or the slope of the function at a specific point.
The process of finding a derivative is called differentiation. For any function, the first derivative gives us the slope of the tangent line at any point, and higher-order derivatives provide further insights into the function, such as finding concavity and inflection points.
  • In our exercise, calculating successive derivatives of the function \[ f(x)=\frac{1}{(x-1)^2} \] was pivotal in forming the Maclaurin series. Each derivative gives us additional terms of increasing degree in the polynomial approximation.
Knowing how to differentiate and use derivations are essential skills for dealing with Taylor and Maclaurin series because they form the building blocks of these expansions.
remainder term
When we expand a function into a Taylor or Maclaurin series, the series is often truncated after a finite number of terms, forming a polynomial approximation. However, there is always some error or 'remainder' left, known as the remainder term.
  • The remainder term quantifies the difference between the actual function and its polynomial approximation.
  • Mathematically, for a Taylor series approximation up to the nth degree, the remainder term \(R_n(x)\) is given by:\[R_n(x) = \frac{f^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!}\]
  • Here, 'c' is some value between 'a' and 'x'.
Essentially, if the rest of the terms beyond a certain degree were left out, the remainder term accounts for what those omitted terms would contribute. Understanding the remainder helps in assessing the accuracy of a series approximation.

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