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Chapter 11: Problem 60

Use the Maclaurin series $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1

Short Answer

Expert verified
Question: Find the Maclaurin series for the function \((x^2 - 4x + 5)^{-2}\) using the given Maclaurin series for \((1+x)^{-2}\). Answer: The Maclaurin series for \((x^2 - 4x + 5)^{-2}\) is given by \(3x^4 - 24x^3 + 74x^2 - 100x + 41 + \cdots\) for \(-1 < (x-2)^2 < 1\).

Step by step solution

01

Write function in the form of (1+x)^{-2}

First, we need to rewrite the given function \((x^2 - 4x + 5)^{-2}\) in the form of \((1+x)^{-2}\). Observe that \((x^2 - 4x + 5) = (1 +(x^2 - 4x + 4))= (1 + (x-2)^2)\). Thus, our function can be rewritten as \([(1 +(x-2)^2]^{-2}}\). Now we have the function in the form \((1+x)^{-2}\), with \(x = (x-2)^2\).
02

Apply the given Maclaurin series

Now that our function is in the desired form, we can substitute \((x-2)^2\) in the Maclaurin series for \((1+x)^{-2}\). $$[(1+(x-2)^2)^{-2}] = 1 - 2(x-2)^2 + 3(x-2)^4 - 4(x-2)^6 + \cdots \text{ for } -1 < (x-2)^2 < 1$$
03

Simplify the expanded series

Next, let's expa薪d and simplify the expression: $$1 - 2(x-2)^2 + 3(x-2)^4 - 4(x-2)^6 + \cdots = 1 - 2(x^2 - 4x + 4) + 3(x^4 - 8x^3 + 24x^2 - 32x + 16)+ \cdots$$ Expanding further and combining like terms, we get: $$1 - 2x^2 + 8x - 8 + 3x^4 - 24x^3 + 72x^2 - 96x + 48 + \cdots$$ This can be simplified as: $$1 - 2x^2 + 8x - 8 + 3x^4 - 24x^3 + 72x^2 - 96x + 48 + \cdots = 3x^4 - 24x^3 + 74x^2 - 100x + 41 + \cdots$$ The Maclaurin series of the given function is: $$(x^2 - 4x + 5)^{-2} = 3x^4 - 24x^3 + 74x^2 - 100x + 41 + \cdots \text{ for } -1 < (x-2)^2 < 1$$

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Key Concepts

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

Series Expansion
Series expansion is a method in mathematics that allows a complicated function to be expressed as the sum of simpler terms. This is helpful when performing calculations or approximations.
  • In the context of this problem, the Maclaurin series is used. It extends functions around the point where the variable, often \(x\), is zero.
  • The basic idea is to represent a function like \((1+x)^{-2}\) through an infinite sum which is easier to manipulate.
In this solution, we took the expression \((x^2-4x+5)^{-2}\). By re-writing it in a form resembling \((1+x)^{-2}\), it became possible to apply the series expansion.through substituting a substitute variable, \((x-2)^2\).This demonstrates how series expansion can function as a tool for simplifying complex expressions for further mathematical work.
Polynomial Approximation
Polynomial approximation allows us to represent complex functions through polynomials, making computations easier. The Maclaurin series is one method which approximates a function around 0 using polynomials.
  • For example, the function \((1+x)^{-2}\) can be approximated by the polynomial \(1 - 2x + 3x^2 - 4x^3 + \cdots \).
  • In the exercise, the expansion yields a polynomial that approximates \((x^2-4x+5)^{-2}\).
This method is beneficial because:
  • Polynomials are easier to evaluate and differentiate compared to exponential or trigonometric functions.
  • It simplifies analysis around points where the approximations provide significant accuracy.
Using polynomial approximations like those derived from the Maclaurin series can greatly simplify complex calculations in calculus and applied mathematics.
Calculus
Calculus is the branch of mathematics focusing on continuous change, utilizing derivatives and integrals. The Maclaurin series connects deeply with calculus concepts by using derivatives to form series expansions.
  • It transforms differentiation of complex functions into manageable segments through polynomial representations.
  • In our problem, every coefficient in the polynomial approximation corresponds to a different derivative of the original function.
Calculus provides the tools to better understand the behavior of functions along with series expansions. These expansions enable us to approximate functions and estimate solutions to differential equations or integrals. From this perspective, the step-by-step approach described in the problem highlights how calculus allows mathematical descriptions and predictions of various phenomena through the elegant utility of derivatives and polynomial expressions.

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Most popular questions from this chapter

Find the function represented by the following series, and find the interval of convergence of the series. (Not all these series are power series.) $$\sum_{k=0}^{\infty} e^{-k x}$$

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Matching functions with polynomials Match functions a-f with Taylor polynomials \(A-F\) (all centered at 0 ). Give reasons for your choices. a. \(\sqrt{1+2 x}\) b. \(\frac{1}{\sqrt{1+2 x}}\) c. \(e^{2 x}\) d. \(\frac{1}{1+2 x}\) e. \(\frac{1}{(1+2 x)^{3}}\) f. \(e^{-2 x}\) A. \(p_{2}(x)=1+2 x+2 x^{2}\) B. \(p_{2}(x)=1-6 x+24 x^{2}\) C. \(p_{2}(x)=1+x-\frac{x^{2}}{2}\) D. \(p_{2}(x)=1-2 x+4 x^{2}\) E. \(p_{2}(x)=1-x+\frac{3}{2} x^{2}\) F. \(p_{2}(x)=1-2 x+2 x^{2}\)
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