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Chapter 11: Problem 19

Find the Maclaurin series for the functions in Exercises \(9-20 .\) $$ x^{4}-2 x^{3}-5 x+4 $$

Short Answer

Expert verified
The Maclaurin series is \(4 - 5x - 2x^3 + x^4\).

Step by step solution

01

Understand the Maclaurin Series

The Maclaurin series is a special case of the Taylor series, where the series is expanded at \(x = 0\). Hence, for a given function \(f(x)\), the Maclaurin series is given by:\[\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]where \(f^{(n)}(0)\) is the \(n\)-th derivative of the function evaluated at \(x = 0\). Our target is to express \(x^4 - 2x^3 - 5x + 4\) as a power series centered at \(x = 0\).
02

Compute the Function Value at Zero

The value of the function at \(x = 0\) is simply the constant term, since by substituting \(x=0\) in the polynomial:\[ (x^4 - 2x^3 - 5x + 4)\big|_{x=0} = 4 \]So, the first term in the Maclaurin series is \(4\).
03

First Derivative and its Value at Zero

Compute the first derivative of the function:\[ f'(x) = 4x^3 - 6x^2 - 5 \]Evaluate at \(x = 0\):\[ f'(0) = -5 \]Thus, the next term in the series is \(-5x\).
04

Second Derivative and its Value at Zero

Compute the second derivative:\[ f''(x) = 12x^2 - 12x \]Evaluate at \(x = 0\):\[ f''(0) = 0 \]This term does not contribute to the series since it is multiplied by \(x^2\).
05

Third Derivative and its Value at Zero

Compute the third derivative:\[ f'''(x) = 24x - 12 \]Evaluate at \(x = 0\):\[ f'''(0) = -12 \]Thus, the term contributed by this derivative is \(-\frac{12}{3!} x^3 = -2x^3 \).
06

Fourth Derivative and its Value at Zero

Compute the fourth derivative:\[ f^{(4)}(x) = 24 \]Evaluate at \(x = 0\):\[ f^{(4)}(0) = 24 \]Thus, the term contributed by this derivative is \(\frac{24}{4!} x^4 = x^4 \).
07

Remaining Derivatives

Compute the fifth derivative:\[ f^{(5)}(x) = 0 \]Since this is a fourth-degree polynomial, all higher derivatives beyond the fourth are zero, contributing nothing to the series.
08

Construct the Series

Combine all the terms to write the Maclaurin series as:\[ f(x) = 4 - 5x - 2x^3 + x^4 \]

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Key Concepts

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

Polynomial Expansion
Polynomial expansion refers to expressing a polynomial in terms of its variables and coefficients that make infinite computation more manageable. In the case of the function given in the exercise, which is a polynomial of degree 4, we have specific powers of the variable, multiplied by coefficients:
  • The term with highest power is the one with the degree of the polynomial: here, it's \(x^4\).
  • It contains constant terms like \(-2x^3\) and \(-5x\), and a constant term like 4.
Polynomial expansion aims to articulate this function explicitly for its ease of use in calculus applications. It lays the foundation for understanding more advanced topics like the power series and deriving specific terms in systems of equations.
Power Series
A power series is a series of the form \(\sum_{n=0}^{\infty} c_n \, x^n\) where \(c_n\) are constants and \(x^n\) are powers of \(x\). It represents functions in the form of an infinite sum of terms based on powers of a variable. In a broader sense, the power series generalizes the concept of polynomial, where the series can potentially have infinitely many terms or coefficients that extend indefinitely.
Here are some key aspects:
  • Each term in a power series has a general format combining coefficients \(c_n\) and powers \(x^n\).
  • Power series are centered at a point; for a Maclaurin series, this point is zero, meaning \(x=0\).
  • It is used to represent functions for interval convergence, offering approximation where simple polynomials cannot convey complex behaviors.
Power series can solve differential equations, perform function approximations, and prove convergence with radius of convergence, which is key in calculus and real analysis.
Derivatives
Derivatives represent the rate of change of a function with respect to one of its variables. Calculating derivatives is essential in the process of finding a Maclaurin series because they determine the coefficients in the series expansion. Here鈥檚 how derivatives come into play in polynomial functions like the one given:
  • The zeroth derivative is simply the function itself and its value at 0 is the first term of the Maclaurin series.
  • The first derivative gives the instantaneous slope of the curve at any point, and its evaluation at 0 gives us the coefficient for the \(x\) term.
  • Further derivatives find finer slopes and curvatures, each evaluated at \(x = 0\), adding to the precision of the power series representation.
In this exercise, we repeatedly differentiate the polynomial to calculate successive terms in the series. Each derivative offers insight into the function鈥檚 behavior, showing how it changes and contributes particular terms in the power series.
Taylor Series
The Taylor series provides a way to expand functions into an infinite sum of terms calculated from the values of the function鈥檚 derivatives at a single point. The Maclaurin series is a specific case of the Taylor series, expanded about zero. Here鈥檚 what makes the Taylor series special:
  • For any function \(f(x)\), its Taylor series approximates \(f(x)\) by summing powers of \((x-a)\) with coefficients based on derivatives.
  • A Maclaurin series is just "Taylor series at zero", which aids in simplifying calculations like the ones in the exercise.
  • Taylor series are instrumental in not only approximating functions but simplifying complex calculations into endless series that are easier to manipulate.
In the exercise, we address the Maclaurin series, a form of Taylor series, showing how a function can be expressed fully through those derivative terms. This allows for more thorough mathematical analysis and approximation in practical applications.

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Most popular questions from this chapter

Show that if \(\Sigma_{n=1}^{\infty} a_{n}\) converges absolutely, then $$\left|\sum_{n=1}^{\infty} a_{n}\right| \leq \sum_{n=1}^{\infty}\left|a_{n}\right|$$

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\left(1+\frac{0.5}{n}\right)^{n} $$

Two complex numbers \(a+i b\) and \(c+i d\) are equal if and only if \(a=c\) and \(b=d .\) Use this fact to evaluate $$ \int e^{a x} \cos b x d x \text { and } \int e^{a x} \sin b x d x $$ from $$ \int e^{(a+i b) x} d x=\frac{a-i b}{a^{2}+b^{2}} e^{(a+i b) x}+C $$ where \(C=C_{1}+i C_{2}\) is a complex constant of integration.

Show that the sum of the first 2\(n\) terms of the series $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\cdots$$ is the same as the sum of the first \(n\) terms of the series $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\cdots$$ Do these series converge? What is the sum of the first \(2 n+1\) terms of the first series? If the series converge, what is their sum?

For what values of \(a,\) if any, do the series in converge? $$\sum_{n=1}^{\infty}\left(\frac{a}{n+2}-\frac{1}{n+4}\right)$$
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