Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=\frac{1}{1+x+x^{2}}\)

Step-by-Step Solution

Answer

The Maclaurin series up to \(x^5\) is \(1 - x + x^3 - 2x^4 - 3x^5\).

Step 1: Identify the Basic Forms

To find the Maclaurin series through terms up to \(x^5\) for \(f(x) = \frac{1}{1 + x + x^2}\), recognize that it may be helpful to use existing series. The geometric series \(\frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots\) is a foundational series that is useful when the denominator is linear.

Step 2: Rewriting the Function

Set \(g(x) = x + x^2\). Then, \(f(x)\) can be expressed as \(\frac{1}{1 + g(x)}\). This allows us to utilize the formula for a geometric series \(\frac{1}{1-u} = 1+u+u^2+u^3+\ldots\). Where \(u = -g(x) = -(x + x^2)\).

Step 3: Apply the Geometric Series Expansion

Plug \(u = -(x + x^2)\) into the geometric series expansion:\[\frac{1}{1 + g(x)} = 1 - (x + x^2) + (x + x^2)^2 - (x + x^2)^3 + \ldots\]Now, calculate and collect terms up to \(x^5\).

Step 4: Calculate Each Expansion Term

Calculate the early terms of the expansion:1. The first term is \(1\).2. The second term (linear term) is \(-(x + x^2) = -x - x^2\).3. The third term (square term) is \((x + x^2)^2 = x^2 + 2x^3 + x^4\).4. The fourth term (cubic term) is \(-(x + x^2)^3 = -(x^3 + 3x^4 + 3x^5 + x^6)\). Disregard \(x^6\) as it's beyond \(x^5\).Now, combine these to collect terms.

Step 5: Combine and Simplify Terms

Combine the terms from the expansions calculated in Step 4:- Constant: \(1\)- Linear: \(-x\)- Quadratic: \(-x^2 + x^2 = 0\)- Cubic: \(-x^3 + 2x^3 = x^3\)- Quartic: \(-3x^4 + x^4 = -2x^4\)- Quintic: \(-3x^5\)Thus, the expansion through \(x^5\) is \(1 - x + x^3 - 2x^4 - 3x^5\).

Step 6: Write Final Series through \(x^5\)

The Maclaurin series for \(f(x) = \frac{1}{1+x+x^2}\) through the \(x^5\) term is:\[f(x) = 1 - x + x^3 - 2x^4 - 3x^5\]

Key Concepts

Geometric Series

A geometric series is a series of terms that have a constant ratio between consecutive terms. It's a powerful tool in mathematics, especially within calculus and series expansions. The classic example of a geometric series is the infinite series \(\frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots\). This series expands as long as the absolute value of \(x\) is less than one.

Understanding geometric series is crucial because it lays the foundation for more complex series like the Maclaurin series. It helps in restructuring functions to make them more manageable for solving or approximating. For instance, the function \(f(x) = \frac{1}{1 + x + x^2}\) in our example isn’t initially in a form suitable for a geometric series. But by reconfiguring it to \(\frac{1}{1 - (-x - x^2)}\), it opens the door to expanding it as a geometric series.

This technique simplifies the derivation of the series expansion by systematically using elementary geometric series results, adapting them to fit more complex scenarios.

Polynomial Expansion

Polynomial expansion is the method of expressing a power or product of variables in a simplified polynomial form. In the context of the Maclaurin series or any series expansion, recognizing how to expand terms like \((x + x^2)^n\) is integral.

When faced with expansions like \( (x + x^2)^2\), each term of the polynomial must be calculated individually, combining powers of \(x\) appropriately:

• Expand this to obtain \( x^2 + 2x^3 + x^4 \).
• Applying similar expansions helps handle higher powers such as \((x + x^2)^3\), where each term is derived methodically.

By expanding these terms, you can break down complex expressions into manageable polynomial components, which are then easy to add to the overall series.

This approach of polynomial expansion aids the simplification and combination of terms in broader series calculations, enhancing computational efficiency.

Series Expansion Techniques

Series expansion techniques involve breaking down functions into their constituent powers and coefficients up to a desired degree. In calculus, these techniques are essential for approximating functions, calculating limits, and solving differential equations.

The Maclaurin series, a specific type of Taylor series, is centered at zero and is defined by taking derivatives of the function at that point. For each section of the series, like in this exercise, terms are drawn from other established expansions to combine seamlessly. This step-by-step process results in an expanded series representing the function over its domain.

• The goal is usually to express a function as a power series up to a specific degree, like \(x^5\) in this exercise.
• You may also involve known series expansions for mathematical functions such as sine, cosine, or exponential functions, employing these as building blocks for more complex series.
• The convergence of these series ensures that as more terms are added, the series approximates the function more accurately.

These techniques make calculus far more tractable, permitting the transformation of complex functions into simpler polynomial forms.

Calculus

Calculus is the mathematical study of continuous change and underpins many series expansion techniques. Within calculus, concepts like differentiation and integration are instrumental in deriving series like the Maclaurin series.

When generating a Maclaurin series, you start with a function and calculate its derivatives at zero. Each derivative value contributes to the term coefficients in the series. This ordered layout of operations allows for the creation of polynomial approximations to represent more complicated expressions.

For calculus applications involving series:

• Derivative calculations help find the series' terms and are essential for ensuring the series' convergence and approximation quality.
• Integration within the series framework allows for area under the curve estimations, involving infinite summation techniques.

The Maclaurin series is vital because it simplifies multi-variable or non-linear equations into forms that are easier to manipulate, analyze, and compute over specific intervals. Understanding calculus principles strengthens the grasp of how series expansions represent the broader scope of mathematical functions.