Question

Find the Maclaurin polynomial of degree \(n\) for the given function. $$f(x)=\frac{1}{1+x}, \quad n=4$$

Short answer

The Maclaurin polynomial is \(P_4(x) = 1 - x + x^2 - x^3 + x^4\).

Step-by-step solution

Understand the Function

The given function is \(f(x) = \frac{1}{1+x}\). A Maclaurin series is a Taylor series expansion of a function about \(x = 0\).

Identify the General Formula

The Maclaurin series for a function \(f(x)\) is given by the sum of \(\frac{f^{(k)}(0)}{k!}x^k\) from \(k = 0\) to \(n\). Here, \(n = 4\).

Determine the Derivatives

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

Evaluate Derivatives at \(x = 0\)

Substitute \(x = 0\) into each of the derivatives:1. \(f(0) = 1\).2. \(f'(0) = -1\).3. \(f''(0) = 2\).4. \(f'''(0) = -6\).5. \(f^{(4)}(0) = 24\).

Construct the Maclaurin Polynomial

Now, substitute the results back into the Maclaurin formula:1. \(P_4(x) = \frac{1}{0!} x^0 - \frac{1}{1!} x^1 + \frac{2}{2!} x^2 - \frac{6}{3!} x^3 + \frac{24}{4!} x^4\).2. Simplify the coefficients: - \(1 - x + x^2 - x^3 + x^4\).

Final Answer

The Maclaurin polynomial of degree 4 for \(f(x) = \frac{1}{1+x}\) is \(P_4(x) = 1 - x + x^2 - x^3 + x^4\).

Key concepts

Taylor series

A Taylor series is a mathematical method for approximating a function with an infinite sum of terms. These terms are based on the function’s derivatives at a single point. The Taylor series can be defined for any function that is infinitely differentiable at a particular point. The general form of a Taylor series centered at a point \(a\) is:

• \(f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots\)

When the series is centered at \(x=0\), it is referred to as a Maclaurin series. It is essentially a special case of the Taylor series. For practical applications, we often truncate the series to a finite number of terms, which makes it a polynomial. This is particularly useful for simplifying complex functions into manageable polynomials.

derivatives

Derivatives represent the rate at which a function changes at any given point. They are fundamental in the process of finding a Taylor or Maclaurin series because each term in these series requires the function to be differentiated one or more times.

• Calculating derivatives involves applying rules like the power rule, product rule, and chain rule, leading to expressions that describe how a function behaves.
• In the process detailed for \(f(x) = \frac{1}{1+x}\), the first derivative is \(f'(x) = -\frac{1}{(1+x)^2}\), the second is \(f''(x) = \frac{2}{(1+x)^3}\), and so on.

By evaluating these derivatives at \(x = 0\), we derive the coefficients required for the Maclaurin series. Proper understanding and calculation of derivatives ensure that the resulting polynomial precisely approximates the behavior of the original function around \(x = 0\).

polynomials

Polynomials are mathematical expressions involving sums of powers in one or more variables multiplied by coefficients. They are straightforward, well-behaved functions that are easy to compute and manipulate, making them incredibly useful in approximations.

• A polynomial can be as simple as \(x^2 + 2x + 1\) or as complex as needed, with many terms and high powers of \(x\).
• In the context of Taylor and Maclaurin series, a polynomial results from truncating the series, which consists of the sum of derivatives of the function at a particular point.
• For the given exercise, the Maclaurin polynomial of degree 4 is \(P_4(x) = 1 - x + x^2 - x^3 + x^4\), which approximates \(f(x) = \frac{1}{1+x}\).

series expansion

Series expansion is a method of expressing a complex function as a sum of simpler terms. This form of representation is especially useful because it allows easier calculations and approximations.

• The series expansion, such as the Taylor or Maclaurin series, breaks down complicated expressions into sums of calculated derivatives of the function at a specific point.
• This method of expansion is significant in calculus and practical applications where function values are either unknown or hard to compute directly.
• The power of this approach is demonstrated when we use it to express \(\frac{1}{1+x}\) in terms of a polynomial, facilitating simpler computation and understanding of the function’s behavior near \(x=0\).

By using series expansion, we can approximate finite and infinite functions to the desired degree of accuracy as displayed in the transformation of \(f(x)\) in the given problem.