Question

Find the Taylor polynomial of degree \(n\), at \(x=c\), for the given function. $$f(x)=\frac{1}{x}, \quad n=5, \quad c=2$$

Short answer

The Taylor polynomial of degree 5 is: \( P_5(x) = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 + \frac{1}{32}(x-2)^4 - \frac{15}{384}(x-2)^5 \).

Step-by-step solution

Understand the Taylor Polynomial Formula

The Taylor polynomial of degree \(n\) for a function \(f(x)\) at \(x = c\) is given by:\[P_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \ldots + \frac{f^{(n)}(c)}{n!}(x-c)^n\]We need to compute the polynomial for \(f(x) = \frac{1}{x}\) with \(n=5\) and \(c=2\).

Calculate Function Value at \(c\)

Calculate \(f(c)\), where \(f(x) = \frac{1}{x}\) and \(c = 2\):\[f(2) = \frac{1}{2}\]

Determine Derivatives of the Function

Find the first few derivatives of \(f(x) = \frac{1}{x}\):- First derivative: \(f'(x) = -\frac{1}{x^2}\)- Second derivative: \(f''(x) = \frac{2}{x^3}\)- Third derivative: \(f'''(x) = -\frac{6}{x^4}\)- Fourth derivative: \(f^{(4)}(x) = \frac{24}{x^5}\)- Fifth derivative: \(f^{(5)}(x) = -\frac{120}{x^6}\)

Evaluate Derivatives at \(c\)

Calculate the derivatives at \(x = 2\):- \(f'(2) = -\frac{1}{4}\)- \(f''(2) = \frac{2}{8} = \frac{1}{4}\)- \(f'''(2) = -\frac{6}{16} = -\frac{3}{8}\)- \(f^{(4)}(2) = \frac{24}{32} = \frac{3}{4}\)- \(f^{(5)}(2) = -\frac{120}{64} = -\frac{15}{8}\)

Substitute into the Taylor Polynomial Formula

Substitute all calculated values into the Taylor polynomial formula for \(n = 5\):\[P_5(x) = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 + \frac{1}{32}(x-2)^4 - \frac{15}{384}(x-2)^5\]

Finalize the Taylor Polynomial Expression

Simplify and write the complete Taylor polynomial: \[P_5(x) = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 + \frac{1}{32}(x-2)^4 - \frac{15}{384}(x-2)^5\]This is the Taylor polynomial of degree 5 for \(f(x) = \frac{1}{x}\) at \(x = 2\).

Key concepts

Maclaurin series

The Maclaurin series is a special case of the Taylor series where the expansion is about zero, that is, the point of expansion, denoted by 'c', is 0. This simplifies the Taylor series formula greatly because all terms with (x-c) just become (x-0) = x. The Maclaurin series is particularly useful in mathematical analysis and calculus because it allows us to express functions as infinite sums of their derivatives at zero.

This series is commonly used with functions like \( e^x \), \( \sin(x) \), \( \cos(x) \), and \( \ln(1+x) \).

The general formula for a Maclaurin series is:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]
The simplicity of Maclaurin series comes from the fact that it is an expansion around zero, making it highly valuable in calculations where you are concerned with behavior around this point.

Higher order derivatives

Higher order derivatives are crucial for creating Taylor polynomials. Each derivative provides additional information on how the function changes.
This is especially true for functions like \( f(x) = \frac{1}{x} \), where subsequent derivatives rapidly increase in complexity, as seen by:

• First derivative: \( f'(x) = -\frac{1}{x^2} \)
• Second derivative: \( f''(x) = \frac{2}{x^3} \)
• Third derivative: \( f'''(x) = -\frac{6}{x^4} \)
• And so forth.

Each derivative gives a new term in the Taylor series and contributes to a more accurate polynomial approximation.
To find a Taylor polynomial, you need to evaluate these derivatives at the point of interest, in this exercise, it's \( x=2 \).
Understanding how to compute and use higher order derivatives enhances your capability to solve calculus problems by providing robust function approximations.

Taylor series approximation

The Taylor series offers a method for approximating complex functions with a polynomial of finite degree. This approximation can be particularly handy when exact solutions are difficult or impossible to derive.
By taking the function's value and its derivatives at a chosen point, such as \( c=2 \), Taylor series transforms these elements into a polynomial - known as the Taylor polynomial.

The benefit is that the polynomial gives us an easier function to work with compared to the original one.

• The degree of the polynomial (\( n \)) dictates the accuracy of the approximation.
• A higher-degree polynomial will typically provide a closer approximation to the actual function.

For instance, in the original problem, the next Taylor polynomial approximates \( \frac{1}{x} \) around \( c=2 \), using a polynomial of degree 5.
The result is a representation of \( \frac{1}{x} \) that is manageable and often sufficient for practical calculation.

Calculus problem solving

Calculus problem solving often necessitates approximating values to solve complex problems without access to exact algebraic solutions. Taylor polynomials provide a tangible tool for this purpose.
Breaking down a function into a polynomial form allows you to estimate function values that may not be easy to calculate directly.

The steps:

• Determine the derivatives of the function.
• Evaluate these derivatives at the specified point.
• Substitute into the Taylor polynomial formula.

This process, highlighted in the original exercise, illuminates how calculus is not just about solving equations but also about understanding function behavior.
This builds on concepts like rates of change and concavity to deliver precise approximations, enhancing both the depth of calculus skills and practical application.