Problem 23 - Calculus

Question

Find a simplified formula for \(P_{5}(x),\) the fifth-degree Taylor polynomial approximating \(f\) near \(x=0\). Use the values in the table. $$\begin{array}{c|c|c|c|c|c} \hline f(0) & f^{\prime}(0) & f^{\prime \prime}(0) & f^{\prime \prime \prime}(0) & f^{(4)}(0) & f^{(5)}(0) \\ \hline-3 & 5 & -2 & 0 & -1 & 4 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

The simplified formula for \(P_5(x)\) is \(-3 + 5x - x^2 - \frac{1}{24}x^4 + \frac{1}{30}x^5\)."

Step 1: Understanding the Taylor Polynomial

To find the fifth-degree Taylor polynomial, we will use the formula for the Taylor polynomial centered at zero (Maclaurin series): \[ P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^k \]For the fifth-degree polynomial, we need terms from \(k=0\) to \(k=5\).

Step 2: Calculate Each Term

Each term of the polynomial is calculated using the derivative values from the table and the factorial of \(k\):- The 0th term: \( \frac{f(0)}{0!} x^0 = -3 \)- The 1st term: \( \frac{f^{\prime}(0)}{1!} x^1 = 5x \)- The 2nd term: \( \frac{f^{\prime\prime}(0)}{2!} x^2 = \frac{-2}{2} x^2 = -x^2 \)- The 3rd term: \( \frac{f^{\prime\prime\prime}(0)}{3!} x^3 = \frac{0}{6} x^3 = 0 \)- The 4th term: \( \frac{f^{(4)}(0)}{4!} x^4 = \frac{-1}{24} x^4 = -\frac{1}{24} x^4 \)- The 5th term: \( \frac{f^{(5)}(0)}{5!} x^5 = \frac{4}{120} x^5 = \frac{1}{30} x^5 \)

Step 3: Combine Terms into Polynomial

Combine all the calculated terms to form the Taylor polynomial:\[ P_5(x) = -3 + 5x - x^2 + 0 - \frac{1}{24}x^4 + \frac{1}{30}x^5 \]This is the simplified fifth-degree Taylor polynomial approximation of \(f\) near \(x=0\).

Key Concepts

Maclaurin series

The Maclaurin series is a special case of the Taylor series. It is used to represent functions as an infinite sum of terms. These terms are calculated from the values of the function's derivatives at a single point, which, in the Maclaurin series, is centered at zero.
It is expressed with the formula:

• \(P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^k\)

When you create a Maclaurin series, you start by finding the derivatives of the function at \(x=0\), then multiply each derivative by powers of \(x\) divided by factorials.

This series helps approximate functions efficiently, especially for complex calculations or when using computational methods.
The more terms you include in the series, the more accurate your approximation will be.

derivatives

Derivatives play a crucial role in forming the Taylor and Maclaurin series. They represent the way a function changes and are defined as the rate at which the function's value changes as its input changes.

In the context of a Maclaurin series, we calculate the function's value and its subsequent derivatives at the point \(x=0\). Each derivative represents a different degree of change in the function:

• The first derivative, \(f'(x)\), reflects the function's slope.
• The second derivative, \(f''(x)\), indicates the rate of change of the slope.
• Higher-order derivatives represent more complex changes.

By evaluating these derivatives at \(x=0\), we can use them to construct the polynomial terms in the series.

factorial

Factorials are a fundamental component of the Maclaurin series formula. A factorial, denoted as \(k!\), is the product of all positive integers less than or equal to \(k\). For example:

• \(0! = 1\)
• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)

Factorials grow rapidly and are used to scale the terms of the series appropriately.

In the Maclaurin series, each term's coefficient includes a division by \(k!\), which helps weigh the derivative appropriately in relation to its position in the polynomial.
It ensures each term contributes correctly to the overall shape and accuracy of the function approximation.

approximation

Approximation with a Maclaurin series allows us to estimate the value of complex functions using polynomials. Polynomials are simpler to compute and analyze, making them useful for approximating more complicated expressions.

For most practical purposes, functions can be approximated up to a certain degree, like the fifth-degree in our exercise, to balance between simplicity and accuracy.
As more terms are added to the series, the approximation becomes closer to the actual function.

When using the Maclaurin series, we can achieve this approximation near the center point (in this case, \(x=0\)), which is particularly useful for calculations around this point. The concept of approximation is widely applied in numerical analysis, engineering, and physics for estimating function values when an exact solution is not feasible or necessary.