Problem 8 - Thomas Calculus

Question

Find the first four terms of the binomial series for the functions. \begin{equation} \left(1+x^{2}\right)^{-1 / 3} \end{equation}

Step-by-Step Solution

Answer

The series is: \( 1 - \frac{1}{3}x^2 + \frac{2}{9}x^4 - \frac{14}{81}x^6 \).

Step 1: Identify the Function and General Binomial Series Formula

The given function is \((1 + x^2)^{-1/3}\). The general formula for a binomial series is \((1 + u)^n = \sum_{k=0}^{ ightarrow \infty} \binom{n}{k} \cdot u^k\), where \( \binom{n}{k} = \frac{n(n-1)(n-2) ext{...}(n-k+1)}{k!} \).

Step 2: Set up the Initial Terms for the Series

Here, \( u = x^2 \) and \( n = -1/3 \). We will substitute these into the binomial series formula and calculate the coefficients individually for the first four terms.

Step 3: Calculate the First Term

For the first term, \( k = 0 \), which gives us \( \binom{-1/3}{0} \cdot (x^2)^0 = 1 \cdot 1 = 1 \). Thus, the first term is 1.

Step 4: Calculate the Second Term

For the second term, \( k = 1 \), use the binomial coefficient: \( \binom{-1/3}{1} = -1/3 \). The term is \( \binom{-1/3}{1} \cdot (x^2)^1 = -\frac{1}{3}x^2 \).

Step 5: Calculate the Third Term

For \( k = 2 \), compute \( \binom{-1/3}{2} = \frac{(-1/3)(-4/3)}{2} = \frac{2}{9} \). The term is \( \binom{-1/3}{2} \cdot (x^2)^2 = \frac{2}{9}x^4 \).

Step 6: Calculate the Fourth Term

For \( k = 3 \), compute \( \binom{-1/3}{3} = \frac{(-1/3)(-4/3)(-7/3)}{6} = -\frac{14}{81} \). The term is \( \binom{-1/3}{3} \cdot (x^2)^3 = -\frac{14}{81}x^6 \).

Step 7: Combine the Terms to Form the Series

The first four terms of the series for \((1 + x^2)^{-1/3}\) are: \( 1 - \frac{1}{3}x^2 + \frac{2}{9}x^4 - \frac{14}{81}x^6 \).

Key Concepts

Binomial Coefficients

Binomial coefficients play a crucial role in binomial series expansions. They are the numerical factors that multiply each term in a series derived from powers of a binomial expression. The binomial coefficient for a general term is denoted as \( \binom{n}{k} \), which is calculated as:

• \( \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \)

This formula helps in determining the cofactor of each term when expanding functions, even with fractional exponents. The coefficient tells us how many times we have a certain combination of elements.
For instance, in the exercise, these coefficients appear in each term when expanding \((1 + x^2)^{-1/3}\). Calculating these coefficients involves identifying the pattern and applying the formula to each term. This knowledge aids in expressing more complex mathematical functions in a manageable form, easing further calculations and interpretations.

Series Expansion

Series expansion allows us to express complex mathematical functions as a sum of simpler terms. For the binomial series, this involves expressing a binomial expression raised to a power as an infinite series. The formula is expressed by

• \((1 + u)^n = \sum_{k=0}^{\infty} \binom{n}{k} \cdot u^k \)

where each term involves a binomial coefficient and a power of the variable of interest.
This series is particularly useful when dealing with fractional or negative exponents, as seen in the function \((1 + x^2)^{-1/3}\). Series expansions help approximate functions, providing a practical method to calculate values and analyze situations that would otherwise be complex.

Calculating Terms

To compute the terms of a binomial series, the first step is setting up the formula given \(u = x^2\) and \(n = -1/3\). Calculating each term means substituting these values into the binomial series and applying it to successive values of \(k\).

• **First Term:** For \(k=0\), the coefficient is \(\binom{-1/3}{0} = 1\). Hence, the term is \((x^2)^0 = 1\).
• **Second Term:** For \(k=1\), compute \(\binom{-1/3}{1} = -1/3\). The term is \(-\frac{1}{3}x^2\).
• **Third Term:** For \(k=2\), \(\binom{-1/3}{2} = \frac{2}{9}\), leading to \(\frac{2}{9}x^4\).
• **Fourth Term:** For \(k=3\), \(\binom{-1/3}{3} = -\frac{14}{81}\). The term is \(-\frac{14}{81}x^6\).

Each step involves applying the coefficient formula and performing some algebraic manipulations to arrive at the terms of the series.

Mathematical Functions

Mathematical functions like \((1+x^2)^{-1/3}\) often appear complex but can be simplified using tools like the binomial series. These functions involve operations that are common in various fields of study, including engineering, physics, and finance.
By using series expansion, we can approximate a function's behavior over a certain range of values. This transformation simplifies calculations and provides insights, allowing us to work with manageable expressions instead of grappling with complex originals.

• Simplifying functions using series expansions is invaluable for numerical solutions.
• It enables deeper analysis of complex systems by breaking them down into simpler parts.

Understanding how to manipulate and expand these functions is crucial for tackling practical problems in different scientific domains.