Problem 9 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The first four terms of the series are: \(1 + \frac{1}{2x} - \frac{1}{8x^2} - \frac{1}{16x^3}\).

Step 1: Identify the Binomial Series Formula

The binomial series formula for \((1 + u)^n\) is given by \(\sum_{k=0}^{\infty} \binom{n}{k} u^k\), where \(\binom{n}{k}\) is the binomial coefficient \(\frac{n!}{k!(n-k)!}\). Here, \(n = \frac{1}{2}\) and \(u = \frac{1}{x}\).

Step 2: Understand the Binomial Coefficients

For our expression, we need to calculate the binomial coefficients for the first four terms: \(\binom{\frac{1}{2}}{0}, \binom{\frac{1}{2}}{1}, \binom{\frac{1}{2}}{2}, \binom{\frac{1}{2}}{3}\). These coefficients are computed as follows: \(\binom{n}{k} = \frac{n (n-1) \ldots (n-k+1)}{k!}\).

Step 3: Calculate the Binomial Coefficients

- \(\binom{\frac{1}{2}}{0} = 1\)- \(\binom{\frac{1}{2}}{1} = \frac{1}{2}\)- \(\binom{\frac{1}{2}}{2} = \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!} = \frac{-1}{8}\)- \(\binom{\frac{1}{2}}{3} = \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!} = \frac{-1}{16}\)

Step 4: Write the First Four Terms

Apply the coefficients to \(\left(\frac{1}{x}\right)^k\):- First term: \(\binom{\frac{1}{2}}{0} \left(\frac{1}{x}\right)^0 = 1\)- Second term: \(\binom{\frac{1}{2}}{1} \left(\frac{1}{x}\right)^1 = \frac{1}{2x}\)- Third term: \(\binom{\frac{1}{2}}{2} \left(\frac{1}{x}\right)^2 = \frac{-1}{8x^2}\)- Fourth term: \(\binom{\frac{1}{2}}{3} \left(\frac{1}{x}\right)^3 = \frac{-1}{16x^3}\)

Step 5: Summarize the Series

Thus, the first four terms of the binomial series are: \(1 + \frac{1}{2x} - \frac{1}{8x^2} - \frac{1}{16x^3}\).

Key Concepts

Binomial Coefficients

Binomial coefficients play a fundamental role in binomial expansions. They are represented by \( \binom{n}{k} \), a notation that represents the number of ways to choose \( k \) elements from a set of \( n \) elements without concern for order. In mathematical terms, it is calculated using the formula:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

In our particular case, we are dealing with non-integer exponents defined by the series, which makes the calculation more interesting yet still useful for approximation. For any general term in a binomial series \( (1+u)^n \), the coefficients are derived using the extended concept:

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

This adaptation allows us to calculate binomial coefficients for fractional and negative values of \( n \). Given our expression \( \left(1 + \frac{1}{x}\right)^{1/2} \), the coefficients for the first four terms are:

• \( \binom{\frac{1}{2}}{0} = 1 \)
• \( \binom{\frac{1}{2}}{1} = \frac{1}{2} \)
• \( \binom{\frac{1}{2}}{2} = \frac{-1}{8} \)
• \( \binom{\frac{1}{2}}{3} = \frac{-1}{16} \)

Calculating these coefficients correctly is crucial because they guide the expansion term values in the binomial theorem.

Binomial Theorem

The binomial theorem is a powerful tool in algebra that gives us a way to expand expressions of the form \((a+b)^n\). This helps in finding terms in the expansion without multiplying repeatedly. For positive integer exponents, the theorem is simply:

• \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)

However, the beauty of the binomial theorem is in its generality. It can be extended to cases where \( n \) is not an integer, through a series that potentially has infinite terms. This expansion is given by:

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

In our example, \( (1 + \frac{1}{x})^{1/2} \), the exponent is \( n = \frac{1}{2} \). By applying the binomial theorem, we expand it and only take the first few terms, as higher-degree terms become negligibly small in many practical applications. The binomial theorem, thus, provides a mechanism to approximate such expressions - a fundamental concept in calculus and algebraic simplifications.

Series Expansion

A series expansion is a way of expressing a function as a sum of terms. For functions that are not simple to analyze directly, series expansions turn them into infinite sums that are often much easier to use. A familiar series expansion is the binomial series, involving terms of powers growing in an orderly fashion based on the binomial coefficients.

• The binomial series expansion of \((1 + u)^n\) takes the form \( \sum_{k=0}^{\infty} \binom{n}{k} u^k \)

This series is particularly useful when the exponent \( n \) is not a whole number, and a direct computation isn't straightforward or when approximations are required. In the given problem, we expanded \((1 + \frac{1}{x})^{1/2}\) for the first four terms, resulting in:

• \( 1 + \frac{1}{2x} - \frac{1}{8x^2} - \frac{1}{16x^3} \)

Each term in this series represents a more refined approximation of the original function, especially when \( x \) is significantly larger than 1. Therefore, series expansions are vital in mathematical modeling and real-world applications. They help by providing approximate solutions where exact ones are either unknown or too complex to derive.