Problem 5 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The first four terms are: \(1 - x + \frac{3}{4}x^2 - \frac{1}{4}x^3.\)

Step 1: Understand the Binomial Series Expansion

The binomial series expansion is a generalization of the binomial theorem, which can be used to expand expressions such as \((1+x)^n\) for any real number \(n\). The series is given by: \[(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots \] applied for small \(x\).

Step 2: Identify Parameters

For the expression \((1+\frac{x}{2})^{-2}\), compare it with the form \((1+x)^n\). Here, \(x\) is replaced by \(\frac{x}{2}\) and \(n = -2\).

Step 3: Calculate the First Term

The first term in the expansion, according to the series, is \(1\). So, for \((1+\frac{x}{2})^{-2}\), the first term is also 1.

Step 4: Calculate the Second Term

The second term is given by \(nx\). Substitute \(n = -2\) and \(x = \frac{x}{2}\) into the formula. So, the second term is \[n\left(\frac{x}{2}\right) = -2\left(\frac{x}{2}\right) = -x.\]

Step 5: Calculate the Third Term

The third term is \(\frac{n(n-1)}{2!}x^2\). Substitute \(n = -2\) and \(x = \frac{x}{2}\):\[\frac{-2(-2-1)}{2}\left(\frac{x}{2}\right)^2 = \frac{-2(-3)}{2}\left(\frac{x^2}{4}\right) = \frac{6}{8}x^2 = \frac{3}{4}x^2.\]

Step 6: Calculate the Fourth Term

The fourth term is \(\frac{n(n-1)(n-2)}{3!}x^3\). Substitute the values, \[\frac{-2(-3)(-4)}{6}\left(\frac{x}{2}\right)^3 = \frac{-24}{6}\left(\frac{x^3}{8}\right) = -2\left(\frac{x^3}{8}\right) = -\frac{1}{4}x^3.\]

Step 7: Compile the First Four Terms

Combine the terms obtained: The first term: \(1\)The second term: \(-x\)The third term: \(\frac{3}{4}x^2\)The fourth term: \(-\frac{1}{4}x^3\)Thus, the first four terms of the binomial series are:\[1 - x + \frac{3}{4}x^2 - \frac{1}{4}x^3.\]

Key Concepts

Binomial Theorem

The binomial theorem is a crucial mathematical tool for expanding expressions that are raised to a power. When given a binomial expression like a\( (1+x)^n\), we can use the binomial theorem to expand it into a series of terms. This theorem essentially uses the concept of binomial coefficients and applies them systematically to each term in the expansion.

Here's the general form of a binomial expansion: \[(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots\] This means you can expand any expression of the form \((1+x)^n\) by continually finding and adding each subsequent term! Each term in this expansion is affected by the power \(n\) and depends on its position in the series (e.g., second, third term, etc.).

Power Series

A power series is a way of expressing a function as an infinite sum of terms defined by a variable's powers. In the context of the binomial theorem, the expansion shown in the previous section is an example of a power series. This allows mathematicians to approximate complex expressions using simpler polynomial terms.

Power series are extremely valuable because they can often transform difficult problems involving functions into more manageable ones, especially when the variable \(x\) is within a certain range. Simple examples of power series appear regularly in calculus and other branches of mathematics to simplify equations and estimate values.

Binomial Coefficient

Binomial coefficients are specific numbers that arise in the expansion of a binomial raised to a power, as seen in the binomial theorem. These coefficients are critical in determining the weight or importance of each term in a binomial expansion.

The symbol \(\binom{n}{k}\) represents the binomial coefficient, which can be calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(n!\) denotes the factorial of \(n\). In simpler terms, the binomial coefficient is the number of ways to choose \(k\) elements from a set of \(n\) elements.

In the binomial series expansion for \((1+x)^n\), each term involves a binomial coefficient that helps adjust the contribution of each power of \(x\), making it an essential part of the expansion process!