Question

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial. $$ \left(1+x^{2}\right)^{-1 / 3} $$

Short answer

The Maclaurin series is: \( 1 - \frac{1}{3}x^2 + \frac{2}{9}x^4 - \frac{7}{81}x^6 + \cdots \).

Step-by-step solution

Recall the General Maclaurin Series

The Maclaurin series is a special case of the Taylor series centered at \(x = 0\). The general formula for a Maclaurin series is \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \].We aim to express \(\left(1 + x^2\right)^{-1/3}\) in the form of this series.

Identify the Function and Binomial Series Expansion

Our function is \((1 + x^2)^{-1/3}\).We can expand a function of the form \( (1 + u)^k \) using the binomial series expansion:\[(1 + u)^k = 1 + k\cdot u + \frac{k(k-1)}{2!}u^2 + \frac{k(k-1)(k-2)}{3!}u^3 + \dots\]Here, \(u = x^2\) and \(k = -\frac{1}{3}\).

Substitute and Expand the Binomial Series

Substitute \( u = x^2 \) and \( k = -\frac{1}{3} \) into the series:\[ (1 + x^2)^{-1/3} = 1 - \frac{1}{3}x^2 + \frac{(-1/3)(-4/3)}{2!}(x^2)^2 - \frac{(-1/3)(-4/3)(-7/3)}{3!}(x^2)^3 + \cdots \]Next, simplify the coefficients for each power of \( x^2 \).

Simplify the Series Terms

Calculate each term's coefficient:- First term: 1- Second term: \(- \frac{1}{3} x^2\)- Third term: \(+ \frac{(-1/3)(-4/3)}{2!} x^4 = + \frac{2}{9} x^4\)- Fourth term: \(- \frac{(-1/3)(-4/3)(-7/3)}{6} x^6 = - \frac{7}{81} x^6\)Thus, the series expansion is:\[ 1 - \frac{1}{3}x^2 + \frac{2}{9}x^4 - \frac{7}{81}x^6 + \cdots \]

Key concepts

Taylor series

The Taylor series is a powerful tool in mathematics for representing functions as infinite series of terms. These terms are calculated by evaluating the function's derivatives at a single point. In the case of the Maclaurin series, that point is always zero. With Taylor series, you can represent almost any function as a sum of polynomial terms. This makes complex functions much easier to work with, especially for approximations. The general form of a Taylor series for a function \( f(x) \) about the point \( a \) is:- \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \)
For the Maclaurin series, we simply center the expansion around \( x = 0 \). This brings us to a simpler form:- \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \)
Here, the process involves calculating the function and its derivatives at zero. Understanding Taylor series helps in recognizing how functions can be decomposed and studied through simpler components.

binomial series expansion

The binomial series expansion is a way to expand expressions of the form \((1 + u)^k\) into a power series. This is especially useful when dealing with terms that might not be easy to expand using elementary algebra methods. The expansion takes the form:- \((1 + u)^k = 1 + ku + \frac{k(k-1)}{2!}u^2 + \frac{k(k-1)(k-2)}{3!}u^3 + \dots\)In this expansion:- \(u\) is a variable or expression, such as \(x^2\) in our problem.- \(k\) is any real number, which in the given problem is \(-\frac{1}{3}\).
This expansion applies even when \(k\) is a fraction, which is the key to solving many calculus problems involving radicals or negative exponents. It helps transform an otherwise difficult expression into a series form, facilitating further operations or approximations.

substitution method

The substitution method is a technique used to simplify complicated series expansions by introducing a new variable. This is especially effective when dealing with functions that are not easily represented in a desired form. In the provided example, we see that the expression \((1 + x^2)^{-1/3}\) does not directly fit a standard power series format. However, by recognizing that it has a form similar to \((1 + u)^k\), we apply the substitution \(u = x^2\). This transforms the function into something that can be directly expanded using the binomial series formula.Benefits of using substitution include:- It makes complex expressions more manageable.- Simplifies the coefficients - easier calculation of each term.- Provides clarity by breaking down the expression into understandable parts.
Once substituted, the expressions can be expanded, making calculation easier, and eventually leading to the realization of the series expansion in a format more usable for further mathematical operations.