Problem 17 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Use your knowledge of the binomial series to find the \(n\) th degree Taylor polynomial for \(f(x)\) about \(x=0 .\) Give the radius of convergence of the corresponding Maclaurin series. One of these "series" converges for all \(x\). $$ f(x)=\sqrt[3]{1+x^{2}}, \quad n=5 $$

Step-by-Step Solution

Answer

The fifth-degree Taylor polynomial for the function \(f(x)=\sqrt[3]{1+x^{2}}\) is \(T_{5}(x) = 1 + \frac{1}{3}x^{2} - \frac{1}{9}x^{4}\) and the radius of convergence is infinite.

Step 1: Rewrite the Function

The function \(f(x)=\sqrt[3]{1+x^{2}}\) can be rewritten as \(f(x)=(1+x^{2})^\frac{1}{3}\), which is more amenable to using the binomial theorem.

Step 2: Apply the Binomial Theorem

Applying the binomial theorem, the function can be rewritten as \(f(x) = 1 + \frac{1}{3}x^{2} - \frac{1}{9}x^{4} + \ldots\), \nwhere each coefficient corresponds to the binomial coefficient.

Step 3: Derive the Fifth-Degree Taylor Polynomial

Given that we are only interested in the fifth-degree Taylor polynomial (n=5), the series can be truncated after the x^{5} term. So, the fifth-degree Taylor polynomial is \(T_{5}(x) = 1 + \frac{1}{3}x^{2} - \frac{1}{9}x^{4}\).

Step 4: Determine the Radius of Convergence

The radius of convergence of a power series is given by the formula R=1/limsup |a_{n}|^{1/n}, where a_{n} are the coefficients of the series. Knowing this series is the Maclaurin series for the function, which is a special type of Taylor series where \(x=0\), and it is known that the Maclaurin series for (1+x)^n converges for all x, so the radius of convergence is infinite.

Key Concepts

Binomial Series

The binomial series is essential for understanding how to expand functions of the form \(1+x)^n\) where \(n\) can be any real number. It's a power series derived from the Binomial Theorem, which originally applied only to positive integer exponents. However, in the binomial series, this is extended to accommodate non-integer exponents, allowing us to express functions with roots as infinite series.

For our exercise, the function \(f(x)=\sqrt[3]{1+x^{2}}\) can be re-expressed as \(f(x)=(1+x^{2})^{\frac{1}{3}}\). To find the \(n\)th degree Taylor polynomial for \(f(x)\) about \(x=0\), we apply the binomial series expansion, taking into account that we only need terms up to the fifth degree for this particular exercise.

Radius of Convergence

When we talk about power series, we often want to know where this series is valid, which is characterized by its radius of convergence. This indicates the range of \(x\) values for which the series converges to a finite sum. The radius of convergence can be calculated using the ratio test or by identifying the limit superior of the absolute values of the coefficients raised to the power of \(1/n\), as given by the formula \(R=1/\limsup |a_{n}|^{1/n}\).

In our example, when we apply this concept to the Maclaurin series for the function \(f(x)\), we discover that this particular series converges for all \(x\) because it's a special case of the binomial series where the exponent is \(\frac{1}{3}\). Therefore, the radius of convergence is infinite, meaning that no matter how far you go in the \(x\)-axis, the series will still converge.

Maclaurin Series

The Maclaurin series is a specific type of Taylor series centered at \(x=0\). It’s an infinite sum that provides a way to represent functions as a series of terms involving derivatives evaluated at zero. For many functions, the Maclaurin series offers a convenient way of approximating functions close to \(x=0\) with a polynomial.

In the exercise, we are tasked with finding the Taylor polynomial about \(x=0\) for the function \(f(x)\), which is essentially finding its Maclaurin series. The series we derive by applying the binomial theorem results in a Maclaurin series because the power series is expanded around the point \(x=0\). This series has the advantageous property of converging for all \(x\) values when \(n\) is in the form of a binomial series, specifically when \(n\) is a fraction like \(1/3\), which leads to an infinite radius of convergence.