Problem 21 - 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 \(x\). $$ f(x)=\frac{x}{\sqrt{4+x}}, \quad n=3 $$

Step-by-Step Solution

Answer

The third degree Taylor polynomial for the function \(f(x)=\frac{x}{\sqrt{4+x}}\) about \(x=0\) is \(P_3(x) = \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{48}x^3\) and the radius of convergence of the corresponding Maclaurin series is \(\infty\).

Step 1: Compute the first few derivatives of \(f(x)\) at \(x=0\)

The goal is to compute the derivatives of \(f(x)\) up to the 3rd degree, evaluated at \(x=0\). The first four derivatives are: \(f'(x) =-\frac{x}{(4+x)^\frac{3}{2}} +\frac{1}{\sqrt{4+x}}\), \(f''(x) =\frac{3x}{(4+x)^\frac{5}{2}}-\frac{1}{(4+x)^\frac{3}{2}}\), \(f'''(x) =-\frac{15x}{(4+x)^\frac{7}{2}}+\frac{3}{(4+x)^\frac{5}{2}}\), \(f^{(4)}(x) =\frac{105x}{(4+x)^\frac{9}{2}}-\frac{15}{(4+x)^\frac{7}{2}}\). At \(x=0\), the values are: \(f(0) =0\), \(f'(0) =\frac{1}{2}\), \(f''(0) =-\frac{1}{8}\), \(f'''(0) =\frac{1}{16}\), \(f^{(4)}(0) =-\frac{5}{128}\).

Step 2: Compute the third degree Taylor polynomial

The Taylor polynomial of degree \(n=3\) is defined as: \(P(x) = f(a) + f'(a)*(x-a) + f''(a)*(x-a)^2/2! + f'''(a)*(x-a)^3/3!\). For \(a=0\), this becomes: \(P_3(x) = f(0) + f'(0)*x + f''(0)*x^2/2! + f'''(0)*x^3/3! = 0 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{48}x^3\).

Step 3: Compute the radius of convergence

The radius of convergence of the Maclaurin series can be determined using the ratio test. For positive value of 'a', the ratio test can be defined as :\(R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right|\). However, since it was mentioned that one of these series converges for all \(x\), and given the nature of the function, we can infer that the radius of convergence is \(\infty\).

Key Concepts

Binomial Series

The binomial series is a powerful tool that lets us expand expressions like \( (1 + x)^n \) into an infinite series. This is especially helpful when dealing with functions that involve radicals or powers. In the given exercise, the function is \( f(x) = \frac{x}{\sqrt{4+x}} \). By clever manipulation, we can express this in a form that resembles a binomial expansion.The binomial series expansion for \( (1 + x)^n \) is given by:

• \( 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots \)

For fractional or negative powers, the series is infinite. In this exercise, employing such expansions allows us to simplify \( \frac{1}{\sqrt{4+x}} \) and obtain derivatives necessary for forming the Taylor polynomial. Understanding how to manipulate binomial series is essential to handle complex functions and derive Maclaurin or Taylor series.

Radius of Convergence

Every power series has a specific set of \(x\)-values for which it converges, known as its interval of convergence. The radius of convergence is a measure that defines how far from the center (in this case, \(x=0\)) the series converges. A valuable technique for finding this radius is the ratio test, which involves examining the limit:\[ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| \]Here, we're handling a Maclaurin series. Our exercise reveals an important feature—the radius of convergence is infinite. This means the series converges for all real numbers \(x\). Many well-known functions, like the exponential function, have this property. The function in our exercise also exhibits such behavior because one of its associated series covers all \(x\). Understanding why this happens can reinforce your grasp of convergence and help tackle complex series.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series where we expand a function around \(x = 0\). It's essentially a way to represent a complicated function as an infinite sum of simpler terms to approximate it around zero. For the Maclaurin series of \( f(x) = \frac{x}{\sqrt{4+x}} \), the focus is on determining the first few derivatives and evaluating them at zero.The general form for a Maclaurin series is:

• \( f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \)

In our exercise, the first three derivatives were calculated and plugged into this formula, yielding:\[ P_3(x) = 0 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{48}x^3 \]The result is a polynomial that approximates the behavior of \( f(x) \) near \( x = 0 \). Therefore, the Maclaurin series grants us a powerful method for approximating functions which might not easily be expressed otherwise. Learning to calculate and apply these series can significantly enhance your problem-solving toolkit.