Problem 30 - Calculus: Early Transcendentals

Question

Suppose you know that \( f^{(n)}(4) = \frac {( - 1)^n n!}{3^n(n + 1)} \) and the Taylor series of \( f \) centered at 4 converges to \( f(x) \) for all \( x \) in the interval of convergence. Show that the fifthdegree Taylor polynomial approximates $ f(5) with error less than 0.0002.

Step-by-Step Solution

Answer

The error of the Taylor polynomial is < 0.0002.

Step 1: Understanding the Taylor Polynomial

The Taylor polynomial of degree 5 for a function \( f \) about \( x = 4 \) is given by:\[T_5(x) = f(4) + f'(4)(x - 4) + \frac{f''(4)}{2!}(x - 4)^2 + \frac{f'''(4)}{3!}(x - 4)^3 + \frac{f^{(4)}(4)}{4!}(x - 4)^4 + \frac{f^{(5)}(4)}{5!}(x - 4)^5\]Here, \( f^{(n)}(4) \) is provided, so we can substitute them directly into the expression.

Step 2: Calculate Individual Derivatives Evaluated at x = 4

Given that \( f^{(n)}(4) = \frac{(-1)^n n!}{3^n(n+1)} \), we calculate each of the needed derivatives:- \( f(4) = \frac{(-1)^0 \, 0!}{3^0(0+1)} = 1 \)- \( f'(4) = \frac{(-1)^1 \, 1!}{3^1(1+1)} = -\frac{1}{6} \)- \( f''(4) = \frac{(-1)^2 \, 2!}{3^2(2+1)} = \frac{4}{27} \)- \( f^{(3)}(4) = \frac{(-1)^3 \, 3!}{3^3(3+1)} = -\frac{2}{27} \)- \( f^{(4)}(4) = \frac{(-1)^4 \, 4!}{3^4(4+1)} = \frac{8}{243} \)- \( f^{(5)}(4) = \frac{(-1)^5 \, 5!}{3^5(5+1)} = -\frac{5}{729} \)

Step 3: Construct the 5th Degree Taylor Polynomial

Substitute the derivatives into the Taylor polynomial for \( x = 5 \):\[T_5(5) = 1 - \frac{1}{6}(5-4) + \frac{4}{27}\frac{(5-4)^2}{2} - \frac{2}{27}\frac{(5-4)^3}{6} + \frac{8}{243}\frac{(5-4)^4}{24} - \frac{5}{729}\frac{(5-4)^5}{120}\]Calculate each term:- Constant: \( 1 \)- First degree: \( -\frac{1}{6} \)- Second degree: \( \frac{4}{54} = \frac{2}{27} \)- Third degree: \( -\frac{2}{162} = -\frac{1}{81} \)- Fourth degree: \( \frac{8}{5832} \)- Fifth degree: \( -\frac{5}{87480} \)

Step 4: Approximate the Taylor Polynomial Value at x = 5

Add the evaluated terms:\[ T_5(5) = 1 - \frac{1}{6} + \frac{2}{27} - \frac{1}{81} + \frac{8}{5832} - \frac{5}{87480} \]Simplify the expression to find the value of the Taylor polynomial at \( x = 5 \).

Step 5: Estimate the Remainder Term (Error)

The error term for the Taylor series is given by the Lagrange remainder:\[ R_5(x) = \frac{f^{(6)}(c)}{6!}(x-4)^6 \]where \( c \) is some number in the interval between 4 and 5. Plug in the given derivative:\[ R_5(5) = \frac{(-1)^6 \, 6!}{3^6(7) \, 720} = \frac{6!}{729 \, 7 \, 720} \]Calculate this value to check if it is less than 0.0002.

Key Concepts

Taylor polynomial

A Taylor polynomial provides an approximation of a function around a specific point by incorporating information about the function's derivatives at that point. In particular, the nth-degree Taylor polynomial of a function \( f \) centered at a point \( a \) is expressed as follows:

• \[ T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n \]

This polynomial utilizes the derivatives \( f^{(k)}(a) \) to capture the behavior of the function around \( a \), up to the degree \( n \).
The Taylor polynomial simplifies the complex nature of functions into a polynomial form, making it easier to analyze and approximate them.
When dealing with higher-degree polynomials, the precision near the point \( a \) generally increases, which is useful in many applications like numerical analysis and engineering.

Lagrange remainder

The Lagrange remainder is an essential concept in Taylor series, describing the error or difference between a function and its nth-degree Taylor polynomial. This error helps determine how accurate a Taylor polynomial approximation is at a particular point.

• For a function \( f \) with a Taylor series expansion, the Lagrange remainder \( R_n(x) \) is given by:\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \]

In this formula, \( c \) is some value between \( a \) (the center of the series) and \( x \) (the point where the approximation is evaluated).
The remainder quantifies the error, offering an upper bound to assess whether the polynomial closely matches the function within the interval of approximation.

Error estimation in Taylor series

Estimating the error in a Taylor series involves determining how closely a Taylor polynomial approximates a function. This often revolves around calculating the Lagrange remainder, as it quantifies the potential error.
Since the remainder is expressed as:

• \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \]

Assessing the error involves determining the possible maximum value of \( f^{(n+1)}(c) \) within the interval. In practical terms, if this remainder is small enough (below a given threshold), then you can trust the Taylor polynomial as a satisfactory approximation.
For many practical applications, ensuring the Lagrange remainder stays within acceptable bounds (such as below 0.0002, as highlighted in the original exercise) allows for a reliable estimation of function values using Taylor polynomials.

Interval of convergence

The interval of convergence for a Taylor series is crucial as it defines where the Taylor series converges to the actual function.
A Taylor series centered at a point \( a \) is given by:

• \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \]

The interval of convergence determines the range of \( x \) within which this series accurately represents the function \( f(x) \).
For many functions, convergence occurs within a specific radius around the center \( a \), resulting in a finite interval.
Accurate knowledge of this interval is important to avoid using the Taylor series in regions where it diverges from the actual function. This underscores the practical utilities and limitations inherent in using Taylor series for approximations.