Problem 30 - Calculus for Biology and Medicine

Question

In Problems 27-30, use the following form of the error term $$ R_{n+1}(x)=\frac{f^{(n+1)}(c)}{(n+1) !} x^{n+1} $$ where \(c\) is between 0 and \(x\), to determine in advance the degree of Taylor polynomial at \(a=0\) that would achieve the indicated accuracy in the interval \([0, x] .\) (Do not compute the Taylor polynomial.) $$ f(x)=\ln (1+x), x=0.1, \text { error }<10^{-2} $$

Step-by-Step Solution

Answer

A degree 1 Taylor polynomial is needed to achieve error < 0.01.

Step 1: Identify the Function and Its Derivatives

We know the function is given as \( f(x) = \ln(1+x) \). To use the error term formula, we need to compute the derivatives of the function. The \((n+1)\)-th derivative of \(f(x)\) is given by \( f^{(n+1)}(x) = \frac{(-1)^n n!}{(1+x)^{n+1}} \). For a Taylor series centered at \(a=0\), this becomes \(f^{(n+1)}(0)= (-1)^n n!\).

Step 2: Substitute Into Error Term Formula

Using the error term formula: \[ R_{n+1}(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}, \] we substitute the derivative into this formula: \[ R_{n+1}(x) = \frac{(-1)^n n!}{(1+c)^{n+1}} \cdot \frac{x^{n+1}}{(n+1)!}. \] Let \(c\) be between 0 and \(x\).

Step 3: Evaluate the Maximum Possible Value of Error

Since \( x = 0.1 \) and \( c \) is between 0 and \(x\), the maximum value of \((1+c)^{-n-1}\) occurs when \( c = 0 \). Thus, \( \frac{1}{(1+c)^{n+1}} \leq 1 \). This simplifies the error bound to: \[ \left|\frac{(-1)^n n!}{(n+1)!} \cdot (0.1)^{n+1}\right| < 10^{-2}. \]

Step 4: Simplify and Solve Inequality

The absolute value of the error becomes: \[ \frac{n!}{(n+1)!} (0.1)^{n+1} < 10^{-2}. \]Simplifying this using \((n+1)! = (n+1) \cdot n!\), we get: \[ \frac{1}{n+1} (0.1)^{n+1} < 10^{-2}. \]Solve for \(n\) by trying different values to find the smallest \(n\) that satisfies this inequality.

Step 5: Trial and Error to Find Suitable Degree

Start plugging in values for \(n\):- For \(n=1\), \(\frac{1}{2} \cdot (0.1)^2 = 0.005 < 0.01\) which satisfies.- Confirm by trying \(n=0\), \(n=0\) produces \(0.1\) which does not satisfy.Thus, the smallest \(n\) that satisfies the inequality is 1. So a degree 1 polynomial (a linear polynomial) is needed.

Key Concepts

Error Analysis

Error analysis is a crucial concept when working with approximations in mathematics. When using a Taylor series to approximate a function, it's important to understand how accurately the approximation represents the function. The "error term" helps us measure this accuracy. It tells us how far off the approximation might be from the actual function value.

The error term for a Taylor series is given by the formula: \[ R_{n+1}(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \]where:

• \( f^{(n+1)}(c) \) is the
• \((n+1)\)-th derivative of the function at some point \(c\) between 0 and \(x\).
• \( (n+1)! \) is the factorial of \(n+1\).
• \( x^{n+1} \) accounts for how the error changes with \(x\).

By assessing this formula, especially how quickly the term \( x^{n+1} \) shrinks for small \(x\), we gain insight into the accuracy of our Taylor polynomial.

In our example, we want the error to be less than \(10^{-2}\). By substituting potential values for \(n\), we determine which degree of Taylor polynomial will keep the error term within this range. Reducing error translates to choosing appropriate \(n\), ensuring our polynomial closely mimics the function over the desired interval.

Derivatives

Derivatives play a significant role in forming the Taylor series and determining the error term. A Taylor series is a sum of derivatives of a function at a specific point, usually known as the center.

In our context, the function given is \( f(x) = \ln(1 + x) \). Its derivatives at a point, typically represented mathematically, are figured out as \( f^{(n+1)}(x) = \frac{(-1)^n n!}{(1+x)^{n+1}} \). When evaluating at \(a = 0\), they simplify to \( f^{(n+1)}(0) = (-1)^n n! \).

• First derivative gives the slope of tangent line at a point.
• Higher order derivatives capture the curvature and other nuanced changes in function behavior.

These derivative terms are components of the Taylor polynomial, dictating how each additional term in the polynomial (as \(n\) increases) improves the accuracy of the approximation.

Understanding derivatives is thus vital. They not only facilitate building the Taylor series but also directly impact the computed error. By properly calculating derivatives, we can optimize our polynomial to approach the function more precisely across an interval.

Polynomial Degree

The degree of a Taylor polynomial is intimately related to its accuracy and the speed with which it approximates a function across an interval. A higher polynomial degree tends to achieve better accuracy because it includes more terms, which results in a closer approximation to the function.

In our task, we approximated \( f(x) = \ln(1+x) \) using a Taylor polynomial at \(a=0\). The goal was to find the degree that keeps the error under \(10^{-2}\) for \( x = 0.1 \).

• The function's error term was simplified, resulting in inequality solving to identify the minimal \(n\) needed for the exactness.
• Trying various trial \(n\) values revealed that \(n=1\) satisfies the inequality, necessitating a first-degree polynomial (linear polynomial).

The process illustrates how increasing the degree from 0 to 1 brought the error below the prescribed limit. So, a careful balance between polynomial degree and computational feasibility is essential in optimization tasks. By doing so, we not only ensure precision but also practical efficiency in approximating functions with Taylor polynomials.