Problem 49 - Thomas Calculus

Question

\begin{equation} \begin{array}{l}{\text { a. Use Taylor's formula with } n=2 \text { to find the quadratic }} \\ {\text { approximation of } f(x)=(1+x)^{k} \text { at } x=0(k \text { a constant) }} \\ {\text { b. If } k=3, \text { for approximately what values of } x \text { in the interval }} \\ {[0,1] \text { will the error in the quadratic approximation be less }} \\ {\text { than } 1 / 100 ?}\end{array} \end{equation}

Step-by-Step Solution

Answer

The quadratic approximation is \( P_2(x) = 1 + 3x + 3x^2 \). For \( k=3 \), the error is less than \( 1/100 \) for \( x \) values between 0 and approximately 0.215.

Step 1: Identify the Function and Derivatives

The given function is \( f(x) = (1+x)^k \). First, compute the derivatives needed for the quadratic approximation: \( f'(x) \) and \( f''(x) \).\[ f'(x) = k(1+x)^{k-1} \] \[ f''(x) = k(k-1)(1+x)^{k-2} \] at \( x = 0 \).

Step 2: Compute Derivatives at x=0

Substitute \( x=0 \) into the derivatives to find their values. For \( f'(x) \) and \( f''(x) \):\[ f'(0) = k(1+0)^{k-1} = k \] \[ f''(0) = k(k-1)(1+0)^{k-2} = k(k-1) \].

Step 3: Construct the Quadratic Approximation

Using Taylor's formula, the quadratic approximation at \( x = 0 \) is:\[ P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2} x^2 \]\[ P_2(x) = 1 + kx + \frac{k(k-1)}{2} x^2 \].

Step 4: Find Error for k=3

With \( k = 3 \), find the quadratic approximation:\[ P_2(x) = 1 + 3x + \frac{3(3-1)}{2} x^2 = 1 + 3x + 3x^2 \]. The error of the approximation is given by the next term in the Taylor series, \( E(x) = \frac{f'''(c)}{6}x^3 \), where \(c\) lies in \((0,x)\).

Step 5: Estimate Error Bounds

For \( k=3 \), \( f'''(x) = 6(1+x)^0 = 6 \). The upper bound for the error is \( |E(x)| \leq \frac{6}{6}x^3 = x^3 \). Set \( x^3 < \frac{1}{100} \) to find the range of \( x \): \[ x < (\frac{1}{100})^{\frac{1}{3}} \approx 0.215 \]. The values of \( x \) for which the error is less than \( \frac{1}{100} \) are approximately in \( [0, 0.215) \).

Key Concepts

Quadratic Approximation

The quadratic approximation is an essential concept in calculus that helps us estimate the value of a function using a polynomial of degree two. This method is particularly useful when the function is too complex to evaluate precisely. We use Taylor's formula to perform this approximation, which requires the calculation of derivatives at a specific point. In the provided exercise, the quadratic approximation for the function \(f(x) = (1+x)^k\) is calculated at \(x=0\).
This approximation is found using a Taylor polynomial of order 2. The Taylor series provides a way to approximate functions as an infinite sum of terms calculated from the values of its derivatives at a single point.
The formula for a quadratic approximation at a point \(a\) is as follows:

• \(P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2\)

Substituting \(x = 0\) into the exercise's function, the quadratic approximation \(P_2(x)\) becomes:

• \(P_2(x) = 1 + kx + \frac{k(k-1)}{2}x^2\)

This polynomial approximates \(f(x)\) around \(x = 0\), allowing us to analyze and estimate its behavior within a local interval.

Error Estimation

Error estimation in quadratic approximation gives us insight into how close our approximation is to the true function. After we construct the quadratic approximation, it is crucial to assess its accuracy. For this, the error term is derived from the Taylor series.
In Taylor series, the error is associated with the remainder of the series beyond the degree of the approximating polynomial. For a quadratic approximation, the error is the next term in the series after the second derivative term. It is given by:

• \(E(x) = \frac{f'''(c)}{6}(x-a)^3\)

where \(c\) is some value between \(a\) and \(x\). In our exercise, to keep the error under 1/100 for \(k=3\), it is determined that the error is less than \(x^3\) by using:

• \(|E(x)| \leq x^3 < \frac{1}{100}\)

Solving the inequality \(x^3 < \frac{1}{100}\) provides the value for \(x\), restricting its range to \, approximately \([0, 0.215)\). This means that for values of \(x\) less than 0.215, our quadratic approximation will have an error less than 1/100.

Derivatives Calculation

To achieve the quadratic approximation, we first need to calculate the derivatives of the function at the point of interest, \(x = 0\). For \(f(x) = (1+x)^k\), we determine its first and second derivatives as follows:

• The first derivative: \(f'(x) = k(1+x)^{k-1}\)
• Evaluated at \(x = 0\): \(f'(0) = k\)
• The second derivative: \(f''(x) = k(k-1)(1+x)^{k-2}\)
• Evaluated at \(x = 0\): \(f''(0) = k(k-1)\)

These calculations are crucial as they provide the coefficients for the polynomial used in the quadratic approximation. Precisely evaluating these derivatives at \(x = 0\) simplifies the expression for the approximation, yielding:

• \(P_2(x) = 1 + kx + \frac{k(k-1)}{2}x^2\)

In our example, this process allows us to approximate \(f(x)\) using these derivatives to simplify and estimate the function’s value within a small interval around \(x = 0\). The derivative calculations form the backbone of constructing the quadratic approximation.