Problem 5 - Thomas Calculus

Question

Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) \(f(x)=1 / x, \quad a=2\)

Step-by-Step Solution

Answer

The Taylor polynomials are: Order 0: \(P_0(x) = \frac{1}{2}\) Order 1: \(P_1(x) = \frac{1}{2} - \frac{1}{4}(x - 2)\) Order 2: \(P_2(x) = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2\) Order 3: \(P_3(x) = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2 - \frac{1}{16}(x - 2)^3\)

Step 1: Understand the problem

We are asked to find Taylor polynomials of orders 0, 1, 2, and 3 for the function \(f(x) = \frac{1}{x}\) at \(a = 2\). This involves calculating derivatives of the function, evaluating them at \(x = 2\), and constructing the polynomials.

Step 2: Calculate the derivatives

Compute the first few derivatives of \(f(x) = \frac{1}{x}\):1. \(f(x) = x^{-1}\)2. \(f'(x) = -x^{-2}\)3. \(f''(x) = 2x^{-3}\)4. \(f'''(x) = -6x^{-4}\)

Step 3: Evaluate derivatives at \(x = 2\)

Substitute \(x = 2\) into each derivative:1. \(f(2) = \frac{1}{2}\)2. \(f'(2) = -\frac{1}{4}\)3. \(f''(2) = \frac{1}{4}\)4. \(f'''(2) = -\frac{3}{8}\)

Step 4: Construct the Taylor Polynomial of Order 0

The 0th order Taylor polynomial is simply the function value at \(a = 2\):\[ P_0(x) = \frac{1}{2} \]

Step 5: Construct the Taylor Polynomial of Order 1

Use the first-order Taylor formula:\[ P_1(x) = f(2) + f'(2)(x - 2) = \frac{1}{2} - \frac{1}{4}(x - 2) \]

Step 6: Construct the Taylor Polynomial of Order 2

Use the second-order Taylor formula:\[ P_2(x) = f(2) + f'(2)(x - 2) + \frac{f''(2)}{2!}(x - 2)^2 = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2 \]

Step 7: Construct the Taylor Polynomial of Order 3

Use the third-order Taylor formula:\[ P_3(x) = f(2) + f'(2)(x - 2) + \frac{f''(2)}{2!}(x - 2)^2 + \frac{f'''(2)}{3!}(x - 2)^3 \ = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2 - \frac{1}{16}(x - 2)^3 \]

Key Concepts

Function Derivatives

In calculus, derivatives represent the rate at which a function changes at any given point. To find the Taylor polynomial, we start with calculating the derivatives of the function.

Considering the given function, - The first derivative, denoted as \(f'(x)\), is the slope of the tangent line at any point on the curve. For \(f(x) = \frac{1}{x}\), the first derivative \(f'(x) = -x^{-2}\).- The second derivative, \(f''(x)\), gives information on the curvature of the function, calculated as \(2x^{-3}\) in this example.- The third derivative, \(f'''(x)\), helps in understanding the change of curvature, calculated as \(-6x^{-4}\).

Derivatives help in forming deeper insights into the behavior and shape of the original function. Calculating these derivatives at a specific point (like \(x = 2\) in the problem) helps in constructing the Taylor polynomial, which approximates the function near that point.

Second-order Taylor Formula

Taylor polynomials approximate functions using derivatives at a particular point. The second-order Taylor Polynomial mainly involves terms up to the second derivative of the function.

The general formula for a second-order Taylor polynomial centered at a point \(a\) is:

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

In our problem, we substitute derivatives calculated at \(a = 2\):

• \(f(2) = \frac{1}{2}\)
• \(f'(2) = -\frac{1}{4}\)
• \(f''(2) = \frac{1}{4}\)

Plugging in these values gives:

• \[ P_2(x) = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2 \]

This polynomial provides a more accurate approximation of the function \(f(x) = \frac{1}{x}\) near \(x = 2\) than the first-order polynomial.

Third-order Taylor Formula

The third-order Taylor polynomial enhances the approximation by including terms up to the third derivative of the function. This involves the third-order Taylor formula.

The standard form of a third-order Taylor polynomial centered at a point \(a\) is:

• \[ P_3(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 \]

For our specific case:

• \(f(2) = \frac{1}{2}\)
• \(f'(2) = -\frac{1}{4}\)
• \(f''(2) = \frac{1}{4}\)
• \(f'''(2) = -\frac{3}{8}\)

Substituting these values results in:

• \[ P_3(x) = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2 - \frac{1}{16}(x - 2)^3 \]

This approximation captures the function's behavior more closely at \(x = 2\), smoothing out predictions with the third derivative's influence.