Problem 8 - Essential Calculus Early Transcendentals

Question

Find the Taylor polynomial \(T_{3}(x)\) for the function \(f\) centered at the number \(a .\) Graph \(f\) and \(T_{3}\) on the same screen. $$f(x)=\tan ^{-1} x, \quad a=1$$

Step-by-Step Solution

Answer

The Taylor polynomial \(T_3(x)\) is \( \frac{\pi}{4} + \frac{1}{2}(x-1) - \frac{1}{4}(x-1)^2 + \frac{1}{48}(x-1)^3 \).

Step 1: Understanding the Taylor Series

The Taylor series for a function \( f(x) \) around a point \( a \) is given by: \[T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]We need the polynomial up to the 3rd degree, so we include terms up to \( f'''(a) \).

Step 2: Calculate Derivatives of \(f(x) = \tan^{-1}(x)\)

The derivatives of \( f(x) = \tan^{-1}(x) \) are:1. \( f'(x) = \frac{1}{1+x^2} \)2. \( f''(x) = \frac{-2x}{(1+x^2)^2} \)3. \( f'''(x) = \frac{-2(1+x^2)^2 + 8x^2}{(1+x^2)^4} \)

Step 3: Evaluate the Derivatives at \(a=1\)

Substitute \( x = 1 \) into the derivatives:1. \( f(1) = \tan^{-1}(1) = \frac{\pi}{4} \)2. \( f'(1) = \frac{1}{1+1^2} = \frac{1}{2} \)3. \( f''(1) = \frac{-2 \cdot 1}{(1+1^2)^2} = \frac{-2}{4} = -\frac{1}{2} \)4. \( f'''(1) = \frac{-2(1+1^2) + 8 \cdot 1^2}{(1+1^2)^4} = \frac{-6 + 8}{16} = \frac{1}{8} \)

Step 4: Form the Taylor Polynomial \(T_3(x)\)

Plug the evaluated derivatives into the Taylor series formula:\[T_3(x) = \frac{\pi}{4} + \frac{1}{2}(x-1) - \frac{1}{4}(x-1)^2 + \frac{1}{48}(x-1)^3\]Simplify each term to construct the polynomial up to the cubic term.

Step 5: Sketch Graphs of \(f(x)\) and \(T_3(x)\)

Plot the function \(f(x) = \tan^{-1}(x)\) and the polynomial \(T_3(x)\) on the same set of axes to compare their shapes around \(x=1\). Notice how the polynomial approximates the function closely near the center \(a=1\).

Key Concepts

Taylor Series

The Taylor Series is a powerful mathematical tool that lets us approximate complex functions using simpler polynomial terms. Imagine trying to approximate a curve by a combination of straight lines and curves. That's exactly what the Taylor Series does, it represents a function as a sum of its derivatives evaluated at a particular point, known as the "center." This provides a polynomial that gets more accurate with more terms.
For instance, the Taylor polynomial of degree 3, denoted as \( T_3(x) \), includes terms up to the third derivative, offering a cubic approximation. This involves calculating derivatives at the centered point \( a \). Thus, a Taylor Polynomial looks like:

• 0th term: \( f(a) \)
• 1st term: \( f'(a)(x-a) \)
• 2nd term: \( \frac{f''(a)}{2!}(x-a)^2 \)
• 3rd term: \( \frac{f'''(a)}{3!}(x-a)^3 \)

By calculating further derivatives and including more terms, a Taylor polynomial can approximate functions even more accurately, especially near the center point \( a \). This helps in various fields like physics, engineering, and economics where understanding behavior near a point is crucial.

Calculus Derivatives

Calculus Derivatives are the building blocks of Taylor Polynomials. Understanding how to find derivatives is crucial since each term in a Taylor polynomial depends on them. Let's consider the derivative of \( f(x) = \tan^{-1}(x) \):

• First derivative: \( f'(x) = \frac{1}{1+x^2} \), gives us the slope or rate of change of the function.
• Second derivative: \( f''(x) = \frac{-2x}{(1+x^2)^2} \), tells us how the rate of change itself is changing.
• Third derivative: \( f'''(x) = \frac{-2(1+x^2)^2 + 8x^2}{(1+x^2)^4} \), adds even more detail to how the curve bends.

Evaluating these derivatives at \( x = 1 \) provides specific values contributing to our Taylor Polynomial. For example, we find \( f'(1) = \frac{1}{2} \), which influences the linear term in our polynomial.
Finding these derivatives involves applying rules like the power rule, product rule, and chain rule in calculus, making it important to understand each rule's application and how it affects the derivative at each step.

Graphing Functions

Graphing Functions is like visualizing how complex functions and their approximations overlap. It's a demonstration that brings mathematical expressions to life. Imagine graphing the original function \( f(x) = \tan^{-1}(x) \) alongside its Taylor Polynomial approximation \( T_3(x) \).
When centered at \( x = 1 \), you would notice that \( T_3(x) \) closely mirrors \( f(x) \) around that point. This is because the polynomial captures the essential shape and behavior of the function where it matters most—the closer you are to the center \( x = 1 \), the more accurate the approximation.
Plotting both graphs can show:

• The matching slopes near the center, thanks to the first derivative.
• How well the polynomial curve hugs the original function, reflecting a snug fit due to higher-order derivatives.

This visualization is key for comprehension, emphasizing how a powerful tool like a Taylor Polynomial can simplify and explain behavior near a specific point with intricate detail.