Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Taylor polynomial of order 3 based at a for the given function. $$ \cot ^{-1} x ; a=1 $$

Step-by-Step Solution

Answer

\( P_3(x) = \frac{\pi}{4} - \frac{1}{2}(x-1) + \frac{1}{4}(x-1)^2 \)

Step 1: Recall the Formula for Taylor Polynomial

The Taylor polynomial of order 3 for a function \( f(x) \) centered at \( a \) is given by:\[P_3(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3\]Our job is to find \( f(a) \), \( f'(a) \), \( f''(a) \), and \( f'''(a) \) for \( f(x) = \cot^{-1}(x) \) at \( a = 1 \).

Step 2: Compute \( f(a) \)

First, we need to evaluate \( f(a) = \cot^{-1}(1) \). \( \cot^{-1}(1) = \frac{\pi}{4} \), because \( \cot \left(\frac{\pi}{4}\right) = 1 \). Thus, \( f(1) = \frac{\pi}{4} \).

Step 3: Compute \( f'(x) \) and \( f'(a) \)

The derivative of \( f(x) = \cot^{-1}(x) \) is \( f'(x) = -\frac{1}{1+x^2} \). Now, evaluate at \( x = 1 \):\[ f'(1) = -\frac{1}{1+1^2} = -\frac{1}{2} \]

Step 4: Compute \( f''(x) \) and \( f''(a) \)

Find the second derivative \( f''(x) \) of \( f(x) = \cot^{-1}(x) \). Start with the first derivative:\[ f'(x) = -\frac{1}{1+x^2} \]Use the quotient rule:\[ f''(x) = \left(\frac{d}{dx} -\frac{1}{1+x^2}\right) = \frac{2x}{(1+x^2)^2} \]Evaluate at \( x = 1 \):\[ f''(1) = \frac{2 \times 1}{(1+1^2)^2} = \frac{1}{2} \]

Step 5: Compute \( f'''(x) \) and \( f'''(a) \)

Now, find the third derivative \( f'''(x) \) using the second derivative.\[ f''(x) = \frac{2x}{(1+x^2)^2} \]The third derivative can be computed using the quotient rule again:\[ f'''(x) = \frac{d}{dx} \left( \frac{2x}{(1+x^2)^2} \right) \]Apply the quotient rule to get:\[ f'''(x) = \frac{2(1+x^2)^2 - 8x^2x}{(1+x^2)^4} \]Simplify it:\[ f'''(x) = \frac{2(1-x^2)}{(1+x^2)^3} \]Evaluate at \( x = 1 \):\[ f'''(1) = \frac{2(1-1^2)}{(1+1^2)^3} = 0 \]

Step 6: Construct the Taylor Polynomial

Now we have all the derivatives evaluated at \( a \):- \( f(1) = \frac{\pi}{4} \)- \( f'(1) = -\frac{1}{2} \)- \( f''(1) = \frac{1}{2} \)- \( f'''(1) = 0 \)Thus, the Taylor polynomial is:\[P_3(x) = \frac{\pi}{4} - \frac{1}{2}(x-1) + \frac{1}{4}(x-1)^2 + 0 \times (x-1)^3\]Simplifying, we obtain:\[P_3(x) = \frac{\pi}{4} - \frac{1}{2}(x-1) + \frac{1}{4}(x-1)^2\]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the actions of the basic trigonometric functions like sine, cosine, and tangent. They are crucial in many areas of mathematics, especially when solving trigonometric equations where one needs to determine the angle given a particular trigonometric ratio. In our problem, the function of interest is the inverse cotangent, denoted as \( \cot^{-1}(x) \). This function gives the angle whose cotangent is \( x \).
For instance, \( \cot^{-1}(1) = \frac{\pi}{4} \) because the cotangent of \( \frac{\pi}{4} \) is 1. Understanding inverse trigonometric functions is essential for working through this exercise, especially when evaluating and differentiating these functions at given points.

Derivative Calculus

Derivative calculus involves finding the rate at which a function changes at any given point; this is known as the derivative of the function. Derivatives are a fundamental tool in calculus, allowing us to understand how functions behave and change over their domain.
For the function \( f(x) = \cot^{-1}(x) \), its first derivative \( f'(x) = -\frac{1}{1+x^2} \) tells us how \( \cot^{-1}(x) \) changes with respect to \( x \). To create the Taylor polynomial, we need derivatives up to the third order:

• First derivative to capture initial rates of change.
• Second derivative, obtained via the quotient rule, reveals curvature information.
• Third derivative, also via the quotient rule, provides further insights into changes in curvature.
  

These calculations are carefully constructed using rules and techniques, such as the quotient rule, to build accurate polynomial approximations.

Quotient Rule

The quotient rule is a method for finding the derivative of the quotient of two functions. It is invaluable in calculus, especially when dealing with functions expressed as ratios. The rule states that if you have a function \( g(x) = \frac{u(x)}{v(x)} \), then its derivative is:
\[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

• This rule was applied to find the second and third derivatives of \( \cot^{-1}(x) \), with \( u(x) \) being constant and \( v(x) \) being a function of \( x \).
• Correct application of this rule results in accurate derivatives, which are essential for constructing Taylor polynomials.
  

Mastering the quotient rule is crucial for handling complicated derivatives like those seen in this exercise.

Polynomial Approximation

Polynomial approximation is a mathematical technique for estimating functions with polynomials. The Taylor polynomial, a type of polynomial approximation, gives a finite series that approximates a function near a specific point. For our task, the Taylor polynomial approximates \( \cot^{-1}(x) \) around \( x=1 \).
We use the formula:
\[P_3(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3\]
By doing this up to third order, we capture not just the function's value at the center point but also information about its slope, curvature, and changes in curvature:

• The first term \( f(a) \) provides the function value.
• The second term involving \( f'(a) \) reflects the slope or first-order change.
• The third term with \( f''(a) \) gives curvature.
• The fourth term, which is zero here, would adjust for further detail where non-zero.
  

This powerful tool can simplify complex functions into more manageable forms, highlighting trends and behavior near particular points.