Problem 24 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=2 x^{3}+x^{2}+3 x-8, \quad a=1\)

Step-by-Step Solution

Answer

The Taylor series is \( -2 + 11(x-1) + 7(x-1)^2 + 2(x-1)^3 \).

Step 1: Define the Taylor Series Formula

The Taylor series of a function \( f \) about \( x = a \) is given by the formula:\[f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]This formula converts a function into an infinite sum of terms calculated from the derivatives of \( f \) at \( a \).

Step 2: Calculate the Function Value at a

Evaluate \( f \, (a) = f(1) \) using the given function:\[f(1) = 2(1)^3 + 1^2 + 3(1) - 8 = 2 + 1 + 3 - 8 = -2\]

Step 3: Find the First Derivative and Evaluate at a

First, calculate the first derivative of \( f(x) = 2x^3 + x^2 + 3x - 8 \):\[f'(x) = 6x^2 + 2x + 3\]Evaluate this derivative at \( x = 1 \):\[f'(1) = 6(1)^2 + 2(1) + 3 = 6 + 2 + 3 = 11\]

Step 4: Find the Second Derivative and Evaluate at a

Calculate the second derivative of \( f(x) \):\[f''(x) = 12x + 2\]Evaluate it at \( x = 1 \):\[f''(1) = 12(1) + 2 = 14\]

Step 5: Find the Third Derivative and Evaluate at a

Calculate the third derivative of \( f(x) \):\[f'''(x) = 12\]Since \( f'''(x) \) is constant, evaluating at \( x = 1 \) also gives:\[f'''(1) = 12\]

Step 6: Construct the Taylor Series

Now, substitute the calculated values into the Taylor series formula:\[f(x) = -2 + \frac{11}{1!}(x-1) + \frac{14}{2!}(x-1)^2 + \frac{12}{3!}(x-1)^3\]Simplify:\[f(x) = -2 + 11(x-1) + 7(x-1)^2 + 2(x-1)^3\]

Step 7: Write the Final Taylor Series

Write down the Taylor series obtained for the function:\[ f(x) = -2 + 11(x-1) + 7(x-1)^2 + 2(x-1)^3 \]

Key Concepts

Derivatives

Derivatives play a crucial role in understanding how a function behaves as its input changes. They measure the rate at which a function's output changes concerning its input. To construct a Taylor series, derivatives of the function at a specific point are used.
At each order (or degree) of the Taylor series, a different derivative is calculated:

• The
  first derivative involves differentiating the original function once, providing information on its slope or rate of change at any point.
• The
  second derivative involves differentiating twice, offering insights into the function's concavity, indicating how the slope itself is changing.
• Higher order
  derivatives, such as the third, fourth, etc., further refine predictions about the function's behavior over intervals.

The process of using derivatives to produce a polynomial that approximates the function at a given point helps ensure that the Taylor series is an effective tool in mathematics.

Polynomial Functions

Polynomial functions are expressions involving sums of powers of variables, with each term consisting of a coefficient, a variable, and an exponent. In the case of Taylor series expansion, polynomial functions serve as the building blocks for approximating more complex functions.

• For example,
  consider a simple polynomial function like \( f(x) = 2x^3 + x^2 + 3x - 8 \). This expression is a sum of terms, each of which involves a power of \( x \).
• These
  polynomial functions are easy to differentiate, a fact that simplifies the process of constructing their Taylor series.

Polynomial functions are vital in creating series expansions because their derivatives result in more polynomial functions, which maintain the simplicity and power of the initial function's form.

Series Expansion

Series expansion, particularly Taylor series expansion, is a method of approximating functions by transforming them into infinite sums of polynomial terms calculated from derivatives at a specific point. This transformation allows mathematicians to analyze functions that may be too complex to study in their original form.
Key aspects of series expansion include:

• The
  Taylor series formula: \[ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]. This provides a systematic method to calculate each term in the series.
• The
  ability to convenience and flexibility of this series to approximate functions near the point \( a \).
• By
  using a finite number of terms, we can closely approximate functions without an endless series and often without significant computational complexity.

Series expansion is a powerful tool that breaks complex functions into more manageable pieces, making it easier to solve, analyze, and understand different mathematical problems.