Problem 26 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=3 x^{5}-x^{4}+2 x^{3}+x^{2}-2, \quad a=-1\)

Step-by-Step Solution

Answer

The Taylor series is \(-7 + 23(x+1) - 41(x+1)^2 + 36(x+1)^3 - 16(x+1)^4 + 3(x+1)^5.\)

Step 1: Evaluating the Function at a

First, we need to evaluate the function at \( a = -1 \). So, substitute \( x = -1 \) into the function \( f(x) = 3x^5 - x^4 + 2x^3 + x^2 - 2 \):\[ f(-1) = 3(-1)^5 - (-1)^4 + 2(-1)^3 + (-1)^2 - 2. \]Calculating each term, we get:\[ f(-1) = 3(-1) - 1 - 2 + 1 - 2 = -3 - 1 - 2 + 1 - 2 = -7. \]

Step 2: First Derivative and Evaluation

Next, find the first derivative of the function and evaluate it at \( a = -1 \).The first derivative is found by differentiating term by term:\[ f'(x) = 15x^4 - 4x^3 + 6x^2 + 2x. \]Now, evaluate \( f'(x) \) at \( x = -1 \):\[ f'(-1) = 15(-1)^4 - 4(-1)^3 + 6(-1)^2 + 2(-1). \]Calculating each term:\[ f'(-1) = 15(1) + 4 + 6 - 2 = 15 + 4 + 6 - 2 = 23. \]

Step 3: Second Derivative and Evaluation

Find the second derivative and evaluate it at \( a = -1 \).The second derivative:\[ f''(x) = 60x^3 - 12x^2 + 12x + 2. \]Now, evaluate \( f''(x) \) at \( x = -1 \):\[ f''(-1) = 60(-1)^3 - 12(-1)^2 + 12(-1) + 2. \]Calculating each term:\[ f''(-1) = -60 - 12 - 12 + 2 = -82. \]

Step 4: Third Derivative and Evaluation

Find the third derivative and evaluate it at \( x = -1 \).The third derivative:\[ f'''(x) = 180x^2 - 24x + 12. \]Evaluate \( f'''(x) \) at \( x = -1 \):\[ f'''(-1) = 180(1) + 24 + 12 = 216. \]

Step 5: Fourth Derivative and Evaluation

Find the fourth derivative and evaluate it at \( x = -1 \).The fourth derivative:\[ f^{(4)}(x) = 360x - 24. \]Evaluate \( f^{(4)}(x) \) at \( x = -1 \):\[ f^{(4)}(-1) = 360(-1) - 24 = -384. \]

Step 6: Fifth Derivative and Evaluation

Find the fifth derivative and evaluate it at \( x = -1 \).The fifth derivative is a constant:\[ f^{(5)}(x) = 360. \]Evaluate at \( x = -1 \):\[ f^{(5)}(-1) = 360. \]

Step 7: Constructing the Taylor Series

Combine all derivatives evaluated at \( x = -1 \) into the Taylor series formula:The Taylor series for the function \( f(x) \) centered at \( a \) is:\[ T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n. \]Substitute the values calculated:\[ T(x) = f(a) + \frac{f'(a)}{1!}(x+1) + \frac{f''(a)}{2!}(x+1)^2 + \frac{f'''(a)}{3!}(x+1)^3 + \frac{f^{(4)}(a)}{4!}(x+1)^4 + \frac{f^{(5)}(a)}{5!}(x+1)^5. \]So, plugging the values:\[ T(x) = -7 + 23(x+1) - \frac{82}{2}(x+1)^2 + \frac{216}{6}(x+1)^3 - \frac{384}{24}(x+1)^4 + \frac{360}{120}(x+1)^5. \]This simplifies to:\[ T(x) = -7 + 23(x+1) - 41(x+1)^2 + 36(x+1)^3 - 16(x+1)^4 + 3(x+1)^5. \]

Key Concepts

Derivative

Derivatives play a critical role in mathematics, especially when dealing with functions.They are essentially tools that tell us the rate at which a function is changing at any point.
Consider the function as a journey and the derivative as the speedometer showing speed changes. For instance, if you have a polynomial function like the one given in the exercise: \[ f(x) = 3x^5 - x^4 + 2x^3 + x^2 - 2 \]Getting its derivative means finding a new function indicating how the above polynomial changes with respect to x.
To find the derivative, apply the power rule: bring down the power as a coefficient, then subtract one from the power. Thus, the derivative of each term of the polynomial is added up to form:\[ f'(x) = 15x^4 - 4x^3 + 6x^2 + 2x. \]The first derivative, \( f'(x) \), tells you the slope of the function at any given point.
Continuing with derivatives, subsequent derivatives (such as second, third, etc.) are found by applying the derivative process again on the previous derivative.
This is why the exercise calculates up to the fifth derivative for the Taylor series expansion.

Polynomial function

Polynomial functions are expressions made up of variables and coefficients composed of terms that are only non-negative integer powers of the variable.
Each term is in the form \( ax^n \) where \( a \) is the coefficient and \( n \) is the power or degree of that term. For example, in the function from the exercise:\[ f(x) = 3x^5 - x^4 + 2x^3 + x^2 - 2 \]The highest degree is 5, making it a fifth-degree polynomial.Polynomials are smooth, continuous functions with no sharp corners or gaps, making them useful for approximation.
They form the basis of calculus operations, such as finding derivatives to determine function behavior, which helps construct Taylor Series.

• Polynomials can be added, subtracted, multiplied, or divided, with the operations performed term by term.
• The simplicity of their structure makes them easy to manipulate mathematically.

Polynomials are the building blocks for much larger systems and concepts in calculus, such as Taylor series and series expansions, as shown in the solution.

Series expansion

Series expansion is a powerful mathematical tool used to express functions as infinite sums of terms. A fundamental branch of this concept is the Taylor Series, which breaks down complex functions into simpler polynomial expressions.
This is achieved by utilizing the function’s derivatives at a specific point, to systematically construct a series. For the function:\[ f(x) = 3x^5 - x^4 + 2x^3 + x^2 - 2 \]and centered at \( a = -1 \), the Taylor series offers a polynomial approximation around \( x = -1 \).
Here's how this works:

• Each term of the Taylor series is derived from the derivatives of the function evaluated at the centered point \( a \).
• The formula for a Taylor series at \( a \) is:\[ T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n. \]
• This involves calculating derivatives, up to 5 in this case, to create the polynomial approximation.

The resulting series expansion mirrors the behavior of the original function closely near the point \( a \) and becomes an invaluable tool for approximating functions, especially in calculus, physics, and engineering applications.