Problem 11 - Essential Calculus Early Transcendentals

Question

Find the Taylor series for \(f(x)\) centered at the given value of \(a\) . [Assume that \(f\) has a power series expansion. Do not show that \(R_{n}(x) \rightarrow 0.1]\) $$f(x)=x^{4}-3 x^{2}+1, \quad a=1$$

Step-by-Step Solution

Answer

The Taylor series for \( f(x) \) centered at \( a = 1 \) is \( 3 - 2(x-1) + 3(x-1)^2 + 4(x-1)^3 + (x-1)^4 \).

Step 1: Understand the Function and the Center

The function given is \( f(x) = x^4 - 3x^2 + 1 \) and we need to find the Taylor series centered at \( a = 1 \). In the Taylor series, each term involves evaluating a derivative of the function at the center.

Step 2: Calculate the First Derivative

Compute the first derivative of \( f(x) \), which is \( f'(x) = 4x^3 - 6x \). This derivative will help in constructing the linear term of the series.

Step 3: Evaluate First Derivative at Center

Evaluate \( f'(x) \) at \( x = 1 \): \( f'(1) = 4(1)^3 - 6(1) = 4 - 6 = -2 \). This gives the coefficient for the linear term \((x-1)\).

Step 4: Calculate the Second Derivative

Compute the second derivative \( f''(x) = 12x^2 - 6 \), which will be used for the quadratic term.

Step 5: Evaluate Second Derivative at Center

Evaluate \( f''(x) \) at \( x = 1 \): \( f''(1) = 12(1)^2 - 6 = 12 - 6 = 6 \). This gives the coefficient for the quadratic term \((x-1)^2\).

Step 6: Calculate the Third Derivative

Compute the third derivative \( f'''(x) = 24x \), which contributes to the cubic term.

Step 7: Evaluate Third Derivative at Center

Evaluate \( f'''(x) \) at \( x = 1 \): \( f'''(1) = 24(1) = 24 \), providing the coefficient for the cubic term \( (x-1)^3 \).

Step 8: Calculate the Fourth Derivative

Compute the fourth derivative \( f^{(4)}(x) = 24 \), used for the quartic term as it is constant.

Step 9: Evaluate Fourth Derivative at Center

The fourth derivative is constant, so \( f^{(4)}(1) = 24 \). This gives the coefficient for the quartic term \( (x-1)^4 \).

Step 10: Write the Taylor Series

The Taylor series centered at \( a = 1 \) is: \[ f(x) = 3 + (-2)(x-1) + \frac{6}{2!}(x-1)^2 + \frac{24}{3!}(x-1)^3 + \frac{24}{4!}(x-1)^4 \]. Simplify to get \( f(x) = 3 - 2(x-1) + 3(x-1)^2 + 4(x-1)^3 + (x-1)^4 \).

Key Concepts

Power Series Expansion

Taylor series is a wonderful example of a power series expansion. Power series are just infinite series that use powers of a variable, like \( x \), to express complex functions as sums of simpler polynomial terms. By centering the series at a point \( a \), we can understand how the function behaves near that point.
The formula for a Taylor series is:

• \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n \)

Here, \( f^{(n)}(a) \) represents the nth derivative of \( f \) evaluated at the center point \( a \), \( n! \) is the factorial of \( n \), and \( (x-a)^n \) are the polynomial terms.
For our specific function \( f(x) = x^4 - 3x^2 + 1 \), the center is \( a = 1 \). This means we are expanding the function around \( x = 1 \), capturing its behavior in terms of integer powers of \( (x-1) \).
Notice how each term involves calculating derivatives of the function at the center. By understanding power series, we unlock a powerful tool to approximate functions and study their properties around specific points.

Derivatives

Derivatives play a key role in creating a Taylor series. They provide the necessary information on how a function changes. The nth derivative at a given center \( a \) reveals the behavior of the function up to that degree. Each derivative gives us a new term in the series, leading to a better approximation.
Here's a refresher on how derivatives are used in this process:
1. **First Derivative**: Tells us the rate of change or slope at a point. For \( f(x) = x^4 - 3x^2 + 1 \), the first derivative is \( f'(x) = 4x^3 - 6x \). Evaluated at \( x = 1 \), it helps form the linear term. 2. **Second Derivative**: Gives us information on the curvature of the function. \( f''(x) = 12x^2 - 6 \) leads to the quadratic term.
3. **Higher-Order Derivatives**: These derivatives like the third, \( f'''(x) = 24x \), and fourth, \( f^{(4)}(x) = 24 \), indicate more subtle changes, forming cubic and higher-order terms.
The calculation of derivatives is fundamental to capturing the essence of a function and contributes directly to the coefficients of the Taylor series.

Polynomial Functions

Polynomial functions are the building blocks of Taylor series, providing a simple structure to express more complex functions. A polynomial is just a sum of powers of \( x \) with constant coefficients.
Understanding polynomials is crucial as each term of a Taylor series can be viewed as a polynomial term centered at \( a \). When the polynomial function is centered, it becomes \( (x-a) \), forming a basic unit for all series terms.
For instance, in our example, the function \( f(x) = x^4 - 3x^2 + 1 \) is already a polynomial of degree 4. This leads to \( f'(1), f''(1), f'''(1), \) and so forth playing critical roles in terms of powers of \( (x-1) \).
Polynomials are beneficial because they are straightforward to work with, offering simple addition, subtraction, and multiplication properties. Analyzing polynomial functions gives us insight into more complex calculus concepts, like convergence and approximation, integral to the Taylor series.