Problem 21 - 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 \( R_n (x) \to 0.\)] Also find the associated radius of convergence. \( f(x) = x^6 - x^4 + 2, \) \( a = 2 \)

Step-by-Step Solution

Answer

The Taylor series is: \(50 + 160(x-2) + 216(x-2)^2 + 152(x-2)^3 + 59(x-2)^4 + 12(x-2)^5 + (x-2)^6\). Radius of convergence is infinite.

Step 1: Compute Derivatives of f(x)

The first step is to compute the derivatives of the function \( f(x) = x^6 - x^4 + 2 \). We need the first few derivatives, at least up to the sixth one, as this is a polynomial of degree 6.- \( f'(x) = 6x^5 - 4x^3 \)- \( f''(x) = 30x^4 - 12x^2 \)- \( f'''(x) = 120x^3 - 24x \)- \( f^{(4)}(x) = 360x^2 - 24 \)- \( f^{(5)}(x) = 720x \)- \( f^{(6)}(x) = 720 \)- \( f^{(7)}(x) = 0 \) (and all higher derivatives are zero)

Step 2: Evaluate Derivatives at a = 2

Next, substitute \( x = 2 \) into each of the derivatives computed in Step 1:- \( f(2) = 2^6 - 2^4 + 2 = 64 - 16 + 2 = 50 \)- \( f'(2) = 6(2)^5 - 4(2)^3 = 192 - 32 = 160 \)- \( f''(2) = 30(2)^4 - 12(2)^2 = 480 - 48 = 432 \)- \( f'''(2) = 120(2)^3 - 24(2) = 960 - 48 = 912 \)- \( f^{(4)}(2) = 360(2)^2 - 24 = 1440 - 24 = 1416 \)- \( f^{(5)}(2) = 720 \times 2 = 1440 \)- \( f^{(6)}(2) = 720 \)

Step 3: Construct Taylor Series

The Taylor series of \( f(x) \) centered at \( a = 2 \) is given by the formula: \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!} (x-2)^n \]Using the derivatives evaluated at \( x=2 \) from Step 2, we have:- \( \frac{f(2)}{0!} = 50 \)- \( \frac{f'(2)}{1!} (x-2) = 160(x-2) \)- \( \frac{f''(2)}{2!} (x-2)^2 = \frac{432}{2} (x-2)^2 = 216(x-2)^2 \)- \( \frac{f'''(2)}{3!} (x-2)^3 = \frac{912}{6} (x-2)^3 = 152(x-2)^3 \)- \( \frac{f^{(4)}(2)}{4!} (x-2)^4 = \frac{1416}{24} (x-2)^4 = 59(x-2)^4 \)- \( \frac{f^{(5)}(2)}{5!} (x-2)^5 = \frac{1440}{120} (x-2)^5 = 12(x-2)^5 \)- \( \frac{f^{(6)}(2)}{6!} (x-2)^6 = \frac{720}{720} (x-2)^6 = (x-2)^6 \)The Taylor series is: \[50 + 160(x-2) + 216(x-2)^2 + 152(x-2)^3 + 59(x-2)^4 + 12(x-2)^5 + (x-2)^6\]

Step 4: Determine the Radius of Convergence

For a polynomial, the radius of convergence is infinite because polynomials are entire functions (analytic everywhere on the complex plane). Therefore, the Taylor series has an infinite radius of convergence.Hence, the Taylor series for \( f(x) \) centered at \( a = 2 \) converges for all \( x \).

Key Concepts

Derivatives

Understanding derivatives is essential to formulating a Taylor series. Derivatives help us describe how a function changes as its input changes. Essentially, they measure the function's rate of change or steepness at any given point. In the context of Taylor series, we use derivatives to approximate functions by polynomials through their behavior at a specific point.

When working on a Taylor series, the first step is to find multiple derivatives of your function, just like in the original exercise for the function \( f(x) = x^6 - x^4 + 2 \). You identify the pattern of change by differentiating repeatedly:

• First derivative \( f'(x) \): Provides the slope of the tangent line at any point \( x \).
• Second derivative \( f''(x) \): Indicates the concavity or convexity of the curve.
• Higher-order derivatives: Capture how the curve oscillates and provides finer approximations for the Taylor polynomial.

Understanding the impact of these derivatives at a point of interest (here, \( x = 2 \)) is crucial as they form the coefficients of your Taylor series, making the entire polynomial approximation reflect the function's behavior perfectly.

Radius of convergence

The "radius of convergence" refers to the range in which a power series (including Taylor series) converges to a function. In layman's terms, it defines how far we can extend our approximation using the Taylor series from the center point before it loses accuracy.

The fascinating aspect of polynomial functions, such as the one in our exercise, is that they have an infinite radius of convergence. This means the Taylor series converges everywhere on the number line, or in complex terms, the entire plane. This happens because polynomials are entire functions, analytic everywhere.

For non-polynomial functions, the radius of convergence takes finite values and is determined by singularities of the function. In mathematical terms, the ratio test or root test helps find this radius when handling more complex functions. However, with polynomials, as any point is analytic, you can rest assured of convergence no matter the chosen \( x \).

Polynomial function

Polynomial functions are expressions composed of variables raised to whole number powers, with coefficients assigned to each term. They are the sum of these terms, such as in our initial function \( f(x) = x^6 - x^4 + 2 \).

Polynomials are among the simplest types of functions and possess several distinguishing properties:

• Smooth and continuous: They have no breaks, jumps, or sharp corners.
• Finite derivatives: They can be differentiated repeatedly until eventually resulting in zero.
• Predictable behavior: As polynomials are characterized by their degree, higher-degree terms prevail at extreme values of \( x \), dictating end behavior.

In the context of Taylor series, polynomials lead to particularly straightforward expressions. Since we can find all derivatives up to the polynomial's degree, gaining an exact representation of the function becomes attainable. As a result, the Taylor series for our function proves exact because we can account for all meaningful changes it undergoes. This reliability makes polynomial functions a fantastic starting block for grasping the concept of Taylor series.