Problem 2 - Thomas Calculus

Question

Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) \(f(x)=\sin x, \quad a=0\)

Step-by-Step Solution

Answer

The Taylor polynomials are: Order 0: 0, Order 1: x, Order 2: x, Order 3: x - \frac{1}{6}x^3.

Step 1: Understand the Taylor Polynomial Formula

The Taylor polynomial of order \( n \) for a function \( f \) centered at \( a \) is given by the formula: \[ T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n \] Since \( a = 0 \), the polynomial simplifies to: \[ T_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \ldots + \frac{f^{(n)}(0)}{n!}x^n \] We will use this to find each Taylor polynomial.

Step 2: Compute Taylor Polynomial of Order 0

For \( n = 0 \), the Taylor polynomial is simply the constant term: \[ T_0(x) = f(0) \] Since \( f(x) = \sin x \), \( f(0) = \sin 0 = 0 \). Thus, \( T_0(x) = 0 \).

Step 3: Compute Taylor Polynomial of Order 1

For \( n = 1 \), we need the first derivative and evaluate it at \( a = 0 \). The derivative of \( \sin x \) is \( \cos x \). So, \( f'(0) = \cos 0 = 1 \). The polynomial is: \[ T_1(x) = f(0) + f'(0)x = 0 + 1 \cdot x = x \] Thus, the first-order Taylor polynomial is \( T_1(x) = x \).

Step 4: Compute Taylor Polynomial of Order 2

For \( n = 2 \), we need the second derivative. The second derivative of \( \sin x \) is \( -\sin x \). So, \( f''(0) = -\sin 0 = 0 \). Thus, the polynomial is: \[ T_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 = 0 + x + 0 = x \] The second-order Taylor polynomial is \( T_2(x) = x \).

Step 5: Compute Taylor Polynomial of Order 3

For \( n = 3 \), we need the third derivative. The third derivative of \( \sin x \) is \( -\cos x \). So, \( f'''(0) = -\cos 0 = -1 \). The third-order polynomial is: \[ T_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 = 0 + x + 0 - \frac{1}{6}x^3 = x - \frac{1}{6}x^3 \] Thus, the third-order Taylor polynomial is \( T_3(x) = x - \frac{1}{6}x^3 \).

Key Concepts

Sinusoidal Functions

Sinusoidal functions are fundamental in trigonometry, with sine and cosine being the most common examples. These functions oscillate between a maximum and minimum value, creating a wave-like graph.
In mathematics, the sine function, denoted as \( \sin(x) \), is periodic with a period of \( 2\pi \). This periodicity means the same pattern repeats at regular intervals of \( 2\pi \). Sinusoidal functions have various applications in physics, engineering, and signal processing.
Understanding these properties is crucial when working with Taylor polynomials, particularly when the function \( f(x) = \sin(x) \) is centered at \( x = 0 \), as seen in the exercise. This setting utilizes the sine function's initial value \( \sin(0) = 0 \), making calculations of consequent Taylor polynomials easier because the initial value at the center point contributes to simpler polynomial expressions.

Derivative Calculations

Derivative calculations involve finding the rate at which a function changes at any point. This concept is critical when constructing Taylor polynomials. Each order of the polynomial requires a calculation of the respective derivative at the center, \( a \).

• The first derivative \( \frac{d}{dx} \sin(x) = \cos(x) \), gives us the slope at any point \( x \), and at \( x = 0 \) it equals 1.
• The second derivative \( \frac{d^2}{dx^2} \sin(x) = -\sin(x) \), provides the concavity, but at \( x = 0 \), it equals 0, simplifying calculations for the Taylor polynomial.
• For the third derivative \( \frac{d^3}{dx^3} \sin(x) = -\cos(x) \), the result at \( x = 0 \) is -1.

Calculating these derivatives helps in constructing the Taylor polynomial by plugging these into the formula, allowing one to approximate the sine function around \( x = 0 \).

Polynomial Approximation

Polynomial approximation refers to the process of using polynomials to estimate more complex functions. Taylor polynomials achieve this by constructing polynomials whose values and derivatives match the target function at one point.
In the exercise, using the Taylor polynomial at \( a = 0 \) results in a series of polynomials: \( T_0(x) = 0 \), \( T_1(x) = x \), \( T_2(x) = x \), and \( T_3(x) = x - \frac{1}{6}x^3 \).
Approximations are particularly valuable near the center point \( a \), where they closely follow the behavior of the original sine function. Expanding higher orders in the polynomial provides better approximations. This means that as the polynomial degree increases, it should match more of the sine function's behavior around 0, improving the accuracy of \( \sin(x) \).

Higher Order Derivatives

Higher order derivatives involve taking successive derivatives of a function. They help in grasping not just the slope or curvature but more complex aspects of the function's graph.
For Taylor series, each successive derivative plays a role in the accuracy of the approximation.

• The zero, first, and second derivatives directly contributed to the results seen in the polynomial \( T_2(x) = x \).
• In the third-order case, the third derivative added another term \(-\frac{1}{6}x^3\), showing the importance of continuing past the first few derivatives for more accurate results.

The higher the derivative, the more "details" about how the function behaves near the approximation point are captured, providing a more complete polynomial expression.