Question

Find the Taylor polynomial of degree \(n\), at \(x=c\), for the given function. $$f(x)=\sin x, \quad n=5, \quad c=\pi / 6$$

Short answer

The 5th degree Taylor polynomial at \(x = \pi/6\) is \(P_5(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{1}{4}(x - \frac{\pi}{6})^2 - \frac{\sqrt{3}}{12}(x - \frac{\pi}{6})^3 + \frac{1}{48}(x - \frac{\pi}{6})^4 + \frac{\sqrt{3}}{240}(x - \frac{\pi}{6})^5\).

Step-by-step solution

Understand Taylor Polynomial Formula

The Taylor polynomial of degree \(n\) for a function \(f(x)\) centered at \(x = c\) is given by the formula:\[P_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \cdots + \frac{f^{(n)}(c)}{n!}(x-c)^n\]

Calculate Derivatives of \(f(x) = \sin x\)

First, we calculate the first five derivatives of \( \sin x \):- \(f(x) = \sin x\)- \(f'(x) = \cos x\)- \(f''(x) = -\sin x\)- \(f'''(x) = -\cos x\)- \(f^{(4)}(x) = \sin x\)- \(f^{(5)}(x) = \cos x\)

Evaluate Derivatives at \(x = \pi/6\)

Now, evaluate the derivatives at \(x = \pi/6\):- \(f(\pi/6) = \sin(\pi/6) = \frac{1}{2}\)- \(f'(\pi/6) = \cos(\pi/6) = \frac{\sqrt{3}}{2}\)- \(f''(\pi/6) = -\sin(\pi/6) = -\frac{1}{2}\)- \(f'''(\pi/6) = -\cos(\pi/6) = -\frac{\sqrt{3}}{2}\)- \(f^{(4)}(\pi/6) = \sin(\pi/6) = \frac{1}{2}\)- \(f^{(5)}(\pi/6) = \cos(\pi/6) = \frac{\sqrt{3}}{2}\)

Substitute into Taylor Polynomial Formula

Substitute the derivatives evaluated at \(x = \pi/6\) into the Taylor polynomial formula:\[P_5(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{1}{2}\frac{(x - \frac{\pi}{6})^2}{2!} - \frac{\sqrt{3}}{2}\frac{(x - \frac{\pi}{6})^3}{3!} + \frac{1}{2}\frac{(x - \frac{\pi}{6})^4}{4!} + \frac{\sqrt{3}}{2}\frac{(x - \frac{\pi}{6})^5}{5!}\]

Simplify the Expression

Simplify the expression to obtain the Taylor polynomial:\[P_5(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{1}{4}(x - \frac{\pi}{6})^2 - \frac{\sqrt{3}}{12}(x - \frac{\pi}{6})^3 + \frac{1}{48}(x - \frac{\pi}{6})^4 + \frac{\sqrt{3}}{240}(x - \frac{\pi}{6})^5\]

Key concepts

Sine Function

The sine function, denoted by \( \sin x \), is a fundamental trigonometric function. It represents the y-coordinate of a point on the unit circle as the angle \( x \) varies. This function is periodic with a period of \( 2\pi \), meaning that its values repeat every \( 2\pi \) units along the x-axis. The sine function is important in mathematics and engineering due to its properties and applications in waveforms.

• The sine of 0 is 0, and it starts at this point when graphing.
• The function reaches its maximum value of 1 at \( \frac{\pi}{2} \) and its minimum value of -1 at \( \frac{3\pi}{2} \).
• It is an odd function, meaning \( \sin(-x) = -\sin(x) \).

Understanding the behavior of the sine function is essential when dealing with calculus concepts such as derivatives and Taylor polynomials.

Derivative Calculation

The process of finding derivatives involves calculating how a function changes as its input changes. When considering Taylor polynomials, it is vital to know how to determine multiple derivatives for a given function.

For the sine function \( f(x) = \sin x \), the derivatives cycle through continuously due to the periodic nature of sine and cosine functions:

• First derivative: \( f'(x) = \cos x \)
• Second derivative: \( f''(x) = -\sin x \)
• Third derivative: \( f'''(x) = -\cos x \)
• Fourth derivative: \( f^{(4)}(x) = \sin x \)
• Fifth derivative: \( f^{(5)}(x) = \cos x \)

These derivatives are necessary to expand a function into a Taylor polynomial and reflect how the original function behaves locally at any point \( x \).

Polynomial Approximation

Polynomial approximation is a method used to estimate complex functions with simpler polynomial expressions. The Taylor polynomial is a specific example of this concept, which approximates a function using its derivatives at a point.

In our example with \( \sin x \), deriving a Taylor polynomial involves evaluating its derivatives at the specified point \( c = \frac{\pi}{6} \), yielding:\

• \( f(\pi/6) = \frac{1}{2} \)
• \( f'(\pi/6) = \frac{\sqrt{3}}{2} \)
• \( f''(\pi/6) = -\frac{1}{2} \)
• \( f'''(\pi/6) = -\frac{\sqrt{3}}{2} \)
• \( f^{(4)}(\pi/6) = \frac{1}{2} \)
• \( f^{(5)}(\pi/6) = \frac{\sqrt{3}}{2} \)

These values are then substituted into the Taylor polynomial formula to generate an approximation. This polynomial closely approximates the behavior of \( \sin x \) in the vicinity of \( \frac{\pi}{6} \).

Series Expansion

Series expansion refers to expressing a function as a series of terms added together. Each term becomes progressively smaller, intending to provide a detailed approximation at a specific point. The Taylor series is one such method where functions are represented as infinite series.

A Taylor polynomial forms the foundation of a Taylor series. In simple terms, it takes the idea of using derivatives (rates of change) at a point to create individual terms of a polynomial. This is done by using:

• The function's value at a point \( c \)
• The first derivative multiplied by \((x-c)\)
• Higher derivatives divided by factorials, each raised to increasing powers of \( (x-c) \)

By limiting the number of terms (in this case, up to the fifth derivative), you obtain the truncated Taylor polynomial. This provides a good approximation while keeping calculations feasible. As shown in the example, the resulting polynomial captures the essence of \( \sin x \) around \( \frac{\pi}{6} \).