Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the Taylor series for \(f(x)\) centered at the indicated value of \(b\). $$ f(x)=\sin x, \quad b=\pi $$

Step-by-Step Solution

Answer

The Taylor series expansion of the function \(f(x) = \sin{x}\) at \(b=\pi\) is \(-(x-\pi)+\frac{(x-\pi)^3}{6}+...\)

Step 1: Define the Function and Identify the Point of Expansion

The function given is \(f(x)=\sin{x}\) and the point of expansion is \(b=\pi\).

Step 2: Compute the First Few Derivatives

The first few derivatives of \(f(x)=\sin{x}\) are: \(f'(x)=\cos{x}\), \(f''(x)=-\sin{x}\), \(f'''(x)=-\cos{x}\) and \(f''''(x)=\sin{x}\).

Step 3: Evaluate the Derivatives at the Point of Expansion

Evaluate these derivatives at \(b=\pi\). We get: \(f(\pi)=0\), \(f'(\pi)=-1\), \(f''(\pi)=0\), \(f'''(\pi)=1\) and \(f''''(\pi)=0\).

Step 4: Form the Taylor Series

The Taylor series expansion of a function \(f(x)\) at a point \(b\) is given by \[f(x)=f(b)+f'(b)(x-b)+\frac{f''(b)(x-b)^2}{2!}+\frac{f'''(b)(x-b)^3}{3!}+...\] Substituting the computed values into this formula, we get: \[f(x) = 0 -1(x-\pi) + \frac{0(x-\pi)^2}{2}+\frac{1(x-\pi)^3}{6}+...\] This simplifies to: \[f(x) = -(x-\pi)+\frac{(x-\pi)^3}{6}+...\]

Key Concepts

Understanding the Sin Function

The sine function, often represented as \( \sin{x} \), is a fundamental trigonometric function that arises frequently in mathematics. It is periodic, meaning it repeats its values in regular intervals of \( 2\pi \). This function maps any real number \( x \) to a range between \(-1\) and \(1\). It is crucial in modeling wave-like phenomena such as sound and light waves in physics.

• The function \( \sin{x} \) has specific symmetry. This symmetry is known as odd, which means \( \sin(-x) = -\sin{x} \).
• The function crosses the x-axis at integer multiples of \( \pi \), that is, at \( 0, \pm\pi, \pm2\pi \), etc.

When dealing with series expansions like the Taylor series, understanding the behavior of the sine function at various points is important, especially in determining how well the approximation represents the function.

The Role of Derivatives in Taylor Series

Derivatives measure the rate at which a function changes. For Taylor series, derivatives give us information about the behavior of the function \( f(x) \) near the point of expansion. The derivatives of the function are used to form the coefficients of the Taylor series.

• The first derivative \( f'(x) \) tells us the slope or rate of change of \( f(x) \).
• Higher-order derivatives, like \( f''(x) \) and \( f'''(x) \), provide similar information about the rate of change of \( f'(x) \) and so forth.

For the sine function, the derivatives alternate between \( \cos{x} \) and \( \sin{x} \), but periodically change signs:
- The first derivative \( \sin{x} \) is \( \cos{x} \). - The second derivative is \( -\sin{x} \). - The third derivative is \( -\cos{x} \).
This cycle repeats every four derivatives, which plays an essential role in forming accurate Taylor series.

Identifying the Point of Expansion

The point of expansion, in this context, is where the Taylor series is centered. For the sine function, we are interested in expanding around \( b=\pi \).

• When a function is expanded around a particular point, it provides an approximation that is particularly accurate near this point.
• The value of the point influences the derivatives' values, as seen when \( f(x)=\sin{x} \) is expanded around \( \pi \).

For our exercise, calculating derivatives at \( b=\pi \) gives numerical coefficients that help construct the Taylor series. These values demonstrate how the function behaves at \( \pi \) and slightly beyond, forming a polynomial that mimics the sine function near \( \pi \).

Forming the Taylor Series Expansion

The Taylor series is a powerful tool used to represent functions as infinite sums of polynomials. A series expansion like the Taylor series can approximate functions with great precision, especially when centered at a specific point.

• The general formula for the Taylor series at a point \( b \) is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(b)}{n!}(x-b)^n \].
• This allows us to write \( f(x) \) as an infinite polynomial, by adding terms involving derivatives evaluated at \( b \).
• The terms \( (x-b)^n \) reflect how far \( x \) is from the point of expansion, modulating the influence of each term as \( x \) moves away from \( b \).

For \( f(x) = \sin{x} \) expanded at \( b=\pi \), we plug in values obtained from earlier steps to achieve:
\[ f(x) = -(x-\pi) + \frac{(x-\pi)^3}{6} + \ldots \] This polynomial provides an approximation of \( \sin{x} \) near \( \pi \), capturing the essential characteristics of the sine function around this point.