Question

Find the remainder term \(R_{n}\) for the nth-order Taylor polynomial centered at a for the given functions. Express the result for a general value of \(n\) $$f(x)=\sin x ; a=0$$

Short answer

The remainder term \(R_n(x)\) for the nth-order Taylor polynomial centered at 0 for the function \(\sin x\) is given by: $$R_n(x) = \sin x - \left[\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right]$$ Where the last term in the sum is the nth term, depending on the parity of n (if n is odd, it is a term with an odd degree, if n is even, it is a term with an even degree).

Step-by-step solution

Find the derivatives of f(x)

Since the function f(x) is given as \(\sin x\), we need to find its derivatives. The first few derivatives are: - \(f^{(1)}(x) = \cos x\) - \(f^{(2)}(x) = -\sin x\) - \(f^{(3)}(x) = -\cos x\) - \(f^{(4)}(x) = \sin x\) Notice that after the 4th derivative, the pattern repeats itself. More generally, we can denote the derivatives as: $$f^{(k)}(x) = \begin{cases} \sin(x), & k \equiv 0 \mod 4 \\ \cos(x), & k \equiv 1 \mod 4 \\ -\sin(x), & k \equiv 2 \mod 4 \\ -\cos(x), & k \equiv 3 \mod 4 \end{cases}$$

Compute the derivatives at a=0

Next, we need to compute the values of the derivatives at the point a = 0: $$f^{(k)}(0) = \begin{cases} 0, & k \equiv 0 \mod 4 \ \text{and} \ k \ne 0\\ 1, & k \equiv 1 \mod 4 \\ 0, & k \equiv 2 \mod 4 \\ -1, & k \equiv 3 \mod 4 \end{cases}$$

Find the remainder term \(R_n\)

Using the remainder term formula, \(R_n(x) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}x^{k}\), substitute the found derivatives and simplify: $$R_n(x) = \sin x - \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}x^{k}$$ Substituting the values of the derivatives at a=0, we get: $$R_n(x) = \sin x - \left[\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right]$$ Where the last term in the sum is the nth term, depending on the parity of n (if n is odd, it is a term with an odd degree, if n is even, it is a term with an even degree). So the remainder term \(R_n(x)\) for the nth-order Taylor polynomial centered at 0 for the function \(\sin x\) is given by: $$R_n(x) = \sin x - \left[\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right]$$

Key concepts

Remainder Term

In the realm of Taylor series, the remainder term is crucial. It tells us how far our Taylor polynomial is from the actual function. For the sine function, when we create a Taylor series around a point, in this case, zero, we want to know how accurate our approximation is. The remainder term, denoted as \( R_n(x) \), provides this insight. It is expressed as:\[ R_n(x) = \, f(x) - P_n(x) \]Here, \( P_n(x) \) is the nth-order Taylor polynomial. By calculating \( R_n(x) \), we can determine how much we miss capturing the true behavior of \( \sin x \) with our polynomial.For the sine function, expressed as \( \sin x \), the remainder term becomes:\[ R_n(x) = \, \sin x - \left[ \frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \right] \]This expression shows the difference between the series expansion and the sine function, providing a clear picture of where the Taylor polynomial deviates from \( \sin x \).

Sine Function

The sine function is a fundamental part of trigonometry, represented as \( \sin x \). It is periodic and oscillates between -1 and 1. This function is incredibly important in various fields, including physics and engineering. The sine function is unique because it looks similar for every period and can be described using trigonometric identities. In the context of Taylor series, the sine function has an interesting property. Its derivatives form a repeating cycle of four distinct patterns:

• 1st derivative, \( \cos x \)
• 2nd derivative, \( -\sin x \)
• 3rd derivative, \( -\cos x \)
• 4th derivative, \( \sin x \)

After the fourth derivative, the pattern repeats. This characteristic is what allows us to build a Taylor series that can approximate \( \sin x \) around a certain point, using only a few terms for reasonable accuracy. Understanding this cycle of derivatives is key to grasping how Taylor series can effectively model the sine function.

Derivatives

Derivatives represent the rate at which a function changes. They are foundational in calculus, giving us insights into the behavior of functions. For the sine function, \( f(x) = \sin x \), derivatives are particularly fascinating due to their cyclical pattern. Let's explore the cycles:- **First derivative**: The derivative of \( \sin x \) is \( \cos x \). This tells us how \( \sin x \) is changing at any point.- **Second derivative**: Taking the derivative of \( \cos x \) gives us \( -\sin x \), indicating how \( \cos x \) is changing.- **Third derivative**: Differentiating \( -\sin x \) results in \( -\cos x \).- **Fourth derivative**: Finally, deriving \( -\cos x \) returns us to \( \sin x \).These derivatives form a repeating sequence every four derivations, which simplifies analyzing functions like sine. Knowing these derivatives privileges us to effectively calculate the Taylor series, as they determine how accurately our polynomial will approximate the function in a specific domain.

Taylor Series Expansion

Taylor series expansion allows us to express complex functions as infinite sums of polynomial terms. This expansion is valuable for approximating functions near a chosen center point, typically denoted as \( a \). For \( \sin x \) centered at \( a = 0 \), the expansion becomes:\[ P_n(x) = \, \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}x^{k} \]Here, each term in the series incorporates the derivative of the function at the center point, divided by the factorial of the derivative's order, scaled by \( x \) raised to the power of that order.For \( \sin x \), the Taylor series is constructed using the cycle of derivatives we discussed:- The non-zero terms occur when \( k \equiv 1 \mod 4 \) and \( k \equiv 3 \mod 4 \), corresponding to \( \cos x \) and \(-\cos x\) respectively when evaluated at \( x=0 \).The result is an alternating series of even-powered terms, providing a path for approximating \( \sin x \) accurately. This construction is fundamental to many applications in both theoretical and practical contexts, as it simplifies otherwise complex function calculations near the center point.