Problem 51 - Calculus Early Transcendentals

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 \(n\) $$f(x)=e^{-x} ; a=0$$

Step-by-Step Solution

Answer

Answer: The remainder term, \(R_n(x)\), for the nth-order Taylor polynomial centered at 0 for the function \(f(x) = e^{-x}\) is given by: $$ R_n(x) = \begin{cases} \frac{e^{-c}}{(n+1)!}x^{n+1} & \mbox{ if } n+1 \mbox{ is even} \\ -\frac{e^{-c}}{(n+1)!}x^{n+1} & \mbox{ if } n+1 \mbox{ is odd} \end{cases} $$ where c is a real number between 0 and x.

Step 1: Calculate the derivatives of f(x) at a = 0

We'll calculate the first few derivatives and then find a general formula for the nth derivative. - \(f(x)=e^{-x}\) - \(f'(x)=-e^{-x}, f'(0)=-1\) - \(f''(x)=e^{-x}, f''(0)=1\) - \(f'''(x)=-e^{-x}, f'''(0)=-1\) Notice that the derivatives alternate between \(e^{-x}\) and \(-e^{-x}\), and the pattern repeats every 2 derivatives. Therefore, the nth derivative can be generalized as: $$ f^{(n)}(x) = \begin{cases} e^{-x} & \mbox{ if } n \mbox{ is even} \\ -e^{-x} & \mbox{ if } n \mbox{ is odd} \end{cases} $$ Now, we evaluate the nth derivatives at a = 0: $$ f^{(n)}(0) = \begin{cases} 1 & \mbox{ if } n \mbox{ is even} \\ -1 & \mbox{ if } n \mbox{ is odd} \end{cases} $$

Step 2: Taylor polynomial

Now, we'll express the nth-order Taylor polynomial using the nth derivatives at a = 0 and the Taylor series formula: $$ P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}(x-0)^k $$ Plugging in the values of the nth derivatives, we get: $$ P_n(x) = \sum_{k=0}^{n} \frac{(-1)^k}{k!}x^k $$

Step 3: Remainder term using Taylor's Remainder theorem

According to Taylor's Remainder Theorem, the remainder term can be expressed as: $$ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-0)^{(n+1)} $$ where c is a real number between 0 and x. The (n+1)th derivative is: $$ f^{(n+1)}(x) = \begin{cases} e^{-x} & \mbox{ if } n+1 \mbox{ is even} \\ -e^{-x} & \mbox{ if } n+1 \mbox{ is odd} \end{cases} $$ Substituting this value in the remainder term expression: $$ R_n(x) = \begin{cases} \frac{e^{-c}}{(n+1)!}x^{n+1} & \mbox{ if } n+1 \mbox{ is even} \\ -\frac{e^{-c}}{(n+1)!}x^{n+1} & \mbox{ if } n+1 \mbox{ is odd} \end{cases} $$ Now, we have expressed the remainder term \(R_n(x)\) for a general value of n for the function \(f(x) = e^{-x}\) centered at a = 0.

Key Concepts

Remainder Term

The remainder term in a Taylor series gives us valuable insight into the error involved when approximating a function using a finite number of terms. For the function \( f(x) = e^{-x} \), centered at \( a = 0 \), we're looking at how well the Taylor polynomial approximates the function for various values of \( n \). The remainder, \( R_n(x) \), tells us how far the Taylor polynomial is from the actual function.

The Taylor's Remainder Theorem provides a formula to find this remainder:

• \( R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-0)^{(n+1)} \)

Here, \( c \) is some value between the center \( a \) and \( x \).

For our specific function, the formula suggests how accurate the approximation is by evaluating the \((n+1)\)-th derivative at some point \( c \), which can vary between \( 0 \) and \( x \). This involves checking whether \( n+1 \) is even or odd, leading to the remainder taking on one of two forms:

• If \( n+1 \) is even: \( R_n(x) = \frac{e^{-c}}{(n+1)!}x^{n+1} \)
• If \( n+1 \) is odd: \( R_n(x) = -\frac{e^{-c}}{(n+1)!}x^{n+1} \)

This alternating sign hints at the alternating nature of the derivatives of the exponential function \( e^{-x} \). estimating the error helps understand the precision of the polynomial approximation.

Exponential Functions

Exponential functions have unique characteristics making them extremely interesting, especially in calculus. The function \( f(x) = e^{-x} \) is a classic example where the base \( e \) represents the natural exponential function. A key property of exponential functions is their consistent growth rate, which remains proportional to their value.

For \( e^{-x} \), the negative exponent modifies its growth, causing the function to decrease as \( x \) increases. These functions are continuously differentiable, which simplifies their analysis in Taylor series and derivatives due to their repetitive derivative patterns.

• The function remains positive even as \( x \) increases, as \( e^x \) never crosses zero.
• They possess a constant rate of growth or decline, which contributes to the simplicity of calculating higher derivatives.

The beauty of exponential functions lies in their ability to represent rapid growth or decay succinctly, making them vital in modeling natural phenomena, compounding processes, and, as we see here, beautifully demonstrating the elegance of calculus through the Taylor series.

Derivative Calculations

Calculating derivatives is an essential technique in understanding how functions change. For the function \( f(x) = e^{-x} \), differentiating produces a series of derivatives that adhere to a simple alternation pattern. This pattern greatly aids in forming the Taylor series.

The derivatives are as follows:

• \( f'(x) = -e^{-x} \)
• \( f''(x) = e^{-x} \)
• \( f'''(x) = -e^{-x} \)

This alternation between \( e^{-x} \) and \(-e^{-x} \) continues, which can be generalized:

• If \( n \) is even: \( f^{(n)}(x) = e^{-x} \)
• If \( n \) is odd: \( f^{(n)}(x) = -e^{-x} \)

When evaluating at \( x = 0 \), these derivatives simplify to either 1 or -1 based on whether \( n \) is even or odd. This repetitive and symmetric nature of the derivatives not only simplifies their calculation but also solidifies the formation of a Taylor series, as it easily integrates into the polynomial approximations for any order \( n \). Moreover, this understanding of derivative calculations highlights the underlying symmetry and elegance of exponential decay functions.