Question

Find a formula for the \(n^{\text {th }}\) term of the Taylor series of \(f(x),\) centered at \(c,\) by finding the coefficients of the first few powers of \(x\) and looking for a pattern. (The formulas for several of these are found in Key Idea 8.8.1; show work verifying these formula.) $$f(x)=e^{-x} ; \quad c=0$$

Short answer

The Taylor series for \(e^{-x}\) centered at 0 is \(\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}x^n\).

Step-by-step solution

Understand Taylor Series

The Taylor series of a function \( f(x) \) centered at \( c \) is a series of the form: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n \] where \( f^{(n)}(c) \) is the \( n^{\text{th}} \) derivative of \( f(x) \) evaluated at \( x = c \). Since we are given \( f(x) = e^{-x} \) and \( c = 0 \), we need to compute the derivatives of \( e^{-x} \) and evaluate them at 0.

Calculate the First Few Derivatives

Compute the derivatives of \( f(x) = e^{-x} \):- \( f(x) = e^{-x} \) so \( f(0) = e^0 = 1 \).- \( f'(x) = -e^{-x} \) so \( f'(0) = -1 \).- \( f''(x) = e^{-x} \) so \( f''(0) = 1 \).- \( f^{(3)}(x) = -e^{-x} \) so \( f^{(3)}(0) = -1 \).Notice a pattern emerging: derivatives alternate in sign, flipping between \( e^{-x} \) and \( -e^{-x} \).

Pattern Recognition and Generalization

The pattern observed in the derivatives is:- Odd derivatives (1st, 3rd, 5th,...) evaluate to \( -1 \).- Even derivatives (2nd, 4th, 6th,...) evaluate to \( 1 \).So, \( f^{(n)}(0) = (-1)^n \). This is the pattern in the coefficients.

Construct the Taylor Series

Utilize the general Taylor series formula with our findings:\[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}x^n \] This series represents the Taylor series of \( e^{-x} \) centered at \( c=0 \).

Verify Using Known Formula

The known Taylor series formula for \( e^x \) is \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \). For \( e^{-x} \), replace \( x \) with \(-x \) in the formula, which gives us: \[ \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!} \] This matches our derived series, confirming its correctness.

Key concepts

Derivative Patterns

Understanding how derivatives behave, especially in a repetitive function like an exponential, can provide valuable insights. In the given problem, we focused on the function \(f(x) = e^{-x}\). When we calculated the derivatives, a pattern emerged, helping streamline the process of finding the Taylor series.

• The first derivative is \(f'(x) = -e^{-x}\), leading to \(f'(0) = -1\).
• The second derivative is \(f''(x) = e^{-x}\), resulting in \(f''(0) = 1\).
• The third derivative is \(f^{(3)}(x) = -e^{-x}\), making \(f^{(3)}(0) = -1\).

This starts a pattern where the sign alternates with each derivative. Notably, the function itself repeats, but with a switch in sign for odd derivatives. Observing these patterns is essential in simplifying and predicting the form of the Taylor series for more complex functions, thereby saving time and reducing the likelihood for errors.

Exponential Function

The exponential function, especially the form \(e^x\), is foundational in calculus and broader mathematical applications. It has a unique property where its derivatives are equivalent to its base function, either positive or negative based on the expression. For \(e^{-x}\),

• The zeroth power or the function itself is \(e^{-x}\).
• First derivative \(-e^{-x}\) reflects the same exponential expression but flipped in sign.

This self-replicating nature makes exponential functions particularly suited for Taylor series expansions, enabling the straightforward application of pattern recognition and derivative evaluation. Understanding how to manipulate and recognize these properties paves the way for tackling more complex problems.

Taylor Series Expansion

A Taylor series serves to approximate functions using polynomials, making them easier to analyze and calculate, especially near a specific point. The general formula for the Taylor series is:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n \]For the exponential function \(e^{-x}\) centered at \(c=0\), the series expansion becomes a simpler expression, given the symmetry and derivative patterns previously identified:\[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}x^n \]This expanded form of the function allows for a powerful approximation tool, essentially turning a complex transcendental function into a polynomial that is much simpler to manipulate, especially when solving practical problems in engineering and physics.

Centered at a Point

In the context of Taylor series, 'centering at a point' refers to the location where the series approximates the function most accurately. The function \(f(x) = e^{-x}\) was centered at \(c=0\) in our problem. When centered at \(c\):

• The series exactly equals the function's value at that center \(x=c\).
• The derivatives are evaluated precisely at \(x=c\), ensuring the series provides a near-perfect approximation around this point.

By centering our series in this way, we not only accommodate for the behavior around the point but also ensure that the computed polynomial retains maximum approximation accuracy, particularly useful when dealing with local behavior of functions.