Problem 11 - Calculus: One and Several Variables

Question

Determine the \(n\) th Taylor polynomial \(P_{n}\) for the function \(f\). $$f(x)=e^{-x}$$

Step-by-Step Solution

Answer

The \(n\)th Taylor polynomial for the function \(f(x) = e^{-x}\) is \(P_{n}(x) = \sum_{k=0}^{n} \frac{(-1)^k x^k}{k!}\).

Step 1: Calculation of \(f^{(k)}(0)\) for k=0 to n

The function \(f(x) = e^{-x}\) has the property that its \(k\)th derivative is \(( -1)^k e^{-x}\). Plugging in \(x = 0\) gives \(f^{(k)}(0) = (-1)^k\).

Step 2: Substituting in the Taylor Series formula.

Substitute \(f^{(k)}(a) = (-1)^k\) and \(a = 0\) into the Taylor series formula to get \(P_{n}(x) = \sum_{k=0}^{n} \frac{(-1)^k x^k}{k!}\).

Step 3: Simplifying the series

The series \(P_{n}(x) = \sum_{k=0}^{n} \frac{(-1)^k x^k}{k!}\) is the \(n\)th Taylor polynomial for the function \(f(x) = e^{-x}\).

Key Concepts

Taylor Series

The Taylor series is a powerful tool in mathematics that allows us to approximate functions using polynomials. It is particularly useful when dealing with complex functions, providing a simpler polynomial form that is easier to work with.

• A Taylor series represents a function as an infinite sum of its derivatives at a particular point.
• The formula for the Taylor series centered at point \(a\) is: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \]
• Every term in the series involves differentials of higher order, scaled by factorial terms.

When creating a Taylor polynomial from a Taylor series, we often choose a finite number \(n\) to determine a polynomial degree. This results in an approximation given by \(P_{n}(x)\), which uses derivatives up to \(f^{(n)}(a)\). This polynomial provides us an approachable way to estimate the function's behavior near \(a\).

Derivatives

Understanding derivatives is crucial, as they are the fundamental building blocks of Taylor series. A derivative represents the rate of change of a function with respect to a variable, often denoted as \(f'(x)\).

• Derivatives help describe how a function changes and are used to calculate slopes of tangent lines.
• Higher-order derivatives, like \(f''(x)\), \(f'''(x)\), and so on, indicate the rate of change of the rate of change, offering further detail on the function's behavior and curvature.

For the exponential function \(f(x) = e^{-x}\), the derivatives have a unique repetitive pattern:

• The first derivative \(f'(x) = -e^{-x}\).
• The second derivative \(f''(x) = e^{-x}\).
• This alternates, making the \(k\)th derivative \(( -1)^k e^{-x}\).

This consistency simplifies deriving Taylor polynomials, as shown in the exercise, where each derivative at \(x = 0\) simply becomes \((-1)^k\).

Exponential Functions

Exponential functions are a key area of study in calculus due to their widespread applications and distinct characteristics. These functions have the form \(f(x) = a^x\) or more commonly \(f(x) = e^x\), where \(e\) is the base of the natural logarithm.

• Exponential functions grow (or decay) at a rate proportional to their current value, making them essential in modeling populations, finance, and decay processes.
• They are unique in that their derivatives reflect their original form, a property that aids in creating Taylor series.

For the specific function \(f(x) = e^{-x}\):

• The function represents exponential decay due to the negative exponent.
• Its Taylor series reveals the unique pattern of its derivatives, simplifying polynomial approximation per the previous step-by-step solution.

Exploring and understanding exponential functions through Taylor polynomials allow us to approximate and analyze these functions in various real-world contexts efficiently.