Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Taylor polynomial of order 3 based at a for the given function. $$ e^{x} ; a=1 $$

Step-by-Step Solution

Answer

The Taylor polynomial of order 3 for \( e^x \) at \( a=1 \) is \( e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3 \).

Step 1: Determine the function and its derivatives

We begin by identifying the given function, which is \( f(x) = e^x \). For a Taylor polynomial of order 3, we need the function and its derivatives up to the third derivative. The function \( f(x) = e^x \) has the same derivative for every order. Thus, the first derivative is \( f'(x) = e^x \), the second derivative is \( f''(x) = e^x \), and the third derivative is \( f'''(x) = e^x \).

Step 2: Evaluate the function and its derivatives at a

Next, we plug the point \( a = 1 \) into the function and its derivatives. Since \( e^x \) is the same for every derivative, we have:- \( f(1) = e^1 = e \)- \( f'(1) = e^1 = e \)- \( f''(1) = e^1 = e \)- \( f'''(1) = e^1 = e \).

Step 3: Apply the Taylor series formula

The Taylor polynomial of order 3 for \( f(x) \) about \( x = 1 \) is given by:\[ T_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 \]Substitute the values we found:\[ T_3(x) = e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3 \]

Step 4: Simplify the polynomial

Combine and simplify the terms to get the final form of the polynomial:\[ T_3(x) = e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3 \]This polynomial represents the Taylor polynomial of order 3 centered at \( a = 1 \) for the function \( e^x \).

Key Concepts

Understanding Derivatives

Derivatives play a crucial role in the study of calculus and in the development of Taylor polynomials. The derivative of a function represents the rate at which the function's value changes at a particular point. For any function, the first derivative gives us the slope of the tangent line at any point, essentially showing how fast the function is increasing or decreasing.

When dealing with the exponential function, specifically, the function and its derivatives are unique in that they remain the same. For instance, if the function is the exponential function \( e^x \), then its derivative is also \( e^x \), and this property holds true for every order of derivative. This feature simplifies the process of finding Taylor series for \( e^x \) significantly.

In essence, derivatives are the building blocks for creating approximations of functions with Taylor and Taylor series polynomials, as they allow us to capture what the function looks like around a particular point by using its tangent, curvatures, and even higher levels if needed.

Demystifying the Exponential Function

The exponential function is one of the most important functions in mathematics, expressed as \( f(x) = e^x \), where \( e \) is a mathematical constant approximately equal to 2.71828. This function is known for its unique property where the rate of increase of the function is proportional to the value of the function itself, making it increase rapidly as \( x \) increases.

One remarkable property of the exponential function is that its value and its derivatives remain the same no matter what order you take. This means that no matter how many times you differentiate \( e^x \), you'll always arrive back at \( e^x \).

• This attribute makes \( e^x \) a favorite choice when constructing Taylor series to approximate other functions around a certain point.

Additionally, the exponential function is extensively used in real-world applications, including compound interest calculations, natural logarithms, and in modeling growth processes in biology and economics. Understanding how it behaves is key to applying calculus concepts effectively.

Unpacking the Taylor Series Formula

The Taylor series is a powerful tool used to approximate complex functions with polynomials, making it easier to analyze and calculate them. At its heart, the Taylor series breaks down a function into an infinite sum of terms calculated from the values of its derivatives at a single point. This series is named after the mathematician Brook Taylor who introduced the concept in the early 18th century.

The Taylor series formula is expressed generally as:

• \( T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n \)

In this formula:

• \( f(a) \) represents the function's value at \( a \).
• The derivatives \( f'(a), f''(a), f'''(a) \) etc., denote the rate of change at the point \( a \).
• \( (x-a)^n \) signifies how far away you are from the point of interest \( a \).

The order of the polynomial \( n \) determines how close the approximation will be to the actual function; a higher-order polynomial typically provides a more accurate approximation. When applied to \( f(x) = e^x \) centered at \( x=1 \), the Taylor polynomial helps in approximating values of the function near \( x=1 \) using derivatives that are all \( e^x \), leading to a compact and symmetric polynomial structure.