Problem 73 - Calculus Early Transcendentals

Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The Taylor polynomials for \(f(x)=e^{-2 x}\) centered at 0 consist of even powers only. b. For \(f(x)=x^{5}-1,\) the Taylor polynomial of order 10 centered at \(x=0\) is \(f\) itself. c. The \(n\) th-order Taylor polynomial for \(f(x)=\sqrt{1+x^{2}}\) centered at 0 consists of even powers of \(x\) only.

Step-by-Step Solution

Answer

Based on the calculations above, we have the following conclusions: Statement A: False - The Taylor polynomial for \(f(x)=e^{-2 x}\) at x = 0 contains both even and odd powers of x. Statement B: False - The 10th-order Taylor polynomial for \(f(x)=x^{5}-1\) centered at \(x=0\) is not equal to the original function \(f(x)\). Statement C: True - The \(n\)th-order Taylor polynomial for \(f(x)=\sqrt{1+x^{2}}\) centered at 0 consists of even powers of \(x\) only.

Step 1: Statement A: Taylor polynomial for \(f(x)=e^{-2 x}\) at x = 0 consists of even powers only

To find Taylor polynomials for \(f(x)=e^{-2 x}\) up to some power \(x^k\) and centered at 0, calculate the polynomial \(P_n(x) = \sum_{n=0}^{k} a_nx^n\), where \(a_n = \frac{f^{(n)}(0)}{n!}\). We have $$ f(x) = e^{-2x}, \, f'(x) = -2e^{-2x}, \, f''(x) = 4e^{-2x}, \, \cdots $$ Evaluate \(f^{(n)}(0)\) for various values of n, and we get $$ \begin{aligned} f(0) &= e^0=1, \\ f'(0) &= -2e^0=-2, \\ f''(0) &= 4e^0=4, \\ f'''(0) &= -8e^0=-8, \\ \cdots \end{aligned} $$ Thus, the corresponding Taylor polynomial is $$ P(x) = 1 - 2x + \tfrac{4}{2!}x^2 - \tfrac{8}{3!}x^3 + \cdots $$ This polynomial contains both even and odd powers of x, disproving the statement. Therefore, Statement A is false.

Step 2: Statement B: The Taylor polynomial of order 10 centered at \(x=0\) for \(f(x)=x^{5}-1\) is \(f\) itself

To find the Taylor polynomial of order 10 for \(f(x)=x^{5}-1\) centered at \(x=0\), calculate the polynomial \(P_n(x) = \sum_{n=0}^{10} a_nx^n\), where \(a_n = \frac{f^{(n)}(0)}{n!}\). We have $$ f(x) = x^{5}-1, \, f'(x) = 5x^{4}, \, f''(x) = 20x^{3}, \\ f'''(x) = 60x^{2}, \, f^{(4)}(x) = 120x, \, f^{(5)}(x) = 120, \, f^{(6)}(x) = 0, \cdots $$ Since all the derivatives for \(n>5\) are zero, the Taylor polynomial of order 10 is equal to the Taylor polynomial of order 5. Thus, $$ P_{10}(x) = P_{5}(x) = (x^5 - 1) + \tfrac{5}{1!}x^{4} - \tfrac{20}{2!} x^3 + \cdots $$ Since \(P_{10}(x)\) is not equal to \(f(x)=x^{5}-1\), Statement B is false.

Step 3: Statement C: The \(n\)th-order Taylor polynomial for \(f(x)=\sqrt{1+x^{2}}\) centered at 0 consists of even powers of \(x\) only

To find the \(n\)th-order Taylor polynomial for \(f(x)=\sqrt{1+x^{2}}\) centered at 0, once again, we use the polynomial \(P_n(x) = \sum_{n=0}^{k} a_nx^n\), where \(a_n = \frac{f^{(n)}(0)}{n!}\). Note the derivative structure that evolves from \(f(x)=\sqrt{1+x^{2}}\): $ \begin{array}{ll} f(x) & = \sqrt{1+x^{2}} \\ f'(x) & = \frac{x}{\sqrt{1+x^{2}}} \\ f''(x) & = \frac{1}{(1+x^{2})^{\tfrac{3}{2}}} \\ f'''(x) &= \frac{-3x}{(1+x^{2})^{\tfrac{5}{2}}} \\ f^{(4)}(x) & = \frac{15 - 12x^{2}}{(1+x^{2})^{\tfrac{7}{2}}} \end{array} $ Notice that for all odd order derivatives, \(x\) appears in the numerator, making \(f^{(n)}(0) = 0\) for all odd n. Thus, the Taylor polynomial will only have even power terms: $$ P_{n}(x) = 1 + Ax^2 + Bx^4 + \cdots $$ Therefore, Statement C is true.

Key Concepts

Maclaurin Series

When learning about Taylor polynomials, we often come across the Maclaurin Series, which is a particular case of the Taylor series centered at zero. The Maclaurin Series is a powerful mathematical tool that allows us to approximate more complex functions using simpler polynomial expressions. This series helps in simplifying calculations and understanding the behavior of functions around the origin.

To form a Maclaurin Series, it involves taking derivatives of the function you're working with at zero, and creating a polynomial using these derivatives. Let's say we have a function \( f(x) \), its Maclaurin Series is given by:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \)

This means that each term in the series is derived by evaluating the derivatives of \( f(x) \) at zero and scaling it by \( n! \), where \( n \) is the order of the derivative.

Using the Maclaurin series, we can effectively represent functions like exponentials, sine, and cosine. It's essential in calculus and applied mathematics, particularly when dealing with function approximations.

Higher-Order Derivatives

Higher-order derivatives play a critical role in constructing Taylor and Maclaurin series. These derivatives are the repeated derivatives of a function, which provide deeper insights into the function’s behavior beyond just slope (first derivative). They help in determining the curvature and changes at an even deeper level.

Calculating higher-order derivatives involves taking the derivative of a function multiple times. Let's say we have a function \( f(x) \), its higher-order derivatives can be thought of sequentially:

• First derivative \( f'(x) \), which represents the slope or rate of change.
• Second derivative \( f''(x) \), often relating to the concavity or acceleration.
• Third derivative \( f'''(x) \), indicating the change in the rate of concavity.
• And so forth...

Higher-order derivatives are essential since each successive derivative contributes a term to the Taylor or Maclaurin series. For Maclaurin series, these calculations are made at \( x = 0 \), while Taylor series could be centered around any point \( a \).

Understanding higher-order derivatives enhances our ability to approximate and understand a function’s full behavior.

Function Approximation

Function approximation is a core concept in calculus, helping us estimate complex functions with simpler polynomial forms. Taylor and Maclaurin polynomials are methods heavily used for this purpose. They provide a way to create an approximation of a function around a specific point, using derivatives to form polynomial expressions.

The essence of function approximation lies in taking a potentially complex function \( f(x) \) and creating a polynomial \( P_n(x) \) that closely estimates \( f(x) \) around a particular value. This comes especially useful in computations where exact values are challenging to attain or unnecessary.

• The idea is to make complex functions approachable, especially when needing quick calculations or graphical schematics.
• Taylor polynomials specifically expand the function in a series using its derivatives at a point to craft a route to simple approximations.

Taking higher-order terms in a polynomial makes the approximation increasingly accurate near the chosen center point.

Hence, function approximation through Taylor and Maclaurin polynomials is invaluable in fields like physics, engineering, and economics, where precise, expedient calculations are often required.