Problem 83 - Calculus: Early Transcendentals

Question

Prove Taylor's Inequality for \( n = 2, \) that is, prove that if \( \mid f''' (x) \mid \le M \text { for } \mid x - a \mid \le d, \) then \( \mid R_2 (x) \mid \le \frac {M}{6} \mid x - a \mid^3 \) for \( \mid x - a \mid \le d \)

Step-by-Step Solution

Answer

Taylor's inequality is proven under the given conditions using the remainder term for a 2nd-order Taylor expansion.

Step 1: Understand Taylor's Remainder Term

The remainder term for a Taylor series expansion of order 2 centered at \( a \) is given by \( R_2(x) = \frac{f'''(c)}{3!}(x-a)^3 \) for some \( c \) between \( a \) and \( x \). This term represents the error in approximating \( f(x) \) using the Taylor expansion up to the second derivative.

Step 2: Apply the Given Condition on the Third Derivative

The problem states that \( \mid f'''(x) \mid \le M \) for \( \mid x - a \mid \le d \). Thus, for any \( c \) between \( a \) and \( x \), it follows that \( \mid f'''(c) \mid \le M \).

Step 3: Estimation of the Remainder Term Magnitude

From Step 1, the magnitude of the remainder term is \( \mid R_2(x) \mid = \left| \frac{f'''(c)}{6} (x-a)^3 \right| \). Given the bound \( \mid f'''(c) \mid \le M \) from Step 2, we have \( \mid R_2(x) \mid \le \frac{M}{6} \mid x-a \mid^3 \).

Step 4: Conclude with the Given Conditions

Since the above inequality holds for all \( x \) such that \( \mid x - a \mid \le d \), Taylor's inequality \( \mid R_2(x) \mid \le \frac{M}{6} \mid x-a \mid^3 \) is satisfied under the given conditions.

Key Concepts

Taylor series

A Taylor series is a way to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It's like making a recipe for a cake where each ingredient is a value from the function or its derivatives. The closer the approximation, the closer your cake tastes like the original recipe.

To construct a Taylor series, we use derivatives of the function evaluated at a point, often denoted as \( a \). This transforms complex functions into a series of simpler polynomial terms. The series is given by:

• \( f(a) \)
• \( f'(a)(x-a) \)
• \( \frac{f''(a)}{2!}(x-a)^2 \)
• ... and so on.

Each term involves higher order derivatives of the function. The farther you go, the more terms you add to increase the accuracy of the approximation. However, this also means more calculations! For simplicity, we often truncate the series to a few terms. This is where the remainder term comes into play.

remainder term

The remainder term in a Taylor series signifies the error or the difference between the function and its Taylor polynomial approximation. Think of it as the gap between your finished cake and the taste of the real recipe if you stopped adding ingredients too soon.

When you're using a Taylor polynomial of order \( n \), the remainder term \( R_n(x) \) is given by an expression involving the \((n+1)\)th derivative. Essentially, it tells us how much the Taylor series fails to capture about the function beyond the nth polynomial term. For example, the remainder term for a second-order approximation (or quadratic Taylor series) is:

• \( R_2(x) = \frac{f'''(c)}{3!}(x-a)^3 \)

This term involves the third derivative of the function, indicating the level of precision the second-order Taylor polynomial has. The variable \( c \) is a mystery player; it lies somewhere between \( a \) and \( x \). Determining \( c \) precisely can be challenging, but Taylor's Inequality helps provide a bound for this remainder term.

third derivative

The third derivative of a function is like the third layer of complexity that measures the rate of change of the rate of change of the rate of change of the function. Each derivative digs deeper into understanding the behavior of the function.

In the context of Taylor's Inequality, the third derivative, \( f'''(x) \), plays a crucial role in determining the remainder term for a second-degree polynomial approximation. The bound \( \mid f'''(x) \mid \leq M \) ensures that the function doesn't behave too erratically.

Using the third derivative in the remainder term allows us to predict how the function might change over a small range around the point \( a \). The tighter this bound, the more confident we are about the accuracy of our polynomial approximation. Essentially, a wild third derivative might mean your cake tastes a bit off from expected! Understanding and estimating this is key in applications like physics and engineering, where approximations often need such precision.

inequality proof

Proving Taylor's inequality involves showing that the remainder term \( \mid R_2(x) \mid \) is bounded as described in the problem. This inequality ensures that despite stopping at a second-order term, the approximation isn't far off.

The steps are straightforward:

• Start with the remainder term \( R_2(x) = \frac{f'''(c)}{3!}(x-a)^3 \)
• Apply the condition \( \mid f'''(x) \mid \leq M \) to get \( \mid f'''(c) \mid \leq M \)
• Substitute into the remainder expression to find \( \mid R_2(x) \mid \leq \frac{M}{6} \mid x-a \mid^3 \)

This inequality emphasizes the bounded nature of the remainder term within the specified interval. It gives clarity on how close the Taylor polynomial is to the actual function within a certain range \( \mid x-a \mid \leq d \). Overall, it reassures that your approximation is trustworthy enough for practical calculations, which is why Taylor's inequality is a fundamental tool in analysis.