Problem 58 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)\) be a function that possesses at least \(n\) derivatives at \(x=a\) and let \(P_{n}(x)\) be the Taylor polynomial of order \(n\) based at \(a\). Show that $$ \begin{aligned} P_{n}(a)=f(a), \quad P_{n}^{\prime}(a)=f^{\prime}(a), \quad P_{n}^{\prime \prime}(a)=f^{\prime \prime}(a), \\ & \ldots, \quad P_{n}^{(n)}(a)=f^{(n)}(a) \end{aligned} $$

Step-by-Step Solution

Answer

The derivatives of the Taylor polynomial match those of the function at \(a\), as shown by consecutive evaluations at \(x=a\).

Step 1: Understanding Taylor Polynomials

The Taylor polynomial of order \(n\) for a function \(f(x)\) at point \(x=a\) is defined as:\[P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n\].This polynomial approximates \(f(x)\) near \(x=a\), using derivatives of \(f\) evaluated at \(a\). For this exercise, we are verifying values of the polynomial and its derivatives at \(x=a\).

Step 2: Evaluating Taylor Polynomial at Base Point

First, substitute \(x=a\) into \(P_n(x)\):\[P_n(a) = f(a) + f'(a)(a-a) + \frac{f''(a)}{2!}(a-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(a-a)^n = f(a)\].All terms except \(f(a)\) vanish because they include \((a-a)\), proving \(P_n(a) = f(a)\).

Step 3: Evaluating Derivative at Base Point

Next, consider the first derivative of the Taylor polynomial:\[P_n'(x) = f'(a) + \frac{f''(a)}{1!}(x-a) + \frac{f'''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{(n-1)!}(x-a)^{n-1}\].Evaluating this at \(x=a\) gives:\[P_n'(a) = f'(a)\], as all terms where \((x-a)\) appears vanish. This shows \(P_n'(a) = f'(a)\).

Step 4: Generalizing for Higher Derivatives

Now, generalize this process to higher order derivatives. The \(k^{th}\) derivative of \(P_n(x)\) at \(x=a\) is:\[P_n^{(k)}(x) = \frac{f^{(k)}(a)}{(k-1)!} + \text{higher order terms in } (x-a)\].When \(x=a\) is substituted, all terms with \((x-a)\) disappear, leaving:\[P_n^{(k)}(a) = f^{(k)}(a)\].Repeat this until \(k=n\) to show equivalence for each derivative up to \(n\).

Step 5: Conclusion

Thus, for the Taylor polynomial \(P_n(x)\) of function \(f(x)\) at \(x=a\), it holds that:\[\begin{aligned}P_n(a) &= f(a), \P_n'(a) &= f'(a), \P_n''(a) &= f''(a), \&\vdots \P_n^{(n)}(a) &= f^{(n)}(a)\end{aligned}\].This validates that the polynomial and its derivatives up to the \(n\)th order match those of \(f(x)\) at \(a\).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, providing information about how a function changes. They are essentially the rate of change or slope of a function at a given point. When we discuss derivatives in the context of Taylor polynomials, we are focusing on how the function behaves around a specific point known as the base point, denoted as \(x = a\).
The first derivative, \(f'(x)\), indicates the rate of change of the function. The second derivative, \(f''(x)\), provides the rate of change of the rate of change, often related to the function's curvature. Higher-order derivatives follow this pattern, describing more complex aspects of the function's behavior.
For the Taylor polynomial \(P_n(x)\), each term is constructed using the derivatives of the function \(f(x)\) evaluated at \(x = a\). This involves calculating derivatives up to order \(n\), denoted as \(f^{(n)}(a)\), and integrating them into the polynomial form.
By finding these derivatives, we can establish the polynomial's ability to closely approximate the behavior of \(f(x)\) near \(x = a\).

Function Approximation

Function approximation is the goal of constructing a mathematical expression that closely resembles a given function, particularly around a specific point. In calculus, Taylor polynomials serve as a powerful tool for approximating functions.
Consider the idea of "zooming in" on the function \(f(x)\) around a base point \(x = a\). The Taylor polynomial \(P_n(x)\) is designed to mimic the behavior of \(f(x)\) by matching it and its derivatives at \(x = a\). This results in an accurate approximation, especially for values of \(x\) close to \(a\).
The polynomial is constructed by summing terms that include the function's derivatives at \(a\) and powers of \((x-a)\). By ensuring that each of these terms reflects the corresponding derivative's influence on the function, the Taylor polynomial captures the essence of \(f(x)\)'s behavior near \(a\).
This approximation is not just about mimicry but also efficiency, as Taylor polynomials often simplify calculations and predictions about a function's behavior.

Mathematical Proof

In mathematics, a proof is a rigorous argument that establishes the validity of a statement. Proving that the Taylor polynomial and the function \(f(x)\) have the same values and derivatives at the base point \(x = a\) involves both logical reasoning and calculation.
The process begins by evaluating \(P_n(x)\) at \(x = a\). This involves substituting \(x = a\) into the polynomial's expression. For the Taylor polynomial, all terms except the constant term \(f(a)\) vanish because they include factors of \((a-a)\), which are zero. This results in the polynomial evaluated at \(a\) equaling \(f(a)\).
Similarly, for derivatives, substituting \(x = a\) into the derivatives of \(P_n(x)\) results in all terms with \((x-a)\) disappearing. This occurs because those terms relate to higher powers of \((x-a)\), which become zero. The remaining portion of each derivative at \(a\) matches the corresponding derivative of \(f(x)\), completing the proof.
Each step of the proof builds on previous logic, ensuring that the polynomial precisely aligns with \(f(x)\), reinforcing the mathematical relationship they share.

Calculus

Calculus is a branch of mathematics that deals with continuous change and includes concepts such as derivatives and integration. It forms the foundation for understanding and applying Taylor polynomials in the context of function approximation.
Through calculus, we gain insight into how functions behave, how they change, and how to quantify these changes. The Taylor polynomial comes into play as a practical application of calculus principles, specifically using the derivatives calculated via differentiation.
The idea of a Taylor polynomial links directly to the core tenets of calculus: describing the instantaneous behavior of functions (via derivatives) and understanding their global behavior as it changes across intervals. Calculus allows us to leverage the stability and predictability offered by Taylor polynomials to approximate functions seamlessly.
Moreover, calculus equips us with the tools to manipulate and analyze these polynomials, ensuring they accurately represent the function \(f(x)\) around the point \(x = a\). Understanding the interplay of these elements in calculus enriches our grasp of Taylor polynomials and function approximation.