Problem 81 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that $$ f(t)=f(a)+\sum_{i=1}^{n} \frac{f^{(i)}(a)}{i !}(t-a)^{i}+\int_{a}^{t} \frac{(t-x)^{n}}{n !} f^{(n+1)}(x) d x $$ provided that \(f\) can be differentiated \(n+1\) times.

Step-by-Step Solution

Answer

The integral term is the remainder in the Taylor series expansion.

Step 1: Understanding the Problem

We need to prove that the given equation represents the Taylor series expansion of a function with a remainder term. The function \( f(t) \) is expressed as a sum of a Taylor polynomial and an integral remainder term. The task is to show that this expression is valid for function \( f \) which can be differentiated \( n+1 \) times.

Step 2: Review the Taylor Series Expansion

The Taylor series expansion of a function \( f \) at point \( a \) is given by:\[f(t) = f(a) + \sum_{i=1}^{n} \frac{f^{(i)}(a)}{i!} (t-a)^i + R_n(t)\]where \( R_n(t) \) is the remainder term for the Taylor series.

Step 3: Examine the Given Expression

The given expression is:\[f(t) = f(a) + \sum_{i=1}^{n} \frac{f^{(i)}(a)}{i!} (t-a)^i + \int_{a}^{t} \frac{(t-x)^n}{n!} f^{(n+1)}(x) \, dx\]Notice that it includes a polynomial part and an integral term, similar to the Lagrange form of the remainder in the Taylor series.

Step 4: Identify the Remainder Term

Compare the given expression with the Taylor series form. The integral term:\[\int_{a}^{t} \frac{(t-x)^n}{n!} f^{(n+1)}(x) \, dx\]acts as the remainder \( R_n(t) \). This is known as the integral form of the Taylor series remainder.

Step 5: Conclude the Verification

The expression matches the Taylor series expansion of a function including its remainder term in integral form. Therefore, the given formula correctly represents the Taylor series and its remainder for a function \( f \) that is \( n+1 \) times differentiable.

Key Concepts

Understanding the Remainder Term

In the Taylor series expansion, the remainder term plays a crucial role. It measures the error between the actual function and its polynomial approximation. When we expand a function using Taylor series, we express it as a sum of derivatives at a specific point raised to increasing powers. However, the series is typically truncated at a certain degree, leading to a remainder term.

• This term is essential because it shows how accurate our approximation is when we cut off the series at the nth degree.
• The integral form of the remainder provides an exact expression for this error, allowing us to quantify it precisely.

Understanding the remainder term is key to assessing the limits and accuracy of polynomial approximations generated through Taylor series.

Differentiability & Its Importance

Differentiability refers to a function's ability to have its derivatives calculated. For a Taylor series expansion to hold, the function in question must be differentiable enough times. Specifically, for a function expanded up to the nth degree, it must be differentiable at least up to the (n+1)th degree.

• This requirement ensures we can evaluate the derivatives needed for the polynomial terms in the expansion.
• Moreover, the differentiability at higher orders is necessary for accurately determining the remainder term, which depends on the (n+1)th derivative.
• Without sufficient differentiability, a Taylor series cannot fit or approximate a function well at given points.

Differentiability is thus vital not just for generating the expansion but also for ensuring the remainder term is meaningful and useful in calculations.

Exploring the Integral Form of the Remainder

The integral form of the remainder term in a Taylor series offers an insightful way to express the error of approximation. Unlike other forms, it uses an integral to fight the remainder as:\[ \int_{a}^{t} \frac{(t-x)^n}{n!} f^{(n+1)}(x) \, dx \]This expression provides an exact measure of how much the polynomial approximation deviates from the true function value.

• It depends on the (n+1)th derivative, integrating the impact of this term over the interval [a, t].
• The limits of integration span from the point of expansion (a) to the point of evaluation (t), covering the entire interval where the approximation is considered.
• The integral form allows us to use properties of integrals and derivatives, giving us flexibility and a deeper understanding of error behavior across continuous intervals.

Using integral forms allows mathematicians and scientists to analyze functions more comprehensively.

Polynomial Approximation

Polynomial approximation is at the heart of the Taylor series. It involves taking a complex function and approximating it by a polynomial of degree n. This method provides a simpler way to evaluate and understand functions within a local region around a point a.

• The polynomial approximation uses derivatives calculated at a to form a polynomial that closely matches the function's value and behavior.
• This approximation is especially useful because polynomials are easier to differentiate, integrate, and evaluate than more complex functions.
• Since the region of accuracy increases with the number of terms used, adding more terms improves the approximation fit.

Despite the complexity reduction, it's important to remember that polynomial approximations are just that—approximations. They are only as good as the degree to which the series is expanded and the remainder term accurately characterizes the leftover error.