Problem 54 - Calculus Early Transcendentals

Question

Find the remainder term \(R_{n}\) for the nth-order Taylor polynomial centered at a for the given functions. Express the result for a general value of \(n\) $$f(x)=1 /(1-x) ; a=0$$

Step-by-Step Solution

Answer

Question: Find the remainder term \(R_{n}(x)\) for the nth-order Taylor polynomial centered at \(a=0\) for the function \(f(x) = \frac{1}{1-x}\). Answer: The remainder term for the nth-order Taylor polynomial centered at \(a=0\) for the given function is \(R_n(x) = \frac{1}{n+1} \cdot \frac{x^{n+1}}{(1-c)^{n+1}}\), where \(c\) is a value between \(0\) and \(x\).

Step 1: Review the Taylor Polynomial and Remainder Formulas

For a given function \(f(x)\), the nth-order Taylor polynomial centered at \(a\) is given by: $$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$$ The remainder term \(R_n(x)\) is given by the Lagrange form: $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$ where \(c\) is a value between \(a\) and \(x\).

Step 2: Find the first derivative of the function

The given function is \(f(x) = \frac{1}{1-x}\). First, we'll find the first derivative of the function: $$f'(x) = \frac{d}{dx} \left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2}$$

Step 3: Identify the pattern for the derivatives

From the first derivative, we can notice that the derivative forms a pattern with increasing powers of \((1-x)\) in the denominator. The second derivative would be: $$f''(x) = \frac{d}{dx} \left(\frac{1}{(1-x)^2}\right) = \frac{2}{(1-x)^3}$$ The pattern suggests that the \((n+1)\)th derivative should be of the form: $$f^{(n+1)}(x) = \frac{(n!)}{(1-x)^{n+1}}$$

Step 4: Apply the Lagrange form and center the Taylor polynomial at \(a=0\)

Now we can substitute this into the Lagrange form of the remainder term: $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-0)^{n+1}$$ Using our expression for the \((n+1)\)th derivative, we have: $$R_n(x) = \frac{\frac{(n!)}{(1-c)^{n+1}}}{(n+1)!}x^{n+1}$$

Step 5: Simplify the remainder term

Finally, we can simplify the remainder term to: $$R_n(x) = \frac{n!}{(n+1)!} \cdot \frac{x^{n+1}}{(1-c)^{n+1}}$$ $$R_n(x) = \frac{1}{n+1} \cdot \frac{x^{n+1}}{(1-c)^{n+1}}$$ And that's the remainder term \(R_n(x)\) for the nth-order Taylor polynomial centered at \(a=0\) for the given function.

Key Concepts

Remainder Term

In the world of calculus, the remainder term in a Taylor polynomial helps quantify the error between the actual function and its polynomial approximation. The expression for the remainder term is crucial when you need to know how accurate the polynomial is in estimating the function. For a function \(f(x)\), the nth-order Taylor polynomial centered around a point \(a\) is given by a sum of terms up to the nth derivative. To calculate how closely this polynomial matches the real function, the remainder term \(R_n(x)\) is used. It helps us understand the magnitude of the deviation as \(n\) increases. This term becomes smaller as we take more terms in the polynomial, thus providing a more precise approximation of the function near the center \(a\). Keep in mind that while the remainder term ensures the accuracy, for practical purposes, an approximation that is 'close enough' might already be sufficient.

Lagrange Form

The Lagrange form of the remainder term gives us an explicit way to express this remainder. The Lagrange form is particularly useful when determining the error's potential size between the Taylor polynomial and the true function. It is expressed as:

• \(R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\)

where \(c\) is some point in the interval between \(a\) and \(x\). Finding this \(c\) gives us a theoretical boundary for the error when using the polynomial. It becomes apparent that if we choose a point \(x\) close enough to \(a\), especially when the increment in derivatives is not very large, the remainder decreases significantly. This makes the Taylor polynomial a very powerful tool in approximating functions, as the Lagrange remainder shows us precisely how the series behaves.

Derivatives

Derivatives play a key role in forming Taylor polynomials. Each term in a Taylor series involves a derivative of the function, starting from the function itself, then its first derivative, second derivative, and so on. For the function \(f(x) = \frac{1}{1-x}\), finding derivatives involves a pattern, where each derivative increases the power of \(1-x\) in the denominator. Initially:

• \(f'(x) = \frac{1}{(1-x)^2}\)
• \(f''(x) = \frac{2}{(1-x)^3}\) and so on.

The pattern continues, where the nth derivative \(f^{(n+1)}(x)\) comes out as \(\frac{n!}{(1-x)^{n+1}}\). Identifying this pattern not only simplifies calculations but also illustrates how the function's behavior influences its polynomial approximation. Understanding these derivatives is key because the Taylor expansion greatly relies on having them correct up to the desired order.

Series Expansion

Series expansion is a major concept distinctively connected to Taylor polynomials. By using a series expansion, we aim to approximate functions with polynomials that can infinitely sum up to the function itself, as \(n\) approaches infinity. A Taylor series is a type of series expansion that utilizes derivatives of a function at a specific point. In the case of the function \(f(x) = \frac{1}{1-x}\), its Taylor series emerges as an infinite geometric series when centered at \(a=0\). The general formula for a Taylor polynomial, therefore, becomes:

• \(P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k\)

This series expansion lets us express complex functions in a simpler polynomial form, which is often easier to compute. As such, Taylor series provide a foundational tool in many calculus applications, from solving differential equations to evaluating limits and more.