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Chapter 10: Problem 6

Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a .\) \(f(x)=1 /(x+2), \quad a=0\)

Short Answer

Expert verified
Order 0: \(T_0(x) = \frac{1}{2}\); Order 1: \(T_1(x) = \frac{1}{2} - \frac{1}{4}x\); Order 2: \(T_2(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2\); Order 3: \(T_3(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2 - \frac{1}{16}x^3\).

Step by step solution

01

Find the function value at a

To find the Taylor polynomials at a point, we begin by evaluating the function at that point. For the given function, evaluate it at \(a = 0\). So, we find \(f(0) = \frac{1}{0+2} = \frac{1}{2}\).
02

Compute the first derivative and its value at a

Take the first derivative of the function \(f(x) = \frac{1}{x+2}\). The derivative is \(f'(x) = -\frac{1}{(x+2)^2}\). Evaluate this at \(a = 0\), yielding \(f'(0) = -\frac{1}{4}\).
03

Compute the second derivative and its value at a

Next, find the second derivative of the function. The second derivative is \(f''(x) = \frac{2}{(x+2)^3}\). Evaluate this at \(a = 0\), giving \(f''(0) = \frac{1}{4}\).
04

Compute the third derivative and its value at a

For the third derivative, differentiate again to get \(f'''(x) = -\frac{6}{(x+2)^4}\). Evaluating at \(a = 0\) gives \(f'''(0) = -\frac{3}{8}\).
05

Form the Taylor polynomial of order 0

The 0th-order Taylor polynomial is simply the function value at \(a\), which is \(f(0) = \frac{1}{2}\).
06

Form the Taylor polynomial of order 1

The first-order Taylor polynomial is \(T_1(x) = f(0) + f'(0)(x-0)\). Plugging the values, \(T_1(x) = \frac{1}{2} - \frac{1}{4}x\).
07

Form the Taylor polynomial of order 2

The second-order Taylor polynomial is \(T_2(x) = f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x-0)^2\). Substitute the values to get \(T_2(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2\).
08

Form the Taylor polynomial of order 3

The third-order Taylor polynomial is \(T_3(x) = f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3\). Plug in the calculated values: \(T_3(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{8}x^2 - \frac{1}{16}x^3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a fascinating branch of mathematics that focuses on change. It's used to understand how things evolve over time and can be broken down into two main parts: **differential calculus** and **integral calculus**.
  • Differential calculus is concerned with rates of change and slopes of curves. It allows us to find derivatives, which tell us how a function behaves as its input changes.
  • Integral calculus deals with accumulation or the total size, such as finding areas under curves.
In the context of Taylor Polynomials, calculus is essential because it provides the tools to approximate complex functions with simpler polynomial expressions. Through derivatives (a part of differential calculus), we can determine the coefficients of a Taylor polynomial and closely estimate a function's behavior around a specific point, known as the center or point of expansion.
Derivatives
Derivatives play a critical role in understanding functions. They give us the rate at which a function's value changes as its input changes. Here's how derivatives were used for our Taylor polynomial example:
  • **First Derivative**: Given a function, its first derivative describes its rate of change or slope. For the function \(f(x) = \frac{1}{x+2}\), the first derivative \(f'(x) = -\frac{1}{(x+2)^2}\) tells us how steep or flat the curve is at any point.
  • **Higher Order Derivatives**: These are simply derivatives of derivatives. So the second derivative \(f''(x)\), gives insight into the function鈥檚 curvature, or how its slope changes. For instance, \(f''(x) = \frac{2}{(x+2)^3}\). The third derivative \(f'''(x)\), provides even more detail about the function's shape around its expansion point.
Understanding derivatives helps us to construct Taylor Polynomials, as the coefficients of the polynomial are derived from evaluating these derivatives at the specific point of expansion.
Polynomial Approximation
Polynomial approximation is a powerful technique where a complex function is approximated using a polynomial. The closer the polynomial is to the function, the better the approximation.In the case of Taylor Polynomials, this approximation depends on:
  • **Order of the Polynomial**: More terms (higher order polynomials) generally provide a better approximation. For example, a third-order Taylor polynomial is typically a closer fit than a first-order polynomial.
  • **Point of Expansion (a)**: This is the point where the function is approximated. In our case, it's \(a = 0\). The polynomial is designed to coincide with the function鈥檚 value and derivatives at this point.
The Taylor series approximates a function by summing its derivatives at the point \(a\), multiplied by powers of \(x-a\), where the coefficients are determined by the function's derivatives. This makes Taylor polynomials highly useful for simplifying calculations and understanding the behavior of functions near a given point.

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Most popular questions from this chapter

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow \infty} x^{2}\left(e^{-1 / x^{2}}-1\right) \end{equation}

Taylor series for even functions and odd functions (Continuation of Section \(10.7,\) Exercise \(59 .\) ) Suppose that \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) converges for all \(x\) in an open interval \((-R, R) .\) Show that \begin{equation} \begin{array}{l}{\text { a. If } f \text { is even, then } a_{1}=a_{3}=a_{5}=\dots=0, \text { i.e., the Taylor }} \\ {\text { series for } f \text { at } x=0 \text { contains only even powers of } x .} \\ {\text { b. If } f \text { is odd, then } a_{0}=a_{4}=a_{4}=\cdots=0, \text { i.e., the Taylor }} \\ {\text { series for } f \text { at } x=0 \text { contains only odd powers of } x .}\end{array} \end{equation}

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so do the following. $$\begin{array}{ll}{\text { a. }} & {\sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. }} & {\sum_{n=1}^{\infty} k a_{n}(k \text { any number })}\end{array}$$

Use the Taylor series for 1\(/\left(1-x^{2}\right)\) to obtain a series for 2\(x /\left(1-x^{2}\right)^{2}\)

The Taylor series generated by \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) is \(\sum_{n=0}^{\infty} a_{n} x^{n}\) A function defined by a power series \(\Sigma_{n=0}^{\infty} a_{n} x^{n}\) with a radius of convergence \(R>0\) has a Taylor series that converges to the function at every point of \((-R, R) .\) Show this by showing that the Taylor series generated by \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) is the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) itself. \begin{equation} \begin{array}{c}{\text { An immediate consequence of this is that series like }} \\ {x \sin x=x^{2}-\frac{x^{4}}{3 !}+\frac{x^{6}}{5 !}-\frac{x^{8}}{7 !}+\cdots} \\ {\text {and}}\end{array} \end{equation} \begin{equation} x^{2} e^{x}=x^{2}+x^{3}+\frac{x^{4}}{2 !}+\frac{x^{5}}{3 !}+\cdots \end{equation} obtained by multiplying Taylor series by powers of \(x,\) as well as series obtained by integration and differentiation of convergent power series, are themselves the Taylor series generated by the functions they represent.
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