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Magazine Chapter 6: Problem 141
Find the Taylor series of the given function centered at the indicated point. $$ 1+x+x^{2}+x^{3} \text { at } a=-1 $$
Short Answer
Expert verified
The Taylor series is \((x+1)^3 - 2(x+1)^2 + 2(x+1)\).
Step by step solution
01
Understand the Problem
We need to find the Taylor series of the function \( f(x) = 1 + x + x^2 + x^3 \) centered at \( a = -1 \). This involves expressing the function as an infinite sum of terms calculated from the derivatives of the function evaluated at the point \( a = -1 \).
02
Compute Derivatives
Compute the first few derivatives of \( f(x) \).- \( f(x) = 1 + x + x^2 + x^3 \)- \( f'(x) = 1 + 2x + 3x^2 \)- \( f''(x) = 2 + 6x \)- \( f'''(x) = 6 \)- \( f^{(4)}(x) = 0 \)Higher-order derivatives starting from the fourth derivative are zero because the original function is a polynomial of degree 3.
03
Evaluate Derivatives at \( a = -1 \)
Find the value of each derivative at \( x = -1 \).- \( f(-1) = 1 - 1 + 1 - 1 = 0 \)- \( f'(-1) = 1 - 2 + 3 = 2 \)- \( f''(-1) = 2 - 6 = -4 \)- \( f'''(-1) = 6 \)- \( f^{(4)}(-1) = 0 \)
04
Write the Taylor Series
Using the formula for the Taylor series:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \]Plug in the values found:- \( f(-1) \) term is \(0\)- \( f'(-1) \) term is \( \frac{2}{1!} (x+1) \)- \( f''(-1) \) term is \( \frac{-4}{2!}(x+1)^2 \)- \( f'''(-1) \) term is \( \frac{6}{3!}(x+1)^3 \)Resulting in:\[ 2(x+1) - 2(x+1)^2 + (x+1)^3 \]
05
Simplify the Series
Combine the terms to simplify the Taylor series:\[ 2(x+1) - 2(x+1)^2 + (x+1)^3 = (x+1)^3 - 2(x+1)^2 + 2(x+1) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Function
The problem begins with the polynomial function, given as \(f(x) = 1 + x + x^2 + x^3\). A polynomial function is an expression consisting of variables and coefficients, constructed using operations of addition, subtraction, multiplication, and non-negative integer exponents. This specific function is a polynomial of degree 3, with terms rising from constant \(1\) to \(x^3\). Each term has a straightforward combination of powers of \(x\). It's essential to distinguish this framework because it influences how derivatives are taken, which is central to constructing a Taylor series.
Derivative Evaluation
To create a Taylor series, derivatives of the polynomial function are critical. Here, the derivatives are calculated at the center of expansion, \(a = -1\).
Derivatives themselves represent the rate of change of a function. They give insight into the behavior and slope of the function graph at a particular point.
Derivatives themselves represent the rate of change of a function. They give insight into the behavior and slope of the function graph at a particular point.
- First derivative: \(f'(x) = 1 + 2x + 3x^2\)
- Second derivative: \(f''(x) = 2 + 6x\)
- Third derivative: \(f'''(x) = 6\)
- Fourth derivative: \(f^{(4)}(x) = 0\)
Series Expansion
The Taylor series is a representation of a function as an infinite sum of terms, which are calculated from the values of its derivatives at a single point. For a polynomial function like \(f(x) = 1 + x + x^2 + x^3\), this sum doesn't extend to infinity since it's finite in degree.
The formula for constructing a Taylor series is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \] In our case, calculating up to the third derivative suffices. Here's how the Taylor series terms are derived:
The formula for constructing a Taylor series is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \] In our case, calculating up to the third derivative suffices. Here's how the Taylor series terms are derived:
- 0th term: \(f(-1) = 0\)
- 1st term: \(\frac{f'(-1)}{1!} (x+1) = 2(x+1)\)
- 2nd term: \(\frac{f''(-1)}{2!}(x+1)^2 = -2(x+1)^2\)
- 3rd term: \(\frac{f'''(-1)}{3!}(x+1)^3 = (x+1)^3\)
Center of Expansion
The center of expansion, denoted by \(a\), is a pivotal component in the Taylor series. In this exercise, the center \(a\) is set at \(-1\).
It indicates the point around which the function is approximated. The choice of this point affects the derivatives' evaluation, making \(-1\) a direct input into calculating higher derivatives.
It indicates the point around which the function is approximated. The choice of this point affects the derivatives' evaluation, making \(-1\) a direct input into calculating higher derivatives.
- The zero term at \(f(-1)\) yields \(0\).
- The changing values for \(f'(-1)\), \(f''(-1)\), and \(f'''(-1)\) alter how each \((x+1)^n\) term appears in the series.
