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Chapter 8: Problem 7

The examples we have considered in this section have all been for Taylor polynomials and series centered at 0 , but Taylor polynomials and series can be centered at any value of \(a\). We look at examples of such Taylor polynomials in this exercise. a. Let \(f(x)=\sin (x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(x=\frac{\pi}{2}\). Then find the Taylor series for \(f(x)\) centered at \(x=\frac{\pi}{2}\). Why should you have expected the result? b. Let \(f(x)=\ln (x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(x=1\). Then find the Taylor series for \(f(x)\) centered at \(x=1\).

Short Answer

Expert verified
For\(\sin(x)\)centered at \(\frac{\pi}{2}\):\(1 -\frac{(x - \frac{\pi}{2})^2}{2} +\frac{(x - \frac{\pi}{2})^4}{24}\). For \(\ln(x)\)centered at 1: \( (x-1) -\frac{(x-1)^2}{2} +\frac{(x-1)^3}{3} -\frac{(x-1)^4}{4}\)

Step by step solution

01

Define the function and derivatives

For part (a), let the function be defined as: \[ f(x) = \sin(x) \]. The derivatives up to the fourth order are: - First derivative: \[ f'(x) = \cos(x) \] - Second derivative: \[ f''(x) = -\sin(x) \] - Third derivative: \[ f'''(x) = -\cos(x) \] - Fourth derivative: \[ f^{(4)}(x) = \sin(x) \].
02

Evaluate the derivatives at the center point

The center point for the Taylor series is at \( x = \frac{\pi}{2} \). Evaluating the derivatives at this point, we get: - \[ f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1 \] - \[ f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0 \] - \[ f''(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1 \] - \[ f'''(\frac{\pi}{2}) = -\cos(\frac{\pi}{2}) = 0 \] - \[ f^{(4)}(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1 \].
03

Construct the Taylor polynomial up to the fourth order for \(f\)

Using the evaluated derivatives, the Taylor polynomial up to the fourth order centered at \( x = \frac{\pi}{2} \) is: \[ P_4(x) = f(\frac{\pi}{2}) + f'(\frac{\pi}{2}) \frac{(x - \frac{\pi}{2})}{1!} + f''(\frac{\pi}{2}) \frac{(x - \frac{\pi}{2})^2}{2!} + f'''(\frac{\pi}{2}) \frac{(x - \frac{\pi}{2})^3}{3!} + f^{(4)}(\frac{\pi}{2}) \frac{(x - \frac{\pi}{2})^4}{4!} \]. Substituting the values, we obtain: \[ P_4(x) = 1 + 0 \cdot (x - \frac{\pi}{2}) - \frac{(x - \frac{\pi}{2})^2}{2} + 0 \cdot \frac{(x - \frac{\pi}{2})^3}{6} + \frac{(x - \frac{\pi}{2})^4}{24} \]. Simplified, this gives: \[ P_4(x) = 1 - \frac{(x - \frac{\pi}{2})^2}{2} + \frac{(x - \frac{\pi}{2})^4}{24} \].
04

Identify the Taylor series for \(f(x) = \sin(x)\) centered at \( x = \frac{\pi}{2} \)

The pattern found in the polynomial suggests a general Taylor series of: \[ \sum_{n=0}^{\infty} (-1)^n \frac{(x - \frac{\pi}{2})^{2n}}{(2n)!} \]. This follows because all odd-power terms are zero and the even-power terms alternate in sign and are divided by the factorial of the order.
05

Define the function and derivatives for part (b)

For part (b), let the function be defined as: \[ f(x) = \ln(x) \]. The derivatives up to the fourth order are: - First derivative: \[ f'(x) = \frac{1}{x} \]. - Second derivative: \[ f''(x) = -\frac{1}{x^2} \]. - Third derivative: \[ f'''(x) = 2\frac{1}{x^3} \]. - Fourth derivative: \[ f^{(4)}(x) = -6\frac{1}{x^4} \].
06

Evaluate the derivatives at the center point for part (b)

The center point for the Taylor series is at \( x = 1 \). Evaluating the derivatives at this point, we get: - \[ f(1) = \ln(1) = 0 \]. - \[ f'(1) = \frac{1}{1} = 1 \]. - \[ f''(1) = -\frac{1}{1^2} = -1 \]. - \[ f'''(1) = 2\frac{1}{1^3} = 2 \]. - \[ f^{(4)}(1) = -6\frac{1}{1^4} = -6 \].
07

Construct the Taylor polynomial up to the fourth order for \(f\)

Using the evaluated derivatives, the Taylor polynomial up to the fourth order centered at \( x = 1 \) is: \[ P_4(x) = f(1) + f'(1) \frac{(x - 1)}{1!} + f''(1) \frac{(x - 1)^2}{2!} + f'''(1) \frac{(x - 1)^3}{3!} + f^{(4)}(1) \frac{(x - 1)^4}{4!} \]. Substituting the values, we obtain: \[ P_4(x) = 0 + 1 \cdot (x - 1) - \frac{(x - 1)^2}{2} + 2 \cdot \frac{(x - 1)^3}{6} - 6 \cdot \frac{(x - 1)^4}{24} \]. Simplified, this gives: \[ P_4(x) = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} \].
08

Identify the Taylor series for \(f(x) = \ln(x)\) centered at \( x = 1 \)

The pattern found in the polynomial suggests a general Taylor series of: \[ \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x - 1)^n}{n} \].

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Key Concepts

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

Taylor polynomials
Taylor polynomials are approximations of functions using polynomials. They are particularly useful for estimating the value of a function near a specific point. A Taylor polynomial of order n centered at a point a for a function f(x) can be expressed as:











































































































































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The Taylor polynomial helps give a good local approximation for functions where calculating the exact value may be complicated.
Taylor series
A Taylor series is an infinite sum of terms calculated from the values of a function's derivatives at a single point. While a Taylor polynomial is a finite approximation, the Taylor series extends this to an infinite number of terms for potentially better accuracy. For a function f(x) centered at point a, the Taylor series is:































































The main advantage of the Taylor series is its ability to represent complex functions with simpler polynomial expressions. This makes it very useful in various fields like physics and engineering.
  • Approximation Accuracy: The more terms you add from the series, the closer you get to the actual function.
  • Summing Up: Taylor series often sum up to give a good representation of a wide range of functions.
Function derivatives
Derivatives measure how a function changes as its input changes. They are an essential part of calculus. Derivatives provide the rates at which functions increase or decrease. The nth-order derivative provides information on the nth rate of change. For a function f(x), the first derivative, denoted as f'(x), represents its rate of change. Higher-order derivatives like the second (f''(x)), third (f'''(x)), and so on provide deeper insights.



















































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Derivatives are essential in constructing Taylor polynomials and series. They form the coefficients of the terms in these polynomials and series, giving an accurate representation of the original function.
  • First Derivative: Tells us about the slope or the rate of change.
  • Second Derivative: Provides information on the concavity and acceleration of the function.
  • Higher-order Derivatives: Offer deeper insights into the behavior of functions beyond mere slope or acceleration.

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Most popular questions from this chapter

In the Limit Comparison Test we compared the behavior of a series to one whose behavior we know. In that test we use the limit of the ratio of corresponding terms of the series to determine if the comparison is valid. In this exercise we see how we can compare two series directly, term by term, without using a limit of sequence. First we consider an example. a. Consider the series \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k}\) We know that the series \(\sum \frac{1}{k^{2}}\) is a \(p\) -series with \(p=2>1\) and so \(\sum \frac{1}{k^{2}}\) converges. In this part of the exercise we will see how to use information about \(\sum \frac{1}{k^{2}}\) to determine information about \(\sum \frac{1}{k^{2}+k} .\) Let \(a_{k}=\frac{1}{k^{2}}\) and \(b_{k}=\frac{1}{k^{2}+k} .\) i) Let \(S_{n}\) be the \(n\) th partial sum of \(\sum \frac{1}{k^{2}}\) and \(T_{n}\) the \(n\) th partial sum of \(\sum \frac{1}{k^{2}+k}\). Which is larger, \(S_{1}\) or \(T_{1}\) ? Why? ii) Recall that $$ S_{2}=S_{1}+a_{2} \text { and } T_{2}=T_{1}+b_{2} $$ Which is larger, \(a_{2}\) or \(b_{2}\) ? Based on that answer, which is larger, \(S_{2}\) or \(T_{2}\) ? iii) Recall that $$ S_{3}=S_{2}+a_{3} \text { and } T_{3}=T_{2}+b_{3} $$ Which is larger, \(a_{3}\) or \(b_{3}\) ? Based on that answer, which is larger, \(S_{3}\) or \(T_{3}\) ? iv) Which is larger, \(a_{n}\) or \(b_{n}\) ? Explain. Based on that answer, which is larger, \(S_{n}\) or \(T_{n}\) ? v) Based on your response to the previous part of this exercise, what relationship do you expect there to be between \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k} ?\) Do you expect \(\sum \frac{1}{k^{2}+k}\) to converge or diverge? Why? b. The example in the previous part of this exercise illustrates a more general result. Explain why the Direct Comparison Test, stated here, works.

For the following alternating series, \(\sum_{n=1}^{\infty} a_{n}=1-\frac{1}{10}+\frac{1}{100}-\frac{1}{1000}+\ldots\) how many terms do you have to go for your approximation (your partial sum) to be within 1e-07 from the convergent value of that series?

In this exercise we investigation the Taylor series of polynomial functions. a. Find the 3 rd order Taylor polynomial centered at \(a=0\) for \(f(x)=x^{3}-2 x^{2}+3 x-1\). Does your answer surprise you? Explain. b. Without doing any additional computation, find the 4 th, 12 th, and 100 th order Taylor polynomials (centered at \(a=0\) ) for \(f(x)=x^{3}-2 x^{2}+3 x-1\). Why should you expect this? c. Now suppose \(f(x)\) is a degree \(m\) polynomial. Completely describe the \(n\) th order Taylor polynomial (centered at \(a=0\) ) for each \(n\).

Compute the value of the following improper integral. If it converges, enter its value. Enter infinity if it diverges to \(\infty\), and -infinity if it diverges to \(-\infty\). Otherwise, enter diverges. $$ \int_{1}^{\infty} \frac{3 d x}{x^{2}+1}= $$ Does the series \(\sum_{n=1}^{\infty} \frac{3}{n^{2}+1}\) converge or diverge? [Choose: converges | diverges to +infinity | diverges to -infinity | diverges]

Represent the function \(\frac{4}{(1-10 x)}\) as a power series \(f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}\) \begin{tabular}{l} \(c_{0}=\) \\ \(c_{1}=\) \\ \(c_{2}=\) \\ \(c_{3}=\) \\ \(c_{4}=\) \\ \hline \end{tabular} Find the radius of convergence \(R=\)
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