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Chapter 10: Problem 25

a. Find the first four nonzero terms of the Taylor series for the given function centered at \(a\) b. Write the power series using summation notation. $$f(x)=\ln x, a=3$$

Short Answer

Expert verified
Question: Find the first four nonzero terms of the Taylor series for the function \(f(x) = \ln(x)\) centered at \(a=3\), and express the power series using summation notation. Answer: The first four nonzero terms of the Taylor series for the function \(f(x) = \ln(x)\) centered at \(a=3\) are: $$\ln(3) + \frac{1}{3}(x-3) -\frac{1}{9}(x-3)^2 + \frac{2}{27}(x-3)^3$$ The power series can be expressed using the summation notation as: $$f(x) \approx \sum_{n=0}^{3} \frac{(-1)^{n-1}(n!)(x-3)^n}{3^n n!}$$

Step by step solution

01

Recall the Taylor series formula

Remember that the Taylor series for a function \(f(x)\) centered at \(a\) is given by: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$ Where \(f^{(n)}(a)\) denotes the nth derivative of the function evaluated at \(a\), and \(n!\) denotes the factorial of \(n\).
02

Compute the first four derivatives of \(f(x)\) at \(a=3\)

We need to compute the first four derivatives of the function \(f(x) = \ln(x)\) evaluated at \(a=3\): 1. \(f(x) = \ln(x)\), so \(f(3) = \ln(3)\). 2. \(f'(x) = \frac{1}{x}\), so \(f'(3) = \frac{1}{3}\). 3. \(f''(x) = -\frac{1}{x^2}\), so \(f''(3) = -\frac{1}{9}\). 4. \(f'''(x) = \frac{2}{x^3}\), so \(f'''(3) = \frac{2}{27}\).
03

Plug in the computed derivatives into the Taylor series formula

Now plug the computed derivatives into the Taylor series formula: $$f(x) \approx \ln(3) + \frac{1}{3}(x-3) -\frac{1}{9}(x-3)^2 + \frac{2}{27}(x-3)^3$$
04

Rewrite the Taylor series in summation notation

We rewrite the Taylor series in summation notation: $$f(x) \approx \sum_{n=0}^{3} \frac{(-1)^{n-1}(n!)(x-3)^n}{3^n n!}$$ Where the coefficients are given as: 1. For \(n=0\), the coefficient is \(\ln(3)\). 2. For \(n=1\), the coefficient is \(\frac{1}{3}\). 3. For \(n=2\), the coefficient is \(-\frac{1}{9}\). 4. For \(n=3\), the coefficient is \(\frac{2}{27}\). The first four nonzero terms of the Taylor series for the function \(f(x) = \ln(x)\) centered at \(a=3\) are: $$\ln(3) + \frac{1}{3}(x-3) -\frac{1}{9}(x-3)^2 + \frac{2}{27}(x-3)^3$$ And the power series can be expressed using the summation notation as: $$f(x) \approx \sum_{n=0}^{3} \frac{(-1)^{n-1}(n!)(x-3)^n}{3^n n!}$$

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Most popular questions from this chapter

Use properties of power series, substitution, and factoring to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Give the interval of convergence for the new series. Use the Taylor series. $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\cdots, \text { for }-1 < x \leq 1$$ $$\sqrt{a^{2}+x^{2}}, a > 0$$

Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) for all \(x\) in the interval of convergence. $$f(x)=\cos x, a=\pi / 2$$

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\cos 2 x+2 \sin x$$

What is the minimum order of the Taylor polynomial required to approximate the following quantities with an absolute error no greater than \(10^{-3}\) ? (The answer depends on your choice of a center.) $$\sin 0.2$$

Use the identity \(\sec x=\frac{1}{\cos x}\) and long division to find the first three terms of the Maclaurin series for \(\sec x\)
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