Problem 25 - Calculus Early Transcendentals

Question

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. $$f(x)=\ln x, a=3$$

Step-by-Step Solution

Answer

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 1: 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 \(n\).

Step 2: 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 \(f'''(3) = \frac{2}{27}\).

Step 3: Plug in the computed derivatives into the Taylor series formula

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

Step 4: 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: $$f(x) \approx \sum_{n=0}^{3} \frac{(-1)^{n-1}(n!)(x-3)^n}{3^n n!}$$