Problem 27 - 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. $$f(x)=2^{x}, a=1$$

Step-by-Step Solution

Answer

Question: Find the Taylor series of the function \(f(x) = 2^x\) centered at \(a = 1\) and write the power series using summation notation. Answer: The first four nonzero terms of the Taylor series for the function \(f(x)=2^x\) centered at \(a=1\) are: $$f(x) \approx 2 + 2\ln(2)(x-1) + \frac{1}{2} 2\ln^2(2) (x-1)^2 + \frac{1}{3!} 2\ln^3(2) (x-1)^3$$ The power series representation of the function \(f(x)=2^x\) in summation notation centered at \(a=1\) is: $$f(x) \approx \sum_{n=0}^{\infty} \frac{2 \ln^n(2)(x-1)^n}{n!}$$

Step 1: Determine the derivatives of the function

We need to find the first four derivatives of the function \(f(x) = 2^x\). 1. First derivative: \(f'(x) = 2^x\ln(2)\) 2. Second derivative: \(f''(x) = 2^x\ln^2(2)\) 3. Third derivative: \(f^{(3)}(x) = 2^x\ln^3(2)\) 4. Fourth derivative: \(f^{(4)}(x) = 2^x\ln^4(2)\)

Step 2: Evaluate derivatives at the center, "a"

Evaluate the derivatives at \(a=1\): 1. \(f(1) = 2^1 = 2\) 2. \(f'(1) = 2^1\ln(2) = 2\ln(2)\) 3. \(f''(1) = 2^1\ln^2(2) = 2\ln^2(2)\) 4. \(f^{(3)}(1) = 2^1\ln^3(2) = 2\ln^3(2)\)

Step 3: Construct the Taylor series

Now, we can construct the Taylor series using the values of the derivatives at \(a=1\): $$f(x) \approx 2 + 2\ln(2)(x-1) + \frac{1}{2} 2\ln^2(2) (x-1)^2 + \frac{1}{3!} 2\ln^3(2) (x-1)^3$$ This is the first four nonzero terms of the Taylor series for the function \(f(x)=2^x\) centered at \(a=1\).

Step 4: Express the Taylor series in summation notation

Finally, we can write the Taylor series in summation notation: $$f(x) \approx \sum_{n=0}^{\infty} \frac{2 \ln^n(2)(x-1)^n}{n!}$$ This is the power series representation of the function \(f(x)=2^x\) in summation notation centered at \(a=1\).