Problem 28 - 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. $$f(x)=10^{x}, a=2$$

Step-by-Step Solution

Answer

Answer: The first four nonzero terms of the Taylor series are \( T_4(x)= 100 + (100\ln(10))(x-2) +\frac{100(\ln(10))^2}{2!}(x-2)^2 +\frac{100(\ln(10))^3}{3!}(x-2)^3 \) and its summation notation representation is \( T_4(x)=\sum_{n=0}^3\frac{100(\ln(10))^n}{n!}(x-2)^n \).

Step 1: Find the first four derivatives of the function

To find the first four derivatives of the function \(f(x)=10^x\), we will rewrite \(10^x\) as \(e^{x\ln(10)}\), and then find the derivatives. 1. First derivative: \(f'(x)=e^{x\ln(10)}\ln(10)\) 2. Second derivative: \(f''(x)=e^{x\ln(10)}(\ln(10))^2\) 3. Third derivative: \(f'''(x)=e^{x\ln(10)}(\ln(10))^3\) 4. Fourth derivative: \(f^{(4)}(x)=e^{x\ln(10)}(\ln(10))^4\)

Step 2: Evaluate the derivatives at a = 2

Now we need to evaluate the four derivatives at \(a=2\): 1. \(f(2)=10^2=100\) 2. \(f'(2)=e^{2\ln(10)}\ln(10)=100\ln(10)\) 3. \(f''(2)=e^{2\ln(10)}(\ln(10))^2=100(\ln(10))^2\) 4. \(f'''(2)=e^{2\ln(10)}(\ln(10))^3=100(\ln(10))^3\)

Step 3: Write the first four non-zero terms of the Taylor series

Now, we can substitute the values of the derivatives evaluated at \(a=2\) into the Taylor series formula: $$T_4(x) = 100 + (100\ln(10))(x-2) +\frac{100(\ln(10))^2}{2!}(x-2)^2 +\frac{100(\ln(10))^3}{3!}(x-2)^3$$

Step 4: Write the power series in summation notation

We can now write the obtained power series in summation notation: $$T_4(x)=\sum_{n=0}^3\frac{100(\ln(10))^n}{n!}(x-2)^n$$ Now we have found the first four nonzero terms of the Taylor series for the given function centered at \(a=2\) and written the power series using summation notation.