Problem 48 - Calculus Early Transcendentals

Question

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\sin 1$$

Step-by-Step Solution

Answer

Answer: The first four nonzero terms of the Taylor series for sin(1) are: 1. \(x = 1\) 2. \(-\frac{1}{6}x^3 = -\frac{1}{6}\) 3. \(\frac{1}{120}x^5 = \frac{1}{120}\) 4. \(-\frac{1}{5040}x^7 = -\frac{1}{5040}\) The series can be approximated as: $$\sin(1) \approx 1 - \frac{1}{6} + \frac{1}{120} - \frac{1}{5040}$$

Step 1: Identify the function and its derivatives

We know that for the sine function, we have the following derivatives: 1. \(f(x) = \sin(x)\) 2. \(f'(x) = \cos(x)\) 3. \(f''(x) = -\sin(x)\) 4. \(f'''(x) = -\cos(x)\) 5. \(f^{(4)}(x) = \sin(x)\) 6. \(f^{(5)}(x) = \cos(x)\) The pattern repeats every four derivatives.

Step 2: Calculate the first four nonzero terms

For the sine function's Taylor series, we need to calculate the first four nonzero terms using the general formula: $$\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$ To find the first four nonzero terms, we need to plug in the values for \(n = 0, 1, 2, 3\): 1. For \(n=0\): \(\frac{(-1)^0}{(2\cdot0+1)!}x^{2\cdot0+1} = \frac{1}{1!}x^1 = x\) 2. For \(n=1\): \(\frac{(-1)^1}{(2\cdot1+1)!}x^{2\cdot1+1} = -\frac{1}{3!}x^3 = -\frac{1}{6}x^3\) 3. For \(n=2\): \(\frac{(-1)^2}{(2\cdot2+1)!}x^{2\cdot2+1} = \frac{1}{5!}x^5 = \frac{1}{120}x^5\) 4. For \(n=3\): \(\frac{(-1)^3}{(2\cdot3+1)!}x^{2\cdot3+1} = -\frac{1}{7!}x^7 = -\frac{1}{5040}x^7\)

Step 3: Substitute \(x=1\) and find the value of the series

Now, we need to substitute \(x=1\) to find the value of the first four nonzero terms: 1. \(x = 1^1 = 1\) 2. \(-\frac{1}{6}x^3 = -\frac{1}{6}(1)^3 = -\frac{1}{6}\) 3. \(\frac{1}{120}x^5 = \frac{1}{120}(1)^5 = \frac{1}{120}\) 4. \(-\frac{1}{5040}x^7 = -\frac{1}{5040}(1)^7 = -\frac{1}{5040}\) So, the first four nonzero terms of the Taylor series for \(\sin(1)\) are: $$\sin(1) \approx 1 - \frac{1}{6} + \frac{1}{120} - \frac{1}{5040}$$