Question

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

Short 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-by-step solution

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.

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 \(n=3\): \(\frac{(-1)^3}{(2\cdot3+1)!}x^{2\cdot3+1} = -\frac{1}{7!}x^7 = -\frac{1}{5040}x^7\)

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: $$\sin(1) \approx 1 - \frac{1}{6} + \frac{1}{120} - \frac{1}{5040}$$