Key Idea 8.8.1 gives the nth term of the Taylor series of common functions. In Exercises 3 – 6, verify the formula given in the Key Idea by finding the first few terms of the Taylor series of the given function and identifying a pattern.
Taylor series of f(x) = sin(x) at = 0 is
\[\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}\]
Taylor series for f(x) at x= a can be found by
\[\ f(x) = \sum_{n=0}^{\infty}\frac{f^{n}(a)}{n!}x^{n}\]
\[\ f(x) = sin x => f(0) = 0\]
\[\ f'(x) = cos x => f'(0) = 1\]
\[\ f''(x) = -sin x => f''(0) = 0\]
\[\ f'''(x) = -cos x => f'''(0) = -1\] \[\ f^{4}(x) = sin x => f^{4}(0) = 0\]
Since \[\ f^{4}(x) = f(x)\], the cycle pattern is {0,1,0,-1}, which shows that every even derivative we get 0, and every odd derivative we get either 1 or -1. Therefore:
\[\ f(x) = \frac{1}{1!}x^{1}+\frac{-1}{1!}x^{3}+\frac{1}{1!}x^{5}+...\]
\[\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}\]