$\begin{align}
 f(x) & = \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2 2 x \\
 & = 1 - \frac{1}{2} \cdot \frac{1 - \cos 4 x}{2} = \frac{3}{4} + \frac{1}{4} \cos 4 x = \left[
 \begin{array}{l}
 y = x - 1 \\ 
 x = y + 1
 \end{array} \right] = \frac{3}{4} + \frac{1}{4} \cos (4 y + 4) \\
 & = \frac{3}{4} + \frac{\cos 4}{4} \cos 4 y - \frac{\sin 4}{4} \sin 4 y \\
 & = \frac{3}{4} + \frac{\cos 4}{4} \sum_{n = 0}^{\infty} (-1)^n \frac{(4 y)^{2 n}}{(2 n)!} - \frac{\sin 4}{4} \sum_{n = 0}^{\infty} (-1)^n \frac{(4 y)^{2 n + 1}}{(2 n + 1)!} \\
 & = \frac{3 + \cos 4}{4} + \sum_{n = 1}^{\infty} \frac{(-1)^n 4^{2 n - 1} \cos 4}{(2 n)!} (x - 1)^{2 n} + \sum_{n = 0}^{\infty} \frac{(-1)^n 4^{2 n} \sin 4}{(2 n + 1)!} (x - 1)^{2 n + 1}
 \end{align}$