91Ó°ÊÓ

First, we will find the general formula for the nth derivative of the function \(\cos 2x\). Since the function is periodic and its derivatives will also be periodic, we can follow a pattern to derive the general formula for the nth derivative. 1st derivative: \(f'(x)=-2\sin 2x\) 2nd derivative: \(f''(x)=-4\cos 2x\) 3rd derivative: \(f'''(x)=8\sin 2x\) 4th derivative: \(f^{(4)}(x)=16\cos 2x\) We notice that the nth derivative oscillates between the sine and cosine functions, along with a power of 2. So we can write the general formula for the nth derivative as follows: $$ f^{(n)}(x)=2^n\begin{cases} \cos 2x, & \textrm{ if } n \equiv 0,2 \pmod{4} \\ -\sin 2x, & \textrm{ if } n \equiv 1,3 \pmod{4} \end{cases} $$

Now, let's find the (n+1)th derivative of the function, using the general formula: $$ f^{(n+1)}(x)=2^{n+1}\begin{cases} \cos 2x, & \textrm{ if } (n+1) \equiv 0,2 \pmod{4} \\ -\sin 2x, & \textrm{ if } (n+1) \equiv 1,3 \pmod{4} \end{cases} $$

By Taylor's theorem, the remainder of the nth-order Taylor polynomial with center a = 0 is given by: $$ R_n(x)=\frac{f^{(n+1)}(0)}{(n+1)!}(x-0)^{n+1} $$ Now, let's use the (n+1)th derivative to compute the remainder: $$ R_n(x)=\frac{2^{n+1}}{(n+1)!}\begin{cases} (0)^{n+1}, & \textrm{ if } (n+1) \equiv 0,2 \pmod{4} \\ -(0)^{n+1}, & \textrm{ if } (n+1) \equiv 1,3 \pmod{4} \end{cases} $$ Since we're evaluating at 0, the remainder will always be zero except if (n+1) is an odd multiple of 2, in which case it becomes: $$ R_n(x)=\frac{2^{n+1}}{(n+1)!}(x)^{n+1}, \textrm{ if } (n+1) \equiv 0,2 \pmod{4} $$ This is the general formula for the remainder of the nth-order Taylor polynomial for the function \(\cos 2x\) with a center at a = 0.