$R_{4}$ for $g(x)=\cos (\pi x)$ on $[0,1]$

Step 1 ：Given the function \(g(x) = \cos (\pi x)\), we are asked to find the fourth degree Taylor polynomial on the interval \([0,1]\).

Step 2 ：The Taylor polynomial of degree \(n\) for a function \(f(x)\) at a point \(a\) is given by: \[P_{n}(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^{2} + \frac{f'''(a)}{3!}(x-a)^{3} + ... + \frac{f^{(n)}(a)}{n!}(x-a)^{n}\]

Step 3 ：In this case, \(f(x) = \cos (\pi x)\), \(a = 0\), and \(n = 4\). So we need to calculate the first four derivatives of \(f(x)\) at \(x = 0\), and then substitute these values into the formula for the Taylor polynomial.

Step 4 ：The first derivative of \(f(x)\) is \(f'(x) = -\pi\sin(\pi x)\), and at \(x = 0\), \(f'(0) = 0\).

Step 5 ：The second derivative of \(f(x)\) is \(f''(x) = -\pi^{2}\cos(\pi x)\), and at \(x = 0\), \(f''(0) = -\pi^{2}\).

Step 6 ：The third derivative of \(f(x)\) is \(f'''(x) = \pi^{3}\sin(\pi x)\), and at \(x = 0\), \(f'''(0) = 0\).

Step 7 ：The fourth derivative of \(f(x)\) is \(f^{(4)}(x) = \pi^{4}\cos(\pi x)\), and at \(x = 0\), \(f^{(4)}(0) = \pi^{4}\).

Step 8 ：Substituting these values into the formula for the Taylor polynomial, we get: \[P_{4}(x) = \frac{\pi^{4}x^{4}}{24} - \frac{\pi^{2}x^{2}}{2} + 1\]

Step 9 ：\(\boxed{P_{4}(x) = \frac{\pi^{4}x^{4}}{24} - \frac{\pi^{2}x^{2}}{2} + 1}\) is the fourth degree Taylor polynomial of the function \(g(x)=\cos (\pi x)\) on the interval \([0,1]\).