I am finding the error bound for the interpolation of the function $f(x) = cos(\pi x) + sin (\pi x) $ on the interval $[-1, 1]$. To do this I need the maximum of the absolute value of the fourth derivative. The fourth derivative is $f^4 = \pi^4cos(\pi x) + \pi^4 sin(\pi x)$.
My first thought is that the absolute value of the maximum of this function is $\pi^4$ which occurs when $x =1$. However this is not correct. Why isn't it? How do you find the correct maximum?
On
Use trig identities to get
$$\pi^4(\cos(\pi x) + \sin(\pi x)) = \pi^4\sqrt{2}\left(\cos(\pi x)\cos\left(\frac{\pi}{4}\right) + \sin(\pi x)\sin\left(\frac{\pi}{4}\right)\right) = \pi^4\sqrt{2}\cos\left(\pi x - \frac{\pi}{4}\right)$$
so the real maximum would have to be $\pi^4\sqrt{2}$ if it occurs on our interval, which it does at $x=\frac{1}{4}$
On
I would normally observe that $|\pi^4 (\sin \pi x + \cos \pi x)| \leq 2\pi^4$ since the two trigonometric functions are bounded by $1$.
However, you have asked for the actual maximum on $[-1,1]$.
Set $g(x) = \pi^4 (\sin \pi x + \cos \pi x)$. This is differentiable on $[-1,1]$ so we can find the absolute maximum through the usual method of finding critical points and comparing with values at endpoints. To facilitate, let's evaluate the endpoints first. $g(-1) = \pi^4(\sin -\pi + \cos -\pi) = \pi^4(0 + -1) = -\pi/4$ and $g(1) = \pi^4(\sin \pi + \cos \pi) = -\pi/4$. So all a point has to give is a value bigger than $-\pi/4$ to be bigger than the endpoints.
We compute $g'(x) = \pi^5(\cos \pi x - \sin \pi x)$. We set this equal to zero to find critical points. So we wish to solve $$ \cos \pi x = \sin \pi x \text{.} $$ Perhaps we recognize that the solutions in the interval are $x = 1/4$ and $x = -3/4$. But if we do not, we can write \begin{align*} 1 &= \frac{\sin \pi x}{\cos \pi x} = \tan \pi x &\text{or}&& \cos \pi x = 0 \end{align*} The first is solved by $ x \frac{1}{\pi} \left( \tan^{-1} 1 + \pi k\right) = \frac{1}{4} + k$ for any integer $k$. As $k$ ranges over the integers, the values landing in the interval $[-1,1]$ are $x = -3/4$ and $x = 1/4$.
We should also check for extraneous and missing solutions since we divided by an expression which can be zero. This is $\cos \pi x = 0$, which is $x = -1/2$ and $x = 1/2$ in the interval. Neither of these satisfies $g' = 0$ nor are they coincident with the solutions found through the tangent, so we need not worry further.
Then $g(1/4) = \pi^4 \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) = \pi^4 \sqrt{2}$ and $g(-3/4) = \pi^4 \left( \frac{-\sqrt{2}}{2} + \frac{-\sqrt{2}}{2} \right) = -\pi^4 \sqrt{2}$. So the desired absolute maximum on $[-1,1]$ is $\pi^4 \sqrt{2}$.
(Notice that this is not so much smaller than $2 \pi^4$ found by a much easier method.)
To find the maximum value of $|f^{(4)}(x)|$, we simply need to check the endpoints and the critical points. Since you have already established that $|f^{(4)}(1)|=|f^{(4)}(-1)|=\pi^4$ is not maximal, then it suffices to test $x_0$ where $f^{(5)}(x_0)=0 \Longleftrightarrow \sin(\pi x_0) = \cos (\pi x_0)$. And of course, $\sin(\pi x_0) = \cos (\pi x_0) = \frac{\sqrt{2}}{2}$ at $x_0 = \frac{1}{4}$. This gives $\left|f^{(4)}\left(\frac{1}{4}\right)\right|=\pi^4\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right)=\pi^4 \cdot \sqrt{2}$ .