Error of the Summed Midpoint-Rule The summed midpoint-rule is defined as:
$(Q_{n})^{0} = \sum_{i=0}^{N-1}(x_{i+1}-x_{i})f(\frac{x_i+x_{i+1}}{2})$
Now the problem is the following: For $f \in C^0([0,1]),  Q_h[f] \\$ with h= 1/N is the summed midpoint rule with $N \in \mathbb N$ with equidistant sampling points in the interval [0,1]. Now I want to show that for $f(x) = x^{\alpha}$ with $0< \alpha < 1$:
$$ | \int_{0}^1f(x)dx-Q_h[f]| = O(h^{\alpha+1})$$, whereas the "O" stands for the corresponding Landau-Notation.
How can I solve this problem? I find it quite hard..
 A: Ok, you were right, it's much trickier than I thought! I can only show that it is a $\mathcal{O}(h^\alpha)$...
Let's assume $c_i = ih + \frac{h}{2}$, and begin with:
$$
\int_0^1 f(u)du - h \sum_{i=0}^{N-1} f(c_i) = \sum_{i=0}^{N-1} \int_{ih}^{(i+1)h} f(u)du - h f(c_i)
$$
This is quite straightforward, I simply split the integral inasmuch pieces as necessary for it to get into the sum. To go further, we will first look at each term of the previous sum for $i>0$, and then take a look at $i=0$ at the end. Assuming $i>0$ implies that $f^{\prime\prime}$ is always bounded on the interval of integration, we will see later why this is useful. Given this, we have:
$$
\int_{ih}^{(i+1)h} f(u)du - h f(c_i) = \int_{ih}^{(i+1)h} \big[ f(u) - f(c_i) - (u-c_i) f^\prime(c_i) \big] du
$$
Indeed, if you verify the equality from right to left, $f^\prime(c_i)$ vanishes because $\frac{(ih + h)^2}{2} -\frac{(ih)^2}{2} = h c_i$. Note that $f^\prime(c_i)$ is always defined because $f^\prime$ is $C^\infty(]0,1])$.
In the previous integrand, we recognize a Taylor-Lagrange expansion of order 2, which we can transform as such:
$$
\forall i > 0,\quad f(u) - f(c_i) - (u-c_i) f^\prime(c_i) = \frac{(u-c_i)^2}{2} f^{\prime\prime}(z_i)
$$
where $z_i\in ]ih,\ (i+1)h[$ for all $i>0$. Integrating each term, we obtain:
$$
\sum_{i=1}^{N-1} \int_{ih}^{(i+1)h} f(u)du - h f(c_i) = 
\sum_{i=1}^{N-1} \frac{ f^{\prime\prime}(z_i) }{ 2 } 
\int_{-\frac{h}{2}}^{\frac{h}{2}} v^2 dv =  \frac{h^3}{24} \sum_{i=1}^{N-1} f^{\prime\prime}(z_i)
$$
where we posed $v = u-c_i$. Finally we can write:
$$
\left| \sum_{i=1}^{N-1} \int_{ih}^{(i+1)h} f(u)du - h f(c_i) \right| \leq \frac{h^3}{24} \big[ (N-1)\sup_{[h,1]} |f^{\prime\prime}| \big]
\leq \frac{h^2}{24} \ \alpha (1-\alpha) h^{\alpha-2} = \mathcal{O}(h^\alpha)
$$
Now, let's not forget the case $i=0$:
$$
\left| \int_0^h f(u)du - h f(c_0) \right| = h^{\alpha+1}\left( \frac{1}{2^\alpha} - \frac{1}{\alpha+1} \right) = \mathcal{O}(h^{\alpha+1})
$$
Unfortunately, this yields an overall precision of $\mathcal{O}(h^{\alpha})$ only, and not $\mathcal{O}(h^{\alpha+1})$ as requested. Maybe it is possible to find a better majoration in the last equation?
