$$y(t) = \frac{1}{B(\alpha)-\alpha+1}[(1-\alpha)t\ E_{\alpha,2} (\frac{-\alpha}{1-\alpha}t^\alpha) + \alpha t^{\alpha + 1}\ E_{\alpha, \alpha+2}(\frac{-\alpha}{1-\alpha}t^\alpha)]$$
(Assume B($\alpha$) = 1, and the E's are Mittag-Leffler functions. $\alpha$ can be either 0.5, 0.7, 0.9, or 0.99.)
My team was assigned to code this equation into MATLAB. He also gave us this numerical approximation of the solution of the above formula:
$$y_{i+1} = y_0 + \frac{1-\alpha}{B(\alpha)}\ f(t_{i},y_{i}) + \frac{\alpha h_N}{B(\alpha)Γ(\alpha+1)} \ \sum_{j=0}^i b_{i+1, j} \ f(t_i,y_j)$$
with
$$b_{i+1,j} = -(i-j)^\alpha + (i - j + 1)^\alpha$$
(B($\alpha$) is still 1, $f(t_i,y_i)$ = $t_i - y(t_i)$ with $y(0) = 0$, $h_N$ is the step size, which is equal to 0.01.)
I'm very sure that my professor wrote the exact equation wrong somehow. When I compare the graph of the exact solution and the numerical approximation, I can see that the numerical approximation is becoming more curved as $\alpha$ increases, which he confirmed was correct. He also told me that for any of the $\alpha$ values mentioned above, the graph of the exact solution and the numerical approximation should be an exact match. However, the exact solution is a straight line no matter the value of $\alpha%$. I rechecked the code multiple times but I am sure of it. Here are the graphs of both at $\alpha$ = 0.99 for reference.
A friend of mine told me that $\alpha t^{\alpha + 1}\ E_{\alpha, \alpha+2}(\frac{-\alpha}{1-\alpha}t^\alpha)$ is causing this linear behavior in the output, even though each component ($\alpha t^{\alpha + 1}$ alone and $E_{\alpha, \alpha+2}(\frac{-\alpha}{1-\alpha}t^\alpha)$ alone) is non-linear on its own. Is there any possible mistake my professor could have done in copying these 2 parts down on paper so that the function is acting linearly even though it shouldn't?
Here is a picture of the full assignment, in case anyone wants to check it out.
Any tips would be greatly appreciated.
