I am reading the paper here and I am running into a few roadblocks. One of them was resolved here and now I am stuck at another. (Pg 177)
Suppose $\alpha\in(0,2)$ and $t_i=ih$ for some fixed $h>0$ and every $i$. It can be shown that $\frac{1}{\Gamma(\alpha)}\int_0^{t_n}((t_{n+1}-\tau)^{\alpha-1})-(t_n-\tau)^{\alpha-1})f(\tau,x(\tau))d\tau\\ =\frac{h}{\Gamma(\alpha)(\alpha-1)}\sum_{i=0}^{t_{n-1}}\int_{t_i}^{t_{i+1}}(z_{n-i}^*(\tau))^{\alpha-2}f(\tau,x(\tau))d\tau $
where $z_j^*(\tau)\in (t_{j-1},t_{j+1})$ for all $j$.
This latter sum is labelled as equation 6 in the paper.
What I now don't understand is the next sentence in the paper:
We can note from (6) that the integrations's kernel $((t_{n+1}-\tau)^{\alpha-1})-(t_n-\tau)^{\alpha-1})$ decays algebraically by the order of $2-\alpha.$
Can someone please explain me the meaning of this. I intutively think it means that as $n$ becomes larger this whole thing should go to zero. But what would be a formal mathematical meaning of this?
Sorry for asking so many questions related to the same topic.
$\endgroup$1 Answer
$\begingroup$I would understand that as $n\to\infty$ it decays as a power of $n$, namely as $n^{-(2-\alpha)}$. To prove it, use the mean value theorem to prove the following inequality: if $1\le\beta$ and $0<x<y$, then $$ y^\beta-x^\beta\le\beta(y-x)y^{\beta-1}. $$ A similar inequality holds if $0<\beta<1$. Apply it to your expression with $y=t_{n+1}-\tau$, $x=t_n-\tau$ and $\beta=\alpha-1$.
The expression algebraic decay is used to contra-position to exponential decay.
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