# Meaning of algebraic decay

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.

-

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 $\beta=\alpha-1$.
Do you mean $n$ instead of $t$? How do I prove this formally in my case? –  Shahab Feb 13 '13 at 13:47
$A_n$ has algebraic decay of order $\beta$ if there exists a constant $C>0$ such that $|A_n|\le C\,n^{-\beta}$. –  Julián Aguirre Feb 13 '13 at 14:57
You misunderstand. I have worked out your answer to $(\alpha-1)h((n+1)h-\tau)^{\alpha-2}$ which proves the result. However this follows from substitution directly in the kernel and does not tell me the need for $z_i^*$. –  Shahab Feb 13 '13 at 15:04