# Orders of convergence

• Oct 3rd 2012, 01:24 AM
alanaj5
Orders of convergence
Hi Guys,

I have a small question on orders of convergence.
I'm happy with the idea that for $\displaystyle$N\in\mathbb{N}$$if, \displaystyle \text{error} \le C_1N^{-\alpha}+C_2N^{-\beta}$$, where $\displaystyle$\alpha<\beta$$, \displaystyle \alpha, \beta \in\mathbb{R}$$, $\displaystyle$\alpha,\beta>0$$, \displaystyle C_1, C_2 \in\mathbb{R}$$ are constants
$\displaystyle$\Rightarrow \text{error} \le (C_1+C_2)N^{-\alpha}$$or \displaystyle \text{error} \le C_1N^{\alpha}+C_2N^{\beta}$$, where $\displaystyle$\alpha<\beta$$, \displaystyle \alpha, \beta \in\mathbb{R}$$, $\displaystyle$\alpha,\beta>0$$,\displaystyle C_1, C_2 \in\mathbb{R}  are constants \displaystyle \Rightarrow \text{error} \le (C_1+C_2)N^{\beta}$$

but for
$\displaystyle$\text{error} \le C_1N^{-\alpha}+C_2N^{\beta}$$, where \displaystyle \alpha, \beta \in\mathbb{R}$$, $\displaystyle$\alpha,\beta>0$$,\displaystyle C_1, C_2 \in\mathbb{R}  are constants. is this the right outcome \displaystyle \text{error} \le (C_1+C_2)N^{\beta}$$ ?
Other than when 0 < N < 1 is there a time when $\displaystyle$N^{-\alpha}$$would win? Thanks :-) • Oct 3rd 2012, 01:50 AM alanaj5 Re: Orders of convergence I think I might have just figured out what I needed to know. I think the size of the coefficients \displaystyle C_1$$ and $\displaystyle$C_2 play an important role here as to which rate dominates the other. Please let me know if you think I'm wrong or I should be considering other factors as well.

Thanks.