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Thread: Orders of convergence

  1. #1
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    May 2008
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    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 :-)
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  2. #2
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    May 2008
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    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.
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