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Proof involving some inequality

  1. Dec 15, 2007 #1
    Difficulty with some inequality

    Suppose that one has three numbers, [tex]v_j^0[/tex] for j=1, 2, and 3, and let their absolute values be less than or equal to 1. Suppose that one has nine continuous functions [tex]c_{jk}[/tex] all of which are bounded, and thus all of which are bounded by an amount [tex]\tfrac{k}{3}[/tex].


    [tex]v_j^{(1)}(s)=v_j^0+\int_0^s \sum_{k=1}^3 c_{jk}(\sigma) v_k^0 d\sigma[/tex]


    [tex]v_j^{(n)}(s)=v_j^0+\int_0^s \sum_{k=1}^3 c_{jk}(\sigma) v_k^{(n-1)}(\sigma) d\sigma[/tex]

    It is perfectly understandable why

    [tex]|v_j^{(1)}-v_j^0| \le ks[/tex],

    but not so clear why

    [tex]|v_j^{(n)}-v_j^{(n-1)}| \le \frac{k^n s^n}{n!}[/tex].

    Somehow, one is supposed to be able to divide by n, but I do not see how is able to.
    Last edited: Dec 15, 2007
  2. jcsd
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