∏ i = 1 k − 1 ( 1 − i 2 N ) ≈ 1 − ( k 2 ) 2 N what assumption are needed for this approximation to be useful.

Question

∏          i      =      1              k      −      1            (    1    −          i              2        N              )    ≈  1  −                                          (                                    k          2                                      )                                      2      N      what assumption are needed for this approximation to be useful.

Abigail Palmer · Accepted Answer

Let   P be the product; then  log  ⁡      P    =      ∑          i      =      1              k      −      1        log  ⁡            (      1      −              i                  2          N                    )      When   N is sufficiently large, then we may use the approximation  log  ⁡      (    1    −    y    )    ≈  −  y and get  log  ⁡      (    1    −    y    )    ≈  −  y  log  ⁡      P    ≈  −      1          2      N            ∑          i      =      1              k      −      1        i  =  −      1          2      N                  k      (      k      −      1      )        2    =  −      1          2      N                          (                    k      2                      )            Thus,  P  ≈      e          −              1                  2          N                                                  (                                    k          2                                      )                                ≈  1  −      1          2      N                          (                    k      2                      )            The above holds when      1    2        1          4              N        2                                          (                            k        2                              )                      2  is sufficiently small.For the 1st &quot;sufficient&quot;, we need      1    2        1          4              N        2                  ∑          i      =      1              k      −      1            i    2    =      1          8              N        2                        (      k      −      1      )      k      (      2      k      −      1      )        6    ≈            k      3              24              N        2            or   k  &amp;lt;      C    1        N          2              /            3      , where       C    1   is a small number. For the second sufficient, we need       k    4        /    (  32      N    2    ) to be small, or   k  &amp;lt;      C    2        N  . For the estimate to hold overall, we need the smaller of the two estimates, or that   k  &amp;lt;      C    2        N  .

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New Questions in Calculus 1