Understanding an Approximation: 1 − ( 1 − μ n ) μ − 1 + ( μ − 1 ) 2 ≈ μ 3 n .

Question

Understanding an Approximation:   1  −            (      1      −              μ        n            )              μ      −      1      +      (      μ      −      1              )        2              ≈            μ      3        n    .

Tatairfzk · Accepted Answer

Note that   μ  −  1  +  (  μ  −  1      )          2        =  μ  (  μ  −  1  )  .. Using the approximation   (  1  +  x      )          α        ≈  1  +  α  x for small   x, we obtain            (      1      −              μ        n            )              μ      −      1      +      (      μ      −      1              )                  2                      ≈  1  −                    μ                  2                    (      μ      −      1      )        n    .Hence  1  −            (      1      −              μ        n            )              μ      −      1      +      (      μ      −      1              )                  2                      ≈                    μ                  2                    (      μ      −      1      )        n    =                    μ                  3                    −              μ                  2                      n    .If we can assume that   μ is sufficiently large (and indeed we assume that   p  &amp;gt;  3            log      ⁡              n              n   so that   μ  =  n  p  &amp;gt;  3  log  ⁡      n  ), then the       μ          2       term pales in comparison to       μ          3      , so the approximation                    μ                  3                    −              μ                  2                      n    ≈            μ              3              n  can also be made. Another way to view this last approximation is to say that   μ  −  1  ≈  μ for large   μ so that       μ          2        (  μ  −  1  )  ≈      μ          2        (  μ  )  =      μ          3        ..

Understanding an Approximation: 1 − <msup> <mrow> (

Answered question

Answer & Explanation

New Questions in Calculus 1