Find the value of m for the following lim n → ∞ ∏ k = 3 n ( 1 − tan 4 ⁡ π 2 k ) = π 3 m

Question

Find the value of m for the following      lim          n      →      ∞            ∏          k      =      3        n        (    1    −          tan      4        ⁡          π              2        k              )    =            π      3        m

EreneDreaceaw · Accepted Answer

Long, but detailed      ∏          k      =      3        n        (    1    −          tan      4        ⁡          π              2        k              )    =      ∏          k      =      3        n        (    1    −          tan      2        ⁡          π              2        k              )        (    1    +          tan      2        ⁡          π              2        k              )    =        ∏          k      =      3        n        (                            cos          2                ⁡                              π                          2              k                                      −                  sin          2                ⁡                              π                          2              k                                                            cos          2                ⁡                              π                          2              k                                            )        (                            cos          2                ⁡                              π                          2              k                                      +                  sin          2                ⁡                              π                          2              k                                                            cos          2                ⁡                              π                          2              k                                            )        =          cos      ⁡              2        x            =              cos        2            ⁡              x            −              sin        2            ⁡              x                    ∏          k      =      3        n              cos      ⁡                        π                      2                          k              −              1                                                          cos        4            ⁡                        π                      2            k                                =            cos      ⁡                        π          4                            cos      ⁡                        π                      2            n                                ⋅      ∏          k      =      3        n        1                  cos        3            ⁡                        π                      2            k                                =              cos      ⁡                        π          4                            cos      ⁡                        π                      2            n                                ⋅      ∏          k      =      3        n                      sin        3            ⁡                        π                      2            k                                                        sin          3                ⁡                              π                          2              k                                          ⋅              cos        3            ⁡                        π                      2            k                                    =          sin      ⁡              2        x            =      2      sin      ⁡              x            cos      ⁡              x                          cos      ⁡                        π          4                            cos      ⁡                        π                      2            n                                ⋅      ∏          k      =      3        n                      sin        3            ⁡                        π                      2            k                                              1                  2          3                    ⋅              sin        3            ⁡                        π                      2                          k              −              1                                            =      2          3      (      n      −      2      )        ⋅            cos      ⁡                        π          4                            cos      ⁡                        π                      2            n                                ⋅      ∏          k      =      3        n                      sin        3            ⁡                        π                      2            k                                              sin        3            ⁡                        π                      2                          k              −              1                                            =        2          3      (      n      −      2      )        ⋅            cos      ⁡                        π          4                            cos      ⁡                        π                      2            n                                ⋅                    sin        3            ⁡                        π                      2            n                                              sin        3            ⁡                        π          4                      =      2          3      n      −      5        ⋅      1          cos      ⁡                        π                      2            n                                ⋅      sin    3    ⁡            π              2        n              =              π      3              2      5        ⋅      1          cos      ⁡                        π                      2            n                                ⋅            (                        sin          ⁡                                    π                              2                n                                                              π                      2            n                              )        3    →            π      3              2      5        ,  n  →  ∞given that  cos  ⁡            π              2        n              →  cos  ⁡      0    =  1  ,  n  →  ∞and            sin      ⁡                        π                      2            n                                      π              2        n              →  1  ,  n  →  ∞

Dwllane4 · Accepted Answer

It is possible to find a closed form for your product, if fact we have for any   k  ≥  3                    1        −                  tan                      4                          ⁡                              (                          π                              2                                  k                                                      )                                              =                              1            −                          tan                              2                                      ⁡                                          (                                  π                                      2                                          k                                                                      )                                                                        cos                              2                                      ⁡                                          (                                  π                                      2                                          k                                                                      )                                                                                      =                              cos            ⁡                                          (                                  π                                      2                                          k                      −                      1                                                                      )                                                                        cos                              4                                      ⁡                                          (                                  π                                      2                                          k                                                                      )                                                                        1        −                  tan                      4                          ⁡                              (                          π                              2                                  k                                                      )                                              =                              8                          sin                              3                                      ⁡                                          (                                  π                                      2                                          k                                                                      )                                                                        sin                              3                                      ⁡                                          (                                  π                                      2                                          k                      −                      1                                                                      )                                                    ×                              cos            ⁡                                          (                                  π                                      2                                          k                      −                      1                                                                      )                                                          cos            ⁡                                          (                                  π                                      2                                          k                                                                      )                                                        Thus, if   n  ≥  3, then :                              ∏                      k            =            3                                n                                                (            1            −                          tan                              4                                      ⁡                                          (                                  π                                      2                                          k                                                                      )                                      )                                              =                  8                      n            −            2                                                (                          ∏                              k                =                3                                            n                                                                                                          sin                                          3                                                        ⁡                                                            (                                              π                                                  2                                                      k                                                                                              )                                                                                                            sin                                          3                                                        ⁡                                                            (                                              π                                                  2                                                      k                            −                            1                                                                                              )                                                                                            )                                3                                    (                      ∏                          k              =              3                                      n                                                                          cos                ⁡                                                      (                                          π                                              2                                                  k                          −                          1                                                                                      )                                                                              cos                ⁡                                                      (                                          π                                              2                                                  k                                                                                      )                                                                                )                                                  =                  2                      3            n            −            5                                                              sin                              3                                      ⁡                                          (                                  π                                      2                                          n                                                                      )                                                          cos            ⁡                                          (                                  π                                      2                                          n                                                                      )

Find the value of m for the following <munder> <mo form="prefix">lim <mrow class="MJX

Answered question

Answer & Explanation

New Questions in Precalculus