The sequence n sin ⁡ ( 4 π 2 n 2 + x 2 ) converges on compacts

Question

The sequence   n  sin  ⁡  (      4          π      2              n      2        +          x      2        ) converges on compacts

Cahokiavv · Accepted Answer

Note that we have for any fixed x                    sin        ⁡        (                  4          π                      n            2                    +                      x            2                          )                            =        sin        ⁡                  (          2          π          n                      1            +                                          (                                  x                                      2                    π                    n                                                  )                            2                                )                                                                =        sin        ⁡                  (          2          π          n                      (            1            +                                          x                2                                            8                π                                  n                  2                                                      +            O            (                          n                              −                4                                      )            )                    )                                                                =                              x            2                                4            π            n                          +        O                  (                      n                          −              3                                )                    whence multiplying by n and letting   n  →  ∞ yields      lim          n      →      ∞        n  sin  ⁡  (      4    π          n      2        +          x      2        )  =            x      2              4      π

Bailee Short · Accepted Answer

Concretely, we have      f    n    (  x  )  =  n    sin  ⁡            (        2  π  n        1    +                  x        2                    4                  π          2                          n          2                                )        .Using the binomial expansion,      1    +                  x        2                    4                  π          2                          n          2                      =  1  +            x      2              8              π        2                    n        2              +  o  (      n          −      4        )  .The sine is  sin  ⁡  t  =  t  −  o  (      t    3    )  .Then                              f          n                (        x        )                            =        n                sin        ⁡                              (                          2        π        n        +                              2            π            n                          x              2                                            8                          π              2                                      n              2                                      +        o        (                  n                      −            3                          )                              )                                                            =        n                sin        ⁡                              (                                                2            π            n                          x              2                                            8                          π              2                                      n              2                                      +        o        (                  n                      −            3                          )                              )                                                            =                              2            π                          n              2                                      x              2                                            8                          π              2                                      n              2                                      −        o        (                  n                      −            2                          )                                          =                              x            2                                4            π                          −        o        (                  n                      −            2                          )                                                    →                      n            →            ∞                                                          x            2                                4            π                              When we consider x only on some compact set, because it&#039;s bounded the convergence can be made uniform, as the estimates will not depend on x (hidden, above in the terms   o  (      n          −      4        ), etc.)

The sequence n sin ⁡ ( <msqrt> 4 π

Answered question

Answer & Explanation

New Questions in Calculus 1