find the Maclaurin series for ln ⁡ 1 + x 1 − x . I found it is x 2 n + 1 2 n + 1 ,is it correct? I got the 1 / 2 outside and solved the maclaurin for the normal log. If it is okay how can I find a relationship between this and the maclaurin series artg.I know it is ( − 1 ) n n of my series but how can I write it?

Question

find the Maclaurin series for   ln  ⁡                    1        +        x                    1        −        x            . I found it is             x              2        n        +        1                    2      n      +      1       ,is it correct? I got the   1      /    2 outside and solved the maclaurin for the normal log. If it is okay how can I find a relationship between this and the maclaurin series artg.I know it is   (  −  1      )    n  n of my series but how can I write it?

Ashlynn Stephens · Accepted Answer

The Taylor series for   log  ⁡                    1        +        x                    1        −        x             at   0 is       ∑          n      =      0        ∞              x              2        n        +        1                    2      n      +      1      , because                    log        ⁡                                            1              +              x                                      1              −              x                                                          =                  1          2                log        ⁡                  (                                    1              +              x                                      1              −              x                                )                                                  =                  1          2                          (          log          ⁡          (          1          +          x          )          −          log          ⁡          (          1          −          x          )          )                                                  =                  1          2                          (          x          −                                    x              2                        2                    +                                    x              3                        3                    −                                    x              4                        4                    +          ⋯          −                      (            −            x            −                                          x                2                            2                        −                                          x                3                            3                        −                                          x                4                            4                        −            ⋯            )                    )                                                  =        x        +                              x            3                    3                +                              x            5                    5                +        ⋯            And the Taylor series for arctan  arctan  ⁡  (  x  ) at   0 is quite similar:       ∑          n      =      0        ∞    (  −  1      )    n              x              2        n        +        1                    2      n      +      1      . That&#039;s because      arctan    ′    ⁡  (  x  )  =      1          1      +              x        2              =  1  −      x    2    +      x    4    −      x    6    +  ⋯

Crancichhb · Accepted Answer

log  ⁡      (                            1          +          x                          1          −          x                      )    =      1    2    log  ⁡      (                  1        +        x                    1        −        x              )    =      1    2        (    log    ⁡    (    1    +    x    )    −    log    ⁡    (    1    −    x    )    )    =      1    2        (          ∑              n        ≥        1              (    −    1          )              n        +        1                            x        n            n        −          ∑              n        ≥        1              (    −    1    )                  x        n            n        )    =      1    2        (          ∑              n        ≥        1              (    −    1          )              n        +        1                            x        n            n        +          ∑              n        ≥        1                            x        n            n        )    =      1    2        (          ∑              n        ≥        1              (    (    −    1          )              n        +        1              +    1    )                  x        n            n        )    =      1    2        (          ∑              n        ≥        1              2                  x                  2          n          −          1                            2        n        −        1              )    =      ∑          n      ≥      1                  x              2        n        −        1                    2      n      −      1        =:  f  (  x  )And the arctan is:      tan          −      1        ⁡  (  x  )  =      ∑          n      ≥      1        (  −  1      )          n      +      1                  x              2        n        −        1                    2      n      −      1        =:  g  (  x  )One relation can be:      tan          −      1        ⁡  (  x  )  =  −  log  ⁡      (                            1          +          i          x                          1          −          i          x                      )  Or alternatively:  g  (  x  )  =  −  f  (  i  x  )

Find the Maclaurin series for ln sqrt(frac(1+x)(1-x))

Answered question

Answer & Explanation

New Questions in Calculus 2