Given this function : s ( x ) = ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! , its derivative c(x)=s'(x). How to show that c'(x)=-s(x) ?This question is related to Relationship Between Sine as a Series and Sine in TrianglesIs c'(x) the double of derivative s ( x ) = ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! ? So I take the derivative of ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! to arrive at c(x) and then take derivative of c(x) to arrive at c'(x) ?

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Given this function :   s  (  x  )  =      ∑          k      =      0        ∞    (  −  1      )    k              x              2        k        +        1                    (      2      k      +      1      )      !      , its derivative c(x)=s&#039;(x). How to show that c&#039;(x)=-s(x) ?This question is related to Relationship Between Sine as a Series and Sine in TrianglesIs c&#039;(x) the double of derivative   s  (  x  )  =      ∑          k      =      0        ∞    (  −  1      )    k              x              2        k        +        1                    (      2      k      +      1      )      !       ? So I take the derivative of       ∑          k      =      0        ∞    (  −  1      )    k              x              2        k        +        1                    (      2      k      +      1      )      !       to arrive at c(x) and then take derivative of c(x) to arrive at c&#039;(x) ?

Tianna Deleon · Accepted Answer

Yes this is exactly the idea.                    s        (        x        )        =                  ∑                      k            =            0                    ∞                (        −        1                  )          k                                      x                          2              k              +              1                                            (            2            k            +            1            )            !                          .            Hence we have that:                    c        (        x        )        =                  s          ′                (        x        )        =                  ∑                      k            =            0                    ∞                (        −        1                  )          k                                      (            2            k            +            1            )                          x                              2                k                                                          (            2            k            +            1            )            !                          =                  ∑                      k            =            0                    ∞                (        −        1                  )          k                                      x                          2              k                                            (            2            k            )            !                          .            Now taking yet another derivative we get:                              c          ′                (        x        )                            =                  ∑                      k            =            1                    ∞                (        −        1                  )          k                                      2            k                          x                              2                k                −                1                                                          (            2            k            )            !                          =                  ∑                      k            =            0                    ∞                (        −        1                  )                      k            +            1                                                2            (            k            +            1            )                          x                              2                (                k                +                1                )                −                1                                                          (            2            (            k            +            1            )            )            !                                                            =        −                  ∑                      k            =            0                    ∞                (        −        1                  )          k                                      x                          2              k              +              1                                            (            2            k            +            1            )            !                          =        −        s        (        x        )        .

Given this function : s ( x ) = <munderover> ∑ <mrow c

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