I would like to calculate the Riemann sum of sin ⁡ ( x ). Fun starts here: R = π n ∑ j = 1 n sin ⁡ ( π n ⋅ j )

Question

I would like to calculate the Riemann sum of   sin  ⁡  (  x  ). Fun starts here:  R  =      π    n        ∑          j      =      1        n    sin  ⁡      (          π      n        ⋅    j    )

owerswogsnz · Accepted Answer

There&#039;s a way to find an expression for the sum      ∑          j      =      1              n        sin  ⁡      (    j    θ    )  by considering instead the geometric sum  1  +  z  +      z    2    +  ⋯  +      z    n    =                    z                  n          +          1                    −      1              z      −      1              for           z      ≠      1      in combination with Euler&#039;s formula by taking   z  =      e          i      θ        =  cos  ⁡      θ    +  i  sin  ⁡      θ   and also using De Moivre&#039;s formula. Then you can find that      ∑          j      =      1              n        sin  ⁡      (    j    θ    )    =            cos      ⁡                        (                      θ            2                    )                    −      cos      ⁡                        (          (          n          +                      1            2                    )          θ          )                            2      sin      ⁡                        (                      θ            2                    )                    This is a standard exercise in most complex analysis books or actually any book that introduces complex numbers. In your case you just have to take   θ  =      π    n

Agostarawz · Accepted Answer

Use  2  sin  ⁡      (          π              2        n              )    sin  ⁡      (          π      n        ⋅    j    )    =  cos  ⁡      (          π              2        n              (    2    j    −    1    )    )    −  cos  ⁡      (          π              2        n              (    2    j    +    1    )    )  Thus the sum telescopes       ∑          j      =      1        n        (    g    (    j    )    −    g    (    j    +    1    )    )    =  g  (  1  )  −  g  (  n  +  1  )      R    n    =      π    n        ∑          j      =      1        n    sin  ⁡      (          π      n        ⋅    j    )    =      π          2      n      sin      ⁡              (                  π                      2            n                          )                  (    cos    ⁡          (              π                  2          n                    )        −    cos    ⁡          (              π                  2          n                    (      2      n      +      1      )      )        )    =      π    n    ⋅      1          tan      ⁡              (                  π                      2            n                          )            The large n limit is easy:      lim          n      →      ∞            R    n    =  2      lim          n      →      ∞            π          2      n        ⋅      1          tan      ⁡              (                  π                      2            n                          )              =  2      lim          x      →      0            x          tan      ⁡      (      x      )        =  2

I would like to calculate the Riemann sum of sin ⁡ ( x ) . Fun starts he

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