Maclaurin series for cos ⁡ 2 x − 1 x 2 using Maclaurin series for cos ⁡ 2 x

Question

Maclaurin series for             cos      ⁡      2      x      −      1              x      2       using Maclaurin series for   cos  ⁡  2  x

Cavalascamq · Accepted Answer

Maclaurin series can be treated as infinite polynomials (most of the time), so we have

\cos 2 x - 1 = - \frac{4 x^{2}}{2!} + \frac{16 x^{4}}{4!} - \frac{64 x^{6}}{6!} + \dots

\frac{\cos 2 x - 1}{x^{2}} = - \frac{4}{2!} + \frac{16 x^{2}}{4!} - \frac{64 x^{4}}{6!} + \dots

= - 2 + \frac{2}{3} x^{2} - \frac{4}{45} x^{4} + \dots

Kamila Frye · Accepted Answer

Since                    c        o        s        (        2        x        )        =        1        −                              4                          x              2                                            2            !                          +                              16                          x              4                                            4            !                          −                              64                          x              6                                            6            !                          .        .        .              cos  ⁡  (  2  x  )  −  1  =            4              x        2                    2      !        +            16              x        4                    4      !        −            64              x        6                    6      !        .  .  .            cos      ⁡      (      2      x      )              x      2        =      4          2      !        +            16              x        2                    4      !        −            64              x        4                    6      !        .  .  .  =  2  +            16              x        2                    4      !        −            64              x        4                    6      !        .  .  .

Maclaurin series for frac(cos2x-1)(x^(2)) using Maclaurin series for cos2x

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