Lagrange Form of the Remainder Term in a Taylor Series

Key Questions

  • By taking the derivatives,

    f(x)=e^{4x}f(x)=e4x
    f'(x)=4e^{4x}
    f''(x)=4^2e^{4x}
    .
    .
    .
    f^{(n+1)}(x)=4^{n+1}e^{4x}

    So,
    R_n(x;3)={f^{(n+1)}(z)}/{(n+1)!}(x-3)^{n+1}={4^{n+1}e^{4z}}/{(n+1)!}(x-3)^{n+1},
    where z is between x and 3.

  • Remainder Term of Taylor Series

    R_n(x;c)={f^{(n+1)}(z)}/{(n+1)!}(x-c)^{n+1},

    where z is a number between x and c.


    Let us find R_3(x;1) for f(x)=sin2x.

    By taking derivatives,

    f'(x)=2cos2x
    f''(x)=-4sin2x
    f'''(x)=-8cos2x
    f^{(4)}(x)=16sin2x

    So, we have

    R_3(x;1)={16sin2z}/{4!}(x-1)^4,

    where z is a number between x and 1.


    I hope that this was helpful.

  • Taylor remainder term

    R_n(x;c)={f^{(n+1)}(z)}/{(n+1)!}(x-c)^{n+1},

    where z is between x and c.

Questions