Constructing a Maclaurin Series

Key Questions

  • How do you find the Maclaurin series of f(x)=cosh(x) ?

    f(x)=coshx=sum_{n=0}^infty{x^{2n}}/{(2n)!}

    Let us look at some details.

    We already know

    e^x=sum_{n=0}^infty x^n/{n!}

    and

    e^{-x}=sum_{n=0}^infty {(-x)^n}/{n!},

    so we have

    f(x)=coshx=1/2(e^x+e^{-x})

    =1/2(sum_{n=0}^infty x^n/{n!}+sum_{n=0}^infty (-x)^n/{n!})

    =1/2sum_{n=0}^infty( x^n/{n!}+(-x)^n/{n!})

    since terms are zero when n is odd,

    =1/2sum_{n=0}^infty{2x^{2n}}/{(2n)!}

    by cancelling out 2's,

    =sum_{n=0}^infty{x^{2n}}/{(2n)!}

    Wataru · 26 · Sep 25 2014
  • How do you use a Maclaurin series to find the derivative of a function?

    The MacLaurin series of a function f is a power series of the form:

    sum_(n=0)^(oo) a_n x^n

    With the coefficients a_n given by the relation

    a_n=(f^((n))(0))/(n!),

    where f^((n))(0) is the nth derivative of f(x) evaluated at x=0.

    Therefore,

    f^((n))(0)=a_n n!

    This reasoning can be extended to Taylor series around x_0, of the form:

    sum_(n=0)^(oo) c_n (x-x_0)^n

    With the relation

    f^((n))(x_0)=c_n n!

    It's important to emphasize that the function nth derivative of f (that is, f^((n)) (x)) cannot be obtained directly from the Taylor/MacLaurin series (only it's value on the point around wich the series is constructed).

    Vinicius M. G. Silveira · 1 · Dec 16 2014

Questions

Power Series