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 #n#th 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 #n#th 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