Constructing a Taylor Series

Key Questions

  • One of the benefits of a Taylor series is the ease of differentiation since it can be done term by term. So, the Talyor series

    #f(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}(x-a)^n#

    can be differentiated as

    #f'(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}[(x-a)^{n}]'#

    by Power Rule,

    #=sum_{n=1}^infty{f^{(n)}(a)}/{n!}[n(x-a)^{n-1}]#

    (Note: #n# starts from #1# since the term is zero when #n=0#)
    by simplifying a little further,

    #=sum_{n=1}^infty{f^{(n)}(a)}/{(n-1)!}(x-a)^{n-1}#

    I hope that this was helpful.

  • Let us solve #y''+y=0# by Power Series Method.

    Let #y=sum_{n=0}^inftyc_nx^n#, where #c_n# is to be determined.

    By taking derivatives,

    #y'=sum_{n=1}^inftync_nx^{n-1} Rightarrow y''=sum_{n=2}^inftyn(n-1)c_nx^{n-2}#

    We can rewrite #y''+y=0# as

    #sum_{n=2}^inftyn(n-1)c_nx^{n-2}+sum_{n=0}^inftyc_nx^n=0#

    by shifting the indices of the first summation by 2,

    #Rightarrow sum_{n=0}^infty(n+2)(n+1)c_{n+2}x^n+sum_{n=0}^inftyc_nx^n=0#

    by combining the summations,

    #Rightarrow sum_{n=0}^infty[(n+2)(n+1)c_{n+2}+c_n]x^n=0#,

    #Rightarrow (n+2)(n+1)c_{n+2}+c_n=0#

    #Rightarrow c_{n+2}=-{c_n}/{(n+2)(n+1)}#

    Let us look at even coefficients.

    #c_2={-c_0}/{2cdot1}=-c_0/{2!}#
    #c_4={-c_2}/{4cdot3}={-1}/{4cdot3}cdot{-c_0}/{2!}=c_0/{4!}#
    #c_6={-c_4}/{6cdot5}={-1}/{6cdot5}cdot c_0/{4!}=-{c_0}/{6!}#
    .
    .
    .
    #c_{2n}=(-1)^n{c_0}/{(2n)!}#

    Let us look at odd coefficients.

    #c_3={-c_1}/{3cdot2}=-{c_1}/{3!}#
    #c_5={-c_3}/{5cdot4}={-1}/{5cdot4}cdot{-c_1}/{3!}={c_1}/{5!}#
    #c_7={-c_5}/{7cdot6}={-1}/{7cdot6}cdot{c_1}/{5!}=-{c_1}/{7!}#
    .
    .
    .
    #c_{2n+1}=(-1)^n{c_1}/{(2n+1)!}#

    Hence, the solution can be written as:

    #y=sum_{n=0}^inftyc_nx^n#

    by splitting into even terms and odd terms,

    #=sum_{n=0}^inftyc_{2n}x^{2n}+sum_{n=0}^inftyc_{2n+1}x^{2n+1}#

    by pluggin in the formulas for #c_{2n}# and #c_{2n+1}# we found above,

    #=c_0sum_{n=0}^infty(-1)^n{x^{2n}}/{(2n)!}+c_1 sum_{n=0}^infty(-1)^n{x^{2n+1}}/{(2n+1)!}#

    by recognizing the power series,

    #=c_0 cosx+c_1 sinx#


    I hope that this was helpful.

  • The Taylor series of a function is a power series, all of whose derivatives match their corresponding derivatives of the function.

    Let us derive the Taylor series of a function #f(x)#, centered at #c#.

    Let

    #f(x)=sum_{n=0}^infty a_n(x-c)^n#

    #=a_0+a_1(x-c)+a_2(x-c)^2+cdots#,

    where coefficients #a_1, a_2, a_3,...# are to be determined.

    By taking the derivatives,

    #f'(x)=a_1+2a_2(x-c)+3a_3(x-c)^2+cdots#

    #f''(x)=2a_2+3cdot2 a_3(x-c)+4cdot3 a_4(x-c)^2+cdots#

    #f'''(x)=3cdot2 a_3+4cdot3cdot2a_4(x-c)+5cdot4cdot3a_5(x-c)^2+cdots#
    .
    .
    .

    By plugging in #x=c#,

    #f(c)=a_0=0! cdot a_0#

    #f'(c)=a_1=1! cdot a_1#

    #f''(c)=2a_2=2! cdot a_2#

    #f'''(c)=3cdot2 a_3=3! cdot a_3#
    .
    .
    .
    #f^{(n)}(c)=n! cdot a_n#

    By dividing by #n!#,

    #a_n={f^{(n)}(c)}/{n!}#

    Hence, we have the Taylor series of #f(x)#, centered at #c#

    #f(x)=sum_{n=0}^infty{f^{(n)}(c)}/{n!}(x-c)^n#.


    I hope that this was helpful.

Questions