Constructing a Taylor Series

Key Questions

  • How do you use a Taylor series to find the derivative of a function?

    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.

    Wataru · · Sep 27 2014
  • How do you use a Taylor series to solve differential equations?

    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.

    Wataru · · Oct 17 2014
  • What is a Taylor series?

    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.

    Wataru · · Oct 18 2014

Questions

Power Series