Power Series Expansion: a review

 

f(x) = n=0S∞ f(n)(a) (x-a)n / n! = f(a) + f '(a) (x-a) + f ''(a) (x-a)2 / 2 + f '''(a) (x-a)3 / 6 + ...

 

            where f ' = df/dx; f '' = d2f/dx2; etc and n! = n * (n-1) * (n-2) * ... * 1

 

The above expansion works exactly for any power series.

Example: f(x) = 5x3 + 8x2 In the expansion if we let a=0, then:

            f(x) = 5(x)3 + 8(x)2 so f(0) = 0

            f '(x) = 15(x)2 + 16(x) so f '(0) = 0

            f ''(x) = 30(x) + 16 so f ''(0) = 16

            f '''(x) = 30 so f '''(0) = 30

            f ''''(x) = 0 and all higher derivatives are also zero.

                        Thus the expansion becomes:

            f(x) = 0 + 0 (x-0) + 16 (x-0)2 / 2 + 30 (x-0)3 / 6 + 0 + 0 + 0 ... = 8x2 + 5x3 .

 

For other than purely power functions, this approximation works well as long as x is close to a.

 

Example #1: f(x) = [1-x]0.5 = Ö [1-x]

            Let's look at the answer close to zero (so a=0):

            f(x) = Ö [1-x] so f(0) = 1

            f '(x) = [1-x]-0.5 (0.5) (-1) so f '(0) = -0.5

            f ''(x) = [1-x]-1.5 (-0.5) (-0.5) (-1) so f ''(0) = -0.25

            f '''(x) = [1-x]-2.5 (-.25) (-1.5) (-1) so f '''(0) = -0.375

                        Thus the expansion becomes:

Ö [1-x] = 1 + -0.5 (x-0) - 0.25 (x-0)2 / 2 + -0.375 (x-0)3 / 6 + ... = 1 - x/2 - x2/8 - 3x3/48 - ...

            For x = 0.1, the "true" answer is f(0.1) = Ö [1-.1] = Ö .9 = 0.94868

            The approximation to zero order gives: f(0.1) = 1 (off by .05132, or 5.41% high)

            The approximation to first order gives: f(0.1) = 1 - (.1)/2 = 0.95 (off by .00132, or 0.139% high)

            The approximation to second order gives: f(0.1) = 1 - (.1)/2 - (.01)/8 = .94875 (off by .00007, or 0.0070% high)

 

Example #2: f(q) = sin(q )

            Let's again look at the answer close to zero (so a=0)

            f(q) = sin(q) so f(0) = 0

            f '(q) = cos(q) so f '(0) = 1

            f ''(q ) = -sin(q ) so f ''(0) = 0

            f '''(q ) = -cos(q ) so f '''(0) = -1

                        Thus the expansion becomes:

            sin(q) = 0 + 1(q-0) + 0(q-0)2/2 + -1(q-0)3/6 + ... = q - q3/6 + ...

            For q =10o = 0.174533 radians, the "true" answer is f(10o) = sin(10o) = 0.173648

            The approximation to zero order gives: f(0.174533) = 0   (off by .173648)

            The approximation to first order gives: f(0.174533) = 0 + (0.174533) = 0.174533   (off by .000885, or 0.51% high)

            The approximation to third order gives:

                        f(0.174533) = 0 + (0.174533) + 0 - (0.174533)3/6 = 0.173647  (off by .000001, or 0.00078% low)

Homework Problem Problem #4: Derive the first 3 non-zero terms of the Taylor series expansion for:
a) cos(x) ;
b) 1/[1+x] ;
c) eax .
            Be sure to look at (at least) two limiting cases for each, and demonstrate with at least two examples how close the first three terms of the series come to giving the "correct" answer.

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