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 – 4.  In the expansion if we let a=0, then:

            f(x)  =  5(x)3 + 8(x)2 – 4,  so f(0) = -4

            f '(x)  =  d[5(x)3 + 8(x)2 – 4]/dx  =  15(x)2 + 16(x),  so f '(0) = 0

            f ''(x)  =  d[f’(x)]/dx  =  d[15(x)2 + 16(x)]  =  30(x) + 16,  so f ''(0) = 16

            f '''(x)  =  d[f’’(x)]/dx = d[30(x) + 16]/dx =  30,  so f '''(0) = 30

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

                        Thus the expansion becomes:

            f(x)  =  -4 + 0(x-0) + 16(x-0)2/2 + 30(x-0)3/6 + 0 + 0 + 0 ...  =  -4 + 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] = [1-x]0.5 ,  so f(0) = 1

            f '(x) =  d[1-x]0.5/dx  =  [1-x]-0.5 (0.5) (-1)  =  -0.5[1-x]-0.5 ,  so f '(0) = -0.5

            f ''(x) =  d[f’(x)]/dx  =  d[ -0.5{1-x}-0.5]/dx  =   (-0.5) [1-x]-1.5 (-0.5) (-1)  =  -0.25[1-x]-1.5 , so f ''(0) = -0.25

            f '''(x) =  d[f’’(x)]/dx  =  d[-0.25{1-x}-1.5 ]/dx  =   (-.25) [1-x]-2.5 (-1.5) (-1)  =  -0.375[1-x]-2.5 , 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) = d[sin(q)]/dq = cos(q),  so f '(0) = 1

            f ''(q ) = d[f’(q)]/dq = d[cos(q)]/dq  =  -sin(q), so f ''(0) = 0

            f '''(q ) = d[f’’(q)]/dq = d[-sin(q)]/dq  =  -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 .
For each case, be sure to demonstrate with at least two values of x how close the first three terms of the series come to giving the "correct" (calculator) answer.  You do not have to worry about limiting cases for this problem.

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