Power Series Expansion: a review f(x) =
 n=0
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'' = d 2 f/dx 2 ; etc and n! = n * (n-1) * (n-2) * ... * 1
The above expansion works exactly for any power series.
Example: f(x) = 5x
3
+ 8x
2
– 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) = df(x)/dx = 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 + 8x 2 + 5x 3 .
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
=

[1-x] Let's look at the answer close to zero (so a=0): f(x) =

[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:

[1-x] = 1 + -0.5 (x-0) - 0.25 (x-0)
2
/ 2 + -0.375 (x-0)
3
/ 6 + ... = 1 - x/2 - x
2
/8 - 3x
3
/48 - ...
For x = 0.1, the "true" answer is f(0.1) =

[1-.1] =

.9 = 0.94868
The approximation to zero order gives: f(0.1) = 1.0000
(off by .05132, or 5.41% high)
The approximation to first order gives: f(0.1) = 1 - (.1)/2 = 0.9500
(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(

) = sin(

)
Let's again look at the answer close to zero (so a=0) f(

) = sin(

), so f(0) = 0 f '(

) = d[sin(

)]/d

= cos(

), so f '(0) = 1 f ''(

) = d[f’(
 d

= d[cos(

)]/d

= -sin(

), so f ''(0) = 0 f '''(

) = d[f’’(
 d

= d[-sin(

)]/d

-cos(

), so f '''(0) = -1
Thus the expansion becomes: sin(

) = 0 + 1(

-0) + 0(

-0)
2
/2 + -1(

-0)
3
/6 + ... =

-
 3
/6 + ...
For

=10 o
= 0.174533 radians, the "true" answer is f(10 o
) = sin(10 o
) = 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
The approximation to third order gives:
(off by .000885) f(0.174533) = 0 + (0.174533) + 0 - (0.174533)
3
/6 = 0.173647
(off by .000001)
Homework Problem: Problem #4: Derive the first 3 non-zero terms of the power series expansion for: a) cos(x) ; b) 1/

[1+x] ; and c) e ax
.
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.