1549901369-Elements_of_Real_Analysis__Denlinger_

6.5 Taylor's Theorem 331

Thus,

!" (0) !"' (0) J C^4 ) (0)

T4(x) = f(O)+ f'(O)(x-O)+~(x-0)^2 +-

3

,-(x-0)^3 +· · ·+-

4

-, -(x-0)^4

• 10 2 - 24 3 24 4


• 3 + 0 · x + 2Tx +
  31
  x + 41x
  10 -24 24
  = 3 + - x^2 + - x^3 + - x^4
  2 6 24
  = 3 + 5x^2 - 4x^3 + x^4.

Notice that in this case, T 4 (x) and f(x) are identical polynomials. D

Now let's take the same function f and find its 4th Taylor polynomial about

a different point, say l.

Example 6.5.4 Find the 4th Taylor polynomial of the function

f(x) = 3 + 5x^2 - 4x^3 + x^4 about l.

Solution. Let f(x) = 3 + 5x^2 - 4x^3 + x^4. Then

JC^0 l(x) = 3 + 5x^2 - 4x^3 + x^4 :::::> JC^0 l(l) = 5

f'(x) = lOx - 12x^2 + 4x^3 :::::> f'(l) = 2

f"(x) = 10 - 24x + 12 x^2 :::::> f"(l) = -2

f"'(x) = - 24 + 24x :::::> f"'(l) = 0

JC^4 l(x) = 24 :::::> JC^4 l(l) = 24

Then,

T4(x) = f(l) + f'(l)(x -1) + --(x-1)f"(l)^2 + --(x-1)f"'(l)^3 + --(JC^4 ) (1) x -1)^4
2! 3! 4!

= 5 + 2(x - 1) + -(-2 x - 1 )2 + -^0 ( x - 1)^3 + -(^24 x - 1)^4

2! 3! 4!

= 5 + 2(x - 1) - (x - 1)^2 + (x - 1)^4.

Notice that in this case, T4 ( x) and f ( x) do not appear to be identical
polynomials. However, it is an easy exercise to "expand" the terms in T 4 (x)
and show that it really is equal to f(x). (Exercise 1) D

Example 6.5.5 Find the nth Taylor polynomial for the function f(x) = ex
about 0.

Solution. This one is easy. For all n E N, the nth derivative of f is

f(n)(x) =ex, and j(n)(O) = l. Thus,