1549901369-Elements_of_Real_Analysis__Denlinger_

332 Chapter 6 • Differentiable Functions

Example 6.5.6 Find the nth Taylor polynomial of f(x) = ln(l + x) about 0.

Solution. We first find J(n) (0) for n = 1, 2, · · · , n.

f (x) =ln(l+x) ==> f (O) =0

1

f'(x) = --= (1 + x)-^1

l+x * f'(O)

= 1 = O!

-1!

f"(x) =-(l+x)-^2 = * f"(O) = -1!

(1 + x)^2

2!

f"'(x) =2(1+x)-^3 = * f"'(O) = 2!

(1 + x)^3

-3'

j<^4 l(x) = -2 · 3(1 + x)-^4 = · =? j<^4 l(O) = -3!

(1 + x)^4

Thus,

n J(k)(O) k n (-l)k+l(k-1)! k n (-l)k+l k

Tn(x) = L - k,-(x-0) = o+ L k! x = L k x

k=O k=l k=l

( 1 r+l

=x-~x^2 +~x^3 -~x^4 + .. · + - xn. 0

n

APPROXIMATION BY TAYLOR POLYNOMIALS

One reason to expect that Tn(x) is a good approximation to f (x) is that at

a, f(x) and Tn (x) have the same value, the same derivative, the same second

derivative, and so on, up to the same nth derivative. The following theorem

makes this explicit.

Theorem 6.5.7 Suppose f has an nth derivative at a. Then Tn(a) = J(a),

T~(a) = f'(a), T;:(a) = f"(a), · · ·, and T~n)(a) = j<nl(a).

Proof. See Exercise 2. •