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Unformatted text preview: 1, 1]. Then so is the function
f (k+1) (x)
g (x) =

(k + 1)!
DE Math 128 331 (B. Forrest)2 4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series The Extreme Value Theorem tells us that g (x) has a maximum on [−1, 1]. Therefore, there
is an M such that
f (k+1) (x)

≤ M
(k + 1)!
for all x ∈ [−1, 1]. Let x ∈ [−1, 1]. Taylor’s Theorem shows that there is a c between x
and 0 such that
f (k+1) (c) k+1
 Rk,0 (x) =
x

(k + 1)!
Therefore,
 f (x) − TK,0 (x)  =  Rk,0 (x) 
= f (k+1) (c) k+1
x

(k + 1)! ≤ M  x k+1
since c is also in [−1, 1].
It follows that
−M  x k+1 ≤ f (x) − Tk,0 (x) ≤ M  x k+1 .
We summarize as follows:
Theorem. [The Approximation Theorem]
Assume that f (k+1) is continuous on [−1, −1]. Then there exists a constant M > 0 such
that
 f (x) − Tk,0 (x) ≤ M  x k+1
or equivalently that
−M  x k+1 ≤ f (x) − Tk,0 (x) ≤ M  x k+1
for each x ∈ [−1, 1].
This theorem is very helpful in calculating many limits.
Example.
cos(x)−1
.
x2
x→0 Calculate lim 2 We know that for f (x) = cos(x) we have T2,0 = 1 − x . Moreover, all...
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