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 [β1, 1]. Therefore, there
is an M such that
f (k+1) (x)

β€ M
(k + 1)!
for all x β [β1, 1]. Let x β [β1, 1]. Taylorβs 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) β TK,0 (x)  =  Rk,0 (x) 
= f (k+1) (c) k+1
x

(k + 1)! β€ M  x k+1
since c is also in [β1, 1].
It follows that
βM  x k+1 β€ f (x) β Tk,0 (x) β€ M  x k+1 .
We summarize as follows:
Theorem. [The Approximation Theorem]
Assume that f (k+1) is continuous on [β1, β1]. Then there exists a constant M > 0 such
that
 f (x) β Tk,0 (x) β€ M  x k+1
or equivalently that
βM  x k+1 β€ f (x) β Tk,0 (x) β€ M  x k+1
for each x β [β1, 1].
This theorem is very helpful in calculating many limits.
Example.
cos(x)β1
.
x2
xβ0 Calculate lim 2 We know that for f (x) = cos(x) we have T2,0 = 1 β x . Moreover, all...
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 Spring '09
 Math, Derivative, Power Series, Taylor Series, Sequences And Series, Taylor's theorem, B. Forrest

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