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Unformatted text preview: f(x) = 0, if and only if lim f ( x) Tk ( x) 0
k lim Rk ( x) 0
k f ( k 1) ( z )( x a ) k 1
0
lim
k (k 1)! f ( n ) (a )( x a ) n
n! Example 1
Show that the Maclaurin Series of ex converges to ex for every possible x:
Using the remainder our limit, we want to make sure that f ( k 1) ( z )( x a ) k 1
lim
0
k (k 1)! e z ( x) k 1
lim
k ( k 1)! We know that the k+1’th derivative evaluated at z will be ez, but z depends on the kth, derivative (depends on k). This means it is very tough to evaluate this limit (perhaps z = k!, then the limit would certainly diverge…). To find a bound that would make this not depend on k, we consider the following:
Say we had x ≥ 0
We know that 0 ≤ z ≤ x which means that by taking e to the inequality, we get:
1 ≤ ez ≤ ex
e z ( x) k 1
e x ( x) k 1
( x) k 1 lim lim
This means that we can write: lim
k k k (k 1)! (k 1)! (k 1)! Example 1 Continued
ex does not depend on k, so we can fa...
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This note was uploaded on 02/07/2014 for the course MATH 1004 taught by Professor Mark during the Summer '00 term at Carleton CA.
 Summer '00
 Mark
 Calculus, Approximation, Taylor Series, Lecture Notes, MATH1004, Kyle Harvey

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