Lesson 28a - Approximating Taylor Series

# F n a x a n n example1 showthatthemaclaurin seriesofex

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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.

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