How do you find n such that |Tn(1) - e| ≤ 10^-5, where Tn is the Maclaurin polynomial for #f (x) = e^x#?

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Answer 1 · 2017-01-18T19:29:04

#abs(T_8(1) - e) < 10^(-5)#

The MacLaurin series for #f(x)=e^x# is:

#e^x = sum_(n=0)^oo x^n/(n!)#

Calculating for #x= 1# and using Lagrange remainders we know that:

#e = sum_(n=0)^N 1/(n!) + R_N(xi)# with #xi in(0,1)#

where:

#R_N = f^((N+1))(xi)/((N+1)!) x^(N+1) = e^xi/((N+1)!) <= e/((N+1)!)#

We therefore have to choose #N# such that:

#e/((N+1)!) < 10^(-5)#

or:

#(N+1)! > 10^5e#

Looking at a table of #(n!)# values we can quickly find that #N=8# is sufficient, and in fact:

#sum_(n=0)^8 1/(n!) = 2.71828#

equals #e# to the fifth decimal.