How do you find the smallest value of #n# for which the Taylor series approximates the function #f(x)=e^(2x)# at #c=2# on the interval #0<=x<=1# with an error less than #10^(-6)#?

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Answer 1 · 2014-11-01T14:15:37

By finding the remainder term,

#|R_n(x;2)|=|{f^{(n+1)}(z)}/{(n+1)!}(x-2)^{n+1}|=|{2^{n+1}e^{2z}}/{(n+1)!}(x-2)^{n+1}|#

#le {2^{n+1}e^{2(2)}}/{(n+1)!}|(0-2)^{n+1}|={2^{2n+2}e^4}/{(n+1)!}#

Since

#{2^{2(21)+2}e^4}/{[(21)+1]!}approx0.000000855 le 10^{-6}#,

we have #n=21#.

I hope that this was helpful.