# 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)?

Nov 1, 2014

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.