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.

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