Question

INSTANT FAN/MEDAL
What degree of the Maclaurin polynomial is required so that the error in the approximation of e^0.6 is less than 0.001?

(no idea how to start)

Answer 1

What do you know about the different error tests that can be used with the series?

Answer 2

@Shirley27 Yes, which do you suggest I apply here and how?

Answer 3

thanks for responding by the way, I appreciate it !

Answer 4

no problem :)
I would suggest using the langrange error bound in this case since the maclaurin series isn't alternating

Answer 5

the actual langrange error bound is through this formula\[error \le \frac{ M \times x ^{n+1} }{ (n+1)! }\]

Answer 6

where \[M \ge f ^{(n+1)}(x)\] and x is a number between 0 and the estimated number that produces the largest derivative

Answer 7

As @Shirley27 mentioned, take the formula of the rest:
\[\frac{f^{(n+1)}(\xi)x^{n+1}}{(n+1)!} \qquad (*)\]
We do not know the value of \(\xi \in (0,x)\). Therefore we look for an upper bound of (*). Since \(f^{(n+1)}(x)\) is an increasing function (check it if necessary), an upper bound of (*) is found at point \(x\): \(f^{(n+1)}(x)x^{n+1}/(n+1)!\) is an upper bound. (\(x=0.6\)).

(If \(f^{(n+1)}\) is a positive decreasing function, the upper bound is found by setting \(\xi=0\).)

Answer 8

An upper bound of (*) is \(\frac{e^{0.6} (0.6)^{n+1}}{(n+1)!}\), for all \(n\). You just have to evaluate this until you obtain a value smaller than 0.001.