Question

INSTANT FAN/MEDAL for answer <3<3

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

Just need correct answer to check my work. Thanks <3

Answer 1

Is it diffferent from http://openstudy.com/study#/updates/570c5cdee4b03d42c8057312 ?

Answer 2

Well.. I don't know if this helps.. but its kind of an example of what your doing :)

Answer 3

Using a degree n Macluarin polynomial, the error is given by
f^(n+1)(c) x^(n+1) / (n+1)! for some c between 0 and x.

In our case, f(x) = e^x and x = 0.3, and
f^(k)(x) = e^x for all positive integers k.

Thus,
|Error| = |e^c (0.3)^(n+1) / (n+1)!| for some c in (0, 0.3).

For x in (0, 0.3), e^c is maximized by e^(0.3) since e^x is increasing for all x.
Moreover, e^(0.3) < 2.
To simplify calculations (though unnecessary), we can use 0.3 < 0.5

Thus,
|Error| <= 2 * (0.5)^(n+1) / (n+1)!.

So, we need to solve
2 * (0.5)^(n+1) / (n+1)! <= .001
<==> (n+1)! / [2 * (0.5)^(n+1)] >= 1000
<==> (n+1)! * 2^n >= 1000.

This is checked to be true when n >= 4.
So, we need the Maclaurin polynomial to have degree at least 4.

I hope this helps!