1. Apr 13, 2008

### bcjochim07

Use Taylor's theorem to determine the degree of the Maclaurin polynomial required for the error in the approximation of the function to be less than .001.

e^.3

So is the procedure to take the derivatives and plug in 0 (since c=0) and find an expression for the n+1 derivative?

f'(c) = 1 f''(c)=1 f'''(c) =1 ......

so the n+1 derivative is 1

So Rn= 1/(n+1)! * (.3) ^(n+1)

Then I set up an equality to find n so that Rn < .001

and n = 3 ???

I want to be sure I am taking the right approach on these problems, so is this the way to do it?

2. Apr 13, 2008

### EngageEngage

Personally I think the best way to do these is to first find the series for the function and then to plug in a few values. The summation for the exponential function is:
$$e^{x} = \sum^{\infty}_{n=0}\frac{x^{n}}{n!}$$

if you compute a few setting x = .3, you will see how many terms you need

$$e^{.3} = \sum^{\infty}_{n=0}\frac{.3^{n}}{n!} = 1+ \frac{.3}{1!}+\frac{.3^{2}}{2!}+...$$

Last edited: Apr 13, 2008
3. Apr 14, 2008

### benorin

Solving the inequality Rn < .001 will give the answer. It is easy to solve this by just plugging in values of n until it is satisfied.