How Many Terms Needed in Maclaurin Polynomial for Error Below 0.001?

  • Thread starter Thread starter bcjochim07
  • Start date Start date
  • Tags Tags
    Polynomial Taylor
bcjochim07
Messages
366
Reaction score
0
Taylor Polynomial Error--Please help!

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?
 
Physics news on Phys.org
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:
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top