Computing e^(-2) Using a Power Series

Date: 04/16/2005 at 10:31:30
From: Anood
Subject: Using power series to compute e^-2

Use power series to compute e^-2 to four decimal place accuracy.  It's
a very difficult question and I don't know how to do it.

Date: 04/18/2005 at 09:20:59
From: Doctor Ricky
Subject: Re: Using power series to compute e^-2

Hey Anood,

This problem isn't as bad as it looks, but you need to understand what 
it's asking.

It's asking you for e^(-2), or in other words, if you are given:

  f(x) = e^(-x)

you want the value of f(x) at x = 2.

Therefore, you are going to find the MacLaurin series expansion for 
e^(-x).  Do you remember the expansion for e^x, which is:

                 x^2   x^3         x^n
  e^x =  1 + x + --- + --- + ... + --- + ...
                  2!    3!          n!

To find the expansion for e^(-x), just substitute -x in for x.  Then
we get:

                    x^2   x^3         [(-1)^n]*(x^n)
  e^(-x) =  1 - x + --- - --- + ... + -------------- + ...
                     2!    3!                n!

Therefore to calculate e^(-2), we just plug the value x = 2 into our 
last equation.  To determine how many terms we need to calculate e^(-
2) to four decimal places accurately, we calculate until the terms get 
small enough to where they are too small to change the value at four 
decimals.

  i.e. | [(-1)^n]*(x^n) |
       | -------------- |  < 10^(-4)
       |       n!       |

We use an absolute value since this is an alternating series.  
Therefore, we can rewrite it as:

  (x^n)       1
  -----  < ------
    n!      1000


Therefore, the number of terms (from zero to n) we will have to 
calculate is when:

          n!
  x^n < ------
         1000


We know x^n will be positive integers greater than one (since x = 2) 
and we can see that n! isn't bigger than 1000 until n = 7.  Therefore 
we know we'll need at least 7 terms.  By finding n! and x^n for larger 
values of n, we see that:

          n!
  x^n < ------
         1000

is true for the first time when n = 10.

Therefore we need to calculate the first 11 terms of the MacLaurin 
series (remember that the first term is when n = 0).  In other words, 
we would plug in our x = 2 into the MacLaurin series and use the first 
11 terms (from x^0 until x^10) to get our answer accurate to four 
decimal places.

If you have any questions or if this was a little confusing, please 
let me know!

- Doctor Ricky, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/19/2005 at 14:22:07
From: Anood
Subject: Thank you (Using power series to compute e^-2)

Dear Doctor Ricky,

Thank you very very verrrrrrrrry much for your help!  :>