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Taylor series expansion

  1. Jan 18, 2012 #1
    Hello all,
    My question is in regards to the Taylor series expansion of
    [tex]f(x)=e^x=1+x+x^2/(2!)+x^3/(3!).......[/tex]
    I calculated the value of
    [tex]e^(-2)[/tex]
    using the first 4 terms, 6 terms, and then the first 8 terms. I then calculated the relative error to compare it to the true value, depcited by my calculator to 6 significant figures. Using the first four terms, I found an error of 2609%. Using the first 6 terms I found an error of 2905%, and lastly using the first 8 terms I found an error of 2952%.

    What can I conclude from this? Does the error increase (at a decreasing rate) until it begins decreasing (at an increasing rate)?
     
  2. jcsd
  3. Jan 18, 2012 #2

    CompuChip

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    How did you get to 2609%?
    If I take 4 terms I get an error of around 46%.

    Are you sure you entered the even powers correctly? One of the most common mistakes is to enter (-2)2 as -22 = -(22).
     
  4. Jan 18, 2012 #3
    Oops, I had a mistake.n However, upon doing it again, I get an error of 346.3.
    relative error=100((e^(-2)-(sum of first four terms))/e^(-2))=100((e^(-2)-(-0.333333))/e^(-2))=346%
     
  5. Jan 18, 2012 #4
    Yes the error after 4 terms is 346%. But in the next 4 terms the errors are 146%, -51%, 15% and 4%; after another 4 terms the error is only 5 parts in a thousand and another four, -2 in 1 million. The error decreases in magnitude with each successive term (except for the second term, x) and converges fast enough for me.
     
    Last edited: Jan 18, 2012
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