1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor expansion

  1. Aug 5, 2005 #1
    Hi

    How do you expand

    (1-exp(-1))^-1

    as Taylor series

    Callisto
     
    Last edited: Aug 5, 2005
  2. jcsd
  3. Aug 5, 2005 #2

    James R

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Try long division to get a Taylor (or McClaurin) series for f(x) = 1/(1-x) for a start.

    Then put x=exp(-1) into that series, and there you go!
     
  4. Aug 5, 2005 #3
    Taylor series for

    1/(1-x) = (1-x)^-1 = 1 + x^2 + x^3 + x^4 +...= SUMi x^i

    Then sub in exp(-1) for x
    That's it?
     
  5. Aug 5, 2005 #4

    James R

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. That's it.
     
  6. Aug 5, 2005 #5
    Cheers James R!
     
  7. Aug 5, 2005 #6
    a constant has a trivial Taylor expansion

    The expression you submitted is a constant, it doesn't depend on x.
    All derivatives of this function with respect to x (the assumed variable) are zero.

    Therefore the Taylor expansion contains only one term (and is exact):

    f(x) = 1/(1-exp(-1)) + 0 + 0 + 0 + 0 + 0 + 0 + ... = 1/(1-exp(-1))

    (unless you did not formulate your problem correctly)
     
    Last edited: Aug 5, 2005
  8. Aug 6, 2005 #7

    James R

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Technically, you're right of course, lalbatros, but your solution doesn't seem to fit the spirit of the question which was asked. Maybe the question was a little unclear, though...
     
  9. Aug 7, 2005 #8
    You are probably right James.
    The question probably was: "how can I get an approximate value for 1/(1-exp(-1)) using Taylor expansions".
    Then, there are numerous ways to choose how to expand.
    For example:

    1/(1-exp(-x)) = 1 /(1-sumi((-x)^i / i!))

    or explicitely:

    1/(1-exp(-x)) = 1/(x-x^2/2+x^3/6-x^4/24+x^5/120 ...)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Taylor expansion
  1. Taylor expansion (Replies: 6)

Loading...