Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor's Theorem Error Bound

  1. Sep 25, 2015 #1
    For the error bound for taylor's theorem, for the n+1 derivative evaluated at some c which maximizes the derivative my textbook says c must be between a and x..but today my teacher said that c must be between absolute value x and negative absolute value x, which is different than I thought.

    An example would be calculating the error of using a second degree taylor polynomial to estimate e^x at x=-1...the n+1 derivative would be e^x, so the question would be do I use 0 because 0 maximizes e^x on [-1,0] or do I use 1 because of absolute value x being 1 and 1 maximizes e^x on [-1,1].

    Hopefully my question makes sense, just to reiterate I am wondering if c is between a and x (which is what textbook says and is what I thought in the past) or between absolute value x and negative absolute value x.

    Additionally I already tried to talk to my teacher to clarify and he insisted it must be between absolute value x and negative absolute value x...but in the past I learned it was x and a which is confirmed by my book.

    Any help on this is appreciated
     
  2. jcsd
  3. Sep 25, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    How did you define a and the variable you call x (despite using it as free variable at the same time)?
     
  4. Sep 25, 2015 #3
    Sorry for not being more clear, a is where the polynomial is centered at and x is where it is being evaluated at... In my example above the polynomial is centered at 0 (forgot to say that sorry) and we are evaluating it at x=-1
     
  5. Sep 25, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Between a and x then.

    If a=0, then "between -|x| and |x|" is a weaker statement. It is not wrong, but sometimes (like here) it leads to a weaker estimate.
     
  6. Sep 25, 2015 #5
    Yes thank you that's what I thought, is there ever a situation where you would need to use abs value x to negative abs value x? Because im trying to figure out why my teacher said that when it seems like a to x works and is more accurate
     
  7. Sep 25, 2015 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Maybe if you want an upper limit for the whole range of the expansion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Taylor's Theorem Error Bound
  1. Taylor's Theorem (Replies: 3)

  2. Error bounds (Replies: 1)

  3. Taylor's Theorem (Replies: 6)

  4. Taylor's Theorem (Replies: 2)

Loading...