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!

Approximating accuracy of Taylor polynomials

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001.

    [tex] e^x ≈ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}[/tex]

    For x < 0

    2. Relevant equations

    Taylor's Theorem to approximate a remainder:

    [tex] |R(x)| = |\frac{f^{n+1}(z)}{(n+1)!}(x-c)^{n+1}|[/tex]

    Where z is some number between c and x, n is the degree of the approximating function, and c is where the function is centered at.

    3. The attempt at a solution

    From the Taylor polynomial given, c = 0 and n = 3. Since f(x) is e^x, the fourth derivative is simply e^x. If I want an error of less than 0.001,

    [tex] |R(x)| = |\frac{e^z}{4!}x^4| < 0.001 [/tex]

    Not too sure about this next part, but I think that since the function above increases as x increases, which means the error increases with an increasing x, I replaced z by x since the maximum error is given by the largest z value, and the largest z value is equal to x, and I'm interested in the error bound anyways.

    [tex] |R(x)| = |\frac{(x^4)(e^x)}{4!}| < 0.001 [/tex]

    ...but I'm still left with the problem that I have to find a value that's stuck in an exponent and outside of one, so I'm assuming I'm doing this wrong and I can't figure out any other way to do it.
     
    Last edited: Nov 7, 2012
  2. jcsd
  3. Nov 7, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    When I saw it I wasn't sure if it was a typo or relevant to your question, but as you have stated the problem it is for ##x<0##, which would help with the ##e^x##.
     
  4. Nov 7, 2012 #3

    lurflurf

    User Avatar
    Homework Helper

    You are on the right track. That is a good estimate (about 7% low). To solve your equation exactly you need the Lambert function. You could also solve it approximately in several ways. One of which is to write it
    x^4e^x=24/1000
    x=fourthroot(24/1000)e^-(x/4)
    and iterate a few times
     
  5. Nov 7, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If are really only worried about x<0 then you have an alternating series. There's an even easier estimate you could use that doesn't even involve e^x. It turns out to be the same error bound that LCKurtz is suggesting you could arrive at by putting in an upper bound for the exponential.
     
    Last edited: Nov 7, 2012
  6. Nov 8, 2012 #5
    Wow, that's very clever.

    I hadn't noticed that the polynomial is the beginning of an alternating series and you can just find the remainder of the alternating series to solve for x.

    I hadn't registered the fact that x < 0 in my mind so I didn't see that you could just put 1 in place of the e^x (and you'd get the same expression as if you had used the alternating series remainder to solve for x)

    That's pretty cool, thanks for the help guys.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Approximating accuracy of Taylor polynomials
Loading...