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Homework Help: 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


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    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


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    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
    and iterate a few times
  5. Nov 7, 2012 #4


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    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.
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