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Justification for upper bound in Taylor polynomial

  1. Aug 6, 2017 #1
    1. The problem statement, all variables and given/known data
    I've been reviewing some Taylor polynomial material, and looking over the results and examples here.

    I'm referring to Example 3 on the page 12 (page numbering at top-left of each page). The question is asking about an upper bound on the error.

    2. Relevant equations

    3. The attempt at a solution
    I was able to get the result in the PDF, but in the discussion of the example, the authors mention that there is a logical issue with using an exact value of e^(0.2) while we're approximating f(x) = e^x . As an alternate bound the authors propose using f(5)(c) < 2 .

    Can anyone tell me where the justification for this value of 2 comes from? Why is this a proposed value for an upper bound?
    Last edited by a moderator: Aug 6, 2017
  2. jcsd
  3. Aug 6, 2017 #2


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    To be honest? I can't see it either. One could take ##0.2 < \frac{1}{2}## and ##f^{(5)}(c) \leq e^{0.2} < e^{0.5} < \sqrt{3} < 2## or something like
    $$ f^{(5)}(c) = 1 + c + \frac{c^2}{2!}+ \frac{c^3}{3!} + \ldots < 1+\frac{1}{5}+\left( \frac{1}{5}\right)^2+ \left( \frac{1}{5}\right)^3+\ldots < 1+\frac{1}{2}+\left( \frac{1}{2}\right)^2 +\ldots = 2$$ but this is a lot of guesswork. Did they eventually had an upper bound in the previous section, or perhaps the ##K## in Rolle?

    Another possibility is to use the fact (if given), that the exponential function is convex. In this case we have
    f^{(5)}(c) < \frac{e^0+e^1}{2} < \frac{1}{2} + \frac{3}{2} = 2
  4. Aug 6, 2017 #3
    Thanks for replying. I will look at the Rolle's theorem possibility.
  5. Aug 6, 2017 #4

    Ray Vickson

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    For ##x > 0## we have ##e^{-x} > 1-x##, so ##e^x < 1/(1-x)##, giving ##e^{0.2} < 1/(1 - 0.2) = 1.25 < 2##.
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