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Homework Help: Am I right or is my book right (Taylor remainder)

  1. Aug 24, 2008 #1
    1. The problem statement, all variables and given/known data
    The approximation [tex] e^{x}=1+x+(x^{2}/2) [/tex] is used when [tex] X [/tex] is small estimate the error when [tex]\left|x \right|<0.1[/tex]


    2. Relevant equations
    [tex]\left|R_{n} \right|<\frac{M(x-a)^{n+1}}{(n+1)!}[/tex]


    3. The attempt at a solution
    Since the Taylor expansion goes to the second power I used the third derivative of [tex]e^{x}[/tex] which is just its self and found the maximum value that it can be between the domain [0,0.1] which is at [tex]e^{0.1}[/tex] then continuing the formula [tex](0.1-0)^{3}[/tex] then i divided it by 3! which gave me an answer of [tex]1.84*10^{-4}[/tex].

    My book on the other hand used [tex]3^{0.1}[/tex] instead of [tex]e^{0.1}[/tex] and as a result the answer in the book was larger than my answer. So which answer is the right answer?
     
  2. jcsd
  3. Aug 24, 2008 #2

    HallsofIvy

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    Science Advisor

    Depends upon what you mean by "estimate". Remember that neither of those answers will be the error- you can't determine the exact error this way. What you are saying is that the error is less than 1.98x10-4 while the text book is saying it is less than a larger value (because e< 3) so both are correct. I don't know if your book has some reason for using "3" rather than "e". e gives a slightly more accurate estimate but they are both estimates.
     
  4. Aug 25, 2008 #3
    The only reason I can imagine is that [tex] 3^{0.1} [/tex] is easier to calculate because it's the solution of [tex] x^{10} = 3 [/tex] and a Newton's method iteration would provide an accurate estimation quickly. The assumption, I suppose, is that if you are approximating [tex] e^x [/tex], you don't have the luxury of calculating it explicitly.
     
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