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Homework Help: Error in Taylor polynomial of e^x

  1. Apr 25, 2008 #1
    Find the Taylor polynomial of degree 9 of

    f(x) = e^x

    about x=0 and hence approximate the value of e. Estimate the error in the approximation.

    I have written the taylor polynomial and evaluated for x=1 to give an approximation of e.
    Its just the error that is confusing me. I have:

    R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1} = \frac{e^z}{10!}

    and I need to find an upper bound for this to give the maximum error of the approximation.
    So far I have

    0 < z < 1, e^0 < e^z < e^1
    and then

    \frac{1}{10!} < \frac{e^z}{10!} < \frac{e}{10!}

    but then the upper bound has an e in it. In a similar example in the book they have just put in 3 instead of e, I guess making the assumption that e < 3. But I'm thinking if i do this I should somehow show that e < 3.

    Thanks for any help :)
  2. jcsd
  3. Apr 25, 2008 #2


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    Have you first approximated e using your Taylor polynomial? That should suggest that e< 3.
  4. Apr 25, 2008 #3
    I have, the approximation less than 3 but that doesn't necessarily mean that the actual value (taylor approximation + error) is < 3, does it?
  5. Apr 25, 2008 #4
    You do not need to show e < 3. Everyone know e is 2.71828 18284 59045 23536... :D

    To be honest, I never knew approximation would give lower value (surprised!), but it should be obvious .. see the red line http://upload.wikimedia.org/wikipedia/commons/5/50/Exp_derivative_at_0.svg

    You are trying to get upper error bound; mine prof said it doesn't matter what you pick (50,100,..) it just have to be above the max function value!

    so e^1 = 1+1+1/2+1/6+.. (you know little more than 2)
    Why should take risk of using something like 2.5..
    simply pick 3!
    Last edited: Apr 25, 2008
  6. Apr 25, 2008 #5
    Thanks for the replies. I had another idea, I could show (by taking the lower sum of small interval slices) that the area under the graph of 1/x between 1 and 3 has to be greater than 1, therefore 3 > e.
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