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Homework Help: Approximation e number using taylors polynomial

  1. Aug 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Well, this problem is quiet similar to the one I've posted before. It asks me to approximate to the e number using taylors polynomial, but in this case tells me that the error must be shorter than 0.0005

    2. Relevant equations

    [tex]R_{n+1}=\displaystyle\frac{f^{n+1}(z)(x-x_0)^{n+1}}{(n+1)!}\leq{\epsilon}[/tex], [tex]0<z<1[/tex]

    [tex]x_0=0[/tex] [tex]x=1[/tex] [tex]f^{n+1}(x)=e^x[/tex] [tex]\epsilon=0.0005[/tex]



    3. The attempt at a solution
    I've tried using the function [tex]f(x)=e^x[/tex] at [tex]x_0[/tex] [tex]\epsilon=0.0005[/tex]



    [tex]P_n(1)=1+1+\displaystyle\frac{1}{2!}+...+\displaystyle\frac{1}{n!}[/tex] which tends to e as n tends to infinity.

    But now I don't know how to get the n, so I get the degree for the polynomial with the error it asks me.

    [tex]R_{n+1}=\displaystyle\frac{f^{n+1}(z)(x-x_0)^{n+1}}{(n+1)!}[/tex] [tex]0<z<1[/tex]

    Lets call [tex]\delta=n+1[/tex]

    [tex]R_{\delta}=\displaystyle\frac{e^z}{\delta!}\leq{0.0005}[/tex] then

    [tex]\displaystyle\frac{e^z}{0.0005}\leq{\delta!}[/tex] I know between which values I can find z, but I don't know how to work the factorial in the inequality.

    Am I proceeding right?

    Bye there.
    Last edited: Aug 12, 2010
  2. jcsd
  3. Aug 12, 2010 #2


    Staff: Mentor

    On the interval [0, 1], e1 < 3.
  4. Aug 12, 2010 #3
    Right, I must use the bigger value on the interval. Then I've got [tex]6000\leq{\delta!}[/tex] So I must use [tex]delta=8[/tex]? but if I use e on the calculator I think I get that seven is a better approximation for it, and I think it works for seven. What you say Mark?

    Sorry, I was wrong. You're right, so [tex]\delta=8[/tex]
  5. Aug 12, 2010 #4


    Staff: Mentor

    I wouldn't introduce another variable, [itex]\delta[/itex]. Just work with n + 1.
  6. Aug 12, 2010 #5
    Thanks Mark ;)
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