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Taylors polynomials

  1. Aug 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Find an approximated value for [tex]\sqrt[ ]{9.03}[/tex] using a Taylors polynomial of third degree and estimate the error.

    2. Relevant equations

    3. The attempt at a solution
    I thought of solving it by using

    [tex]f(x)=\sqrt[]{x}[/tex] centered at [tex]x_0=9[/tex]



    Then I evaluated it at x=9.03, so I get:


    I don't know if this is right, I've tried with the calculator and it gives 3.00500.... Now, how do I estimate the error? just by resting to the value the calculator gives the one I get?
  2. jcsd
  3. Aug 11, 2010 #2


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    9.03-9 = 0.03, not 0.3. :)
  4. Aug 11, 2010 #3
    ****, thanks haha
  5. Aug 11, 2010 #4


    Staff: Mentor

    For the error estimate, there's a formula for the error in approximating a Taylor series by the first n terms (terms up through degree n). If my memory is correct, it looks something like this.

    [tex]R_n(x)= \frac{f^{n + 1}(c)}{(n + 1)!}(x - a)^{n + 1}[/tex]

    Since it is usually impossible to determine c (which is between x and a), in practice, one finds an upper bound on |f(n+1)(x)|.
  6. Aug 12, 2010 #5
    Thank you Mark. I think I get it.

    Here it is:

    [tex]R_{n+1}(x)=\displaystyle\frac{-15(x-9)^4}{4!16\sqrt[ ]{\alpha}}[/tex]

    [tex]\alpha=9+0.03\theta[/tex], [tex]\theta\in{(0,1)}[/tex]
    Last edited: Aug 12, 2010
  7. Aug 12, 2010 #6
    I could be wrong, but shouldn't this be an inequality? That's how I remember it.
  8. Aug 12, 2010 #7


    Staff: Mentor

    No, it shouldn't be an inequality. Checking with a calculus text, the formula for the remainder is correct.
  9. Aug 12, 2010 #8
    It can be thought as inequality like this:

    [tex]|R_n(x)|= |\frac{f^{n + 1}(c)}{(n + 1)!}(x - a)^{n + 1}|\leq{}\displaystyle\frac{(x-a)^{n+1}}{(n+1)!} k[/tex]
    Where [tex]k=max{f(b) \textsf{ such that b} \in{[x,a]}[/tex]
    Last edited: Aug 12, 2010
  10. Aug 12, 2010 #9


    Staff: Mentor

    The remainder is an equality. Usually it suffices to get an upper bound on that derivative - then you have an inequality.
  11. Aug 12, 2010 #10


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    The various forms of the Taylor remainder are generalizations of the mean value theorem to higher derivatives and higher degree polynomials -- the remainder is exactly equal to the given expression for some point c in a known interval.

    Typical practical applications find the maximum over the entire interval, and then use that as an upper bound on the true remainder.
  12. Aug 12, 2010 #11
    Thank you both, Mark and Hurkyl for clearing things up!
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