Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

    So

    [tex]P_n(x)=3+\dysplaystyle\frac{(x-9)}{6}-\dysplaystyle\frac{(x-9)^2}{216}+\dysplaystyle\frac{3(x-9)^3}{3888}[/tex]

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

    [tex]P_n(x)=3+\dysplaystyle\frac{(0.3)}{6}-\dysplaystyle\frac{(0.3)^2}{216}+\dysplaystyle\frac{3(0.3)^3}{3888}\approx{3.049604167}[/tex]

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

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

    Mark44

    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

    Mark44

    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

    Mark44

    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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook