1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding K for integral approximation errors help

  1. Sep 4, 2014 #1

    LBK

    User Avatar

    Hello, I am having a hard time getting my errors to come out to what the book says the answers should be. My approximations are correct, so I think I'm just misunderstanding how to find K.
    Q.a) Find the approximations T8 and M8 for ∫(0 to 1) cos(x2)dx
    I found these to be T8=0.902333 and M8=0.905620

    b) Estimate the errors in the approximations. And here's where my numbers don't match. What I did:

    2nd deriv f''(x)=-2x*sin(x2)-4x2cos(x2)

    since the graph is bounded by |1| I thought the max. would be at f''(0) but that would make f''(0)=0
    So I'm really confused on where to go from here.
    FYI--the book answer gives a value of ET=<or=0.0078 and EM=<or=0.0039

    I have the formula for error for trapezoidal as <or= [K(b-a)3] / 12n2
    In this case then, I should have (b-a)3=1 and 12n3=768
    Working backward from the book's correct answer that would make 0.0078=K/768 and K=5.99
    So I'm not seeing where that would come from. Any help, please? I'm really struggling in this class and very confused.
     
    Last edited: Sep 4, 2014
  2. jcsd
  3. Sep 4, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    What do you mean by ##T_8## and ##M_8##?
     
  4. Sep 4, 2014 #3

    LBK

    User Avatar

    Oh, I'm sorry that would be integral approximation by trapezoidal rule and by midpoint rule, respectively with n=8

    *also original post corrected for error formula, denominator should be 12n^2 not cubed
     
  5. Sep 4, 2014 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    For your function ##f(x) = \cos(x^2)## the second derivative should be
    [tex] f''(x) = -2 \sin(x^2)-4x^2 \cos(x^2),[/tex]
    which is a bit different from what you wrote. If you plot ##f''(x)## you will see that it is negative and strictly decreasing on the interval ##0 \leq x \leq 1##. Thus ##|f''(x)|## is strictly increasing.

    Anyway, for Trapezoidal, the error bound is
    [tex] |E| \leq \frac{K(b-a)^3}{12 n^2} = \frac{K}{12\times 8^2}= \frac{K}{768}, \: K \equiv \max_{0 \leq x \leq 1} |f''(x)|[/tex]
    In your case, ##K = |f''(1)| \doteq 3.84415##, so ##|E| \leq 0.0050054 \doteq 0.005.## I cannot get the book's answer.

    I think I can see where you went wrong: the actual error formula is
    [tex] E = -\frac{f''(\xi) (b-a)^3}{12 n^2} = - \frac{f''(\xi)}{768}[/tex]
    In this case, ##f''(\xi) < 0## so the error is ##>0##. To bound the error we want the least value of ##f''##, that is the most negative value. That is at ##\xi = 1##.
     
  6. Sep 4, 2014 #5

    LBK

    User Avatar

    I worked on it just now and I got .005 too. A classmate said he couldn't get the book answer either, so I feel much better now. And for the record, your explanation is clearer than my teacher and the book, so thank you again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding K for integral approximation errors help
Loading...