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Numerical Analysis: Composite Trapezoidal Sum Rule

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the composite trapezoidal sum rule to evaluate

    I = integral from 0 to 1 of: (exp(x)-1)/x

    At x = 0 the integrand evaluates to 1.
    Select the step size h in order to guarantee an approximation error less than
    10e-5.
    Carry out your calculation with at least 10 decimal significant figures.
    Note: the exact answer rounds to
    1:31790215145440389486:

    2. Relevant equations

    I know that i need to find the max of f''(x), but it goes to infinity as x goes to zero. How do i find this max value?

    3. The attempt at a solution

    Using the Trapezoidal Sum rule with n=1, the error using the Newton Cotes Method is: (10e-5)=(1/12)*f''(x)*h^2

    So h = sqrt((12*10e-5)/f''(x))

    f''(x) = ((x^2 -2x+2)*exp(x)-1)/(x^3)

    I tried L'Hopitals rule, but that gave me 1/3 as x goes to zero, which is incorrect.
     
  2. jcsd
  3. May 28, 2009 #2
    Check you L'Hopitals evaluation again.

    f''(x) does not go to infinity as x approaches zero.

    Check your f''(x) expression - your "-1" in expression for f''(x) should be "-2" (but this does not effect your L'Hopitals evaluation, but it's incorrect)

    Another way you can look at this is by looking at the Taylor Series expansion of exp(x) at x=0. Plug that into the expression for f(x) and then take derivatives. Should get the same answer, f''(0) = 1/3.
     
  4. May 28, 2009 #3
    Trying the Taylor Series:

    f = (exp(x)-1)/x = sum (n=0 to infinity) (x^n)/(n+1)!

    f'' =[n(n-1)x^(n-2)]/(n+1)!

    How do I evaluate this to find the max value?
     
  5. May 29, 2009 #4
    You don't (directly). It was presented to allow you to evaluate f'' at x=0.

    However, if you can show that f'' is positive throughout the interval and uniformly increasing in the interval of intrest [0,1], then you know the max of f'' is at the right domain endpoint (x=1). Or, if it is positive throughout the interval and uniformly decreasing, then you know the max of f'' is at the left domain endpoint (x=0).

    Now, take another look at that series expansion and what can you say about f''(x) in the interval [0,1]?
     
  6. May 29, 2009 #5
    well, expanding the series out, it appears that it is indeed positive throughout the interval and uniformly increasing, so x_max =1.
    Then f''(x)_max = f''(1) = ((1^2-2+2)*exp(1)-1)/(1^3) = 1.718281828 ?

    Thus, h = 0.0083568657
    1/h = 119.66 ~120
    So the # of intervals N = 120? (ie 0 to 0.0083333333, 0.008333333 to 0.0166666667... all the way to 1?)

    Then, using the composite sum rule,

    (h/2)*(f0 +f1 + f1 + f2 + f2 + f3 + f3 +f4...+f120)?

    That does not seem correct, as doing this by hand would take a long time
     
    Last edited: May 29, 2009
  7. Jun 1, 2009 #6
    As I said, I believe you have an error in your expression for f''(x). I came up with the following:

    [tex]f''(x) = \frac{(q(x) \cdot e^x-2)}{x^3}[/tex]

    with

    [tex]q(x) = x^2-2x+2[/tex]

    This would yield [itex]f''(1) = e-2[/itex] = 0.71828...

    Hence, I compute the minimum required n = 25 via the expression below where I've taken [itex]\epsilon[/itex] = 10E-5 = [itex]10^{-4}[/itex] and [itex]\zeta = 1[/itex].

    [tex]n \ge \sqrt{\frac{{(b-a)}^3}{12 \cdot \epsilon} \cdot f''(\zeta)}[/tex]
     
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