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Help with Newton-Raphson Method

  1. Feb 13, 2008 #1
    Hello All,
    I am new to this forum, and was wondering if anyone could help me with a homework problem I am having.

    1. Consider the Integral, int(h(x)dx)=c, integrated from lower limit "a" and upper limit "b".
    where "a" is unknown and 0<a<b, c>0, h(x)>0, a<=x<=b.

    Task: Develop a general expression for determining "a" by a Newton-Raphson procedure, assuming that b,c,h(x) are known.

    2. Relevant equations

    I know I need to derive the NR equation below in order to solve for the unknown lower limit.

    A(p+1)=A(p) -[f(Ap)/f ' (Ap)]

    I am just not sure how to apply the integral of a function to the NR method. Any help will be greatly appreciated!
  2. jcsd
  3. Feb 13, 2008 #2
    You want to set it up so you're finding the root of an equation. So you want to find [tex]a[/tex] to solve
    [tex] \int_a^b h(x)dx - c = 0 [/tex]
    So if you know h(x), you could solve
    [tex] F(b) - F(a) - c = 0 [/tex]
    where F is the anitderivative of h. Here you get the derivative for free because the derivative of F is h, which you know. Or if you don't know F, you could do the integral numerically. For example, using the trapezoidal rule you get something like
    [tex] \frac{h(b)}{2} + \sum_j h(x_j) + \frac{h(a)}{2} - c = 0 [/tex]
    In this case you would be computing derivatives of h(a) with respect to a.
    I don't know if any of those are correct, but they're some suggestions to start with.
  4. Feb 14, 2008 #3
    Okay, thanks for the input.

    I am not given b,c, or h(x), I am just supposed to assume that they are provided, and then use that to develop an expression for determining "a" using the Newton-Raphson method.

    So if I say that F(b)-F(a) -c = 0, can I also say that

    [tex]\int[/tex] h*(b - a) - c = 0 ? (integral of h multiplying b subtracted by integral of h multiplying a)

    I think I need to use Leibnitz's Rule, but I am a little confused how to apply this equation to the rule. Thanks!
    Last edited: Feb 14, 2008
  5. Feb 16, 2008 #4
    My take on it


    [tex] F(x) = \int_a^b {h(x) dx} = c [/tex]

    which is also

    [tex] F(b) - F(a) = c [/tex]

    Then, for our zero function, let

    [tex]G(a) = F(b) - F(a) - C = 0 [/tex]

    Note that here, our variable is "a". Thus, our Newton search will be expressed as:

    [tex] \[
    a_{\nu+1} = a_{\nu} - \frac{G(a_{\nu})}{G'(a_{\nu})}

    Here we note that the prime on G indicates a derivative wrt a

    For the derivative, we have (assuming b and c are constants)

    G'(a) = \frac{d G(a)}{da} = -h(a)

    Thus, the iteration scheme is

    [tex] \[
    a_{\nu+1} = a_{\nu} + \frac{\int^b_{a_{\nu}} {h(x) dx} - c}{h(a_{\nu})}
  6. Feb 21, 2008 #5
    Looks like you want to write your own code. Try looking up a book called Numerical Recipies in C. It's free online.
  7. Feb 21, 2008 #6
    hi all, i've had this same question. and when i go to apply the method it seems to make sense that the f(a)/f'(a) term should be smaller than the a term in the equation such that there is actual convergence.

    well, no matter what i do i can't get f(a)/f'(a) to be smaller than my inital guess. i'm ripping my hair out. anyone have any suggestions?
  8. Feb 21, 2008 #7
    I've had it work for some contrived examples. But you have to be careful as, just like all Newton-Raphson implementations, the starting guess must be sufficiently close to the root.

    Also, if a numerical procedure is being used to evaluate the integral, care must be exercised such that the evaluation is performed with sufficient accuracy to allow convergence. Of course, for the iteration process, you don't want to go too crazy here - pick an efficient solver. A delicate balance here.

    Can you provide some more detail on the actual problem you are trying to solve and methods you are using?

    P.S. I don't think the Numerical Recipies one line books are free anymore. However, efficient solvers are readily available - may need to select based on problem, but Simp's Rule may be adequate for most depending on the precision of "a" one is trying to solve for.
  9. Feb 21, 2008 #8
    the function is x^2*exp(x)

    integrated from a to b=3 set =c which is 97.704

    then we have to find a convergence to a

    so i integrated, and solved for the subsequent function of a=0

    I plotted the function and saw a root near 1. so it seems i'm on the right track. Now i just need to figure out how to make a loop in matlab....
  10. Feb 21, 2008 #9
    Yep, sounds like you're on the right track.

    GSXtuner21 - did you have any other questions? (you're OP after all).
  11. Feb 21, 2008 #10
    I think you guys have sufficiently answered my questions and then some, so thank you very much for the help!

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