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Math Help - integration matlab qn

  1. #1
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    integration matlab qn

    1. The problem statement, all variables and given/known data
    Let us consider an approximation to an integral. Let f(x) be some continuous function on
    [a, b]. We wish to find an approximation for the integral
    I = int from a to b of f(x)dx
    in the following manner:
    Subdivide the interval into N intervals of length h = (b−a)/N. Let xi = ih for i = 0, . . . ,N.
    Let
    Ij = int from 0 to h of f(xj+t)dt

    Find a cubic polynomial Pj (x) that goes through (xj , f(xj)), (xj + h/3,f(xj + h/3)),(xj+2h/3, f(xj+2h/3) and (xj+1,f(xj+1))
    We form an approximation for the integral by letting
    I=sum(j=0 to N-1) of w0*f(xj)+w1*f(xj+h/3)+w2*f(xj+2h/3)+w3*f(xj+1)
    Find these weights, wi.
    In 2 peices of code, plot the first three Bessel functions, J0(x), J1(x) and J2(x), on the
    interval [0, 20]. The first peice of code should be a MATLAB function BJ(x, n) outputing
    the approximation for the integral representation of Jn, given by
    Jn(x) =(1/pi)int from 0 to pi of cos(nt − x sin t)dt
    using the above method for 100 subdivisions of [0, pi]. The second peice of code should call
    the function an produce the required plots with 2000 subdivisions of [0, 20].

    Im just gobsmacked with this qn.. as i only started using matlab a couple of days, ago and have no programming experience.

    what i have done so far is really no good, but i have no idea.

    function [Jn]=BJ(x,n)
    N=100;
    b=pi;
    a=0
    h=(b-a)/N;
    x=a:h:b;
    xj=i*h;
    Ij=0;
    J0=cos(n*xj-x*sin(xj));
    J1=cos(n*(xj+h/3)-x*sin(xj+h/3));
    J2=cos(n*(xj+(2*h/3))-x*sin(xj+(2*h/3)));
    J3=cos(n*(xj+1)-x*sin(xj+1));
    for i=0:N;
    Ij=Ij+w0*J0+w1*J1+w2*J2+w3*J3;
    J=(1/pi)*Ij;
    end

    can someone please help me
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  2. #2
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    my lecturer told me this... when i asked for help... but have no idea what it means, or what he has done

    The trick about finding the weights is to consider the Polynomial of
    degree 3 of the form

    P(x) = f(xj) P_0(x) + P_1(x)f(xj + h/3) + P_2(x)f(xj + 2h/3) P_3(x)f(xj + h)

    dont try ax^3 + bx^2 + cx + d, that derivation is many many pages
    long, and we require a trick from polynomial interpolation to avoid
    complications. It is a trick we used in one of the derivations of
    simpsons rule.

    Now, we can transform this integral to one over 0 to h, which just
    simplifies the calculations to a page or two less. P_i are all of
    degree 3 and
    P(0) = f(xj)
    P(h/3) = f(xj+h/3)
    P(2h/3) = f(xj+2h/3)
    P(h) = f(xj + h)
    this defines the polynomials completely as
    P_0(0) = 1 and P_1(0) = P_2(0) = P_3(0) = 0
    P_1(h/3) = 1 and P_0(h/3) = P_2(h/3) = P_3(h/3) = 0
    P_2(2h/3) = 1 and P_0(2h/3) = P_1(2h/3) = P_3(2h/3) = 0
    P_3(h) = 1 and P_0(h) = P_1(h) = P_3(h) = 0
    these define the P_i completely since you know each polynomials roots.
    Now w_i is the integral of P_j over 0 to h. This is exactly like what
    we did for Simpsons rule.
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  3. #3
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    ok I looked through it and i got..

    P_0=(0,1),(h/3,0),(2h/3,0),(h,0)
    P_1=(0,0),(h/3,1),(2h/3,0),(h,0)
    P_2=(0,0),(h/3,0),(2h/3,1),(h,0)
    P_3=(0,0),(h/3,0),(2h/3,0),(h,1)

    is that right?
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  4. #4
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    is it something like Simpson's 3/8 rule?? i just wiki'd and since its interpolated with a cubic instead of a quadratic, it mite be that? But how do i derive the rule from what i have? or even better code it?
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  5. #5
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    ok from Pj(x).. and integrating over 0 to h i got

    h/8*P_0+3h/8*P_1+3h/8*P_2+h/8*P_3 this is really i close i think.. but not sure, please can anyone help?
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