MATLAB Bisection method and numerical integration

Click For Summary
The discussion focuses on using the composite Simpson's rule and the bisection method in MATLAB to find the upper limit of integration, ##x_l##, such that the integral equals 170. The user is encountering errors in their code, particularly in calculating the integral at each midpoint of the bisection method. Suggestions include using MATLAB's built-in numerical integrators like `quadl`, `quadgk`, or `integral` for the integration, and `fzero` for root finding, which can simplify the process. A correct implementation using `fzero` is provided, demonstrating how to achieve the desired result. The conversation emphasizes the importance of accurately defining the problem and utilizing MATLAB's built-in functions for efficiency.
roam
Messages
1,265
Reaction score
12
In Matlab I am trying to use the composite Simpson's rule to find ##x_l## so that

$$170=\int^{x_l}_0 \sqrt{1+(y')^2} dx = \int^{x_l}_0 \sqrt{1+\left( \frac{x^2}{68000} \right)^2} dx $$

For convenience this can be written as

$$I(x) = 170 - \int^x_0 \sqrt{1 + (\frac{x^2}{68000})} dx$$

The limits of integration would be from ##0## to ##x=170##. Now to find the zero of this function I want to employ the bisection method while using Simpsons rule to evaluate the integral involved in evaluating ##I(x)## at each step.

Here's my code so far:

Code:
a=0; b=170;

for x=[0:]

f = sqrt(1+((x.^2)./68000).^2);

   %Simpson's rule
   if numel(f)>1
    n=numel(f)-1; h=(b-a)/n;
    I= h/3*(f(1)+2*sum(f(3:2:end-2))+4*sum(f(2:2:end))+f(end));
        else
            h=(b-a)/n; xi=a:h:b;
            I= h/3*(f(xi(1))+2*sum(f(xi(3:2:end-2)))+4*sum(f(xi(2:2:end)))+f(xi(end)));
   end
  f = 170 - I;tol=1e-6;
while abs(b-a) > tol
    x = (a+b)/2;
    y = ff(x);

    if y == 0
        rv = x;
        break
    end

    if ff(a)*y < 0
        b = x;
    else
        a = x;
    end
end

rv=(a+b)/2;

But I get lots of errors and the code does not run. I think it is because I need to get the Simpson's rule to calculate new value for the function at each midpoint of the bisection method. I'm not really sure how to fix this. :confused:

Alternatively, if bisection does not work with Simpson's method, I would appreciate it if anyone could show me how to exactly incorporate built in root finders into my code.

Any help would be greatly appreicated.
 
Last edited:
Physics news on Phys.org
I'd use regula-falsi for the root finding part of this problem. If memory serves me correctly, MATLAB has a simpson's rule built in...
 
  • Like
Likes roam
The question is confusing because first you say you are trying to find the upper limit of integration ##x_l## (such that the value of the integral is 170), but then later you say that the upper limit of integration is 170.

If you are trying to find the upper integration limit such that the value of the definite integral is 170, the following will work. quadl is a numerical integrator that comes with MATLAB, but you could also use quadgk or integral to the same effect. Also, fzero is the built-in root finder.

Code:
>> f = @(xl) quadl(@(x)sqrt(1+(x.^2/68000).^2),0,xl);
>> R = fzero(@(x) 170-f(x),4)
R =
  167.2340
>> quadl(@(x)sqrt(1+(x.^2/68000).^2),0,R)
ans =
  170.0000
 
Last edited:
  • Like
Likes roam and DrClaude

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K