Implementing the Trapezoidal Method

Hi. The aim is to utilize Trapezoidal Method for an integral with no variables, so we will get the x after integration.

where U is a constant number and u is a set of 37 experimental data. here is what I have tried:

input= [18.83183;18.85426;18.84941;18.81932;18.68937;18.35632;18.01761;17.59767;16.89197;16.18128;15.58579;14.55176;13.93881;12.94744;12.06294;11.14389;10.32951;9.8233;9.33477;8.77343;8.83133;8.52088;9.06752;9.56366;10.71528;11.46361;12.7156;13.63052;14.39822;15.24683;16.28031;17.07743;17.68488;18.46132;18.78772;18.87824;18.90421];
f=@ (x) (1-(input/18.73835 ).^2 ); 
a=-0.05;                % Lower limit
b=0.05;                 % Upper limit
h=0.05;                 % Increment
n=(b-a)/h;              % Number of subintervals
sum=0;
for k=1:1:n-1
  x(k)=a+k*h;
  y(k)=f(x(k));                 %%%%%% line 31
  sum=sum+y(k);
end
% Formula:  (h/2)*[(y0+yn)+2*(y1+y2+y3+..+yn-1)]
result = h/2*(f(a)+f(b)+2*sum);
fprintf('\n Friction coefficient is: %f',result);

Error message:

Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

Error in iki_cd (line 31)
  y(k)=f(x(k));
  

I need your help to fix the code. Thanks.

[Jan]

f(x) replies a constant vector independent from the argument x. Then f(x(k)) is a vectir, but y(k) is a scalar.

If your function is constant, there is no need to apply the trapezoidal method to get the integral. If you calculate the integral over a vector, the output is a vector also. It looks like you want to compute a scalar result. So maybe you want to get the mean value and no integral?

[Turgut Ataseven]

@Jan Thanks for the reply.

I want to calculate the result of the integral, which is 0.1*[1-u^2/U^2], and then put the "input" values to u one by one, thus get a single scalar value, "result".

[Torsten]

"u" in your integral is the array "input", I guess.

You have 37 array values - what are the corresponding x-values in the interval [-0.05:0.05] ?

Equidistant points

x = linspace(-0.05,0.05,37)

?

Then

result = trapz(x,1-(input/18.73835 ).^2)

But if "input" are velocities, should it be highest in the middle, i.e. at x=0 (input(19)) ?

Answer 1

[Ravi]

Hi Turgut Ataseven,

I understand that you are facing an issue in implementing trapezoidal method for computing the numeric integration value.

I have noticed a small error in the code where you are computing the “f” function value. Instead of passing “x” as the parameter, “f” is taking “input” as parameter in computing the function. On replacing “input” with “x” you would obtain the desired results.

Please find the below modified code.

input= [18.83183;18.85426;18.84941;18.81932;18.68937;18.35632;18.01761;17.59767;16.89197;16.18128;15.58579;14.55176;13.93881;12.94744;12.06294;11.14389;10.32951;9.8233;9.33477;8.77343;8.83133;8.52088;9.06752;9.56366;10.71528;11.46361;12.7156;13.63052;14.39822;15.24683;16.28031;17.07743;17.68488;18.46132;18.78772;18.87824;18.90421]; 
f=@ (x) (1-(x/18.73835 ).^2 );  
a=-0.05;                % Lower limit 
b=0.05;                 % Upper limit 
h=0.05;                 % Increment 
n=(b-a)/h;              % Number of subintervals 
sum=0; 
for k=1:1:n-1 
    x(k)=a+k*h; 
    y(k)=f(x(k));                 %%%%%% line 31 
    sum=sum+y(k); 
end 
% Formula:  (h/2)*[(y0+yn)+2*(y1+y2+y3+..+yn-1)] 
result = h/2*(f(a)+f(b)+2*sum); 
fprintf('\n Friction coefficient is: %f',result); 

 Friction coefficient is: 0.100000

I hope this modification resolves the issue you are facing.

Thanks,

Ravi Chandra