How to use the trapezium rule

Hi,

I've been trying to figure out how to use the trapezium rule and completely failing, since I'm a bit of a simpleton at Matlab.

I'm integrating the product of two functions and trying to find the value of the definite integral using the trapezium rule.

This is my code:

syms x f=(((x.^2)-1)^n);

Pn=(1/((2.^n).*factorial(n))).*diff(f,x,n); Pm=(1/((2.^m).*factorial(m))).*diff(f,x,m);

P=Pn.*Pm

x = -1 :0.1: 1;

y=P;

trapz(y,x)

I keep getting these errors:

??? Error using ==> trapz at 59 LENGTH(X) must equal the length of the first non-singleton dimension of Y.

Error in ==> Question7b at 11 trapz(y,x)

Thanks.

Answer 1

[Roger Stafford]

Pm would be the m-th Legendre Polynomial except that, as you have defined it, it appears to involve the m-th derivative of the n-th power of x^2-1, not its m-th power. I see no way for the m value to automatically be substituted for n in that formula. You may have to define another function, g, which uses m for this purpose.

Also I see no place where you have given n and m specific values before calling on 'trapz'. It does not know how to do its computation without being given specific numerical values for all its variables. As far as I know, it is incapable of producing a symbolic answer.

In fact n and m need to be defined before matlab can produce a function into which x values could be substituted. The form of the functions depends on the values of n and m.

I don't have an updated symbolic toolbox to check this with, but your error message indicates that the vector length of y is incorrect. You had better chose a small n and m and manually check out the forms of the Pn and Pm functions. Also check out that the proper substitution has been made of the x numerical values in getting y. With your x you should get a vector length for y of 21. If not, you need to find out why not.

Finally I notice that your call on 'trapz' is erroneous. It should be

 trapz(x,y)

not

 trapz(y,x).

(Note that you can also use 0.1*trapz(y) with your x sub-intervals all that same length.)

[Mike]

Hi Roger,

I was asked to create a function which computes the integral of P (which is equivalent to Pm*Pn) for any given value of n and m, using the trapezium rule. So far, as you can see I haven't figure out how to make it a function, and have neither figured out how to solve my errors. My code is relatively unchanged:

     syms x
       n=1;
       m=1;
        f=(((x.^2)-1)^n);
        Pn=(1/((2.^n).*factorial(n))).*diff(f,x,n);
        Pm=(1/((2.^m).*factorial(m))).*diff(f,x,m);
        P=Pn.*Pm
        x = -1 :0.1: 1;
    y=P;
trapz(x,y)
end

I get the following errors:

??? Undefined function or method 'max' for input arguments of type 'sym'.

Error in ==> trapz at 43 perm = [dim:max(ndims(y),dim) 1:dim-1];

Error in ==> Question7b at 12 trapz(x,y)

Thanks for your help, and sorry if I wasn't clear.

Answer 2

[Mike]

Ok so I edited the code to try and include the function legendrepolynomial:

 function y=legendrepolynomial(x)

    n=1;
    m=1;

    f=(((x.^2)-1)^n);
    g=(((x.^2)-1)^m);

    Pn=(1/((2.^n).*factorial(n))).*diff(f,x,n);
    Pm=(1/((2.^m).*factorial(m))).*diff(g,x,m);
    P=Pn.*Pm

 y=P;
 trapz(x,y)
 end

In the command window I input

 x=-1:0.1:1
 legendrepolynomial(x)

and the following errors show up:

??? Error using ==> mpower Inputs must be a scalar and a square matrix. To compute elementwise POWER, use POWER (.^) instead.

Error in ==> legendrepolynomial at 6 f=(((x.^2)-1)^n);

Thanks again.