Please help me to look at the code. Thx

I don't know what is wrong with that code. Here is the screenshot of the issue and the code:

function intGL=gausslegendre (a,b,f,M,varargin)
y = [-1/ sqrt (3),1/ sqrt (3)];
H2 = (b-a)/(2*M);
z = [a:2*H2:b];
zM = (z(1:end -1)+z(2:end ))*0.5;
x = [zM+H2*y(1), zM+H2*y(2)];
f = f(x,varargin {:});
intGL = H2*sum(f);
return

Answer 1

[per isakson]

This returns a numerical result, but it's not an integer.

fun = @(x) exp(-x.^2./2);
int = gausslegendre( 0, 2, fun, 1 )
function intGL=gausslegendre (a,b,f,M,varargin)
y = [-1/sqrt(3),1/sqrt(3)];
H2 = (b-a)/(2*M);
z = [a:2*H2:b];
zM = (z(1:end -1)+z(2:end ))*0.5;
x = [zM+H2*y(1), zM+H2*y(2)];
f = f(x,varargin{:});
intGL = H2*sum(f);
end

Comments

• sqrt (3) (with a space) works at the command line but not in a function (R2018b)

[Tianlan Yang]

That really helps. Thank you !!

Answer 2

[Walter Roberson]

inf(3)

ans = 3×3

Inf Inf Inf Inf Inf Inf Inf Inf Inf

[-1./inf (3)]

ans = 1×2

0 3

If inf(3) returns a 3 x 3 array of inf, then why doesn't [-1./inf (3)] take -1 divided by a 3 x 3 array of inf?

The answer is that there is a space between the inf and the (3), and within [], space is the concatenation operator unless there is an operator before the space

[1 2]

ans = 1×2

1 2

[1-2]

ans = -1

[1 -2]  %space before operator but no space between operator and number --> unary operator

ans = 1×2

1 -2

[1 - 2] %space before and after operator --> subtraction

ans = -1

So your line

[-1/ sqrt (3),1/ sqrt (3)];

is being treated as

[-1/ sqrt() (3),1/ sqrt() (3)];

which in turn is

[(-1/ sqrt()), (3), (1/ sqrt()), (3)];

and that fails because sqrt() cannot be invoked with no parameters. My example with inf did not complain because inf can be invoked with no parameters (and usually is.)