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

# Please help me to look at the code. Thx

2 views (last 30 days)

Show older comments

Tianlan Yang
on 19 Oct 2021

Commented: Tianlan Yang
on 19 Oct 2021

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

##### 0 Comments

### Accepted Answer

per isakson
on 19 Oct 2021

Edited: per isakson
on 19 Oct 2021

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

### More Answers (1)

Walter Roberson
on 19 Oct 2021

inf(3)

[-1./inf (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]

[1-2]

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

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

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.)

##### 0 Comments

### See Also

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!