Operations on anonymous functions in MATLAB

Question

My code is for approximating functions using Haar wavelet for different levels of approximation. However when I try to calculate the error using the formula squareroot(integral(squareof(u(t)-uk(t)))) over the limits 0 to 1 where u(t) is the function f to be approximated uk(t) is the Haar approximation of the function until k terms However I'm not able to calculate the error properly due to some rogue matlab operation.Here's my code. The code for phi is

function a=phi(x)
if(0 <= x & x< 1)
a=1;
else
a=0;
end

function approxx(j)
f=@(x)sin(x);
b=@(j,t,k)phi((power(2,j)*t)-k);
a=@(x,j,k)(f(x).*b(j,x,k));

sum=@(x)0;
for k=0:power(2,j)
   ak=integral(@(x)a(x,j,k),power(2,-j)*k,power(2,-j)*(k+1));
   c=@(x)ak*power(2,j)*phi((power(2,j)*x)-k);
   sum=@(x)(sum(x)+c(x));    
end
fplot(f,[0,1],'r');hold on;
fplot(sum,[0,1]);

er=power(integral(@(x)power(f(x)-sum(x),2),0,1),0.5);
disp(er);
end

What is phi? Also: What output do you get, what would you expect instead? — knedlsepp
– knedlsepp, Commented Apr 9, 2015 at 17:42
i have updated my code check out the definition of phi in it.. — Hemanshu Sethi
– Hemanshu Sethi, Commented Apr 9, 2015 at 18:05
actually this is the haar approximation of a function and when we increase the value of j in the function that we have written the error(er in code) should decrease but instead it shows some random values being calculated and also it's increasing. — Hemanshu Sethi
– Hemanshu Sethi, Commented Apr 9, 2015 at 18:11

knedlsepp · Accepted Answer · 2015-04-09 18:55:33Z

1

Your function phi is not properly vectorized.

The call

integral(@(x)power(f(x)-sum(x), 2), 0, 1)

uses a vectorized integration method to compute the integral, i.e. it evaluates the function for a vector instead of a single value. sum(x) for a vector x = linspace(0,1,100) however returns a single double value instead of a vector of length 100. MATLAB thinks you want to subtract a single double value sum(x)==0 from the vector f(x) and doesn't throw an error, but computes the wrong values instead.

You should replace phi with it's correctly vectorized version:

function a=phi(x)
a = double(0<=x & x<1);
end

BTW: Consider directly computing the values instead of using nested anonymous functions. Those will get really really slow for large depth of nesting. Also sum is a built-in function; you should refrain from naming your variables like built-ins.

edited Apr 9, 2015 at 18:55

answered Apr 9, 2015 at 18:43

knedlsepp

6,1043 gold badges22 silver badges43 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

knedlsepp Over a year ago

@HemanshuSethi: If this solved your problem, please consider accepting my answer by clicking the grey check mark next so it turns green.

Collectives™ on Stack Overflow

Operations on anonymous functions in MATLAB

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related