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integral with vectoric varying coeficient

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fima v
fima v on 6 May 2020
Commented: Ameer Hamza on 7 May 2020
Hello , i have a basic function
exp(-x.^2).*log(x).^2
which i integrate in a certain interval.
i want to multiply my basic function with a vectoric coefficient called coef_vec that varies with the interval.
so if the integral is at 5 my basic function whould be multiplied with coef_vec(5).
i know that its some how turning the integral into a loop.
Is it possible in matlab?
Thanks.
coef_vec=linspace(1,10,100)
fun = @(x)coef_vec*exp(-x.^2).*log(x).^2;
q = integral(fun,1,10);

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Accepted Answer

Ameer Hamza
Ameer Hamza on 6 May 2020
Edited: Ameer Hamza on 6 May 2020
You need to consider that value of x is not always an integer, it can be a floating-point too. In that case, coef_vec(x) will fail. You need to interpolate your vector to give value on that range. Try this code
coef_vec = linspace(1,10,100);
coef_fun = @(x) interp1(1:numel(coef_vec), coef_vec, x);
fun = @(x) coef_fun(x).*exp(-x.^2).*log(x).^2;
q = integral(fun,1,10);

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fima v
fima v on 7 May 2020
Hello Ameer, but i need a continues formula to use in my integral.
vq = interp1(x,v,xq)
how do i convert my vq data into a contines formula?
Thanks.
Ameer Hamza
Ameer Hamza on 7 May 2020
I have defined coef_fun as a function handle using interp1 so that it can be continually evaluated at all the points. If you just use interp1 alone, then you will again get a finite vector. The function handle allows the integral to evaluate interp1 at an arbitrary point.
Following code shows the difference between interp1 and polyfit
x = 0:0.1:10;
coef_vec = gensig('square', 2, 10, 0.1) + 0.1*rand(size(x)).';
coef_fun = @(xq) interp1(x, coef_vec, xq);
xq = linspace(0, 10, 200);
p = polyfit(x, coef_vec, 10);
figure;
plot(x, coef_vec, '+-', xq, polyval(p, xq), '-o')
title('Polyfit')
figure;
plot(x, coef_vec, '+-', xq, coef_fun(xq), '-o')
title('interp1')
Ameer Hamza
Ameer Hamza on 7 May 2020
Try this
data = load('Default Dataset4.csv');
x = data(:,1);
y = data(:,2);
[x, index] = unique(x);
coef_fun = @(xq) interp1(x, y(index), xq);
xq = linspace(3.5,23,100000);
plot(xq, coef_fun(xq))
title('interp1')

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More Answers (1)

David Hill
David Hill on 6 May 2020
Why not do it after computing the integral since it is a constant.
fun = @(x)coef_vec*exp(-x.^2).*log(x).^2;
q = integral(fun,1,10);
q=q*linspace(1,10,100);

  1 Comment

fima v
fima v on 6 May 2020
because i want the coeffecient to be tied to the interval
so it needs to be during the integration.

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