integral with vectoric varying coeficient

1 回表示 (過去 30 日間)
fima v
fima v 2020 年 5 月 6 日
コメント済み: Ameer Hamza 2020 年 5 月 7 日
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);

採用された回答

Ameer Hamza
Ameer Hamza 2020 年 5 月 6 日
編集済み: Ameer Hamza 2020 年 5 月 6 日
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);
  5 件のコメント
Ameer Hamza
Ameer Hamza 2020 年 5 月 7 日
編集済み: Ameer Hamza 2020 年 5 月 7 日
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 2020 年 5 月 7 日
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')

サインインしてコメントする。

その他の回答 (1 件)

David Hill
David Hill 2020 年 5 月 6 日
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 件のコメント
fima v
fima v 2020 年 5 月 6 日
because i want the coeffecient to be tied to the interval
so it needs to be during the integration.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeCreating and Concatenating Matrices についてさらに検索

Community Treasure Hunt

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

Start Hunting!

Translated by