determining the number of divisions in riemann sums
2 ビュー (過去 30 日間)
古いコメントを表示
Hi, since riemann sum is all about adding smaller divided rectangles below the graph. I developed a code which calculates the difference between present sum and previous sum, when the difference is greater than 1e-5, the code must store the number of rectangles(terms) added to reach the difference and store the number of terms in n. c is the final area, the problem is that my code is not giving me the number of divisions nor the difference between sums, please help me find the (1).difference between sums and (2).the number of rectangles/divisions when the difference is greater than 1e-5
format longG
syms x
f = 2*pi*(0.16*(0.25-(x-1)^2)+44.52)*(1+(-0.32*x - 0.32)^2)^1/2;
a = int(f,5,17.688);
b = sym(a);
c = double(b)
if diff(c) > 0.00001
n = length(c)
end
0 件のコメント
採用された回答
David Hill
2020 年 4 月 26 日
You did not say if you wanted left, right, middle riemann sum. This is left.
f = @(x)2*pi*(0.16*(0.25-(x-1).^2)+44.52).*(1+(-0.32*x - 0.32).^2).^1/2;
a=1;%whatever bounds you want
b=3;%whatever bounds you want
Rs(1) = f(a)*(b-a);
Rs(2) = sum(f(linspace(a,b,2)).*((b-a)/2));
n=2;
while abs(Rs(n)-Rs(n-1))>1e-5
n=n+1;
Rs(n) = sum(f(linspace(a,b,n)).*((b-a)/n));
end
1 件のコメント
その他の回答 (0 件)
参考
製品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!