Mid point value of a function

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Mubashara Wali
Mubashara Wali 2021 年 2 月 5 日
回答済み: Jan 2021 年 2 月 7 日
I need to find the sum with value of T at mid of every subinteval
for j= 1:1:n+1
Rsum2=Rsum2+bb(j+1)*T(j+1/2)
end
but get an error
"Array indices must be positive integers or logical values."
How can I find value of T at midpoint of every interval in this loop?
  3 件のコメント
Mubashara Wali
Mubashara Wali 2021 年 2 月 7 日
How can I find this sum in a loop?

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回答 (2 件)

Jan
Jan 2021 年 2 月 5 日
T(j+1/2) is not valid, when T is a vector. You need integer values as indices.
Perhaps you mean:
(T(j) + T(j+1)) / 2

Jan
Jan 2021 年 2 月 7 日
interval is [0,1] with n subintervals and I need to find the sum of the function at midpoint of every subinterval.
n = 27;
interval = linspace(0, 1, n);
% If T is vectorized:
Result = sum(T(interval(1:n-1) + interval(2:n)) / 2)
% If T works for scalars only:
Result = 0;
for k = 1:n-1
Result = Result + T(0.5* (interval(k) + interval(k+1)));
end

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