how to calculate cauchy product in summation
8 次查看(过去 30 天)
显示 更早的评论
a(1)=1;
b(1)=1;
c(1)=1;
S=0;
Z=0;
for i=1:20
for p=1:i-1
a(i+1)=(1/(i+1)).*(a(i)-b(i));
b(i+1)=(1/(i+1)).*(a(i)-S-b(i));
c(i+1)=(1/(i+1))*(Z-c(i));
S=S+a(p)*c(i-p);
Z=Z+a(p)*b(i-p);
disp([S Z])
end
end
S ,Z are cauchy product how to calculate
2 个评论
Ankit
2022-3-4
what problem you are facing ?.. inside the loop you are trying to access the element which is not defined
a(i+1)=(1/(i+1)).*(a(i)-b(i));
回答(1 个)
Jan
2022-3-4
The shown code cannot run. In the 1st iteration of the "for i" loop, the inner loop is not entered: "for p = 1:i-1". In the 2nd iteration i=2 the innerloop is entered and in the line
a(i+1)=(1/(i+1)).*(a(i)-b(i));
% ^^^ ^^^
The not existing elements a(2) and b(2) are requested.
Seeing only the failing code does not allow to suggest an improvement, because the readers have no chance to guess, what you want to do instead.
3 个评论
Jan
2022-3-4
编辑:Jan
2022-3-4
@shiv gaur: You have been asked repeatedly to use a proper code formatting. Please read an consider this: https://www.mathworks.com/matlabcentral/answers/help/rtc#rtc_summary . Thanks.
"pl see the runing program" - the code still stops with an error and we still cannot guess, what it should calculate.
"Resolve the problem" is not a polite way to ask for help, but a rough command. Refusing to post your code in a formatted way and posting not running code repeatedly let it look, like it is your intention to provoke the community.
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!