quintuple summation using a for loop

3 次查看(过去 30 天)
Ive tried setting the sum to an empty cell and filling it and not declaring it and I do not know what to do or whats wrong and why it will not give me an answer different than the inital declaration.
syms i j k l p q
U = ((-1)^j)*a' - ((-1)^i)*a;
V = ((-1)^j)*a' - ((-1)^i)*a;
W = ((-1)^j)*a' - ((-1)^i)*a;
r = sqrt(U*U+V*V+W*W);
Phi = -U*W*log(r-U)-V*W*log(r-V)+U*V*atan((U*V)/(r*W))-r*W;
for i = 0:1
for j = 0:1
for k = 0:1
for l = 0:1
for p = 0:1
for q = 0:1
sum = sum + (((-1)^(i+j+k+l+p+q))*Phi);
disp(sum)
end
disp(sum)
end
disp(sum)
end
disp(sum)
end
disp(sum)
end
disp(sum)
end
disp(sum)
  4 个评论
Wyatt Panaccione
Wyatt Panaccione 2020-6-23
%% Inputs
a = 0.035;%length of up/low magnet
b = 0.014;%width of up/low magnet
c = 0.007;%height of up/low magnet
am = .03;%length of middle magnet
bm = .01;%width of middle magnet
cm = 0.01;%height of middle magnet
%% Negative K
syms iw jw k l p q
U = ((-1)^jw)*am - ((-1)^iw)*a;
V = ((-1)^l)*bm - ((-1)^k)*b;
W = ((-1)^q)*cm - ((-1)^p)*c;
r = sqrt(U*U+V*V+W*W);
Phi = -U*W*log(r-U)-V*W*log(r-V)+U*V*atan((U*V)/(r*W))-r*W;
sum =[0];
for iw = 0:1
for jw = 0:1
for k = 0:1
for l = 0:1
for p = 0:1
for q = 0:1
sum = sum + (((-1)^(iw+jw+k+l+p+q))*Phi);
disp(sum)
end
end
end
end
end
end
disp(sum)

请先登录,再进行评论。

回答(1 个)

David Goodmanson
David Goodmanson 2020-6-23
编辑:David Goodmanson 2020-6-23
Hello Wyatt,
The sum comes out zero because it is, in fact, zero. Each term in the desired sum is
((-1)^(iw+jw+k+l+p+q))*constant;
Each of the indices is +-1. Their sum is either even or odd.
When it's even, the term contributes (-1)^even = +constant to the desired sum.
When it's odd, the term is contributes (-1)^odd = -constant to the desired sum.
There are 2^6 instances of (iw+jw+k+l+p+q), and half of those sums are even, half odd as you can check. So the desired sum ends up as zero.
It would be better if you went with Walter's recommendation and used something other than 'sum' for the sum variable. 'Sum' would work fine.

类别

Help CenterFile Exchange 中查找有关 Logical 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by