Sum of Maple to MATLAB
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Hi All,
There is a saying "A pee to an ant is flooding".... I guess my question appears simple but being new to MATLAB, couldn't find my way. Seeking some help...
Lm = [1 2 3]';
Um = [4 5 6]';
Maple:
sum(Lm[p,1] * sum(Lm[q,1] * (Um[p,1] - Um[q,1]),q=1..3), p=1..3)
What will be the equivalent MATLAB for the above?
Thank you.
1 个评论
Matt Fig
2011-5-14
I would have to know what that Maple command did first! Since this is a MATLAB site, it is possible that nobody on here knows what that Maple command does...
采纳的回答
Walter Roberson
2011-5-14
The most direct equivalent would be
Lm(1,1) * (Lm(2,1) * (Um(1,1) - Um(2,1)) + Lm(3,1) * (Um(1,1) - Um(3,1))) + Lm(2,1) * (Lm(1,1) * (Um(2,1) - Um(1,1)) + Lm(3,1) * (Um(2,1) - Um(3,1))) + Lm(3,1) * (Lm(1,1) * (Um(3,1) - Um(1,1)) + Lm(2,1) * (Um(3,1) - Um(2,1)))
Which works out to be 0
If we generalize to
sum(Lm[p,1] * sum(Lm[q,1] * (Um[p,1] - Um[q,1]),q=1..N), p=1..N) %MAPLE
where N is the length of Um and Lm, then:
breaking it down,
T(p) := sum(Lm[q,1] * (Um[p,1] - Um[q,1]),q=1..N) %Maple
would be
T(p) = sum(Lm .* (Um(p)-Um)) %MATLAB
and when Lm and Um are column vectors, that could be written
T(p) = Lm.' * (Um(p)-Um) %MATLAB
Then,
sum(Lm[p,1] * T[p,1],p=1..N) %Maple
would be
T = zeros(length(Lm),1); %MATLAB
for p=1:T
T(p) = Lm.' * (Um(p)-Um)
end
sum(Lm .* T)
but the sum(Lm .* T) could also be written as Lm.' * T leading to
T = zeros(length(Lm),1); %MATLAB
for p=1:T
T(p) = Lm.' * (Um(p)-Um)
end
Lm.' * T
In turn, in MATLAB that could be written as
Lm.' * arrayfun(@(p) Lm.' * (Um(p)-Um),1:length(Lm)).'
Note here that arrayfun is returning a row vector instead of a column vector, so the .' after the arrayfun is needed to switch back to a column vector for the array multiplication.
更多回答（3 个）
Matt Fig
2011-5-14
It would help, for those of us who do not have Maple and consequently do not know what the result of that Maple command is, if you would show the expected output. Even if you have to make a simple example and create the output by hand, show inputs and expected outputs.
The best I can guess would be:
S = 0; % This is the result you want at the end.
for p = 1:3
tmp = 0;
for q = 1:3
tmp = tmp + Lm(q) * (Um(p)-Um(q));
end
S = S + tmp*Lm(p);
end
S % Look at the final result...
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