MATLAB: Sum of Maple to MATLAB

maplemaple to matlabsum

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.

Best Answer

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.
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