sum of series. Vectorised (no loop)
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Respuesta aceptada
Khalid Mahmood
el 27 de Abr. de 2021
Editada: Khalid Mahmood
el 27 de Abr. de 2021
% To reduce 2 more lines
function s=vsum(n)
if nargin<1
s=1; return
end
if mod(n,2)==0, n=n-1;end
s=1+sum(1./[3:2:n] -1./[2:2:n])
3 comentarios
DGM
el 28 de Abr. de 2021
Editada: DGM
el 28 de Abr. de 2021
It's perfect if you want the wrong answer 50% of the time.
This is demonstrable. Just test it.
function s=vsum(n)
if nargin<1
s=1; return
end
if mod(n,2)==0, n=n-1;end
s=1+sum(1./[3:2:n] -1./[2:2:n]);
end
Test with odd argument:
s1 = vsum(5)
s2 = 1 - 1/2 + 1/3 - 1/4 + 1/5
results match
s1 =
0.7833
s2 =
0.7833
Test with even argument:
s1 = vsum(4)
s2 = 1 - 1/2 + 1/3 - 1/4
results don't match
s1 =
0.8333
s2 =
0.5833
This whole thing looks like an attempt to make the vector lengths match when they shouldn't.
s2 = sum(1./(1:2:nt))-sum(1./(2:2:nt))
is simpler and actually correct.
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