Vectors and Sum operation

1 visualización (últimos 30 días)
bondpen
bondpen el 19 de Mzo. de 2018
Comentada: bondpen el 19 de Mzo. de 2018
Trying to use sum to figure about resistor formula, 1/[(1/r1)+1rn)] where n can be any number.
What i have come up with:
function R = scirpt(y)
R = 1/sum(1/(y));
end
Well be entering this:
scirpt([100,200,300]) should output 54.5455
if true
% code
end

Respuesta aceptada

John D'Errico
John D'Errico el 19 de Mzo. de 2018
Editada: John D'Errico el 19 de Mzo. de 2018
R = 1/sum(1./y);
y is a vector. You want to divide EACH of the elements of y into 1. So you use the ./ operator, which is designed to solve element-wise operations. (Just like the .* and .^ operators.)
The other divide is a scalar/scalar operation, so / works fine there, although ./ would also be acceptable. That is:
R = 1./sum(1./y);
would also work.
Finally, there was no need to put a parens around y in your original line. So while it is fine to add additional parens if you are not sure you need them in some spot, too many parens that need not be there will only make things confusing, hard to read and debug.
  1 comentario
bondpen
bondpen el 19 de Mzo. de 2018
Thanks for explaining why you need ./ operator. Thanks.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Stair Plots en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by