Taking the sum of exponentials
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L'O.G.
le 22 Mar 2022
Réponse apportée : Star Strider
le 22 Mar 2022
To do the calculation,
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/937934/image.png)
is the following code correct? Is the (:) necessary? I think that makes a column vector, but I don't think it's necessary. I don't think cumsum would be useful here. Could somebody please advise? If it matters,
. And a is a vector of
.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/938124/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/937949/image.png)
x = linspace(0.01,100,10000);
f_x = sum(exp(-x./a(:)));
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Star Strider
le 22 Mar 2022
‘If it matters,
. And a is a vector of
.’
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/937999/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/938004/image.png)
I don’t see ‘t’ defined anywhere, so it depends on what relationship ‘t’ has to ‘x’ or a.
However, since as @_ noted, since a is already a column vector by definition, the (:) is not necessary. The important aspect is that with respect to MATLAB coding, ‘x./a’ must be a matrix in order that sum produces the row vector necessary for the result.
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Paul
le 22 Mar 2022
Break it up with simple inputs to see what's going on:
x = [1 2]; % row vector
a = [1 2 3]; % row vector
x./a(:) % implicit expansion makes a 3 x 2 matrix
exp(x./a(:)) % element wise exp
sum(exp(x./a(:))) % sum down the columns
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