Exponential approximation for vector input

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Eduardo
Eduardo 2023-1-31
评论： Eduardo 2023-2-1
Ran in:
I was double checking the behaviour of a sigmoid function used in my Simulink model and I noticed that I was getting incorrect approximations when I made the computation for a vector of values
vect = [-5.0000 -5.0000 -5.0000 1.0000 0.9000 0.8000 0.7000 -5.0000 -5.0000];
y_vect = 1/(1+exp(-2*(vect'-1)));
% Value calculated using the vector
y_vect(4)
ans = 0
% Value calculated alone
y_val = 1/(1+exp(-2*(vect(4)-1)))
y_val = 0.5000
This approximation in my case causes great confussion due to the magnitude of the quantity expected.
Is there any way to solve this?

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采纳的回答

Sulaymon Eshkabilov
Ran in:
You have overlooked one dot. Here is the corrected commands:
vect = [-5.0000 -5.0000 -5.0000 1.0000 0.9000 0.8000 0.7000 -5.0000 -5.0000];
y_vect = 1./(1+exp(-2*(vect-1)));
% Value calculated using the vector
y_vect(4)
ans = 0.5000
% Value calculated alone
y_val = 1/(1+exp(-2*(vect(4)-1)))
y_val = 0.5000
  1 个评论
Eduardo
Eduardo 2023-2-1
Oh nice to know!
I wrongly thought the broadcasting would be done automatically since we just had a scalar in the numerator

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更多回答（1 个）

Voss
Voss 2023-1-31
Ran in:
vect = [-5.0000 -5.0000 -5.0000 1.0000 0.9000 0.8000 0.7000 -5.0000 -5.0000];
Using / (matrix right division), as you have it now:
y_vect = 1/(1+exp(-2*(vect'-1)));
disp(y_vect)
1.0e-05 * 0.6144 0 0 0 0 0 0 0 0
Using ./ (element-wise right division):
y_vect = 1./(1+exp(-2*(vect'-1)));
disp(y_vect)
0.0000 0.0000 0.0000 0.5000 0.4502 0.4013 0.3543 0.0000 0.0000
https://www.mathworks.com/help/matlab/matlab_prog/matlab-operators-and-special-characters.html

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