Octave/Matlab: Difference between e^(-1*z) and exp(-1*z)

Question

I am new with Octave and I have a problem.I thought the following codes were the same, but they produces different results. What is the difference? Thanks

Octave/Matlab: Difference between e^(-1*z) and exp(-1*z)

g = 1./(1 + e^(-1*z));

g = 1./(1 + exp(-1*z));

Where z is a vector, element or matrix

Answer 1

In Octave

exp(1) equals e where e is Euler's number.

There are 4 operations/functions that are to be noted here:

e^x is same as expm(x) and e.^(x) is same as exp(x).

• e^x and expm(m) represent e raise to the matrix x.
• e.^(x) and exp(x) represent exponential e^x for each element in matrix x.

If x is a scalar then all (e^x, expm(x), e.^x and exp(x)) are mathematically equal.
For your case, z is a matrix and hence you get different results.

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In MATLAB,

e is not defined in MATLAB. exp(x) and expm(x) have same definitions in MATLAB as those that are described for Octave above.

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_{PS: e or E are also used for E-notation in both MATLAB and Octave but that's a different thing.}

Answer 2

In Octave, it is important to note that e^x and exp(x), where x is a double precision scalar variable, are not necessarily the same.

For instance:

>> a = e ^ 2
a =  7.3891

>> b = exp (2)
b =  7.3891

>> b - a
ans = 8.8818e-16

The reason is that exp (2) uses a dedicated algorithm to compute the exponential function, while e ^ 2 actually calls the function e () to get the value of e, and then squares it:

>> c = realpow (e (), 2)
c =  7.3891

>> c - a
ans = 0

Another reason why e ^ x and exp (x) differ is that they compute completely different things when x is a square matrix, but this has already been discussed in Sardar's answer.