Help evaluating complex function in form m+ni?

  • #1
NotASmurf
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Hey all, I need the complex version of the sigmoid function in standard form, that is to say $$f(\alpha) =\frac{1}{1+e^{-\alpha}} , \hspace{2mm}\alpha = a+bi , \hspace{2mm} \mathbb{C} \to \mathbb{C}$$ in the simplified form: $$f = m+ni$$ but found this challenging, for some reason i assumed there was an identity for $$e^{e^{x} }, \hspace{2mm} x \in \mathbb{C}$$, so wasted my time with
$$e^{-a-bi}= e^{e^{tan^{-1}\frac{b}{a}i + ln[\sqrt{ a^{2} + b^{2} }]}}$$ and tried from there, (just showing I did make an attempt, no matter how abysmal). Any help appreciated as I am not too familiar with complex numbers outside of the basics needed for transformation matrices.
 
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  • #2
NotASmurf said:
so wasted my time with
e−a−bi=eetan−1bai+ln[√a2+b2]e−a−bi=eetan−1bai+ln[a2+b2]​
e^{-a-bi}= e^{e^{tan^{-1}\frac{b}{a}i + ln[\sqrt{ a^{2} + b^{2} }]}} and tried from there
No need to go that far.
Try multiplying ##f(\alpha) =\frac{1}{1+e^{-\alpha}}## with ##\frac{1+e^{-\alpha^*}}{1+e^{-\alpha^*}}## and then use Euler formula.
 
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  • #3
Thanks :D ,Comes to $$ \frac{1}{2cos(b) e^{-a}}[1+e^{-\overline{\alpha}]}$$ right (before further simplification)?
Then $$ =[ \frac{1}{2cos(b) e^{-a}} +\frac{1}{2}] + [\frac{1}{2} tan(b)]i $$ ? Or have I screwed up? (can't exactly substitute in numbers as easily in this case to test)
 
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  • #4
[itex]\frac{1}{1+e^{-\alpha}}=\frac{1}{1+e^{-a}(cosb-isinb)}=\frac{1+e^{-a}(cosb+isinb)}{(1+e^{-a}cosb)^2+(e^{-a}sinb)^2}=\frac{1+e^{-a}(cosb+isinb)}{1+2e^{-a}cosb+e^{-2a}}[/itex]

[itex]\alpha[/itex] and a look alike in itex.
 
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  • #5
Thanks mathman but since my previous answer is more computationally efficient (has to run vast iterations for the program) my previous answer correct?
 
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  • #6
Turns out this was a fruitless exercise since $$\frac{\partial u }{\partial x} \neq \frac{\partial v }{\partial y} $$ , so it doesn't conform with the Cauchy -Riemann equations D: (It needs to be differentiable)
 
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  • #7
NotASmurf said:
Turns out this was a fruitless exercise since $$\frac{\partial u }{\partial x} \neq \frac{\partial v }{\partial y} $$ , so it doesn't conform with the Cauchy -Riemann equations D: (It needs to be differentiable)
I think you should check that again. 1/(1+e) is analytic in the complex plane except where (1+e) = 0.
 
  • #8
well when I reworked it I got $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} $$ BUT $$\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} $$ when it should be equal to the negative does that mean its diffentiable, but only for certain regions? (havn't had to do complex differentiation before), also quick question , the Riemann - Cauchy equations essentially say $$\frac{\partial f}{\partial z} = 0 $$ must be true for it do be differentiable, this confuses me, can someone elucidate on the intuition here?
 
  • #9
NotASmurf said:
$$\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} $$ when it should be equal to the negative
I think there must be a sign problem somewhere.
The series of operations α => -α => e => 1 + e gives an entire function (analytic in the entire complex plane)
Then 1/(1 + e) is analytic for α in ℂ except where it is a division by 0.
the Riemann - Cauchy equations essentially say $$\frac{\partial f}{\partial z} = 0 $$ must be true for it do be differentiable,
This is not right. I'm not familiar with the Wirtinger derivatives (https://en.wikipedia.org/wiki/Wirtinger_derivatives), but apparently this should be the partial wrt z conjugate (see equation 3 of https://en.wikipedia.org/wiki/Cauchy–Riemann_equations )
 
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  • #10
NotASmurf said:
Thanks mathman but since my previous answer is more computationally efficient (has to run vast iterations for the program) my previous answer correct?
I doubt it. It is different from mine - unlikely to be correct. Check out the denominator.
 
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  • #11
NotASmurf said:
the Riemann - Cauchy equations essentially say
∂f∂z=0​
\frac{\partial f}{\partial z} = 0 must be true for it do be differentiable, this confuses me, can someone elucidate on the intuition here?
No, you are wrong. An analytic function is characterized by [itex] \frac{\partial f}{\partial \bar{z}}=0[/itex] (you need to define [itex]\frac{\partial f}{\partial \bar{z}}[/itex] in a sensible manner, but that should not be too hard).
 
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