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I Help evaluating complex function in form m+ni?

  1. Aug 16, 2016 #1
    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.
     
  2. jcsd
  3. Aug 16, 2016 #2

    blue_leaf77

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    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.
     
  4. Aug 16, 2016 #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)
     
    Last edited: Aug 16, 2016
  5. Aug 16, 2016 #4

    mathman

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    [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.
     
  6. Aug 16, 2016 #5
    Thanks mathman but since my previous answer is more computationally efficient (has to run vast iterations for the program) my previous answer correct?
     
    Last edited: Aug 16, 2016
  7. Aug 16, 2016 #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)
     
    Last edited: Aug 16, 2016
  8. Aug 17, 2016 #7

    FactChecker

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    I think you should check that again. 1/(1+e) is analytic in the complex plane except where (1+e) = 0.
     
  9. Aug 17, 2016 #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?
     
  10. Aug 17, 2016 #9

    FactChecker

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    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.
    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 )
     
  11. Aug 17, 2016 #10

    mathman

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    I doubt it. It is different from mine - unlikely to be correct. Check out the denominator.
     
  12. Aug 28, 2016 #11

    Svein

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    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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