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Thread: How to simply sigmoid function?

  1. #1
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    Question How to simply sigmoid function?

    Is it possible to simply the combination of two sigmoid functions

    $$\sigma\big(k(t_1-t)\big) - \sigma\big(k(t_2-t)\big)$$
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  2. #2
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    Re: How to simply sigmoid function?

    I'm not sure what you are asking. Simply is not a verb. If you are asking if it is possible to simplify, it depends on which sigmoid function you are using and what you mean by simplify. I'm gonna go with probably not.
    Last edited by SlipEternal; Aug 24th 2017 at 02:50 AM.
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  3. #3
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    Re: How to simply sigmoid function?

    Sorry for the typo, yes I meant simplify.

    Is it possible to express

    $$x = \sigma\big(k(t_1-t)\big) - \sigma\big(k(t_1-t)\big)$$

    in a simpler form?
    Last edited by brianx; Aug 24th 2017 at 09:00 AM.
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    Re: How to simply sigmoid function?

    Quote Originally Posted by brianx View Post
    Sorry for the typo, yes I meant simplify.

    Is it possible to express

    $$x = \sigma\big(k(t_1-t)\big) - \sigma\big(k(t_1-t)\big)$$

    in a simpler form?
    Yes. $x = 0.$

    $$x = \sigma\big(k(t_1-t)\big) - \sigma\big(k(t_1-t)\big) = 0$$ is simpler.
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  5. #5
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    Re: How to simply sigmoid function?

    Quote Originally Posted by brianx View Post
    Sorry for the typo, yes I meant simplify.

    Is it possible to express

    $$x = \sigma\big(k(t_1-t)\big) - \sigma\big(k(t_1-t)\big)$$

    in a simpler form?
    An example sigmoid function is: $\sigma(z) = \dfrac{1}{1+e^z}$. If we were to plug in:

    $\dfrac{1}{1+e^{k(t_1-t)}} - \dfrac{1}{1+e^{k(t_2-t)}} = \dfrac{e^{k(t_2-t)}-e^{k(t_1-t)}}{\left( 1+e^{k(t_1-t)}\right) \left( 1+e^{k(t_2-t)} \right)}$

    I do not see that as simpler. So, it depends on which sigmoid function you are using and what you mean by "simpler".
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