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How to find a tangent (not differentiating) Newton Raphson maybe? watch

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    Ok, I have an increasing logistic function that passes through 0,0. X, the variable, is time, so the logistic starts at x=o. The problem is that since logistics have two horizontal asymptotes, when it passes through the point (0,0) its derivative is not 0.

    I need to start at 0,0 with 0 as derivative, since it represents velocity and it begins at rest. But, I also need to create some kind of quadratic or other function that has a minimum at 0,0 and that can be a tangent to the logistic at some point.

    Anyone know how to setup this problem? Is it differential equations? I dont want any direct solutions, but maybe some hints or starting points.
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    This is what I did, ax^2=logistic. Then I need to solve for a to give me the correct function right? I dont know how to solve for a that way, for that equation to have only one answer at the desired place, so I iterated with wolframalpha.com until i got a=4.7996103028305. The question is, how can i solve it algebraically?
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    Love to help, but I've no idea what a logistic function is or it properties.
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    Hmmm
    the problem here is that Logistic functions never have a derivative of Zero....
    starting from the D.E. defn. dv/dt = v(1-v), one can see that zero's only occur if v = 0,1 ... which are of course the asymptotes to which you refer...
    What is the actual question/ problem that you are working on .... I will have a think..
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    (Original post by Mrm.)
    Hmmm
    the problem here is that Logistic functions never have a derivative of Zero....
    starting from the D.E. defn. dv/dt = v(1-v), one can see that zero's only occur if v = 0,1 ... which are of course the asymptotes to which you refer...
    What is the actual question/ problem that you are working on .... I will have a think..
    Thanks Mrm. A logistic function is basically like this: 1/(1+e^(-x)) The minus gives a positive function.
    This is my actual problem, with simplified numbers: Ax^2=1/(1+e^(-x))
    I need to find an A, which will give a minimum/only one root in the equation: Ax^2 - 1/(1+e^(-x))= 0 I talked to my maths teacher he said I had to use Newton-Raphson but that its not necessary at my level... He told me my solution from iterations were fine the way they were (I rounded A to 8 decimals). He also told me algebraic solutions for polynomials and exponentials simply dont happen lol. Anyone here think its ok to do it with Newton-Raphson method? Any tips appreciated
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    (Original post by personlis69)
    This is my actual problem, with simplified numbers: Ax^2=1/(1+e^(-x))
    I need to find an A, which will give a minimum/only one root in the equation: Ax^2 - 1/(1+e^(-x))= 0 I talked to my maths teacher he said I had to use Newton-Raphson but that its not necessary at my level... He told me my solution from iterations were fine the way they were (I rounded A to 8 decimals). He also told me algebraic solutions for polynomials and exponentials simply dont happen lol. Anyone here think its ok to do it with Newton-Raphson method? Any tips appreciated
    It's still not at all clear what you're actually trying to do here. "minimum" and "only one root" don't mean the same thing, surely?

    Do you mean that you want to find the value of A so that \displaystyle Ax^2-\frac{1}{1+e^{-x}} = 0 has exactly one root? I have to say, I'm not seeing how you'd even find an approximate solution using N-R in that case, but maybe I'm just being stupid.

    What exactly have you calculated?
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    Considering the graph of the logistic function, it doesn't even seem possible.
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    (Original post by Scipio90)
    Considering the graph of the logistic function, it doesn't even seem possible.
    You're right - it's not possible. Thanks.
 
 
 
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