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    kidos Posts: 5, Reputation: 1
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    #1

    Feb 27, 2012, 10:17 AM
    Math, building a simple non-linear function
    Hi

    I“m trying to build a function which is the inverse function of f= -1/(1 exp(b)) 1/(1 exp(b-2*b*h^n)). Where b and n are constants to fit the function to a curve of data. The look of function f should be attached. So the range of the plot is f(0)=0 , f(0.5)= close to 0.5 and f(1)=1 , and simple constants to shift and fit it well. I did get something close looking to the inverse function of f but its way to complex and long, that“s exp(ln((1/2)*(b-ln(-(h h*exp(b)-exp(b))/(h h*exp(b) 1)))/b)/n). Help would be great!

    kidos
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    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Feb 27, 2012, 10:42 AM
    kidos - pelase do not cust and paste the equations from another program - as you can see some of the important characters get dropped.

    I think what you wrote is this:



    Is that right? And you want to express h as a function of f(h)?
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    #3

    Feb 27, 2012, 11:11 AM
    Hi ebaines

    Yes that“s right for f(h), and yes I want to express h as a function of f(h). Sorry for the confusing paste straight from Matlab. I“ve attached the nasty complex one inverse I“ve had so far, trying to get a nice simpler one.

    Regards

    kidos
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    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Feb 27, 2012, 11:29 AM
    Sorry, I still don't understand what you're trying to do. Specifically - why are there 'h' terms in your inverse equation? If you start with:



    then I would expecty the inverse to be of the form


    In any event - one way to simplify your equation is to use the fact that

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    #5

    Feb 27, 2012, 02:06 PM
    Hi ebaines

    Thanks for the responses. I apologize for being unclear in my explanations. The matter of case is that I had few points of results (the blue curve in my first post) and I wanted to make a function model fitting the curve of the data where h was a vector. I had tried polyfit in Matlab but that did not fit well enough so I had worked on a function which was getting close, the f(h) function but then realized that I wanted to plot the h terms on the y-axis but not on the x-axis, so I was getting my function plotting the curve mirrored through x=y line to what I wanted. It was not working to change the order of terms in my plots because the vectors values of the function fit model and results will be off (hard to explain this :S) when real values of forces and height (h) are being compared with the model values. So I was just trying to find a similar simple function as f(h) but inverse, that should fix my problem.
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    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Feb 27, 2012, 02:53 PM
    Here's the inverse function I get:



    yields:
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    #7

    Feb 28, 2012, 04:45 AM
    Hi ebaines

    I have now a function which is simple and close to what I was looking for (see attached figure and function) but this function is f(0)= -Inf and f(1)= Inf , and I am trying to get f(0)=0 and f(1)=1.
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    ebaines Posts: 12,131, Reputation: 1307
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    #8

    Feb 28, 2012, 06:51 AM
    Perhaps a cubic would work for your purposes. You can curve fit a third order polynomial to hit the end points:

    f(x) = 2x^3 - 3x^2 +2x

    **EDIT A more general cubic equation that fits your needs and lets you adjust the slope of the line at x=1/2 (let's call this value 's') is:

    y = Ax^3 + Bx^2 + Cx

    where A = 4(1-s)
    B = -3A/2
    C = A/2+1

    The attached figure shows how the function looks for s = 0.5 and s = 0.3.
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    #9

    Feb 28, 2012, 02:33 PM
    Thanks for the help ebaines. Any thoughts on the GOP primaries? I am fan of Ron Paul :)

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