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Math Help - Finding the inverse of an implicit polynomial function

  1. #1
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    Ulsan
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    Finding the inverse of an implicit polynomial function

    I have read various responses to finding an inverse function in the pre-calculus section and decided to post this question here instead since it isn’t really a ‘pre-calculus’ type of question.. The solutions to finding an inverse of a function look reasonably straight forward but I was wondering whether anyone might be able to assist me in finding an inverse function for the following;

    y= ax + bx^c

    Clearly it is not possible to find 'x' in terms of 'y' since the equation is implicit, hence the format for the explicit examples I've seen can't be used, so how does one approach finding the inverse of this function? Note that I have found the variables a, b, and c and they are not integers for the purposes of my investigation.

    Now, having stated the initial problem, it gets a little more complicated as I also have to find the integral to this inverse function but just cannot get past the first step using the examples already provided in the forum. Any helpful tips to get me started would be greatly appreciated.


    In order to lay out the problem completely I have presented it in the terms I am presently using.;

    If
    \tau = \left(\frac{\delta Pr}{2L} \right) = a\gamma+b\gamma^c


    and


    v=\frac{1}{R^2}\int_{0}^R r^2f^{-1}\left(\frac{\delta Pr}{2L} \right)dr


    Then I imagine that by guessing a value for
    \delta P I can use an iterative process can find a value for v and compare that to known values for v in order to modify my initial guess. I already have defined parameters for L and R

    I am sure you can now see why I need to find the inverse ! What do you think?
    For now if you've got any ideas on how to approach this little problem, I'd be pleased to know.

    Cheers!
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  2. #2
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    I'm not entirely sure about this. But let me try to offer what I think may be a route to it's analytic solution and there may be problems if the function is not strictly one-to-one.

    We want:

    \int_{0}^R r^2f^{-1}\left(\frac{\delta Pr}{2L} \right)dr

    given that f(x)=ax+bx^c

    First let k=\frac{\delta P}{2L} and v=kr then:

    \int_{0}^R r^2f^{-1}\left(\frac{\delta Pr}{2L} \right)dr=\frac{1}{k^3}\int_0^{Rk} v^2 f^{-1}(v)dv

    Now let u=f^{-1}(v) then v=au+bu^c and \frac{dv}{du}=a+\frac{bc}{u} u^c then:

    \frac{1}{k^3}\int_0^{Rk} v^2 f^{-1}(v)dv=\frac{1}{k^3}\int_{p_0}^{p_1} (au+bu^c)^2 u (a+\frac{bc}{u}u^c) du

    where:

    p_0=\text{Solve}(0==au+bu^c,u)

    p_1=\text{Solve}(Rk==au+bu^c,u)

    What I would do is start simple with just y=x^2 go through it, test it numerically, add complexity to it, test it again numerically, and then finally test everything I did above numerically.
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  3. #3
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    Ulsan
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    Dear Shawsend,
    Fistly, my sincere thanks to you for your interest and your clear suggestion. This does indeed seem to be a way forward. I tried following a method suggested in "The Mathematical Gazette" (March 2000) which provided a formula for integrating inverse functions but I ended up getting lost in a quagmire of difficulties when it came to the limits. Your method however is well worth exploring as it directly addresses this aspect.You have shed a ray of light and a glimmer of hope on this problem for me. Thank you. If you are beset by the ironies of life then consider a euphemism in the context of the problem I posed to a total stranger that your help has indeed been one-to-one.
    Best regards, Alex.
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  4. #4
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    Hello Alex. Not life in general, just mine. It's from the Last Samurai with Tom Cruise. Hey, it's not hard to check that as we can numerically integrate the inverse directly: Just do a table of {f(x),x}, generate an "interpolation" function on that table, bingo-bango (the code is Mathematica):

    Code:
    In[83]:=
    f[x_] := (1/2)*x + (3/4)*x^(1/2); 
    mytable = Table[{f[x], x}, 
        {x, 0, 10, 0.1}]; 
    myinverse = Interpolation[mytable]; 
    rval = 4; 
    kval = 2; 
    i1 = NIntegrate[t^2*myinverse[kval*t], 
       {t, 0, rval}]
    p1 = First[x /. Solve[8 == (1/2)*x + 
            (3/4)*x^(1/2), x]]; 
    i2 = NIntegrate[(1/kval^3)*
        ((1/2)*u + (3/4)*u^(1/2))^2*u*
        (1/2 + 3/8/u^2^(-1)), {u, 0, p1}]
    
    Out[88]=
    167.62359195110528
    
    Out[90]=
    167.62360953565704
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