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Math Help - inverse function

  1. #1
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    inverse function

    Hello,

    I have a function to inverse, I proved that thise function is inversible but I didn't manage to find the expression of the inverse:

    here is the function:
    y= ((P1*A*r^(-alpha))/(N+(P2*A*(r^2-2*R*cos(theta)*r+R^2)^(-alpha/2))))

    the function is decreasing for r< R/cos(theta)

    so I have to find the expression of 'r' in function of y,P1, P2, alpha, N, R.

    all the parameters here are positive; and r<R/cos(theta).


    I used matlab (with 'solve' function) but it didn't manage to inverse this function.

    Thanks for your precious help
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by stanfordia View Post
    here is the function:
    y= ((P1*A*r^(-alpha))/(N+(P2*A*(r^2-2*R*cos(theta)*r+R^2)^(-alpha/2))))
    does your function look like this:

    y=\frac{P_1 A r^{-\alpha}}{N+P_2 A(r^2-2Rcos\theta+R^2)^{-\frac{\alpha}{2}}}
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  3. #3
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    yes exactly

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  4. #4
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    And what is the independent variable? In other words, to find the inverse function, what should we solve for? r? \theta?

    If the problem is to solve <br />
y=\frac{P_1 A r^{-\alpha}}{N+P_2 A(r^2-2Rcos\theta+R^2)^{-\frac{\alpha}{2}}} for \omega, this is relatively straight forward. Multiply both sides by the denominator: y(N+P_2 A(r^2- 2Rcos\theta+ R^2)^\frac{\alpha}{2}= P_1 A r^{-\alpha}.

    Take the 2/\alpha power of both sides: y^{\frac{2}{\alpha}}(N+ P_2 A(r^2- 2Rcos(\theta)+ R^2))= P_1^{\frac{2}{\alpha}}A^{\frac{2}{\alpha}}r^-2

    Multiply both sides by y^{\frac{\alpha}{2}}: N+ P_2 A(r^2- 2Rcos(\theta)+ R^2)= y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^{\frac  {2}{\alpha}}r^-2

    Subtract N from both sides: P_2 A(r^2- 2Rcos(\theta)+ R^2)= y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^{\frac  {2}{\alpha}}r^-2- N

    Divide both sides by P_2 A: r^2- 2Rcos(\theta)+ R^2= \frac{y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^  {\frac{2}{\alpha}}r^-2- N}{P_2 A}

    Subtract r^2+ R^2 from both sides: - 2Rcos(\theta) = \frac{y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^  {\frac{2}{\alpha}}r^-2- N}{P_2 A}- r^2- R^2

    Finally, divide both sides by -2R: cos(\theta) = -\frac{\frac{y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alp  ha}}A^{\frac{2}{\alpha}}r^-2- N}{P_2 A}- r^2- R^2}{2R}
    and take the arccosine of both sides:
    : \theta = arccos\left(-\frac{\frac{y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alp  ha}}A^{\frac{2}{\alpha}}r^-2- N}{P_2 A}- r^2- R^2}{2R}\right)

    If you want to solve for r, having both r^2 and r^{-\alpha} makes it much harder.
    Last edited by HallsofIvy; July 4th 2009 at 09:13 AM.
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  5. #5
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    There is an error in your equation : the N is not is the exponent part


    <br />
y (N+P_2 A(r^2-2Rcos\theta+R^2)^{-\frac{\alpha}{2}}) =P_1 A r^{-\alpha}<br />

    and TO REPLY ti your question I need to solve r
    Last edited by mr fantastic; July 4th 2009 at 02:09 PM. Reason: Merged posts
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