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Thread: 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:

    $\displaystyle y=\frac{P_1 A r^{-\alpha}}{N+P_2 A(r^2-2Rcos\theta+R^2)^{-\frac{\alpha}{2}}}$
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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? $\displaystyle \theta$?

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

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

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

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

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

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

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

    If you want to solve for r, having both $\displaystyle r^2$ and $\displaystyle r^{-\alpha}$ makes it much harder.
    Last edited by HallsofIvy; Jul 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


    $\displaystyle
    y (N+P_2 A(r^2-2Rcos\theta+R^2)^{-\frac{\alpha}{2}}) =P_1 A r^{-\alpha}
    $

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