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October 23rd, 2018, 02:50 AM  #1 
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1  Inverse of function with 3 variables
So, I do have a formula that generates a point, to do so, it will compute with 3 variables (u, v, a) and I want to find out what the variables are for a given point. Here's the function: $\displaystyle \binom{u}{v}+\frac{u*\sin(e)v*\cos(e)}{sin(a)}*\binom{\cos(a+e)}{\sin(a+e)} + \frac{u*\cos(e+d+0.5a)+v*\sin(e+d+0.5a)}{\sin(a)}*\binom{\sin(e+0.5a)}{\cos(e+0.5a)} $ UV is a vector and a,d and e are angles, although e & d are set parameters. The formula works, but I don't know how I inverse it. Do I write the formula down like: f(u,v,a)=[...] or can I split it up f(u)*f(v)*f(a) or is it a chain, I'm completely lost here, so any insight or link would help me. 
October 23rd, 2018, 04:37 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 
you've got $\begin{pmatrix}x \\y\end{pmatrix} = f(u,v,a)$ do you want $\begin{pmatrix}u\\v\\a\end{pmatrix} = f^{1}(x,y) ?$ I suspect there are infinite combinations of $(u,v,a)$ that produce a given $(x,y)$ and thus no inverse exists. Where does $f$ arise from? 
October 23rd, 2018, 05:08 AM  #3  
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1 
Hey romsek, thanks for replying. Quote:
Quote:
$\begin{pmatrix}u\\v\\a\end{pmatrix} = f^{1}(x,y,r)$ Now I explain why r is important. The formula is describing a centerpoint of a circle. that circle has a radius: $r = \frac{b*\cos(a/2)}{\sin(a)}$ There there is only one combination of u,v,a for x,y,r.  
October 23rd, 2018, 05:48 AM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
\[f(u,v,a) = f(u,v,g(u,v)) \] and locally the level set will be the graph of $g$. This means level sets are 1dimensional differentiable manifolds near any regular point. As you have noticed, if you fix a value of $r = r_0$, then level sets for this in the domain of $f$ are also 1dimensional differentiable manifolds. Generically, two 1dimensional manifolds will intersect in $\mathbb{R}^2$ at a single point (locally) which is the unique value you are describing. See the following link for a more thorough introduction. https://en.wikipedia.org/wiki/Inverse_function_theorem Last edited by SDK; October 23rd, 2018 at 06:18 AM.  
October 23rd, 2018, 06:53 AM  #5  
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1  Quote:
as someone who is not studying in a mathrelated field, missing knowledge of calculus 1 or 2, I see lot's of things I haven't heard or used before, so going from trigonometry to Inverse function theorem seems ambitious. Here is, how I would approach it though: $\displaystyle \begin{pmatrix}u\\v\\a\end{pmatrix}=f^{1}(x,y,r) \\ f(u,v,a) = \begin{pmatrix}x\\y\\r\end{pmatrix} $ $\displaystyle x = u+\frac{u*\sin(e)v*\cos(e)}{\sin(a)}*\cos(a+e)+\frac{u*\cos(e+d+a/2)v*\sin(e+d+a/2)}{\sin(a)}*\sin(e+a/2) $ $\displaystyle y = v+\frac{u*\sin(e)v*\cos(e)}{\sin(a)}*\sin(a+e)+\frac{u*\cos(e+d+a/2)v*\sin(e+d+a/2)}{\sin(a)}*\cos(e+a/2) $ $\displaystyle r = \frac{b*\cos(a/2)}{\sin(a)} $ now I would ask: what is u,v,a for $x=x_1$? what is u,v,a for $y=y_1$? what is u,v,a for $r=r_1$? and the overlap of these answers would be the parameters that create my unique circle. for $x_1$ $u_1=[...]$ $v_1=[...]$ $a_1=[...]$ for $y_1$ $u_2=[...]$ $v_2=[...]$ $a_2=[...]$ for $r_1$ $u_3=[...]$ $v_3=[...]$ $a_3=[...]$ then I would: $u_1=u_2=u_3=u$ $v_1=v_2=v_3=v$ $a_1=a_2=a_3=a$ Do you think that would work? Last edited by theHeavyLobster; October 23rd, 2018 at 06:55 AM. Reason: bad wording  
October 26th, 2018, 09:32 AM  #6 
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1  ditched the parameter d
So I chose to get rid of $d$ as it is 0 in this case and simplified the function to the following one: $\displaystyle \begin{bmatrix} x \\ y \\ \end{bmatrix} = \binom{u/2}{v/2}+\binom{v}{u}*\frac{1+cos(a)}{2*sin(a)} $ $\displaystyle \begin{bmatrix} r \end{bmatrix} = \frac{b*cos(a/2)}{sin(a)} $ I'll try to inverse that now, any help is appreciated as always. 
October 26th, 2018, 09:46 AM  #7  
Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315  Quote:
$\Large \begin{align*} &u = \frac{\sqrt{4 r^2b^2}}{b}+2 y \\ \\ &v = \frac{2 b \sqrt{4 r^2b^2} (xy)+b^24 r^2}{b^24 r^2}\\ \\ &a = 2 \left(\tan^{1}\left(\frac{\sqrt{4 r^2b^2}}{r},\frac{b}{r}\right)\right) \end{align*}$ The $\tan^{1}(x,y)$ function being used here takes two arguments, the first treated as the xcoordinate, and the second is the y coordinate and returns the angle in the proper quadrant. See if these formulas work for you.  
October 26th, 2018, 10:24 AM  #8 
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1 
I'll check those thanks, for a I did get: a = 2*Asin(b/2r) which seems to work 
October 27th, 2018, 03:29 AM  #9  
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1  Quote:
I had "a" figured out for "r" already, so I just plugged it into "x" and "y": $\displaystyle a = 2*sin^{1}(b/2*r) \\ $ $\displaystyle x = \frac{u}{2} + v* \frac{1+cos(a)}{2*sin(a)} \\ $ $\displaystyle x = \frac{u}{2} + v* \frac{1+cos(2*sin^{1}(b/2*r))}{2*sin(2*sin^{1}(b/2*r))} \\ $ $\displaystyle u = \frac{x}{2} + v*\frac{1 \pm sqrt(1(b/2*r)^2)}{2b/r} $ $\displaystyle y = \frac{v}{2} + u* \frac{1+cos(a)}{2*sin(a)} \\ $ $\displaystyle v = \frac{y}{2} + u*\frac{1 \pm sqrt(1(b/2*r)^2)}{2b/r} $ Now I plugged "v" into "u": $\displaystyle u = \frac{x}{2} + (\frac{y}{2} + u*\frac{1 \pm sqrt(1(b/2*r)^2)}{2b/r} )*\frac{1 \pm sqrt(1(b/2*r)^2)}{2b/r} $ Now I just need to get all the u's to the left and I've got the formula for u. Is that correct? I did this already, but the result was also not working, although it almost works, which is, why I think there's probably a little mistake in the simplificationprocess, let me show you: The green circle is fed directly by the parameters x,y,r and the red one goes through the inverse of f and f itself. If you find any mistake, let me know Last edited by theHeavyLobster; October 27th, 2018 at 03:44 AM.  
October 27th, 2018, 06:30 AM  #10 
Member Joined: Jul 2018 From: Germany Posts: 37 Thanks: 1  it is working!
Alright, Now I've checked my approach another time and have seen a couple of mistakes, now it's a matter of simplifications again, but u,v & a are set! Thanks for the help again! 

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