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In my math assignment I have to find the inverse of

$$ f(x_1, x_2) = \left(\ln \left(\frac{x_2}{x_1}\right), x_1^2 + x_2^2\right) $$

Now I already have looked into this, and came up with the following:

I split f into $$ f_1(x_1,x_2) = \ln \left(\frac{x_2}{x_1} \right)\\ \text{and} \\ f_2(x_1,x_2) = x_1^2+x_2^2 $$

solving $f_1$ for $x_2$ resulted in $$e^{f_1} \cdot x_1 = x_2 $$

and after using this $x_2$ in $f_2$ $$ x_1 = \sqrt{\frac{f_2}{1+e^{2f_1}}} $$

which I then used in $x_2$ again and led to:

$$ x_2 = e^{f_1} \cdot \sqrt{\frac {f_2}{1+e^{2f_1}}} $$

So according to what I have done the inverse should be:

$$ f^{-1}(f_1, f_2) = \left( \sqrt{\frac{f_2}{1+e^{2f_1}}} \ ,\ e^{f_1} \cdot \sqrt{\frac {f_2}{1+e^{2f_1}}} \right) $$

But after looking at another example on this site, it appears to be wrong. So I'd like to ask whether this is wrong or not, and if it is where I made the mistakes.

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  • $\begingroup$ Should be $e^{2f_1}$, not $e^{2}f_1$, in the last occurrence near the bottom. Don't know if this was the mistake tho.. $\endgroup$ – Dzoooks May 14 at 22:32
  • $\begingroup$ @Dzoooks You're right I'd missed a bracket. thank you $\endgroup$ – DoingItSideways May 14 at 22:34
  • $\begingroup$ Looks fine to me, as long as you don't run into problems with the signs for the arguments of square root or logarithms $\endgroup$ – Andrei May 14 at 23:17
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$$ f^{-1}(f_1, f_2) = \left( \sqrt{\frac{f_2}{1+e^{2f_1}}} \ ,\ e^{f_1} \cdot \sqrt{\frac {f_2}{1+e^{2f_1}}} \right) $$

was indeed correct.

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