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.

  • $\begingroup$ 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

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

was indeed correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.