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Let $f(x,y)=(x+e^{2y}-1, \sin({x^2+y}))$ and let $h(x,y)=(1+x)^5-e^{4y}$. Show that there exists a continuously differentiable function $g(x,y)$ defined in a neighborhood of $(0,0)$ such that $g(0,0)=0$ and $g\circ f=h$. Compute $\frac{\partial g}{\partial y}(0,0)$.

Any hint on how to procced? Thanks

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    $\begingroup$ Notice that if you had such a function, and if you managed to show that $f$ is invertible in a neighborhood of $(0,0)$, then necessarily you'd have $g = h\circ f^{-1}$ in a neighborhood of $(0,0)$. Does this help? $\endgroup$
    – user2093
    Jun 24 '12 at 21:20
  • $\begingroup$ Yes it helps, thanks William. $\endgroup$
    – KWO
    Jun 24 '12 at 21:24
  • $\begingroup$ So it actually has nothing to do with inverse or implicit function theorem? $\endgroup$
    – KWO
    Jun 24 '12 at 21:25
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    $\begingroup$ Well, it has everything to do with the inverse function theorem. For example, how would you show that $f$ is invertible in a neighborhood of $(0,0)$? One easy way is to differentiate $f$ at $(0,0)$ and use the inverse function theorem. $\endgroup$
    – user2093
    Jun 24 '12 at 21:26
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    $\begingroup$ You can write up the solution yourself at the bottom. Best way for a question to get answered IMO. $\endgroup$
    – MGN
    Jun 24 '12 at 22:14
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As William mentioned, you can construct your function $g$ if $f$ has a local inverse about the origin. Hence we must use the inverse function theorem. First note that $f$ is continuously differentiable everywhere. Moreover, we can compute the Jacobian $J$:

$J(x,y) = \begin{bmatrix} 1 & 2e^{2y} \\ 2x\cos(x^2+y^2) & \cos(x^2 + y) \end{bmatrix}$.

Note that at $(0,0)$, $\det(J(0,0)) = \left| \begin{matrix} 1 & 2 \\ 0 &1\end{matrix}\right| = 1$. So we have satisfied the hypotheses of the inverse function theorem, thus we can construct such a $g$. But the inverse function theorem is even stronger in that it gives you a formula for the derivative of $f^{-1}$ locally in terms of $f'$. You can probably work it out from there by observing $f^{-1}(0,0) = (0,0)$.

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    $\begingroup$ Thanks I have done it with $\frac{\partial g}{\partial y}(0,0)=-14$. $\endgroup$
    – KWO
    Jun 24 '12 at 22:34
  • $\begingroup$ Wonderful! Cheers. $\endgroup$
    – MGN
    Jun 24 '12 at 22:35

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