Linear approximation

**Killer** · Feb 4th 2011, 04:13 PM

Let $f:\mathbb R^2\to\mathbb R$ so that $f(x,y)=(s,t)=\left( x+\dfrac{1}{2}\arctan y,y+\dfrac{1}{2}\arctan x \right)$

Find the linear approximation of $f^{-1}$ on a neighborhood of $(s_0,t_0)=f(0,1).$

Don't worry about the invertibility, that was another question which I solved, but now I need help with this one, I don't get how to solve it.

**Jose27** · Feb 4th 2011, 06:21 PM

By the chain rule, we get that $Df(0,1)\circ Df^{-1}(s_0,t_0)=I$ .

**Killer** · Feb 4th 2011, 06:47 PM

Okay... but, I don't know how to relate it to find the linear approximation.

If you could help me a bit more.

**Jose27** · Feb 4th 2011, 06:51 PM

It's pretty straightforward: Since that relation gives that the derivatives involved are invertible we get $Df^{-1}(s_0,t_0)=Df(0,1)^{-1}$

**Killer** · Feb 4th 2011, 07:19 PM

Oh, now I get, so I just need to compute the inverse but do I need to multiply it by a vector or something?

**HallsofIvy** · Feb 5th 2011, 06:47 AM

The linear approximation to g(x, y) at (x_0, y_0) is g(x_0, y_0)+ D_g(x_0, y_0)(x, y) where the last term is the product of the matrix D_g(x_0, y_0) with the vector (x, y).

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