       Re: Inverse Function

• To: mathgroup at smc.vnet.net
• Subject: [mg121472] Re: Inverse Function
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Fri, 16 Sep 2011 05:48:42 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201109150840.EAA29206@smc.vnet.net>

```On 09/15/2011 03:40 AM, Giuseppe Benedetti wrote:
> Hello to everyone,
> I have a function of two variables g[k,s] and I want to compute (symbolically) its inverse with respect to the first variable k (ie keeping s constant). I am therefore looking for a function G[g,s], which I then need to differentiate (twice) with respect to the second variable s. Does anybody know how to do it???
>
> Giuseppe.
>

G[g[k, s], s] - k == 0

Differentiate once and twice with respect to k and likewise with respect
to x. All these derivatives are zero. Use that to solve for D[0,2][G] in
terms of derivatives of g.

In:= Derivative[0, 2][G][g[k, s], s] /.
Solve[{D[G[g[k, s], s] - k, k] == 0,
D[G[g[k, s], s] - k, {k, 2}] == 0, D[G[g[k, s], s] - k, s] == 0,
D[G[g[k, s], s] - k, {s, 2}] == 0}, {Derivative[2, 0][G][g[k, s],
s], Derivative[0, 2][G][g[k, s], s],
Derivative[1, 0][G][g[k, s], s], Derivative[0, 1][G][g[k, s], s]}]

Out= {-2*Derivative[0, 1][g][k, s]*
Derivative[1, 1][G][g[k, s], s] -
(Derivative[0, 2][g][k, s] - (Derivative[0, 1][g][k, s]^2*
Derivative[2, 0][g][k, s])/Derivative[1, 0][g][k, s]^2)/
Derivative[1, 0][g][k, s]}

In principle I would have thought we'd also need the second order mixed
derivative, but it seems to have worked without that.

Daniel Lichtblau
Wolfram Research

```

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