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MathGroup Archive 2007

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Re: Inverse of arbitrary functions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74582] Re: [mg74497] Inverse of arbitrary functions
  • From: "Chris Chiasson" <chris at chiasson.name>
  • Date: Tue, 27 Mar 2007 04:00:20 -0500 (EST)
  • References: <200703240006.TAA16019@smc.vnet.net>

It is worth pointing out that /; and -> are not symbols. They are more
like infix operators, such as +, -, /, etc... The main difference is
that /; and -> only join two other pieces of syntax, while + can be
used to join two or more pieces of syntax.

After transformation to an expression, the result of 1+a+b is
Plus[1,a,b], whcih you can check by using FullForm. The result of 1/;a
is Condition[1,a].

On 3/23/07, Dave Rudolf <dave.rudolf at usask.ca> wrote:
> Hey all,
>
> Is there some way to get Mathematica to find the symbolic inverse of a
> user-defined function? I know that I can use InverseFunction to find the
> inverse of simple functions, like so:
>
>         InverseFunction[Log] -> Exp
>
> However, I would like to do something like
>
>         f[ x_ ] = x + sin[ x + cos[ x ] ]
>
>         InverseFunction[ f ]
>
> I did run across a similar newsgroup post that said to use the Solve
> function, like so:
>
>         inverse[f_, x_] /; PolynomialQ[f, x] && Not[FreeQ[f, x]] :=
>                  Module[{y}, Solve[f == y, x][[1, 1, 2]] /. y -> x]
>
> However, it doesn't seem to work as is, and I don't understand the
> syntax of some of the stuff. Like, what's with the -> symbol? And \; for
> that matter?
>
> Anyway, let me know if there is some other way to pull this off.
>
> Thanks.
>
> Dave
>
>


-- 
http://chris.chiasson.name/


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