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Re: Inverse function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg115564] Re: Inverse function
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sat, 15 Jan 2011 04:42:49 -0500 (EST)

Do the same

Reduce[y == x^x, x]

C[1] \[Element] Integers && y != 0 &&
 x != 0 && ((2 I \[Pi] C[1] + Log[y] !=
      0 && ((Im[ProductLog[-1, 2 I \[Pi] C[1] + Log[y]]] > -\[Pi] &&
         x == E^ProductLog[-1, 2 I \[Pi] C[1] + Log[y]]) ||
       x == E^ProductLog[
         2 I \[Pi] C[1] +
          Log[y]] || (Im[
           ProductLog[1, 2 I \[Pi] C[1] + Log[y]]] <= \[Pi] &&
         x == E^ProductLog[1, 2 I \[Pi] C[1] + Log[y]]))) || (y == 1 &&
      x == 1))

Simplify[Reduce[{y == x^x, x > 0}, x, Reals], x > 0]

(y == 1 && x == 1) || (Inequality[E^(-E^(-1)), LessEqual, y, Less,
    1] &&
      (E^ProductLog[Log[y]] == x ||
     E^ProductLog[-1, Log[y]] == x)) ||
   (y > 1 && E^ProductLog[Log[y]] == x)

If an inverse does not exist, then use numeric techniques.

f[y_?NumericQ] := x /. FindRoot[y == x^x, {x, 1}]

Plot[f[y], {y, E^(-1/E), 2}]


Bob Hanlon

---- "=D0=90=D1=80=D0=BD=D0=BE=D0=BB=D1=8C=D0=B4 =D0=91=D0=BE=D0=B3=D0=BE=
=D0=BF=D0=BE=D0=BB=D1=8C=D1=81=D0=BA=D0=B8=D0=B9" <sender999ster at gmail.com> wrote:

==========================
And how to be in a case y==x^x?
Thanks.

2011/1/14 Bob Hanlon <hanlonr at cox.net>

> Solve[y == x^2, x]
>
> {{x -> -Sqrt[y]}, {x -> Sqrt[y]}}
>
> Reduce[y == x^2, x]
>
> x == -Sqrt[y] || x == Sqrt[y]
>
>
> Bob Hanlon
>
> ---- Arnold <sender999ster at gmail.com> wrote:
>
> ==========================
> Function y=x^2 is given. Give, please, a code in WM-7 with which help it is
> possible to receive function, return the given. That is, x=g (y).
> Thanks.
>



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