Re: finding inverses of functions
- To: mathgroup at smc.vnet.net
- Subject: [mg126441] Re: finding inverses of functions
- From: "djmpark" <djmpark at comcast.net>
- Date: Thu, 10 May 2012 05:00:52 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <4767427.124602.1336550815297.JavaMail.root@m06>
An interesting problem. I'm going to take your "e" to be Exp. Let's define
the y function:
Clear[x, y];
y[x_] := 3 x^3 + 2 E^(2 x);
The function looks to be monotonically increasing.
Plot[y[x], {x, -2, 2}]
And we can verify it.
ForAll[x, D[y[x], x] > 0]
Resolve[%, Reals]
...
True
We will solve a differential equation and need an initial condition:
y[0]
2
Write the inverse slope in terms of x[y] and solve the differential
equation.
Clear[x]
x'[y] == 1/(D[y[x], x] /. x -> x[y]);
xsol = DSolve[%, x, y][[1, 1]]
x -> Function[{y}, InverseFunction[2 E^(2 #1) + 3 #1^3 &][y + C[1]]]
Solve the initial condition for C[1].
x[2] == 0 /. xsol;
Solve[%, C[1]][[1, 1]]
C[1] -> 0
We can now define the x function. It is in terms of an InverseFunction but
easily evaluable.
x[y_] = x[y] /. (xsol /. C[1] -> 0)
InverseFunction[2 E^(2 #1) + 3 #1^3 &][y]
We can check numerically that the function is inverse, at least for small or
exact values of x.
x[y[x]] == x;
% /. x -> 10
True
(I wish I knew a method to show this symbolically.)
We can plot the y function, inverse x function and their composition to
check, at least numerically, the proper relation.
Show[
{Plot[y[x], {x, -2, 2}, PlotStyle -> Black],
Plot[x[y], {y, -20, 20}, PlotStyle -> Blue],
Plot[y[x[z]], {z, -10, 10}, PlotStyle -> Red]},
AspectRatio -> Automatic,
PlotRange -> 10,
Axes -> False,
Frame -> True]
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/index.html
From: John Accardi [mailto:accardi at accardi.com]
Hello, I am trying to get Mathematica to find the inverse of:
y = 3x^3 + 2e^(2x) (which I know is invertible)
InverseFunction only seems to give inverses of built-ins, like Sine.
I tried:
Solve[ y == 3x^3 + 2e^(2x), x ] but get a message that Solve does not have
methods suitable. (Solve works for simpler functions, however.)
Any ideas? Thanks.